First slide

Capacitance

Question

A network of four capacitors of capacity equal to $C_{1} = C, C_{2} = 2 C, C_{3} = 3 C, C_{4} = 4 C$ are connected to a battery as shown in the Fig. 3.81. The ratio of the charges on $C_{2} and C_{4}$ is :

Easy

A

$\frac{7}{4}$
B

$\frac{22}{3}$
C

$\frac{3}{22}$
D

$\frac{4}{7}$

Solution

$Q_{4} = (4 C) V = 4 CV$
$Q_{2} = (\frac{6}{11} C) V = \frac{6}{11} CV \Rightarrow \frac{Q_{2}}{Q_{4}} = \frac{6 CV}{11} x \frac{1}{4 CV} = \frac{3}{22}$

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