The diagram given below shows two bulbs rated 40 W, 12 V each.

Given,

• Power rating of each bulb (P) = 40 W
• Voltage rating of each bulb (V) = 12 V
• E.m.f. of the battery = 12 V

(a) As,

$$\begin{aligned}\text P = \frac{\text V^2}{\text R} \\[1em] \Rightarrow \text R = \frac{\text V^2}{\text P} = \frac{12^2}{40} = \frac{144}{40} = 3.6\thinspace\text{Ω}\end{aligned}$$

Hence, the resistance of each bulb is 3.6 Ω.

(b) 1. When S₁ is closed and S₂ and S₃ are open :

The current flows from the battery through the first bulb, then through the closed switch S₁ and then through the second bulb. Hence the two bulbs are connected in series across the 12 V battery.

Total resistance (R_s) = 3.6 + 3.6 = 7.2 Ω

Total power consumed = $\frac{\text V^2}{\text R_s} = \frac{12^2}{7.2} = \frac{144}{7.2}$ = 20 W

Hence, the total power consumed by the circuit is 20 W.

2. When S₁ is open and S₂ and S₃ are closed :

The closed switch S₃ provides a path from one terminal of the battery directly to the second bulb, and the closed switch S₂ provides a path from the first bulb directly to the other terminal of the battery. Hence each bulb now gets connected directly across the 12 V battery, i.e., the two bulbs are connected in parallel.

So, total resistance:

$$\begin{aligned}\frac{1}{\text R_p}=\frac{1}{3.6}+ \frac{1}{3.6}\\[1em] \frac{1}{\text R_p} = \frac{2}{3.6}\\[1em] \text R_p= \frac{3.6}{2}\\[1em] \text R_p=1.8\thinspace\text{Ω}\end{aligned}$$

Total power consumed = $\frac{\text V^2}{\text R_p} = \frac{12^2}{1.8} = \frac{144}{1.8}$ = 80 W

Hence, the total power consumed by the circuit is 80 W.