Question

In the circuit shown in figure three ideal sources have $\text{EMF}$, $E_1=1\text{V}$ and $E_2=2.5\text{V}$ and the resistances have the values $R_1=10\Omega$ and $R_2=20\Omega$. Find the potential differences $V_A-V_B$ between the plates $\text{A}$ and $\text{B}$ of the capacitor $\text{C}$ in steady state.

• A. $-0.5\text{V}$
• B. $-1\text{V}$
• C. $-1.5\text{V}$
• D. $-2\text{V}$

Answer 1

The correct option is A $-0.5\text{V}$

As we know that in steady state no current flows through the branch in which capacitor is connected, so in the given circuit current only flows in the outer loop as shown in figure.


Writing $KVL$ equation for the outer loop $\text{POMLKNP}$ gives,

$20i+10i=2.5-1.0$

$\Rightarrow 30i=1.5$

$\Rightarrow i=\frac{1.5}{30}=0.05\text{A}$

Writing equation of potential drop from capacitor terminal $A$ to $B$ for the loop $\text{NMOPN}$ gives,

$V_B+20i-2.5+1=V_A$

$\Rightarrow V_B-V_A=1.5-20i$

$\Rightarrow V_B-V_A=1.5-20\times 0.05$

$\Rightarrow V_B-V_A=0.5\text{V}$

$\therefore V_A-V_B=-0.5\text{V}$

Hence, option $\left(a\right)$ is correct answer.