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(B) Let the resistance of each bulb be $R$.Case $(i)$: All six bulbs are glowing. Section $A$ has $3$ bulbs in parallel,and section $B$ has $3$ bulbs

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$9:4$

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(B) Let the resistance of each bulb be $R$.Case $(i)$: All six bulbs are glowing. Section $A$ has $3$ bulbs in parallel,and section $B$ has $3$ bulbs in parallel. These two sections are in series. The equivalent resistance is $R_{eq_1} = (R/3) + (R/3) = 2R/3$. The power consumed is $P_1 = E^2 / R_{eq_1} = 3E^2 / 2R$.Case $(ii)$: Two bulbs from section $A$ are in parallel (resistance $R/2$) and one bulb from section $B$ is in series with this combination. The equivalent resistance is $R_{eq_2} = R/2 + R = 3R/2$. The power consumed is $P_2 = E^2 / R_{eq_2} = 2E^2 / 3R$.The ratio of power consumption is $P_1 : P_2 = (3E^2 / 2R) : (2E^2 / 3R) = 9 : 4$.

Six similar bulbs are connected as shown in the figure with a $DC$ source of $emf\; E$,and zero internal resistance. The ratio of power consumption by the bulbs when $(i)$ all are glowing and $(ii)$ in the situation when two from section $A$ and one from section $B$ are glowing,will be

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