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. $\frac{14}{3} Ω$
Now, match the given columns and select the correct option from the codes given below

Moderate

Solution

As $3 Ω$ and $7 Ω$ are in series across 6 v and in series potential divides in proportion to resistance.
So $V_{A} - V_{D} = \frac{R \times X}{(R + S)} = \frac{3 \times 6}{3 + 7} = \frac{18}{10} = 1 .8 V$
Hence (i) $\to$ (q)
Similarly, $V_{A} - V_{B} = \frac{P \times V}{(P + Q)} = \frac{2 \times 6}{2 + 4} = 2 V$
$(V_{A} - V_{B}) - (V_{A} - V_{D}) = V_{D} - V_{B} = 2 - 1 .8 = 0 .2 V$
Hence (ii) $\to$ (r)
Energy stored in the capacitor will be zero if V_B = V_D
i.e., $\frac{P}{Q} = \frac{R}{S} \Rightarrow \frac{2}{Q} = \frac{3}{7} or Q = \frac{14}{3} Ω$
So (iii) $\to$ (s)
Hence, [i-q], [ii-r], [iii-s] and [iv-p]

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

SCAN ME

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.	$\frac{14}{3} Ω$