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(D) Let the resistance of each bulb be $R$ and the voltage of each battery be $V$.For circuit $(P)$: Three bulbs are in series with one battery. Total

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$P < R < Q < S$

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(D) Let the resistance of each bulb be $R$ and the voltage of each battery be $V$.For circuit $(P)$: Three bulbs are in series with one battery. Total resistance $= 3R$. Power $P_P = \frac{V^2}{3R} \approx 0.33 \frac{V^2}{R}$.For circuit $(Q)$: Three bulbs are in parallel with one battery. Total resistance $= R/3$. Power $P_Q = \frac{V^2}{R/3} = \frac{3V^2}{R} = 3 \frac{V^2}{R}$.For circuit $(R)$: One bulb is connected to one battery. Total resistance $= R$. Power $P_R = \frac{V^2}{R} = 1 \frac{V^2}{R}$.For circuit $(S)$: One bulb is connected to two batteries in series. Total voltage $= 2V$. Total resistance $= R$. Power $P_S = \frac{(2V)^2}{R} = \frac{4V^2}{R} = 4 \frac{V^2}{R}$.Comparing the values: $0.33 \frac{V^2}{R} Therefore,the order of increasing power consumption is $P < R < Q < S$.

The following figures show different combinations of identical bulb$(s)$ connected to identical battery(ies). Which option is correct regarding the total power dissipated in the circuit?

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