Electrostatic Potential and Capacitance - Result Question 36
38. A network of four capacitors of capacity equal to $C_1=C, C_2=2 C, C_3=3 C$ and $C_4=4 C$ are connected to a battery as shown in the figure. The ratio of the charges on $C_2$ and $C_4$ is: [2005]
(a) $\frac{4}{7}$
(b) $\frac{3}{22}$
(c) $\frac{7}{4}$
(d) $\frac{22}{3}$
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Answer:
Correct Answer: 38. (b)
Solution:
- (b)
Equivalent capacitance for three capacitors $(C_1, C_2 & C_3)$ in series is given by $\frac{1}{C _{\text{eq. }}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}=\frac{C_2 C_3+C_3 C_1+C_1 C_2}{C_1 C_2 C_3}$
$\Rightarrow C _{\text{eq. }}=\frac{C_1 C_2 C_3}{C_1 C_2+C_2 C_3+C_3 C_1}$
$\Rightarrow C _{\text{eq. }}=\frac{C(2 C)(3 C)}{C(2 C)+(2 C)(3 C)+(3 C) C}=\frac{6}{11} C$
$\Rightarrow$ Charge on capacitors $(C_1, C_2 & C_3)$ in series
$=C _{\text{eq }} V=\frac{6 C}{11} V$
Charge on capacitor $C_4=C_4 V=4 C V$
$\frac{\text{ Charge on } C_2}{\text{ Charge on } C_4}=\frac{\frac{6 C}{11} V}{4 C V}=\frac{6}{11} \times \frac{1}{4}=\frac{3}{22}$
