Question

Consider the circuit diagram shown.

Column I		Column II
i	Potential difference across $A$ and $D$ in steady state is	p	2 V
ii	Potential difference across capacitor in steady state is	q	1.8 V
iii	Value of $Q$ for which no energy is stored across capacitor is	r	0.2 V
iv	Potential difference across $A$ and $B$ in steady state is	s	$\frac{14}{3} \Omega$

Now, match the given columns and select the correct option from the codes given below. Codes:

• A. i - (p), ii - (r), iii - (s), iv - (p)
• B. i - (r), ii - (p), iii - (q), iv - (s)
• C. i- (q), ii -(q), iii - (s), iv - (r)
• D. i - (p), ii - (p), iii - (q), iv - (s)

Answer 1

Answer: (A)

As $3 \,\Omega$ and $7 \,\Omega$ are in series across $6 \,V$ and $1 \, n$ series potential divides in proportion to resistance.
So $V_{A}-V_{D}=\frac{R \times V}{(R+S)}$
$=\frac{3 \times 6}{3+7}=\frac{18}{10}=1.8 \, V$
Hence (i) $\rightarrow$ (q)
Similarly, $V_{A}-V_{B}=\frac{P \times V}{(P+Q)}$
$=\frac{2 \times 6}{2+4}=2 \,V$
$\left(V_{A}-V_{B}\right)-\left(V_{A}-V_{D}\right)=V_{D}-V_{B}$
$=2-1.8=0.2 \,V$
Hence (ii) $\rightarrow(r)$
Energy stored in the capacitor will be zero if $V_{B}=V_{D}$
i.e., $\frac{P}{Q}=\frac{R}{S} $
$\Rightarrow \frac{2}{Q}=\frac{3}{7}$ or $Q=\frac{14}{3} \Omega$
So (iii) $\rightarrow( s )$
Hence, [i-q],[ii-r],[iii- s]and [iv- $p ]$