Question

Calculate the steady state current in the $2\Omega$ resistor and steady state charge on the capacitor in the circuit?

• A. $0.9\text{A};0.36\mu \text{C}$
• B. $0.7\text{A};0.27\mu \text{C}$
• C. $0.5\text{A};0.20\mu \text{C}$
• D. $0.2\text{A};2\mu \text{C}$

Answer 1

The correct option is A $0.9\text{A};0.36\mu \text{C}$

Let us distribute potentials in the circuit as shown in figure.



As in steady state, no current flows through the branch having capacitor. Thus, in the middle branch of this circuit no current flows, and we can consider zero potential drop across $4\Omega$ resistance as shown.

Writing $\text{KCL}$ equation for the node $x$,

$\frac{x}{2.8}+\frac{x-6}{2}+\frac{x-6}{3}=0$

$\Rightarrow 3x+4.2x-25.2+2.8x-16.8=0$

$\Rightarrow 10x=42$

$\Rightarrow x=\frac{42}{10}=4.2\text{V}$

The current in $2\Omega$ resistance is,

$I_{2\Omega }=\frac{6-x}{2}=\frac{6-4.2}{2}=\frac{1.8}{2}=0.9\text{A}$

Steady state charge on the capacitor is,

$q=CV=C(6-x)$

$=0.2\times {10}^{-6}\times 1.8$

$=0.36\times {10}^{-6}=0.36\mu \text{C}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.