Question

Find the charge and potential drop across the capacitor $C_4$ in the given circuit ?

• A. $\frac{40}{7}\mu \text{C},\frac{20}{7}\text{V}$
• B. $\frac{80}{7}\mu \text{C},\frac{40}{7}\text{V}$
• C. $\frac{30}{7}\mu \text{C},\frac{15}{7}\text{V}$
• D. $\frac{20}{7}\mu \text{C},\frac{40}{7}\text{V}$

Answer 1

The correct option is A $\frac{40}{7}\mu \text{C},\frac{20}{7}\text{V}$

We can solve this by potential method, by assigning potential as shown in the figure.


Using $\text{KVL}$ for isolated system $-\text{I}$,

$(V_1-5)4+(V_1-0)2+(V_1-V_2)4=0$

$\Rightarrow 4V_1-20+2V_1+4V_1-4V_2=0$

$\Rightarrow 10V_1-4V_2=20....\left(1\right)$

Using $\text{KVL}$ for isolated system $-\text{II},$

$(V_2-5)2+(V_2-0)4+(V_2-V_1)4=0$

$\Rightarrow 2V_2-10+4V_2+4V_2-4V_1=0$

$\Rightarrow 10V_2-4V_1=10...\left(2\right)$

Solving eqn (1) and (2)

$V_1=\frac{20}{7}\text{V}$ and

$V_2=\frac{15}{7}\text{V}$

Charge and potential drop across capacitor $C_4$ is,

$Q_4=C_4(V_1-0)=2\times (\frac{20}{7}-0)=\frac{40}{7}\mu \text{C}$

$V_4=\frac{Q_4}{C_4}=\frac{40}{\frac{7}{2}}=\frac{20}{7}\text{V}$

Hence, option (a) is the correct answer.