Bulb A rated 160 W, 40 V and Bulb B rated 40 W, 40 V are connected as shown in the diagram.

Given,

• Power rating of the bulb A ($\text P_\text A$) = 160 W
• Voltage rating of the bulb A ($\text V_\text A$) = 40 V
• Power rating of the bulb B ($\text P_\text B$) = 40 W
• Voltage rating of the bulb B ($\text V_\text B$) = 40 V
• Supply voltage ($\text V$) = 40 V

(a) Resistance of bulb A is given by,

$$\begin{aligned}\text R_\text A = \frac{\text V_\text A^2}{\text P_\text A} \\[1em] = \frac{40^2}{160} \\[1em] = \frac{1600}{160} \\[1em] = 10\ \text Ω\end{aligned}$$

Similarly,

Resistance of bulb B is given by,

$$\begin{aligned}\text R_\text B = \frac{\text V_\text B^2}{\text P_\text B} \\[1em] = \frac{40^2}{40} \\[1em] = \frac{1600}{40} \\[1em] = 40\ \text Ω\end{aligned}$$

From Ohm's law,

Potential drop (V) = Current (I) x Resistance (R)

Since the bulbs are in series, so the same current flows through both bulbs and the potential difference divides in the ratio of their resistances i.e.,

V₁ : V₂ = $\text R_\text A : \text R_\text B$ = 10 : 40 = 1 : 4

Hence, the ratio V₁ : V₂ = 1 : 4.

(b) False because if bulb A fuses, the circuit becomes open, so no current can flow through the circuit. Hence the current becomes zero, not the same.