Three capacitors C_{1} = 2 , mu F,C_{2} = 6 , | vedclass

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(C) Let the potential at point $A$ be $V$ and at point $B$ be $0$. Let the potential at point $D$ be $V_{D}$.Since the capacitors $C_{2}$ and $C_{3}$

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$1: 2: 2$

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(C) Let the potential at point $A$ be $V$ and at point $B$ be $0$. Let the potential at point $D$ be $V_{D}$.Since the capacitors $C_{2}$ and $C_{3}$ are in series,the charge on them is the same. Using the principle of conservation of charge for the isolated plate system at node $D$:$(V_{D} - V) C_{2} + (V_{D} - 0) C_{3} = 0$$(V_{D} - V) 6 + (V_{D} - 0) 12 = 0$$6V_{D} - 6V + 12V_{D} = 0$$18V_{D} = 6V \implies V_{D} = \frac{V}{3}$Now,calculate the charges:$q_{1} = C_{1} V = (2 \, \mu F) V = 2V \, \mu C$$q_{2} = C_{2} (V - V_{D}) = 6 \, \mu F (V - \frac{V}{3}) = 6 \, \mu F (\frac{2V}{3}) = 4V \, \mu C$$q_{3} = C_{3} (V_{D} - 0) = 12 \, \mu F (\frac{V}{3} - 0) = 4V \, \mu C$The ratio of charges is $q_{1} : q_{2} : q_{3} = 2V : 4V : 4V = 2 : 4 : 4 = 1 : 2 : 2$.

Three capacitors $C_{1} = 2 \, \mu F$,$C_{2} = 6 \, \mu F$,and $C_{3} = 12 \, \mu F$ are connected as shown in the figure. Find the ratio of the charges on capacitors $C_{1}$,$C_{2}$,and $C_{3}$ respectively.

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