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(B) First,calculate the resistance of each bulb using $R = V^2 / P$.For bulb $A$: $R_A = (24)^2 / 48 = 576 / 48 = 12\ \Omega$.For bulb $B$: $R_B = (24

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When the switch is open,$A$ delivers $9\ W$ and $B$ delivers $12\ W$ approximately.

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(B) First,calculate the resistance of each bulb using $R = V^2 / P$.For bulb $A$: $R_A = (24)^2 / 48 = 576 / 48 = 12\ \Omega$.For bulb $B$: $R_B = (24)^2 / 36 = 576 / 36 = 16\ \Omega$.Case $1$: Switch is closed.Each bulb is connected to its own $12\ V$ battery. The power delivered is $P = V^2 / R$.$P_A = (12)^2 / 12 = 144 / 12 = 12\ W$.$P_B = (12)^2 / 16 = 144 / 16 = 9\ W$.Case $2$: Switch is open.The two bulbs are in series with a total voltage of $24\ V$.The total resistance is $R_{eq} = R_A + R_B = 12 + 16 = 28\ \Omega$.The current in the circuit is $I = V / R_{eq} = 24 / 28 = 6 / 7\ A$.The power delivered by $A$ is $P_A = I^2 R_A = (6/7)^2 	imes 12 = (36/49) 	imes 12 \approx 8.8\ W \approx 9\ W$.The power delivered by $B$ is $P_B = I^2 R_B = (6/7)^2 	imes 16 = (36/49) 	imes 16 \approx 11.7\ W \approx 12\ W$.Thus,option $B$ is correct.

Two light bulbs in the circuit have ratings $A\,[24\ V, 48\ W]$ and $B\,[24\ V, 36\ W]$ as shown in the figure.

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