Three identical bulbs B 1 , B 2 and B 3 each of power rating 18 W, 12 V are connected to a battery of 12 V.

Given,

Battery emf (ε) = 12 V

Voltage rating of eac bulb (V) = 12 V

Power rating of eac bulb (P) = 18 W

(a)

1. Let resistance of each bulb be R.

Then,

$$\begin{aligned}\text P = \frac {\text V^2}{\text R} \\[1 em] ⇒ \text R = \frac {\text V^2}{\text P} = \frac {12^2}{18} = \frac {144}{18}\\[1 em] ⇒ \text R = 8\ \text Ω\end{aligned}$$

2. As, B₂ and B₃ are in parallel combination then

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

Here, R₁ = R₂ = 8 Ω

$$\begin{aligned}⇒ \frac{1}{\text R_\text P} = \frac{1}{8} + \frac{1}{8} = \frac{2}{8}\\[1 em] ⇒ \text R_\text P = \frac{8}{2} = 4\ \text Ω\end{aligned}$$

Now this combination is in series with B₁ then

R_S = R_P + R = 4 + 8 = 12 Ω

So,

$$\begin{aligned}\text {Current drawn} = \frac{\text {Battery emf}}{\text {Total resistance}} = \frac{\text ε}{\text R_\text S} \\[1 em] =\frac{12}{12} = 1\ \text A\end{aligned}$$

Hence, current drawn from the battery is 1 A.

(b) When B₃ is removed, the circuit becomes :

Now, the equivalent resistance = 8 + 8 = 16 Ω.

Since, the resistance increases, current in the circuit decreases.

Hence, brightness of B₁ decreases.