Question

A network consisting of three resistors, three batteries, and a capacitor is shown in the figure


$\begin{array}{ccc}& \text{Column I}& \text{Column II}\\ \left(A\right)& \text{Current in branch EB is}& \text{(P)}10\mu C\\ \left(B\right)& \text{Current in branch CB is}& \text{(Q)}0.5\mu C\\ \left(C\right)& \text{Current in branch ED is}& \text{(R)}1.5\mu C\\ \left(D\right)& \text{Charge on capacitor is}& \text{(S)}5\mu C\\ & & \text{(T)}7.5\mu C\end{array}$
Which of the following option has the correct combination considering column - I and Column - II.

• A. $A\to Q$
• B. $B\to S$
• C. $C\to R$
• D. $D\to P$

Answer 1

The correct option is D $D\to P$

$A\to R;B\to Q;C\to Q;D\to P$
When a steady state is reached, no current passes through the capacitor or the branch CE.


Considering the loop $\text{ABEFA}$
$5\times (i_1+i_2)=10$ or $i_1+i_2=2\text{A}...\left(1\right)$
Considering the loop $\text{BCDEB},$
$4i_2=12-10=2\Rightarrow i_2=0.5\text{A}$
So, $i_1=2-0.5=1.5\text{A}$
To find the charge on capacitors, we must know potential difference across the plates.
Consider the loop $\text{CEDC}:$
$-12+4i_2+3\times 0-v_c+8=0$
$v_c=-2\text{V}$ .
Potential difference across capacitor is $2\text{V}$
So, the charge on capacitor $Q=CV=10\mu C$