Current Electricity

6.25 x 10¹⁸

Reason — The number of electrons in 1 C charge is 6.25 x 10¹⁸.

ammeter, series

Reason — By connecting the ammeter in series, the entire current flowing through the circuit also flows through the ammeter, allowing it to measure the current accurately.

I = $\frac{\text {ne}}{\text t}$

Reason — If n electrons pass through a cross-section of a conductor in time t, then current in conductor is I = $\frac{\text {ne}}{\text t}$

V = W/Q

Reason — If W joule of work is done in bringing a test charge Q coulomb from infinity to the point P, then electric potential V at point P is V = W/Q .

voltmeter, parallel

Reason — Connecting the voltmeter in parallel ensures that it measures the potential difference accurately because it doesn't affect the current flow in the circuit.

All of the above

Reason — According to Ohm's Law : V = IR

Hence, I ∝ V and V/I = constant

Hence, all are correct.

volt/ampere

Reason — 1 ohm = volt/ampere

ohm^-1

Reason — The unit of conductance is ohm^-1.

solar cell

Reason — Nichrome, Copper sulphate solution with copper electrodes and dil sulphuric acid, Silver, all the three follow Ohm's Law i.e., V = IR, hence they are ohmic conductors whereas a solar cell is a non-ohmic conductor.

all of the above

Reason — The three factors on which the resistance of wire depends are —

1. Dependence on length of the wire — the resistance of a wire is directly proportional to the length of the wire.
   R ∝ l
2. Dependence on the thickness of the wire — the resistance of a wire is inversely proportional to it's area of cross section (a) normal to the direction of flow of current.
   R ∝ $\bold{\frac{1}{a}}$
3. Dependence on the temperature of the wire — the resistance of conductor increases with an increase in it's temperature.

Same

Reason — Specific resistance of an object is an intrinsic property which is independent of it's dimensions but does depend upon nature of material and temperature. Since both rod and thin wire are of same material i.e., aluminium so both will have same specific resistance.

9R

Reason — When the wire is stretched to thrice it's length, it's area of cross section becomes 1/3 and it's length becomes 3l.

Let, a be the area of initial cross section and ρ be the specific resistance of the material of wire.

Then,

• length = l,
• Resistance = R Ω,
• New length = 3l,
• New area = $\frac{\text a}{3}$

From relation

R = ρ $\frac{\text l}{\text a}$ = ρ $\frac{\text l}{\text {πr}^2}$

Initial resistance (R₁) = R = ρ $\frac{\text l}{\text a}$    [Equation 1]

New resistance (R_n) = ρ $\frac{3\text l}{\frac{\text a}{3}}$ = ρ $\frac{9\text l}{\text a}$    [Equation 2]

On dividing eqn (ii) by (i), we get,

$$\begin{aligned}\frac{\text R_\text n}{\text R} = \frac{\text ρ \frac{9\text l}{\text a}}{\text ρ \frac{\text l}{\text a}} \\[1em] = 9 \\[1em] \Rightarrow \text R_\text n = 9\text R\end{aligned}$$

Hence, the new resistance = 9R.

nichrome wire

Reason — The conductors which obey the Ohm's law are called the ohmic resistors or linear resistances. A Nichrome wire obeys Ohm's law hence it is an ohmic resistor.

carbon

Reason — Resistance decreases with increase in temperature for carbon.

Constantan

Reason — Standard resistors are made of constantan as it has high specific resistance and the effect of change in temperature on its resistance is negligible.

Both its high resistivity and its increase in resistance with rise in temperature

Reason — Nichrome wire is commonly used in heating appliances due to its following properties:

1. Specific resistance is high — Nichrome wire has a relatively high specific resistance, which means it resists the flow of electric current efficiently, resulting in heating when current passes through it.
2. Resistance increases with an increase in temperature — This property is desirable for heating elements because it helps maintain a consistent level of heating even as the temperature rises.

both zero resistance and infinite conductance

Reason — A superconductor exhibits both infinite conductance and zero resistance, allowing electric current to flow through it without any loss of energy due to resistance.

The expression is —

R = ρ$\frac{\textbf l}{\textbf a}$

where

• R = resistance of wire
• ρ = specific resistance of the material of wire
• l = length of wire
• a = area of cross section of wire

The order of specific resistance is as follows:

1. The specific resistance is very low for metals, because it allows most of current to pass through it.
2. The specific resistance is low for semiconductor.
3. The specific resistance is very high for insulators, as the current won't pass through it.

The substance whose specific resistance remains almost unchanged with increase in temperature is manganin.

(i) The material used for the filament of an electric bulb is a tungsten wire because it has a high melting point.

(ii) The material used for the heating element of a room heater is nichrome because the specific resistance of nichrome is high and it's resistance increases to a great extent with the increase in temperature.

A substance of zero resistance (or infinite conductance) at a very low temperature is called a superconductor.

Current is defined as the rate of flow of charge. The S.I. unit of charge is coulomb and therefore current is measured in coulomb per second which has been given the name ampere (A).

The electric potential at a point is defined as the amount of work done per unit charge in bringing a positive test charge from infinity to that point. The S.I. unit of electric potential is volt.

The potential difference (p.d.) between two points is equal to the work done per unit charge in moving a positive test charge from one point to the other. It's S.I. unit is volt. It is a scalar quantity.

The potential difference between two points is said to be 1 volt if the work done in moving 1 coulomb charge from one point to other is 1 joule i.e.,

1 volt = $\frac{\text{1 joule}}{\text{1 coulomb}}$ = 1 J C^-1

(a) Current is a scalar quantity.

By stating the direction of current in a conductor we mean that the direction of motion of electrons is opposite to it.

(b) Potential is a scalar quantity.

When the potential is positive at a point in the vicinity of a positive charge since work has to be done on the positive test charge against the repulsive force due to the positive charge in bringing it from infinity, while it is negative at a point in the vicinity of a negative charge since the attractive force on test charge due to the negative charge does work by itself.

The obstruction offered to the flow of current by the conductor (or wire) is called it's resistance.

The S.I. unit of resistance is Ohm.

(a) The particles which are responsible for the flow of current in a metallic wire are free electrons.

(b) In a metallic wire, the moving charges are the free electrons which constitute the current.

If n electrons pass through the cross section of a conductor in time t, then total charge passed through the conductor is given as

Q (charge) = n × e

and the current in conductor is

$\text{I} = \frac{\text{Q}}{\text{t}} = \frac{\text{ne}}{\text{t}}$

(c) When the ends of a metal wire are connected to a cell, the electrons inside it experience a force in direction from the end at negative potential to positive, due to which they began to move as shown in the diagram below.

Due to force, their speed increases but during the movement they collide with the fixed positive ions and loose some of their kinetic energy due to which their speed decreases. This lost energy heats up the wire.

After the collision, they are again accelerated towards the positive potential due to the existing potential difference so their speeds again increases and then again in collision with the positive ions, their speed decreases. This process continues. As a result, the electrons do not move in bulk with a continuously increasing speed, but their is a drift of electrons towards the positive terminal. Thus, a metal wire offers some resistance to the flow of electrons through it.

(a) The law which relates the potential difference and current in a conductor is known as Ohm's law.

Ohm's law states that the current flowing through the conductor is directly proportional to the potential difference across it's ends provided that the physical conditions and the temperature of the conductor remain constant.

(b) The necessary condition for a conductor to obey Ohm's law is that the physical conditions and the temperature should remain constant.

Graph (a) is non-ohmic resistor and

Graph (b) is ohmic resistor.

The I-V graph for (b) is a straight line or linear while for (a), the graph is a curve.

Resistance of a wire is directly proportional to the length of the wire.

R ∝ l

In a long conductor, the number of collisions of free electrons with the positive ions will be more as compared to a shorter one. Therefore, a longer conductor offers more resistance. Hence, the resistance of a conductor is directly proportional to the length of the wire.

With the increase in temperature of a conductor, the random motion of electrons increases. As a result, the number of collisions of electrons with the positive ions increase. Hence, the resistance of a conductor increases with an increase in it's temperature.

The iron wire will have more resistance.

Iron wire has more resistance as compared to copper because specific resistance of iron is more than that of copper.

The three factors on which the resistance of wire depends are —

1. Dependence on length of the wire — the resistance of a wire is directly proportional to the length of the wire.
   R ∝ l
2. Dependence on the thickness of the wire — the resistance of a wire is inversely proportional to it's area of cross section (a) normal to the direction of flow of current.
   R ∝ $\bold{\frac{1}{a}}$
3. Dependence on the temperature of the wire — the resistance of conductor increases with an increase in it's temperature.

Specific resistance of a material is the resistance of a wire of that material of unit length and unit area of cross section.

The S.I. unit of specific resistance is ohm × metre or (Ω m)

(a) Two factors on which the specific resistance of a wire depends are —

1. Material of the substance — It is a characteristic property of the substance. It is different for different substances.
2. Temperature of the substance — It increases with the increase in temperature for metals, but it decreases with the increase in temperature for the semiconductor.

(b) Both the wires will have the same specific resistance because the specific resistance is a characteristic property of the material and as both the wires are of copper hence both will have same specific resistance.

The specific resistance of a semi-conductor decreases with the increase in temperature.

(a) Dependency of resistance of a wire is as follows:

1. Resistance of a wire is directly proportional to length of the wire.
2. Resistance of a wire is inversely proportional to the square of radius of the wire.

(b) Dependency of specific resistance of a wire is as follows:

1. Specific resistance of a wire does not depend on length of the wire as it is a characteristic property of the material of wire.
2. Specific resistance of a wire does not depend on radius of the wire as it is a characteristic property of the material of wire.

(a) Copper or aluminium materials are used for making connection wires because their specific resistance is very low and hence they possess least possible resistance.

Due to low (or negligible) resistance of connection wires, the current in circuit remains unaffected, and the loss of energy due to heating is prevented. Hence, they are made of materials such as copper or aluminium, whose specific resistance is very small.

(b) The connection wires are made thick so that their resistance becomes low.

Standard resistors are made of magnanin, constantan etc. for which the specific resistance is high and the effect of change in temperature on their resistance is negligible.

The material used for making fuse wire is an alloy of lead and tin because it's melting point is low and it's specific resistance is more than that of copper or aluminium so that the resistance of a short and thin fuse wire is more than that of the connecting wire. It permits current upto it's safe limit to pass through it, but an excessive current melts it so that it blows off and the circuit is broken.

A superconductor is a substance of zero resistance (or infinite conductance) at a very low temperature.

Example — Mercury below 4.2 K, lead below 7.25 K and niobium below 9.2 K

Resistances of these substances decrease tremendously with the decrease in temperature and become almost zero in the low temperature range near absolute zero. Zero resistance of a superconductor means it's infinite conductivity (i.e., once a current starts flowing in a superconductor, it persists even when there is no potential across it).

According to Ohm's law, the current flowing in a conductor is directly proportional to the potential difference applied across it's ends provided that the physical conditions and the temperature of the conductor remain constant.

If a current I flows in a conductor when the potential difference across it's ends is V and R is the resistance, then according to Ohm's law V = IR

Below circuit diagram shows the setup for verifying Ohm's law:

(a) V-I graph for a conductor obeying Ohm's law is shown below:

(b) Slope of V-I graph for a conductor represents resistance.

I-V graph for a linear resistor

Slope of I-V graph — The slope of I-V graph is $\frac{\textbf {ΔI}}{\textbf {ΔV}}$ which is the reciprocal of resistance of the conductor i.e.,

Slope = $\frac{Δ\text I}{Δ\text V}$ = $\frac{1}{\text{resistance of the conductor}}$

The conductors which obey the Ohm's law are called the ohmic resistors or linear resistances. Examples are all metallic conductors (such as Silver, Aluminium, Copper, Iron etc.)

Current–Voltage relationship of an ohmic resistor is shown in the below graph:

Resistance is determined in the form of slope from the graph.

The conductors which do not obey the Ohm's law are known as the non-ohmic resistors (or non-linear resistances). Examples — LED, solar cell, junction diode, etc.

Current-Voltage relationship of non-ohmic resistors is shown in the graph below:

V-I graph of a conductor at two different temperatures T₁ and T₂ is shown below:

Figure shows two straight lines A and B on the V-I graph for a conductor at two different temperatures T₁ and T₂ (T₁ > T₂) respectively.

The straight line A is more steeper than the line B because the resistance of conductor is more at high temperature T₁ than at low temperature T₂

(a) In a thick conductor, electrons get a larger area of cross section to flow as compared to a thin conductor, therefore a thick wire offers less resistance. The resistance of a conductor is inversely proportional to it's area of cross section (a), normal to the direction of flow of current i.e.,

R ∝ $\frac{1}{\text a}$

Hence,

R ∝ $\frac{1}{\text {πr}^2}$

where r is the radius of the wire.

Hence, resistance of a wire is inversely proportional to the square of the radius of the wire.

(b) As we have seen above that the resistance of a conductor is inversely proportional to it's area of cross section (a), normal to the direction of flow of current. Hence, the thin wire, will offer more resistance as it's area of cross section is less.

Given,

• Number of electrons flowing (n) = 6.25 × 10¹⁶
• Time taken (t) = 2 s
• e = 1.6 × 10^-19 C

Current in conductor is given by,

$\text I = \frac{\text{ne}}{\text{t}}$

Substituting the values in the formula above we get,

$$\begin{aligned}\text I = \frac{6.25 \times 10^{16} \times 1.6 \times 10^{-19}}{2} \\[1em] = \frac{6.25 \times 1.6 \times 10^{-3}}{2} \\[1em] = 6.25 \times 0.8 \times 10^{-3} \\[1em] = 5 \times 10^{-3}\ \text A\end{aligned}$$

Hence, the current flowing through the conductor is 5 mA from B to A.

Given,

• Current (I) = 3.2 mA = 3.2 × 10^-3 A
• Charge of one electron = -1.6 × 10^-19 coulomb
• t = 1 s

Charge flowing through the conductor is given by,

Charge (Q) = current (I) x time (t)

Substituting the values in the formula above we get,

Q = 3.2 x 10^-3 x 1 = 3.2 x 10^-3 C

Hence,

$$\begin{aligned}\text{No. of electrons} = \frac{\text{Total charge flowing}}{\text{Charge on one electron}} \\[1em] = \frac{3.2 \times 10^{-3}}{1.6 \times 10^{-19}} \\[1em] = 2 \times 10^{16}\end{aligned}$$

Therefore, the number of electrons that will pass each second through the cross section of that conductor is 2 x 10¹⁶.

Given,

• Current (I) = 300 mA = 0.3 A
• Resistance (R) = 20 Ω
• Potential difference (V) = ?

From Ohm's law

V = IR

Substituting the values in the formula above, we get,

V = 0.3 × 20

V = 6 V

Hence, the required potential difference is 6 V.

Given,

• Potential difference (V) = 12 V
• Current (I) = 2 A
• Resistance = ?

From Ohm's law

V = IR

Substituting the values in the formula above, we get,

12 = 2 x R

⇒ R = $\frac{12}{2}$

⇒ R = 6 Ω

Hence, the resistance of the filament of the bulb is 6 Ω.

The resistance will be less when the bulb is not glowing.

V-I graph for the readings is shown below:

(a) We can observe from the graph that when current is 0.5 A then potential difference is 1.25 V.

(b) We can observe from the graph that when potential difference is 0.75 V, then current is 0.3 A.

(c) Slope of the graph gives the resistance, and as the graph is linear

So, if potential difference (V) = 1.0 V

current (I) = 0.4 A

From Ohm's law

V = IR

Substituting the values in the formula above, we get,

1.0 = 0.4 x R

⇒ R = $\frac{1.0}{0.4}$

⇒ R = $\frac{10}{4}$

⇒ R = 2.5 Ω

Hence, the resistance in circuit = 2.5 Ω

(i) R = ρ$\frac{\text l}{\text a}$ = ρ$\frac{\text l}{\text {πr}^2}$

Resistance for wire of radius 1 mm,

R₁ = ρ$\frac{l}{\text π\times 1^2}$ = ρ$\frac{l}{\text π}$

Resistance for wire of radius 2 mm,

R₂ = ρ $\frac{l}{\text π \times 2^2}$ = ρ $\frac{l}{4\text π}$

Hence, ratio between the two is :

$$\begin{aligned}\frac{\text R_1}{\text R_2}= \frac{\text ρ\frac{\text l}{\text π}}{\text ρ\frac{\text l}{4\text π}} \\[1em] = \frac{4}{1} \\[1em] = 4 : 1\end{aligned}$$

Hence, the ratio of resistance between the two wires = 4 : 1

(ii) The specific resistance of the two wires will be same because the specific resistance is a characteristic property of the material, hence it does not change with radius.

Therefore, ratio of specific resistance between the two wires = 1 : 1

When the wire is stretched to double it's length, it's area of cross section becomes half and it's length becomes double.

Let, a be the area of initial cross section and ρ be the specific resistance of the material of wire.

Then,

• Initial length = l
• R = 1 Ω
• New length = 2l
• New area = $\frac{a}{2}$

From relation

R = ρ $\frac{\text l}{\text a}$ = ρ $\frac{\text l}{\text {πr}^2}$

Initial resistance (R₁) = 1 = ρ$\frac{\text l}{\text a}$    [Equation 1]

New resistance (R₂) = ρ $\frac{2\text l}{\frac{\text a}{2}}$ = ρ$\frac{4\text l}{\text a}$    [Equation 2]

On dividing eqn (2) by (1), we get,

$$\begin{aligned}\frac{\text R_2}{1} = \frac{\text ρ \frac{4\text l}{\text a}}{\text ρ \frac{\text l}{\text a}} \\[1em] \Rightarrow \text R_2 = 4\ \text Ω\end{aligned}$$

Hence, the new resistance = 4 Ω.

Given,

• Resistance (R) = 3 Ω
• Length (l) = 10 cm

Let, a be the area of initial cross section and ρ be the specific resistance of the material of wire.

Then,

• New length = 30 cm
• New area a = $\frac{\text a}{3}$

From relation

R = ρ $\frac{\text l}{\text a}$ = ρ $\frac{\text l}{\text {πr}^2}$

Initial resistance (R₁) = 3 = ρ$\frac{10}{\text a}$    [Equation 1]

New resistance (R₂) = ρ$\frac{30}{\frac{\text a}{3}}$ = ρ $\frac{90}{\text a}$    [Equation 2]

On dividing eqn (1) by (2), we get,

$$\begin{aligned}\frac{3}{\text R_2} = \frac{\text ρ \frac{10}{\text a}}{\text ρ \frac{90}{\text a}} \\[1em] \frac{3}{\text R_2} = \frac{10}{90} \\[1em] \frac{3}{\text R_2} = \frac{1}{9} \\[1em] \Rightarrow \text R_2 = 3 \times 9 \\[1em] \Rightarrow \text R_2 = 27\ \text Ω\end{aligned}$$

Hence, the new resistance = 27 Ω

Given,

• Resistance (R) = 9 Ω

When the wire is tripled on itself, it's area of cross section becomes thrice and it's length becomes $\frac{\text l}{3}$.

Let, a be the area of initial cross section and ρ be the specific resistance of the material of wire.

Then,

• Length = 30 cm,
• New length = $\frac{\text l}{3}$ = $\frac{30}{3}$ = 10 cm,
• New area a₂ = 3a

From relation

R = ρ $\frac{\text l}{\text a}$

Initial resistance (R₁) = 9 = ρ$\frac{30}{\text a}$    [Equation 1]

New resistance (R₂) = ρ$\frac{10}{3\text a}$    [Equation 2]

On dividing eqn (2) by (1), we get,

$$\begin{aligned}\frac{\text R_2}{9} = \frac{ρ \frac{10}{3\text a}}{ ρ \frac{30}{\text a}} \\[0.5em] \frac{\text R_2}{9} = {\frac{10}{90}} \\[1em] \frac{\text R_2}{9} = {\frac{1}{9}} \\[1em] \text R_2 = 1\ \text Ω\end{aligned}$$

Hence, the new resistance = 1 Ω.

Given,

• Specific resistance (ρ) = 1.7 × 10^-8 Ω m
• Radius (r) = 1 mm = 10^-3m
• Resistance (R) = 2 Ω
• Length (l) = ?

From relation

R = ρ$\frac{\text l}{\text a}$ = ρ$\frac{\text l}{\text {πr}^2}$

Substituting the values in the formula above we get,

$$\begin{aligned}2 = 1.7 \times 10^{-8} \frac{\text l}{\text π(10^{-3})^2} \\[1em] 2 = 1.7 \times 10^{-8} \frac{\text l}{3.14 \times (10^{-3})^2} \\[1em] 2 = \frac{1.7 \times 10^{-2} \times \text l}{3.14} \\[1em] 2 = \frac{1.7 \times \text l}{314} \\[1em] \Rightarrow \text l = \frac{314 \times 2}{1.7} \\[1em] \Rightarrow \text l = 369.4 \text{ m}\end{aligned}$$

Hence, the length of the copper wire is 369.4 m.

Case 1 —

• Current (I) = 100 m A = 0.1 A
• Potential Difference (V) = 0.2 V

From Ohm's law

V = IR

Substituting the values in the formula above, we get,

0.2 = 0.1 x R

⇒ R = $\frac{0.2}{0.1}$

⇒ R = 2 Ω

Hence, resistance of filament of bulb = 2 Ω

Case 2 —

• Current (I) = 400 m A = 0.4 A
• Potential Difference (V) = 1.0 V

From Ohm's law

V = IR

Substituting the values in the formula above, we get,

1.0 = 0.4 x R

⇒ R = $\frac{1.0}{0.4}$

⇒ R = 2.5 Ω

Hence, resistance of filament of bulb = 2.5 Ω

Therefore, we observe that with increase in temperature resistance of the wire increases.

Hence, resistance of filament increases with the increase in temperature.

(i) The best conductor is silver because it has the lowest specific resistance among the three given materials.

(ii) Specific resistance (resistivity) indicates how strongly a material opposes the flow of electric current. A lower value of resistivity means less opposition to current flow and hence better conductivity.

Since silver has the smallest resistivity compared to copper and constantan, it allows electric current to pass most easily, making it the best conductor.

Given,

• I₁ = 2 A
• V₁ = 50 V
• I₂ = 3 A
• V₂ = 60 V

To determine whether the conductor is ohmic, we check if it obeys Ohm’s law, which requires the ratio $\frac{\text V}{\text I}$ to remain constant.

On applying Ohm's law i.e, V = IR,

At 50 V and 2 A :

$$\begin{aligned}\text R_1 = \frac{\text V_1}{\text I_1} \\[1em] = \frac{50}{2} \\[1em] = 25\ \text Ω\end{aligned}$$

At 60 V and 3 A :

$$\begin{aligned}\text R_2 = \frac{\text V_2}{\text I_2} \\[1em] = \frac{60}{3} \\[1em] = 20\ \text Ω\end{aligned}$$

Since the resistance is not constant (25 Ω ≠ 20 Ω), the conductor does not obey Ohm’s law.

Therefore, the conductor is non-ohmic.

the material of electrodes

Reason — The emf of a cell depends on :

1. the material of electrodes and
2. the electrolyte used in the cell.

equal to, greater than

Reason — Emf of a cell is equal to the terminal voltage when cell is not in use and when the current starts to flow in the circuit, then there is a voltage drop due to internal resistance, thus we get, V = E - Ir. Hence, Emf is greater than the terminal voltage when cell is in use.

$\text I = \frac{\text ε}{(\text R + \text r)}$

Reason — The current drawn from the cell

$\text I = \frac{\text { e.m.f. of cell}}{\text { total resistance}} = \frac{\text ε}{(\text R + \text r)}$

current is same in each resistance

Reason — Current is same in each resistance in series combination of resistances. The current has a single path to flow, hence, same current passes through each resistor.

p.d. is the same across each resistance

Reason — Potential difference is same across each resistor in parallel combination of resistances, which is equal to the potential difference across the terminals of the battery (or source).

All of the above

Reason — For parallel combination of resistances :

1. On I-V graph, the slope of line is more because resistance in parallel combination is less than series combination.
2. The potential difference across each resistance is same because, the ends of each resistor are connected to the ends of the same source of potential.
3. According to Ohm's Law V = IR. Hence, the current in a resistor is inversely proportional to the resistance.

assertion is false but reason is true..

Explanation

Assertion (A) is false. E.m.f. is greater than the terminal voltage only when the cell delivers the current to the external circuit.

Reason (R) is true. When a heavy current is drawn from the cell, a large number of charge carriers flow through the electrolyte and hence more work is done. This results in more voltage drop v, and hence less terminal voltage V. Thus, terminal voltage V of a cell depends on the amount of current I drawn from it.

(a) Equivalent resistance R of three resistors R₁, R₂ and R₃ joined in parallel is given by —

$\frac{1}{\text R}$ = $\frac{1}{\text R}_1$ + $\frac{1}{\text R_2}$ + $\frac{1}{\text R_3}$

(b) Equivalent resistance R of three resistors R₁, R₂ and R₃ joined in series is given by —

R = R₁ + R₂ + R₃

(a) When same current flows in each resistor, then the two resistors are joined in series

(b) When potential difference is same across each resistor, then the two resistors are joined in parallel

(c) When equivalent resistance is less than either of the two resistances, then the two resistors are joined in parallel

(d) When equivalent resistance is more than either of the two resistances, then the two resistors are joined in series

(a) In the circuit, two resistors of 2 Ω each are connected in parallel. In parallel, the equivalent resistance is R_P, then

$$\begin{aligned}\frac{1}{\text R_\text P} = \frac{1}{2} + \frac{1}{2} \\[1em] \frac{1}{\text R_\text P} = \frac{2}{2} \\[1em] \Rightarrow \text R_\text P = 1\ \text Ω\end{aligned}$$

Hence, R_P = 1 Ω

(b) In the circuit, three resistors of 1 Ω, 1 Ω and 2 Ω are connected in parallel. In parallel, the equivalent resistance is R_P, then

$$\begin{aligned}\frac{1}{\text R_\text P} = \frac{1}{1} + \frac{1}{1} + \frac{1}{2} \\[1em] = \frac{2 + 2 + 1}{2} \\[1em] \Rightarrow \text R_\text P = \frac{2}{5} Ω \\[1em] \Rightarrow \text R_\text P = 0.4\ \text Ω\end{aligned}$$

Hence, R_P = 0.4 Ω

(c) In the circuit two resistors of 1 Ω each are connected in parallel. In parallel, the equivalent resistance is R_P, then

$$\begin{aligned}\frac{1}{\text R_\text P} = \frac{1}{1} + \frac{1}{1} \\[1em] \frac{1}{\text R_\text P} = \frac{1 + 1}{1} \\[1em] \frac{1}{\text R_\text P} = \frac{2}{1} \\[1em] \text R_\text P = \frac{1}{2} \\[1em] \Rightarrow \text R_\text P = 0.5\ \text Ω\end{aligned}$$

Hence, R_P = 0.5 Ω

(d) In the circuit there are three parts. In the first part two resistor of 1 Ω each are connected in series. In series, the equivalent resistance is R'_S, then

R'_S = (1 + 1) Ω = 2 Ω

In the second part two resistor of 1 Ω each are connected in series. In series, the equivalent resistance is R''_S, then

R''_S = (1 + 1) Ω = 2 Ω

In the third part,R'_S and R''_S are in parallel, the equivalent resistance is R_P, then

$$\begin{aligned}\frac{1}{\text R_\text P} = \frac{1}{2} + \frac{1}{2} \\[1em] \frac{1}{\text R_\text P} = \frac{1 + 1}{2} \\[1em] \frac{1}{\text R_\text P} = \frac{2}{2} \\[1em] \Rightarrow \text R_\text P = 1\ \text Ω\end{aligned}$$

Hence, R_P = 1 Ω

Therefore, on observing the values we can say that (a) and (d) have same equivalent resistances i.e., 1 Ω.

The factors on which the internal resistance of a cell depends are —

1. The surface area of the electrodes — larger the surface area of electrodes, less is the internal resistance.
2. The distance between the electrodes — more the distance between the electrodes, greater is the internal resistance.

Given,

• e.m.f. = ε
• internal resistance = r
• external resistance = R

(a) The total resistance of circuit = R + r

(b) The current drawn from the cell

$\text I = \frac{\text {e.m.f. of cell}}{\text {Total resistance}} = \frac{\text ε}{(\text R + \text r)}$

(c) the p.d. across the cell = V = IR

Substituting from above,

$\text V = \bold{\frac{\text ε}{(\text R + \text r)}}\times \text R$

(d) Voltage drop inside the cell V = Ir

Substituting from above,

$\textbf V = \bold{\frac{\textbf ε}{(R + r)}}\times \textbf r$

(a) When a cell is used to send current to an external circuit, it's terminal voltage V is less than it's e.m.f.

Hence, terminal voltage < e.m.f.

(b) The emf of the cell is equal to it's terminal voltage when no current is drawn.

The slope of V-I graph gives the resistance. Since the straight line A is less steeper than B, so the straight line A represents small resistance. In parallel combination, the equivalent resistance is less than in series combination so A represents the parallel combination.

e.m.f — e.m.f of a cell is defined as the energy spent (or the work done) per unit charge in taking a positive test charge around the complete circuit of the cell. (i.e., in the circuit outside the cell as well as in the electrolyte inside the cell).

Terminal voltage — The terminal voltage of a cell is defined as the work done per unit charge in carrying a positive test charge around the circuit connected across the terminals of the cell.

Internal resistance — The resistance offered by the electrolyte inside the cell, to the flow of current, is known as the internal resistance of the cell.

In a closed circuit, the current flows through the circuit. There is a fall of potential across the internal resistance of the cell. Hence, the p.d. across the terminals in a closed circuit is less than the p.d. across the terminals in an open circuit by an amount equal to the potential drop across the internal resistance of the cell.

Hence, the p.d. across the terminals of a cell is more in an open circuit and reduced in a closed circuit.

Below diagram shows two resistors connected in series:

If current I is drawn from the battery, the current through each resistor will also be I.

By Ohm's law,

p.d. between A and B is V₁ = V_A - V_B = IR₁

p.d. between B and C is V₂ = V_B - V_C = IR₂

Adding these we get,

V = V₁ + V₂

= V_A - V_B + V_B - V_C

= V_A - V_C

= IR₁ + IR₂

= I (R₁ + R₂)    [Equation 1]

If the equivalent resistance between the points A and C is R_S, then the potential difference between the points A and C is

V = V_A - V_C = IR_S    [Equation 2]

Therefore from equation 1 and 2,

IR_S = I (R₁ + R₂)

⇒ R_S = R₁ + R₂

Thus, in the series combination, the equivalent resistance is equal to the sum of the individual resistances.

Below diagram shows two resistors connected in parallel:

Let I₁ and I₂ be the currents through the resistances R₁ and R₂ respectively, then total current drawn from the battery is

I = I₁ + I₂    [Equation 1]

If potential difference between the two ends A and B is V, then by Ohm's law

current in R₁ is I₁ = $\frac{\text V}{\text R_1}$

current in R₂ is I₂ = $\frac{\text V}{\text R_2}$

On adding these,

I₁ + I₂ = $\frac{\text V}{\text R_1}$ + $\frac{\text V}{\text R_2}$    [Equation 2]

If the equivalent resistance of the combination between the points A and C is R_p, then total current drawn from the source is

I = $\frac{\text V}{\text R_\text p}$    [Equation 3]

Substituting the values of I and I₁ + I₂ from equation 3 and 2 in 1, we get,

$$\begin{aligned}\frac{\text V}{\text R_\text p} = \text V\Big(\frac{1}{\text R_1} + \frac{1}{\text R_2}\Big) \\[1em] \Rightarrow \frac{1}{\text R_\text p} = \frac{1}{\text R_1} + \frac{1}{\text R_2}\end{aligned}$$

Thus, in the parallel combination, the reciprocal of the equivalent resistance is equal to the sum of the reciprocals of the individual resistances.

Given,

• e.m.f. (ε) = 4 volt
• Internal resistance (r) = 2 ohm
• External resistance (R) = 8 ohm

(i) When the key is open then no current is flowing in the circuit and hence the ammeter reading = 0

From relation,

Voltage (V) = ε – Ir

Substituting the values in the formula we get,

V = 4 – (0 × 2)

V = 4 volt

Hence, voltmeter reading is 4 volt.

(ii) When key is closed, current drawn from the cell,

$\text{I} = \frac{\text{e.m.f. of cell}}{\text{total resistance}} = \frac{\text{ε}}{\text{R} + \text{r}}$

Substituting the values in the formula above we get,

I = $\frac{4}{8 + 2}$

⇒ I = $\frac{4}{10}$

⇒ I = 0.4 ampere

Thus, ammeter reading = 0.4 ampere

From relation,

Voltage (V) = ε – Ir

Substituting the values in the formula we get,

V = 4 - (0.4 x 2) = 3.2 V

Hence, voltmeter reading is 3.2 volt.

Given,

• E.m.f. (ε) = 6.0 V
• Current (I) = 3 A
• Potential difference (V) = 5.4 V
• Internal resistance (r) = ?

From relation,

V = ε – Ir

Substituting the values in the formula above we get,

5.4 = 6 - (3r)

⇒ r = 6 - 5.4

⇒ 3r = 0.6

⇒ r = $\frac{0.6}{3}$

⇒ r = $\frac{6}{30}$

⇒ r = $\frac{2}{10}$

⇒ r = 0.2 Ω

Hence, internal resistance of the battery is 0.2 Ω.

(a) Given,

• e.m.f. (ε) = 1.8 V
• Internal resistance (r) = 2 Ω
• I = ?

Total resistance of arrangement = 2 + 0.7 + 4.5 = 7.2 Ω

From relation,

$\text I = \frac{\text ε}{\text R}$

Substituting the values in the formula above we get,

I = $\frac{1.8}{7.2}$

⇒ I = 0.25 A

Hence, reading of ammeter = 0.25 A

(b) Current (I) = 0.25 A

total resistance (excluding internal resistance) = 4.5 + 0.7 = 5.2 ohm

Using ohm's law

V = IR

Substituting the values in the formula above we get,

V = 0.25 × 5.2

V = 1.3 V

Hence, potential difference across the terminals of the battery is 1.3 V.

(a) Given,

• I = 400 mA = 400 x 10^-3 = 0.4 A
• V = 12 V

From Ohm's law,

V = IR

Substituting the values in the formula we get,

12 = 0.4 x R

⇒ R = $\frac{12}{0.4}$

⇒ R = 30 Ω

Hence, the resistance of the music system is 30 Ω.

(b) Given,

• I = 320 mA = 320 x 10^-3 A = 0.32 A
• R = 30 Ω

From Ohm's law,

V = IR

Substituting the values in the formula we get,

V = 0.32 x 30 = 9.6V

Hence, the battery voltage when the music system stops playing is 9.6 V.

Given,

• e.m.f. = ε
• Internal resistance = r
• Current (I) = 1.0 A
• External resistance (R) = 1.9 ohm

Case 1 :

From relation,

ε = I(R + r)

Substituting the value in the formula above we get,

ε = 1 x (1.9 + r)

ε = 1.9 + r    [Equation 1]

Case 2 :

• I = 0.5 A
• R = 3.9 Ω

Substituting the value in the formula above we get,

ε = 0.5 x (3.9 + r)

ε = 1.95 + 0.5r    [Equation 2]

Equating 1 and 2 we get,

$$\begin{aligned}1.9 + \text r = 1.95 + 0.5\text r \\[1em] \text r - 0.5 \text r = 1.95 - 1.9 \\[1em] 0.5 \text r = 0.05 \\[1em] \Rightarrow \text r = \frac{0.05}{0.5} \\[1em] \Rightarrow \text r = 0.1\ \text Ω\end{aligned}$$

Now, substituting the value of r in equation 1, we get,

ε = 1.9 + 0.1

ε = 2 V

Hence, ε = 2 V , r = 0.1 Ω

In order to get a total resistance less than 1 Ω, the three resistors of values 2 Ω, 3 Ω and 5 Ω, should be connected in parallel as shown in the diagram below:

In parallel, if the equivalent resistance is R_P, then

$$\begin{aligned}\frac{1}{\text R_\text P} = \frac{1}{\text R_1} + \frac{1}{\text R_2} + \frac{1}{\text R_3} \\[1em] = \frac{1}{2} + \frac{1}{3} + \frac{1}{5} \\[1em] = \frac{15 + 10+ 6}{30} \\[1em] = \frac{31}{30} \\[1em] = 0.97\ \text Ω\end{aligned}$$

Hence, the three resistors should be connected in parallel and total resistance is 0.97 Ω.

When two resistors (2 Ω each) in parallel combination are connected to the third resistor (2 Ω) in series connection then the resultant resistance is 3 Ω. The below diagram shows this arrangement of resistors:

To verify,

Total resistance of two resistors in parallel,

In parallel, if the equivalent resistance is R_P, then

$$\begin{aligned}\frac{1}{\text R_\text P} = \frac{1}{\text R_1} + \frac{1}{\text R_2} \\[1em] \frac{1}{\text R_\text P} = \frac{1}{2} + \frac{1}{2} \\[1em] \frac{1}{\text R_\text P} = \frac{2}{2} \\[1em] \Rightarrow \text R_\text P = 1\ \text Ω\end{aligned}$$

Thus, R_P = 1 Ω

In series connection,

Equivalent resistance = R_E = R_P + R₃ = 1 + 2 = 3 Ω

Hence, equivalent resistance = 3 Ω

Let the resistors be R₁, R₂, R₃, R₄ each of 2.0 Ω.

In parallel, if the equivalent resistance is R_P, then

$$\begin{aligned}\frac{1}{\text R_\text P} = \frac{1}{\text R_3} + \frac{1}{\text R_4} \\[1em] \frac{1}{\text R_\text P} = \frac{1}{2} + \frac{1}{2} \\[1em] \frac{1}{\text R_\text P} = \frac{2}{2} \\[1em] \Rightarrow \text R_\text P = 1\ \text Ω\end{aligned}$$

Thus, R_P = 1 Ω

In series connection,

Equivalent resistance = R_E = R₁ + R₂ + R_P = 2 + 2 + 1 = 5 Ω

Hence, equivalent resistance = 5 Ω

Let the four set of resistors be R₁, R₂, R₃, R₄ .

As each set consists of three resistors of 2 Ω each in series:
Resistance of each set = 2 + 2 + 2 = 6 Ω

As four similar sets of equivalent resistance 6 Ω are arranged in parallel, so the equivalent resistance (R_P) will be:

$$\begin{aligned}\frac{1}{\text R_\text P} = \frac{1}{\text R_1} + \frac{1}{\text R_2} + \frac{1}{\text R_3} + \frac{1}{\text R_4} \\[1em] \frac{1}{\text R_\text P} = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} \\[1em] \frac{1}{\text R_\text P} = \frac{4}{6} \\[1em] \text R_\text P = \frac{6}{4} \\[1em] \Rightarrow \text R_\text P = 1.5\ \text Ω\end{aligned}$$

Hence, equivalent resistance = 1.5 Ω

In the circuit, there are three parts. In the first part, resistors of 4 Ω and 8 Ω are connected in series. If the equivalent resistance of this part is R'_s then

R'_s = 4 + 8 = 12 Ω

In the second part, resistors of x Ω and 5 Ω are connected in series. If the equivalent resistance of this part is R''_s then

R''_s = (x + 5) Ω

In the third part, the two parts of resistance R'_s = 12 Ω and R''_s = (x + 5) Ω are connected in parallel. If the equivalent resistance between points A and B is R_p then

$$\begin{aligned}\frac{1}{\text R_\text p} = \frac{1}{\text R'_\text s} + \frac{1}{\text R''_\text s} \\[1em] = \frac{1}{12} + \frac{1}{x + 5}\end{aligned}$$

But it is given that equivalent resistance between points A and B is 4 Ω
∴ R_p = 4 Ω

Putting value of R_p in the equation above and solving for x:

$$\begin{aligned}\frac{1}{4} = \frac{1}{12} + \frac{1}{\text x + 5} \\[1em] \frac{1}{4} = \frac{\text x + 5 +12}{12(\text x + 5)} \\[1em] \frac{1}{4} = \frac{\text x + 17}{12(\text x + 5)} \\[1em] \frac{12(\text x + 5)}{4} = \text x + 17 \\[1em] 3(\text x + 5) = \text x + 17 \\[1em] 3\text x + 15 = \text x + 17 \\[1em] 3\text x - \text x = 17 - 15 \\[1em] 2\text x = 2 \\[1em] \Rightarrow \text x = 1 Ω\end{aligned}$$

Hence, the value of x = 1 Ω

In the circuit, there are four parts. In the first part, three resistors of 1 Ω each are connected in series. If the equivalent resistance of this part is R'_s then

R'_s = (1 + 1 + 1) Ω = 3 Ω

In the second part, three resistors of 2 Ω each are connected in series. If the equivalent resistance of this part is R''_s then

R''_s = (2 + 2 + 2) Ω = 6 Ω

In the third part, the two parts of resistance R'_s = 3 Ω and R''_s = 6 Ω and 2 Ω are connected in parallel. If the equivalent resistance is R_p then

$$\begin{aligned}\frac{1}{\text R_\text p} = \frac{1}{\text R'_\text s} + \frac{1}{2} + \frac{1}{\text R''_\text s} \\[1em] = \frac{1}{3} + \frac{1}{2} + \frac{1}{6} \\[1em] = \frac{2 + 3 + 1}{6} \\[1em] = \frac{6}{6} \\[1em] \Rightarrow \text R_\text p = 1\ \text Ω \\[1em]\end{aligned}$$

∴ R_p = 1 Ω

In the fourth part 1 Ω, (R_p = 1 Ω) and 1 Ω are connected in series in between points A and B.

Hence, the equivalent resistance is (1 + 1 + 1) Ω = 3 Ω

Given,

• Uniform wire with a resistance of 27 Ω is divided into three equal pieces. Hence, resistance of each piece = 9 Ω.

Three such pieces are joined in parallel.

In parallel, if the equivalent resistance is R_P, then

$$\begin{aligned}\frac{1}{\text R_\text P} = \frac{1}{\text R_1} + \frac{1}{\text R_2} + \frac{1}{\text R_3} \\[1em] = \frac{1}{9} + \frac{1}{9} + \frac{1}{9} \\[1em] = \frac{3}{9} \\[1em] = \frac{1}{3} \\[1em] \Rightarrow \text R_\text P = 3\ \text Ω\end{aligned}$$

Hence, equivalent resistance = 3 Ω

Below diagram shows the arrangement of resistors:

In the circuit, there are two parts. In the first part, two resistors of 6 Ω and 12 Ω are connected in parallel. If the equivalent resistance is R_p then

$$\begin{aligned}\frac{1}{\text R_\text P} = \frac{1}{12} + \frac{1}{6} \\[1em] = \frac{1+2}{12} \\[1em] = \frac{3}{12} \\[1em] = \frac{1}{4} \\[1em] \Rightarrow \text R_\text P = 4\ \text Ω\end{aligned}$$

Thus, R_P = 4 Ω

In the second part, 1 Ω and (R_P = 4 Ω) are connected in series. If the equivalent resistance is R_s then

R_s = 1 + R_P = 1 + 4 = 5 Ω

Hence, total resistance of the circuit = 5 Ω

In the circuit, there are two parts. In the first part, resistors of 12 Ω, 6 Ω and 4 Ω are connected in parallel. If the equivalent resistance of this part is R'_p then

$$\begin{aligned}\frac{1}{{\text R'}_\text P} = \frac{1}{12} + \frac{1}{6} + \frac{1}{4} \\[1em] = \frac{1 + 2 + 3}{12} \\[1em] = \frac{6}{12} \\[1em] = \frac{1}{2} \\[1em] \Rightarrow \text R'_\text P = 2\ \text Ω\end{aligned}$$

Thus, R'_P = 2 Ω

In the second part, resistors of 2 Ω, (R'_p = 2Ω) and 5 Ω are connected in series. If the equivalent resistance of this part is R_s then

R_s = (2 + 2 + 5) Ω = 9 Ω

Hence, effective resistance between A and B = 9 Ω

In the circuit, there are three parts. In the first part, two resistors of 3 Ω, 2 Ω are connected in series. If the equivalent resistance is R'_s then

R'_s = 3 + 2 = 5 Ω

In the second part, two resistors of 6 Ω, 4 Ω are connected in series. If the equivalent resistance is R''_s then

R''_s = 6 + 4 = 10 Ω

In the third part R'_s, 30 Ω and R''_s are connected in parallel. If the equivalent resistance is R_p then

$$\begin{aligned}\frac{1}{\text R_\text P} = \frac{1}{\text R'_\text s} + \frac{1}{30} + \frac{1}{\text R''_\text s} \\[1em] = \frac{1}{5} + \frac{1}{30} + \frac{1}{10} \\[1em] = \frac{6 + 1 + 3}{30} \\[1em] = \frac{10}{30} \\[1em] = \frac{1}{3} \\[1em] \Rightarrow \text R_\text P = 3\ \text Ω\end{aligned}$$

Hence, R_P = 3 Ω

In the circuit, there are two parts. In the first part, three resistors of 2 Ω each are connected in series. If the equivalent resistance of this part is R'_s then

R'_s = (2 + 2 + 2) Ω = 6 Ω

In the second part, the resistance of first part (R'_s = 6 Ω ) and 2 Ω are connected in parallel. If the equivalent resistance of this part is R_p then

$$\begin{aligned}\frac{1}{\text R_\text p} = \frac{1}{6} + \frac{1}{2} \\[1em] = \frac{1 + 3}{6} \\[1em] = \frac{4}{6} \\[1em] \Rightarrow \text R_\text p = 1.5\ \text Ω\end{aligned}$$

∴ R_p = 1.5 Ω

(b) In the circuit, there are three parts. In the first part, two resistors of 2 Ω each are connected in series. If the equivalent resistance of this part is R'_s then

R'_s = (2 + 2) Ω = 4 Ω

In the second part, two resistors of 2 Ω each are connected in series. If the equivalent resistance of this part is R''_s then

R''_s = (2 + 2) Ω = 4 Ω

In the third part, the two parts of resistance R'_s = 4 Ω and R''_s = 4 Ω are connected in parallel. If the equivalent resistance between points A and C is R_p then

$$\begin{aligned}\frac{1}{\text R_\text p} = \frac{1}{4} + \frac{1}{4} \\[1em] = \frac{2}{4} \\[1em] = \frac{1}{2} \\[1em] \Rightarrow R_p = 2\ \text Ω\end{aligned}$$

∴ Equivalent Resistance between A and C = 2 Ω

(a) In the circuit, there are two parts. In the first part, two resistors of 3 Ω each are connected in series. If the equivalent resistance of this part is R'_s then

R'_s = (3 + 3) Ω = 6 Ω

In the second part, resistance R'_s = 6 Ω and 3 Ω are connected in parallel. If the equivalent resistance between points P and Q is R_p then

$$\begin{aligned}\frac{1}{\text R_\text p} = \frac{1}{6} + \frac{1}{3} \\[1em] = \frac{1 + 2}{6} \\[1em] = \frac{3}{6} \\[1em] \Rightarrow \text R_\text p = 2\ \text Ω\end{aligned}$$

∴ Equivalent resistance between the points P and Q = 2 Ω

(b) In the circuit, 3 Ω, R_p = 2 Ω and 3 Ω are connected in series. If the equivalent resistance of this part is R_s then

R_s = (3 + 2 + 3) Ω = 8 Ω

∴ Equivalent resistance between the points X and Y = 8 Ω

(a) Labelled diagram of the arrangement is shown below:

(b) Two resistors of 2 Ω and 3 Ω are connected in parallel. If the equivalent resistance of this part is R_p then

$$\begin{aligned}\frac{1}{\text R_\text p} = \frac{1}{2} + \frac{1}{3} \\[1em] = \frac{3 + 2}{6} \\[1em] = \frac{5}{6} \\[1em] \Rightarrow \text R_\text p = \frac{6}{5} \\[1em] = 1.2\ \text Ω\end{aligned}$$

So,

• Equivalent resistance = 1.2 Ω
• Current (I) = 0.5 A
• Potential Difference (V) = ?

From Ohm's law

V = IR

Substituting the values in the formula above, we get,

V = 0.5 x 1.2 = 0.6 V

Current through 2 Ω = ?

From Ohm's law

V = IR

Substituting the values in the formula above, we get,

0.6 = I x 2

⇒ I = $\frac{0.6}{2}$ = 0.3 A

Thus, current through 2 Ω resistor = 0.3 A

Current through 3 Ω = ?

From Ohm's law

V = IR

Substituting the values in the formula above, we get,

0.6 = I x 3 ⇒ I = $\frac{0.6}{3}$ = 0.2 A

Hence, current through 3 Ω resistor = 0.2 A

(a) Given,

• R = 1 Ω
• V = 2 V
• Current through 1 Ω = ?

From Ohm's law

V = IR

Substituting the values in the formula above, we get,

2 = I x 1

⇒ I = 2 A

So, current through Resistor A = 2 A

(b) Given,

• R = 2 Ω
• V = 2 V
• Current through 2 Ω = ?

From Ohm's law

V = IR

Substituting the values in the formula above, we get,

2 = I x 2
⇒ I = $\frac{2}{2}$ = 1 A

Hence, current through Resistor B = 1 A

(a) Given,

• V = 4 V
• I = 0.4 A

From Ohm's law

V = IR

4 = 0.4 x R

⇒ R = $\frac{4}{0.4}$ = 10 Ω

Hence, total resistance of the circuit = 10 Ω

(b) As total resistance of the circuit is equal to 10 Ω and the two resistors R and 20 Ω are connected in parallel, hence we get,

$$\begin{aligned}\frac{1}{10} = \frac{1}{\text R} + \frac{1}{20} \\[1em] \Rightarrow \frac{1}{\text R} = \frac{1}{10} - \frac{1}{20} \\[1em] = \frac{2 - 1}{20} \\[1em] = \frac{1}{20} \\[1em] \text R = 20\ \text Ω\end{aligned}$$

Hence, the value of R = 20 Ω

(c) current flowing in R = ?

• V = 4 V
• R = 20 Ω
• I = ?

From Ohm's law

V = IR

⇒ 4 = I x 20

⇒ I = $\frac{4}{20} = \frac{1}{5}$ = 0.2 A

Hence, the current flowing in R = 0.2 A

(a) Resistance of 1 m of wire = 3 Ω.

Resistance of 1.5 m of wire = 3 x 1.5 = 4.5 Ω. As three such wires are joined in parallel and if the equivalent resistance of this part is R_p then

$$\begin{aligned}\frac{1}{\text R_\text p} = \frac{1}{4.5} + \frac{1}{4.5} + \frac{1}{4.5} \\[1em] = \frac{3}{4.5} \\[1em] \Rightarrow \text R_\text p = \frac{4.5}{3} \\[1em] = 1.5\ \text Ω\end{aligned}$$

Hence, total resistance of circuit = 1.5 Ω

(b) I = 2 A

From Ohm's Law

V = IR

Substituting the values in the formula above we get,

V = 2 x 4.5 = 9 V

Hence, potential difference = 9 V

(c) R = 3 Ω for 1 meter wire

Therefore, for 5 m

R = 3 x 5 = 15 Ω

Here the area is twice and resistance is inversely proportional to area.

Thus, resistance becomes half

R = $\frac{15}{2}$ = 7.5 Ω

Hence, resistance = 7.5 Ω

(i) Given,

Two resistors each of 2 Ω connected in parallel. If the equivalent resistance of this part is R_p then

$$\begin{aligned}\frac{1}{\text R_\text p} = \frac{1}{2} + \frac{1}{2} \\[1em] = \frac{2}{2} \\[1em] \Rightarrow \text R_\text p = 1\ \text Ω\end{aligned}$$

So, equivalent resistance R_p = 1 Ω

Given,

• I = 1.2 A

From relation,

ε = I(R + r)

= 1.2 x (1 + r)

= 1.2 + 1.2r    [Equation 1]

When the resistors are connected in series, it supplies a current of 0.4 A,

If the equivalent resistance of this part is R_s then

R_s = 2 + 2 = 4 Ω

Thus, equivalent resistance R_s = 4 Ω

Given,

• I = 0.4 A

From relation,

ε = I(R + r)

= 0.4 x (4 + r)

= 1.6 + 0.4r    [Equation 2]

Equating 1 and 2, we get,

1.2 + 1.2r = 1.6 + 0.4r

⇒ 1.2r – 0.4r = 1.6 - 1.2

⇒ 0.8r = 0.4

⇒ r = $\frac{0.4}{0.8}$

⇒ r = 0.5 Ω

Hence, internal resistance r = 0.5 Ω

(ii) Substituting the value in equation 1 we get,

ε = 1.2(1 + r)
= 1.2 x (1 + 0.5)
= 1.2 x 1.5
= 1.8 V

Hence, e.m.f. of the cell = 1.8 V

(a) Given,

• e.m.f. = 16 V
• Internal resistance = 2 Ω
• Current through battery = ?

If R_p is the equivalent resistance of resistors 3 Ω and 6 Ω connected in parallel, then

$$\begin{aligned}\frac{1}{\text R_\text p} = \frac{1}{3} + \frac{1}{6} \\[1em] = \frac{2 + 1}{6} \\[1em] = \frac{3}{6} \\[1em] = \frac{1}{2} \\[1em] \Rightarrow \text R_\text p = 2 Ω\end{aligned}$$

From relation,

ε = I(R + r)

Substituting the value in the formula above we get,

16 = I(2 + 2)

⇒ 16 = I x 4

⇒ I = $\frac{16}{4}$ = 4 A

Hence, current through the battery = 4 A

(b) Potential difference between the terminals of the battery = ?

Using Ohm's law

V = IR

• R = 2 Ω
• I = 4 A

Substituting the values in the formula above we get,

V = 4 x 2 = 8 V

Hence, potential difference between the terminals of the battery = 8 V

(c) Current in 3 Ω resistor = ?

Using Ohm's law

V = IR

• R = 3 Ω
• V = 8 V
• I = ?

Substituting the values in the formula above we get,

8 = I × 3

I = $\frac{8}{3}$

⇒ I = 2.66 A

Hence, current in 3 Ω resistor is 2.66 A

(d) Current in 6 Ω resistor = ?

Using Ohm's law

V = IR

• R = 6 Ω
• V = 8 V
• I = ?

Substituting the values in the formula above we get,

8 = I × 6
⇒ I = $\frac{8}{6} = \frac{4}{3}$ = 1.333 A

Hence, current in 6 Ω resistor is 1.34 A

(a) Given,

• R = 4 Ω
• I = 0.25 A
• The p.d. across the 4 Ω resistor (V) = ?

Using Ohm's law

V = IR

Substituting the values in the formula above we get,

V = 0.25 x 4 = 1 V

Hence, the p.d. across the 4 Ω resistor (V) = 1 V

(b) Given,

• Internal resistance = 3 Ω
• I = 0.25 A
• The p.d. across the internal resistance of the cell = ?

Using Ohm's law

V = IR

Substituting the values in the formula above we get,

V = 0.25 x 3 = 0.75 V

Hence, the p.d. across the internal resistance (V) = 0.75 V

(c) Potential difference across R Ω or 2 Ω

V = V_net - V_{across 4 Ω} - V_{across 3 Ω}

Hence, we get,

V = 2 - 1 - 0.75 = 0.25 V

(d) The p.d. across resistor of R Ω = 0.25 V

Let the equivalent resistance of the resistors of 2 Ω and R Ω connected in parallel be R'_p

$$\begin{aligned}\frac{1}{\text R'_\text p} = \frac{1}{\text R} + \frac{1}{2} \\[1em] \Rightarrow \frac{1}{\text R'_\text p} = \frac{2 + \text R}{2\text R} \\[1em] \Rightarrow \text R'_\text p = \frac{2\text R}{2 + \text R}\end{aligned}$$

Using Ohm's law

V = IR

0.25 = 0.25 x R'_p

Substituting the value of R'_p from above:

$$\begin{aligned}0.25 = 0.25 \times \Big(\frac{2\text R}{2 + \text R}\Big) \\[1em] \Rightarrow \frac{2\text R}{2 + \text R} = 1 \\[1em] \Rightarrow 2\text R = 2 + \text R \\[1em] \Rightarrow 2\text R - \text R = 2 \\[1em] \Rightarrow \text R = 2\ \text Ω\end{aligned}$$

Hence, value of R = 2 Ω

(a) In the circuit, there are two parts. In the first part, resistors of 2.0 and 4.0 Ω are connected in series. If the equivalent resistance of this part is R_s then

R_s = 2 + 4 = 6 Ω

In the second part, R_s = 6.0 and resistor of 6.0 Ω are connected in parallel. If the equivalent resistance of this part is R_p then

$$\begin{aligned}\frac{1}{\text R_\text p} = \frac{1}{6} + \frac{1}{6} \\[1em] = \frac{1 + 1}{6} \\[1em] = \frac{2}{6} \\[1em] \Rightarrow \text R_\text p = \frac{6}{2} \\[1em] = 3.0\ \text Ω\end{aligned}$$

Hence, the effective resistance of the circuit = 3 Ω

(b) The reading of ammeter = ?

• R = 3 Ω
• V = 6.0 V

Using Ohm's law,

V = IR

Substituting the values in the formula above we get,

6 = I x 3

⇒ I = $\frac{6}{3}$ = 2 A

Hence, the reading of ammeter = 2 A

(a) In the circuit, there are three parts. In the first part, resistors of 10 Ω and 40 Ω are connected in parallel. If the equivalent resistance of this part is R'_p then

$$\begin{aligned}\frac{1}{\text R'_\text p} = \frac{1}{10} + \frac{1}{40} \\[1em] = \frac{4 + 1}{40} \\[1em] = \frac{5}{40} \\[1em] \Rightarrow \text R'_\text p = \frac{40}{5} \\[1em] = 8\ \text Ω\end{aligned}$$

In the second part, resistors of 30 Ω, 20 Ω and 60 Ω are connected in parallel. If the equivalent resistance of this part is R''_p then

$$\begin{aligned}\frac{1}{\text R''_\text p} = \frac{1}{30} + \frac{1}{20} + \frac{1}{60} \\[1em] = \frac{2 + 3 + 1}{60} \\[1em] = \frac{6}{60} \\[1em] \Rightarrow \text R''_\text p = \frac{60}{6} \\[1em] = 10\ \text Ω\end{aligned}$$

In the third part, resistors R'_p and R''_p are connected in series. If the equivalent resistance of this part is R_s then

$$\begin{aligned}\text R_\text s = 8 + 10 \\[1em] = 18\ \text Ω\end{aligned}$$

Hence, the total resistance of the circuit = 18 Ω

(b) Given,

• E.m.f. = 1.8 V
• Effective resistance of the circuit = 18 Ω
• I = ?

From Ohm's law

V = IR

Substituting the values in the formula above, we get,

1.8 = I x 18

I = $\frac{1.8}{18}$

⇒ I = 0.1 A

Hence, the reading of ammeter = 0.1 A

(a) Given,

• e.m.f. = 2 V
• I = ?

In the circuit, there are two parts. In the first part, resistors of 4.5 Ω and 9 Ω are connected in parallel. If the equivalent resistance of this part is R_p then

$$\begin{aligned}\frac{1}{\text R_\text p} = \frac{1}{4.5} + \frac{1}{9} \\[1em] = \frac{9 + 4.5}{4.5 \times 9} \\[1em] = \frac{13.5}{40.5} \\[1em] \Rightarrow \text R_\text p = \frac{40.5}{13.5} \\[1em] = 3\ \text Ω\end{aligned}$$

In the second part, 1.2 Ω, 0.8 Ω and R_p = 3 Ω are connected in series. If the equivalent resistance of this part is R_s then

$$\begin{aligned}\text R_\text s = 1.2 + 0.8 + 3 \\[1em] = 5\ \text Ω\end{aligned}$$

So, the effective resistance of the circuit = 5 Ω

Using Ohm's law,

V = IR

2 = I x 5

⇒ I = $\frac{2}{5}$

⇒ I = 0.4 A

Hence, the reading of the ammeter = 0.4 A

(b) The potential difference across the ends of the cells = ?

• ε = 2 V
• I = 0.4 A
• r = 1.2 Ω

From relation,

Voltage (V) = ε – Ir

Substituting the values in the formula we get,

V = 2 - (0.4 x 1.2)

⇒ V = 2 - 0.48

⇒ V = 1.52 V

Hence, potential difference across the terminals of the cell = 1.52 V

(c) The potential difference across the 4.5 Ω resistor = ?

Current flowing is I = 0.4 A.

Now the current I divides in two parts.

Let the current in 4.5 Ω resistor be I₁ and in 9 Ω resistor be I₂.

So,

I = I₁ + I₂

And

I₁ x 4.5 = I₂ x 9

On solving,

$$\begin{aligned}\text I_1 = \frac{9}{4.5 + 9} \times \text I \\[1em] = \frac{9}{4.5 + 9} \times 0.4 \\[1em] = 0.2667\ \text A\end{aligned}$$

$$\begin{aligned}\text {P.d. across the 4.5 Ω resistor} = \text I_1 \times 4.5 \\[1em] = 0.2667 \times 4.5 \\[1em] = 1.2\ \text V\end{aligned}$$

Alternate Method:

V_4.5Ω = V_cell - V_ammeter

$$\begin{aligned}\text V_{4.5Ω} = 1.52 - \text {IR} \\[1em] = 1.52 - (0.4 \times 0.8) \\[1em] = 1.52 - 0.32 \\[1em] = 1.2\ \text V\end{aligned}$$

Hence, p.d. across the 4.5 Ω resistor = 1.2 V

• Emf of the battery (ε) = 12 V
• Internal resistance of the battery (r) = 5.0 x 10^-2 Ω
• Current drawn (I) = 90 A

As, terminal voltage (V) of the battery is given by,

$$\begin{aligned}\text V = \text ε - \text {Ir} \\[1em] = 12 - 5 \times 10^{-2} \times 90 \\[1em] = 12 - 450 \times 10^{-2} \\[1em] = 12 - 4.5 \\[1em] = 12 - 4.5 \\[1em] = 7.5\ \text V\end{aligned}$$

Hence, the teminal voltage of battery is 7.5 V.

All of the above

Reason — We know,

• W = QV

As, Q = It

Hence, substituting we get

• W = VIt

And By Ohm's Law : V = IR

Hence, substituting we get

• W = (IR)It = I²Rt

3600 J

Reason — 1 watt-hour = 1 watt x 1 hour = 1 W x (60 x 60 s) = 3600 J

Hence, 1 WH = 3600 J

both the resistance of the wire and the duration for which current flows

Reason — The amount of heat produced in a wire on passing current through it, depends on the following three factors.

1. The amount of current passing through the wire — The amount of heat H produced in a wire is directly proportional to the square of current I passing through the wire, i.e., H ∝ I²
2. The resistance of wire — The amount of heat H produced in the wire is directly proportional to the resistance R of the wire, i.e., H ∝ R.
3. The time for which current is passed in the wire — The amount of heat H produced in a wire is directly proportional to the time t for which current is passed in the wire i.e., H ∝ t

144 Ω

Reason — Resistance of an electrical appliance (rating 100 W, 120 V) is :

$\text R_A = \frac{\text V^2}{\text P_\text A}$

Substituting the values in the formula above we get,

$$\begin{aligned}\text R_\text A = \frac{120^2}{100} \\[1em] \Rightarrow R_A = 144\ \text Ω\end{aligned}$$

Hence, the resistance of an electrical appliance is 144 Ω.

nearly 25 W

Reason — Given,

• Power (P) = 100 W
• Voltage (V) = 220 V

We know that,

Power (P) = VI

Substituting the values in the formula above, we get,

100 = 220 x I

⇒ I = $\frac{100}{220}$

⇒ I = 0.45 A

Therefore, current through the lamp = 0.45 A

Power consumed when voltage is 110 V = ?

$\text {Resistance of lamp (R)} = \frac{\text V^2}{\text P}$

Substituting the values in the formula, we get,

$$\begin{aligned}\text R = \frac{220^2}{100} \\[1em] = 484\ \text Ω\end{aligned}$$

From relation,

$\text P = \frac{\text V^2}{\text R}$

Substituting the values in the formula, we get,

$$\begin{aligned}\text P = \frac{110^2}{484} \\[1em] = 25\ \text W\end{aligned}$$

Hence, the power consumed is 25 W.

The brightness of bulb A increases but that of C decreases.

Reason — When switch S is open, bulb B is out of the circuit, so bulbs A and C are in series. When the switch is closed, bulbs B and C form a parallel combination. This reduces the equivalent resistance of that part of the circuit, so the total resistance of the circuit decreases.

As a result, the current in the main circuit increases. Since bulb A is in series with the whole circuit, it gets more current and glows brighter.

But after closing the switch, the current in the right-hand part divides between bulbs B and C. Therefore, the current through bulb C decreases, so its brightness decreases.

Hence, the brightness of bulb A increases while that of bulb C decreases.

W₁ < W₂ < W₃

Reason — The bulbs B₁ and B₂ are connected in series in the upper branch, while bulb B₃ is connected in a parallel branch directly across the 240 V supply.

For bulbs of the same voltage rating,

$\text R=\frac{\text V^2}{\text P}$

So, resistance is inversely proportional to power.

Thus,

• 100 W bulb has the least resistance
• 60 W bulb has greater resistance
• 40 W bulb has the greatest resistance

Hence,

R₁ < R₂ < R₃

Since B₁ and B₂ are in series, the same current flows through both. Power in a series circuit is given by:

P=I²R

Therefore, the bulb with greater resistance in that branch will consume more power.

So,

W₂ > W₁

Bulb B₃ is connected directly across the mains, so it gets the full 240 V and works at its rated power, i.e. 40 W.

But B₁ and B₂, being in series, share the voltage, so each consumes less power than B₃.

Therefore, the output powers are in the order : W₁ < W₂ < W₃.

assertion is true but reason is false.

Explanation

In a series circuit, the same current flows through all the bulbs.

For bulbs of the same voltage rating,

$\text R=\frac{\text V^2}{\text P}$

So, the bulb of lower power rating has higher resistance.

Now, the power consumed in a series circuit is:

P=I²R

Since the current I is the same through each bulb, the bulb having greater resistance consumes more power and therefore glows brighter.

Hence, the lower power bulb emits more light than the higher power bulb in series. So, the Assertion is true.

The Reason is false because the lower power bulb does not get more current. In series combination, current is the same through every bulb.

Hence, assertion is true but reason is false.

both A and R are true and R is the correct explanation of A

Explanation

Assertion (A) is true because in a parallel combination, each appliance operates independently and draws its own current. The total current supplied by the source is the sum of currents in all branches. Since total power is given by P = VI, the total power consumed becomes the sum of the individual powers of all appliances.

Reason (R) is true because in a parallel circuit, each appliance is connected directly across the source, so the same potential difference (voltage) is applied across each appliance.

Hence, both A and R are true and R is the correct explanation of A.

The expression for electrical energy spent in flow of current through an electrical appliance in terms of current, resistance and time is:

W = I²Rt

(a) Expression for electrical power spent in flow of current through a conductor in terms of resistance and potential difference is —

$\text W = \frac{\text V^2}{\text R}$

(b) Expression for electrical power spent in flow of current through a conductor in terms of current and resistance is —

$\text W = \text I^2\text R$

The S.I. unit of electrical power is volt x ampere (VA) or watt (W) or J s^-1

One watt is the electric power consumed when a current of 1 ampere flows through a circuit having a potential difference of 1 volt.

(i) The physical quantity measured in kW is electrical power.

(ii) The physical quantity measured in kWh is electrical energy. kWh is commercial unit of electrical energy.

(iii) The physical quantity measured in Wh is electrical energy. Wh is commercial unit of electrical energy.

$$\begin{aligned}\text{(a) 1 kWh} = \frac{1 \text{ volt} \times 1 \text{ ampere} \times \boldsymbol{1 \space hour} }{1000} \\[1em] \text{(b) 1 kWh} = \boldsymbol{3.6 \times 10^6 J}\end{aligned}$$

For circuit (I) :

• Net Resistance (R) = 3 Ω
• Voltage (V) = 6 V

$\text {Heat Dissipated} = \text P = \frac{\text V^2}{\text R} = \frac{6\times6}{3} =\frac{36}{3} = 12\ \text W$ .......... (1)

For circuit (II) :

As resistances are in parallel combination and also have equal value then

$\frac{1}{\text R} = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$

So, net resistance (R) = $\frac{3}{2}$ Ω

Voltage (V) = 6V

Heat Dissipated = P = $\frac{\text V^2}{R} = \frac{6\times6}{\frac{3}{2}} =\frac{36\times2}{3} = 12\times2=24$ W .......... (2)

For circuit (III) :

As resistances are in series combination then

$\text R = 2 + 3 = 5\ \text Ω$

So, net resistance (R) = 5 Ω

Voltage (V) = 6V

Heat Dissipated = P = $\frac{\text V^2}{R} = \frac{6\times6}{5} =\frac{36}{5} = 7.2$ W .......... (3)

From (1), (2) and (3) it is clearly visible that in circuit (iii) heat dissipation is minimum.

(a) In the expression,

$\text P = \frac{\text Q \times \text V}{\text {time}}$

Symbol Q represents charge and the symbol V represents voltage.

(b) The expression for power P in terms of current and resistance is:

$\text P = \text I^2\text R$

where,

• I represents current and
• R represents resistance.

The S.I. unit of electrical energy is Joule.

The relation between Wh and Joule is as follows —

1 Wh = 3600 J

The statement 'the power of an appliance is 100 W' means that 100 J of electrical energy is consumed by the appliance in 1 second.

(i) The household unit of electricity is Kilowatt hour (kWh).

One kilowatt-hour (kWh) is the electrical energy consumed by an electrical appliance of power 1 kilowatt when it is used for 1 hour.

(ii) The voltage of the electricity that is generally supplied to a house is 220 volt.

(iii) The electrical energy is consumed by various appliances in our houses (or industries) and it's cost is paid to the electrical company.

One kilowatt-hour (kwh) is the electrical energy consumed by an electrical appliance of power 1 kilowatt when it is used for 1 hour.

Value of kilowatt-hour (kWh) in S.I. unit is —

1 kWh = 3.6 × 10⁶ J

Kilowatt is the unit of electrical power. i.e., one kilowatt is the electric power consumed when a current of 1 ampere flows through a circuit having a potential difference of 1 volt.

Kilowatt-hour is the unit of electrical energy i.e., One kilowatt - hour (kwh) is the electrical energy consumed by an electrical appliance of power 1 kilowatt when it is used for 1 hour.

An electrical appliance such as electric heater, geyser etc is rated with it's power and voltage.

For example, an electric bulb is rated as 100 W - 220 V. It means that if the bulb is lighted on a 220 V supply, the electric power consumed by it is 100 W (i.e., 100 J of electric energy is consumed by the bulb in 1 s).

(a) Resistance of element of appliance while in use is —

$\text R = \frac{\text V^2}{\text P}$

(b) The safe limit of current in it, while in use is —

$\text I = \frac{\text P}{\text V}$

If current exceeds this value, the power supplied at voltage V will exceed the rated power of the appliance and the appliance may get damaged. So this value of current is called the safe current which can flow through the appliance at voltage V.

When an electric bulb is rated '100 W, 250 V' it means that if the bulb is lighted on a 250 V supply, the electric power consumed by it is 100 W (i.e., 100 J of electrical energy is consumed by the bulb in 1 s).

(i) Energy consumed by fluorescent tube in one day = 0.24 kWh

Hence, energy consumed in 30 days = 0.24 x 30 = 7.2 kWh.

(ii) Energy consumed by television set in one day = 0.36 kWh

Hence, energy consumed in 30 days = 0.36 x 30 = 10.8 kWh.

(iii) Energy consumed by room heater in one day = 2 kWh

Hence, energy consumed in 30 days = 2 x 30 = 60 kWh.

Resistance of lamp A (rating 220 V, 50 W) is —

$\text R_\text A = \frac{\text V^2}{\text P_\text A}$

Substituting the values in the formula above we get,

$$\begin{aligned}\text R_\text A = \frac{220^2}{50} \\[1em] = 968\ \text Ω\end{aligned}$$

Resistance of lamp B (rating 220 V, 100 W) is —

$\text R_\text B = \frac{\text V^2}{\text P_\text B}$

Substituting the values in the formula above we get,

$$\begin{aligned}\text R_\text B = \frac{220^2}{100} \\[1em] = 484\ \text Ω\end{aligned}$$

From formula, we get, P = I² R

As we observe that R_A > R_B and same current is flowing in both the lamps as both are connected in series therefore P_A > P_B.

Hence, 50 W lamp consumes more power than 100 W lamp.

The amount of heat produced in a wire on passing current through it, depends on the following three factors.

1. The amount of current passing through the wire — The amount of heat H produced in a wire is directly proportional to the square of current I passing through the wire, i.e., H ∝ I²
2. The resistance of wire — The amount of heat H produced in the wire is directly proportional to the resistance R of the wire, i.e., H ∝ R.
3. The time for which current is passed in the wire — The amount of heat H produced in a wire is directly proportional to the time t for which current is passed in the wire i.e., H ∝ t

(a) Given,

• Resistance (R) = 500 Ω
• Current (I) = 0.4 A
• Power = ?

From Ohm's law,

V = IR

Substituting the values in the formula above, we get,

V = 0.4 x 500 = 200 V

Power (P) = VI

Substituting the values in the power formula we get,

P = 200 x 0.4 = 80 W

Hence, the power of bulb = 80 W

(b) The potential difference at its end is 200 V.

(a) Given,

• Current (I) = 3 A
• Resistance (R) = 75 Ω
• Time (t) = 2 min = 120 s
• Heat produced = ?

Using

H = I²Rt

Substituting the values in the formula above, we get,

H = 3² x 75 x 120 = 81000 J

Hence, heat energy produced = 81000 J

(b) Charge passed through the resistance = ?

From relation,

Q = It

Substituting the values in the formula above we get,

Q = 3 x 120 = 360 C

Hence, charge passed through the resistance = 360 C

Given,

• Power (P) = 60 W
• Voltage (V) = 250 V

We know that,

Power (P) = VI

Substituting the values in the formula above, we get,

60 = 250 x I

⇒ I = $\frac{60}{250}$

⇒ I = 0.24 A

So, current through the lamp = 0.24 A

Power consumed when voltage is 200 V = ?

Resistance of lamp (R) = $\frac{\textbf V^2}{\textbf P}$

Substituting the values in the formula, we get,

$$\begin{aligned}\text R = \frac{250^2}{60} \\[1em] = 1041.6\ \text Ω\end{aligned}$$

From relation,

P = $\frac{\textbf V^2}{\textbf R}$

Substituting the values in the formula, we get,

$$\begin{aligned}\text P = \frac{200^2}{1041.6} \\[1em] = 38.4\ \text W\end{aligned}$$

Hence, the power consumed reduces to 38.4 W.

Given,

• Power (P) = 100 W
• Voltage (V) = 250 V

From relation,

Power (P) = VI

Substituting the values in the formula above, we get,

100 = 250 x I

⇒ I = $\frac{100}{250}$

⇒ I = 0.4 A

Hence, current drawn = 0.4 A

(a) Given,

• P = 100 W
• V = 220 volt

We know that,

Power P = $\frac{\textbf V^2}{\textbf R}$

Substituting the values in the formula above, we get,

$$\begin{aligned}100 = \frac{220^2}{\text R} \\[1em] \Rightarrow \text R = \frac{220^2}{100} \\[1em] = 484\ \text Ω\end{aligned}$$

Hence, resistance of electric bulb = 484 Ω

(b) From relation,

P = VI

Safe current (I) = $\frac{\text P}{\text V}$

Substituting the value we get,

I = $\frac{100}{220}$

⇒ I = 0.45 A

Hence, safe current limit = 0.45 A

Given,

• Power (P) = 60 W
• time (t) = 12.5 h for 30 days

As energy consumed is given by,

E = P × t

= 60 x 12.5

= 750 Wh

Therefore, the electrical energy consumed in one day is 750 Wh.

Electrical energy consumed in 30 days = 750 x 30 = 22500 Wh = 22.5 kWh

Hence, the electrical energy consumed in 30 days is 22.5 kWh.

Given,

• Power (P) = 750 W
• time (t) = 16 h

As energy consumed is given by,

E = P × t

= 750 x 16

= 12000 Wh

= 12 kWh

Hence, the electrical energy consumed in 16 hours is 12 kWh.

(i) Given,

• Resistance (R) = 200 Ω
• Voltage (V) = 200 volt
• Time (t) = 5 min = 300 sec

As we know,

$\textbf {Energy (E)} = \frac{\textbf V^2\textbf t}{\textbf R}$

Substituting the values in the formula above we get,

$\text E = \frac{200^2 \times 300}{200}$ = 60,000 J

Hence, energy consumed = 60,000 J

(ii) In kWh = ?

As we know,

1 kWh = 3.6 x 10⁶ J

⇒ 1 J = $\frac{1}{3.6 \times 10^6}$ kWh

⇒ 60,000 J = $\frac{60,000}{3.6 \times 10^6}$ kWh = 0.0167 kWh

Hence, energy consumed = 0.0167 kWh

(i) Given,

• Voltage (V) = 12 V
• Power (P) = 24 W
• Current (I) = ?

From relation,

Power (P) = VI

Substituting the values in the formula above, we get,

24 = 12 x I

⇒ I = $\frac{24}{12}$

⇒ I = 2 A

Hence, the current flowing through it is 2 A.

(ii) Energy (E) = P × t

• Time (t) = 20 min = 1200 sec

E = 24 x 1200 = 28,800 J

Hence, energy consumed = 28,800 J

(i) Given,

• Current (I) = 0.2 A
• Potential difference (V) = 20 V

From Ohm's law

V = IR

Substituting the values in the formula above, we get,

20 = 0.2 x R

⇒ R = $\frac{20}{0.2} = \frac{200}{2}$ = 100 Ω

Hence, the resistance of the wire = 100 Ω

(ii) Given,

• Time (t) = 60 sec
• heat energy produced in 1 minute = ?

From relation,

Heat energy (H) = I²Rt

Substituting the values in the formula above, we get,

H = 0.2² x 100 x 60 = 240 J

Hence, the heat energy produced in 1 minute is 240 joules.

Given,

• Voltage (V) = 240 V
• Power (P) = 60 W
• Resistance (R) = ?

From relation

$\textbf P = \frac{\textbf V^2}{\textbf R}$

We get,

$$\begin{aligned}60 = \frac{240^2}{\text R} \\[1em] \Rightarrow \text R = \frac{240^2}{60} \\[1em] = 960\ \text Ω\end{aligned}$$

So, resistance of the electric lamp = 960 Ω

From the relation for current through the element, we know,

$\textbf I = \frac{\textbf P}{\textbf V}$

Substituting the values in the formula above, we get,

$$\begin{aligned}\text I = \frac{60}{240} \\[1em] = 0.25\ \text A\end{aligned}$$

When one lamp is connected across the mains, it draws 0.25 A current. If two such lamps are connected in series across the mains, current through each bulb becomes —

$$\begin{aligned}\frac{240 \text V}{(960 + 960) \text Ω} \\[1em] = 0.125\ \text A\end{aligned}$$

i.e., current is halved.

Hence, heating (= I²Rt) in each bulb becomes one fourth, so each bulb appears less bright.

Given,

• Voltage (V₁) = 220 V
• Voltage (V₂) = 110 V
• P = 60 W (both bulbs have same power)
• Resistance (R) = ?

From relation

$\textbf P = \frac{\textbf V^2}{\textbf R}$

Case 1 :

$$\begin{aligned}60 = \frac{220^2}{\text R_1} \\[1em] \Rightarrow \text R_1 = \frac{220^2}{60}\ \text Ω\end{aligned}$$

Case 2 :

$$\begin{aligned}60 = \frac{110^2}{\text R_2} \\[1em] \Rightarrow \text R_2 = \frac{110^2}{60}\ \text Ω\end{aligned}$$

Ratio between the two is —

$$\begin{aligned}\text R_1 : \text R_2 = \frac{220^2}{60} : \frac{110^2}{60} \\[1em] = \frac{\frac{220^2}{60}}{\frac{110^2}{60}} \\[1em] = \frac{\frac{220^2}{\cancel{60}}}{\frac{110^2}{\cancel{60}}} \\[1em] = \frac{220^2}{110^2} \\[1em] = \frac{2^2}{1^2} \\[1em] = \frac{4}{1}\end{aligned}$$

Hence, the ratio of resistances of the two bulbs is 4 : 1.

(i) Given,

• Power (P) = 250 W
• Voltage (V) = 230 V
• Time (t) = 1 h = 3600 sec
• Energy (E) = ?

From relation,

E = P × t

Substituting the values in the formula above, we get,

E = 250 x 3600 = 9 x 10⁵ J

Hence, energy consumed in one hour = 9 x 10⁵ J

(ii) Let t be the time in which the bulb will consume 1.0 kWh energy when connected to 230 V mains.

P = 1.0 kWh = 1000 Wh

Substituting the value in the formula, we get,

1000 = 250 x t

⇒ t = $\frac{1000}{250}$ = 4 h

Hence, time = 4 h

(i) Given,

• Power (P) = 250 W
• Voltage (V) = 100 V
• Current through each heater (I) = ?

As

P = VI

Substituting the values in the formula above we get,

250 = 100 x I

⇒ I = 250 / 100 = 2.5 A

So, current through each heater = 2.5 A

And

Current for three heaters = 3 x 2.5 = 7.5 A

Hence, total current taken from supply = 7.5 A

(ii) Resistance for each heater (R) = ?

Since,

V = IR

Substituting the values in the formula above we get,

100 = 2.5 x R

⇒ R = $\frac{100}{2.5}$ = 40 Ω

Hence, resistance of each heater = 40 Ω

(iii) Energy supplied to three heater in 5 h = ?

Energy (E) = P × t

Substituting the values in the formula above, we get,

E = 250 x 5 = 1250 Wh = 1.25 kWh

Now,

Energy for three heaters = 3 × 1.25 = 3.75 kWh

Hence, energy supplied to the three heaters = 3.75 kWh

(i) Given,

• Voltage (V) = 4 V
• Internal resistance = 2.5 Ω
• Current (I) = 0.5 A
• time (t) = 10 min = 600 sec

Energy supplied by the battery (E) = V x I x t

Substituting the values in the formula, we get,

E = 4 x 0.5 x 600 = 1200 J

Hence, total energy supplied = 1200 J

(ii) Resistance of the bulb = ?

From relation,

V = I (R + r)

We get,

4 = 0.5 (R + 2.5)
⇒ 4 = (0.5 R) + (0.5 x 2.5)
⇒ 4 = (0.5 R) + 1.25
⇒ 0.5R = 4 - 1.25
⇒ R = $\frac{2.75}{0.5}$ = 5.5 Ω

Hence, resistance of the bulb = 5.5 Ω

(iii) Energy dissipated in the bulb in 10 min = ?

From relation

E = I²Rt

Substituting the values in the formula above, we get,

E = 0.5 x 0.5 x 5.5 x 600 = 825 J

Hence, energy dissipated = 825 J

(i) Given,

• Resistance, R_A = 4 Ω
• Resistance, R_B = 6 Ω

Let equivalent resistance of the two resistors connected in parallel be R_P

$$\begin{aligned}\frac{1}{\text R_\text P} = \frac{1}{4} + \frac{1}{6} \\[1em] = \frac{3 + 2}{12} \\[1em] = \frac{5}{12} \\[1em] \text R_\text P = \frac{12}{5} \\[1em] = 2.4\ \text Ω\end{aligned}$$

Thus, equivalent resistance = 2.4 Ω

From relation

$\textbf P = \frac{\textbf V^2}{\textbf R}$

We get,

$$\begin{aligned}\text P = \frac{6^2}{2.4} \\[1em] = 15\ \text W\end{aligned}$$

Hence, power supplied = 15 W

(ii) Power dissipation across each resistor = ?

From relation,

P = VI

Current across A (4 Ω) resistor —

I_A = $\frac{6}{4} = \frac{3}{2}$ = 1.5 A

Power dissipation across A = VI_A

Substituting we get,

P = 6 x 1.5 = 9 W

Similarly,

Current across B (6 Ω) resistor —

I_B = $\frac{6}{6}$ = 1 A

Power dissipation across B resistor = VI_B

Substituting the values in the formula above, we get,

P = 6 x 1 = 6 W

Hence, the power dissipated across A is 9 W and across B is 6 W.

(i) Given,

• e.m.f. (V) = 15 V
• r = 2 Ω
• R₁ = 4 Ω
• R₂ = 6 Ω

Electrical energy = ?

As the battery and resistors are connected in series, equivalent resistance is given by

R = 2 + 4 + 6 = 12 Ω

From Ohm's law —

V = IR

Substituting the values in the formula above, we get,

15 = I x 12

⇒ I = $\frac{15}{12}$ = 1.25 A

So, current in circuit = 1.25 A

Now,

Voltage across 6 Ω = IR = 1.25 × 6 = 7.50 V

Voltage across 6 Ω = 7.5 V

Time (t) = 1 min = 60 sec

We know,

$\textbf E = \frac{\textbf V^2 \times \textbf t}{\textbf R}$

Substituting the values in the formula above, we get,

E = $\frac{7.5^2 \times 60}{6}$

= 7.5² x 10

= 562.5 J

Hence, energy spent across the 6 Ω resistor = 562.5 J

Given,

• V = 220 V
• time (t) = 5 min = 300 sec

From relation

$\textbf P = \frac{\textbf V^2}{\textbf R}$

and

Heat produced (H) —

H = P x t

⇒ H = $\frac{\text{V}^2}{\text{R}}$ x t

Case 1 : For V = 220 V

$\text H = (\frac{220^2}{\text R}) \times 300 \quad \quad \quad \textbf {[Equation 1]}$

Case 2 : For V = 200 V

$\text H = (\frac{200^2}{\text R}) \times \text t \quad \quad \quad \textbf {[Equation 2]}$

Equating 1 and 2 we get,

$$\begin{aligned}(\frac{220^2}{\text R}) \times 300 = (\frac{200^2}{\text R}) \times \text t \\[1em] \Rightarrow \text t = (\frac{220^2}{200^2}) \times 300 \\[1em] = 363\ \text s \\[1em] \Rightarrow \text t = 6.05 \text { min}\end{aligned}$$

Hence, time taken = 6.05 min

Given,

• I = 8 A
• V = 220 V
• t = 2 h
• Rate of electricity = 4.50 per kWh
• Cost = ?

From relation,

E = V x I x t

we get,

E = 220 x 8 x 2 = 3520 Wh = 3.52 kWH

As cost of energy for 1 kWh = ₹ 4.50

Therefore, cost of energy for 3.52 kWh = 4.5 x 3.52 = ₹15.84

Hence, cost of electrical energy = ₹15.84

Given,

• Power (P) = 2 kW
• Voltage (V) = 220 V
• Time (t) = 7 x 3 = 21 h

From relation,

E = P × t

Substituting the values in the formula above, we get,

E = 2 x 21 = 42 kWh

Cost for 1 kWh of energy = ₹ 4.25

∴ Cost for 42 kWh of energy = 4.25 x 42 = ₹ 178.50

(i) Given,

• P = 1500 W
• V = 250 volt

Current drawn (I) = $\frac{\text P}{\text V}$

Substituting the values in the formula, we get,

I = $\frac{1500}{250}$

⇒ I = 6 A

Hence, current drawn = 6 A

(ii) energy consumed in 50 hours = ?

• t = 50 h

From relation,

E = P × t

Substituting the values in the formula above, we get,

E = 1500 x 50 = 75000 Wh = 75 kWh

Hence, energy consumed in 50 hours = 75 kWh

(iii) cost of energy consumed at ₹ 4.20 per kWh = ?

• Cost per unit of energy = ₹ 4.20

Cost for 75 kWh of energy = 4.2 x 7.5 = ₹ 315

Hence, cost of energy consumed = ₹ 315