Question

The circuit shown is in steady state. Find the charge in capacitor $C_1$.

• A. $20\mu \text{C}$
• B. $30\mu \text{C}$
• C. $40\mu \text{C}$
• D. $10\mu \text{C}$

Answer 1

The correct option is D $10\mu \text{C}$

In steady state, Capacitors in a circuit are open.

So, the given circuit can be simplified as shown below.


Using $\text{KVL}$ in the loop $ABCDA$

$45-5I-30I-10I=0$

$\Rightarrow I=1\text{A}$



Let $V$ be the potential at point $A$

Therefore, Charge present on capacitor $C_1$ is $Q_1=C_1(V-0)$

Charge present on capacitor $C_2$ is $Q_2=C_2(V-10)$

Charge present on capacitor $C_3$ is $Q_3=C_3(V-40)$

Applying $\text{KCL}$ at point $A$ we get, $Q_1+Q_2+Q_3=0$

$\Rightarrow 0.5V-20+2V-60+1.5V=0$

$\Rightarrow V=20\text{volts}$

Thus, $Q_1=0.5\times 20=10\mu \text{C}$

Hence, option (d) is the correct answer.