An electric bulb rated $220 \, V, 60 \, W$ is working at full efficiency. $(i)$ State the resistance of the coil of the bulb. Another identical bulb is connected in series with the first one and the system is connected across the mains as shown below. $(ii)$ State the rate of conversion of energy in each bulb. $(iii)$ Calculate the total power. $(iv)$ What will be the total power if the bulbs are connected in parallel?

(N/A) Given: Rated voltage $V = 220 \, V$,Rated power $P = 60 \, W$.
$(i)$ Using the formula $P = \frac{V^2}{R}$,the resistance $R$ of the bulb is:
$R = \frac{V^2}{P} = \frac{(220)^2}{60} = \frac{48400}{60} \approx 806.67 \, \Omega$.
$(ii)$ When two identical bulbs are connected in series,the total resistance $R_T = R + R = 2R = 2 \times 806.67 = 1613.34 \, \Omega$.
The current $I$ flowing through the circuit is $I = \frac{V}{R_T} = \frac{220}{1613.34} \approx 0.136 \, A$.
The rate of energy conversion (power consumed) in each bulb is $P' = I^2 R = (0.136)^2 \times 806.67 \approx 15 \, W$.
$(iii)$ Total power consumed in series connection is $P_{total} = I^2 R_T = (0.136)^2 \times 1613.34 \approx 30 \, W$.
$(iv)$ If the bulbs are connected in parallel,the total power is $P_{total} = P_1 + P_2 = 60 + 60 = 120 \, W$.