For the circuit shown in the figure, all the three batteries are ideal. Match the column
Column-I Column-II
(A) The value of R such that current through 3 V battery is zero. (p) $1 Ω$
(B) The value of R such that current through 1 V battery is zero. (q) $2 Ω$
(C) The value of R such that current current through 2 V battery is zero (r) $3 Ω$
(D) The value of R for which current through 3 V battery is 5/8 A (s) Not possible for any value of R

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$(A) \to q; (B) \to q; (C) \to p; (D) \to s$

$(A) \to s; (B) \to s; (C) \to s; (D) \to p$

$(A) \to s; (B) \to s; (C) \to p; (D) \to q$

$(A) \to p; (B) \to p; (C) \to q; (D) \to s$

answer is B.

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Detailed Solution

Let current through various branches is as shown in the figure.
Question Image
$\begin{array}{l} i_{1} + i_{2} + i_{3} = 0... (i) \\ V_{A} - V_{B} = V_{D} - V_{C} = V_{E} - V_{F} \\ 1 - 1 i_{1} = 2 - i_{2} = 3 - R i_{3} .... (i i) \end{array}$
$\begin{array}{l} F r o m E q s . (i) a n d (i i), w e g e t \\ i_{1} + \frac{2 - 1 + i_{1}}{2} + \frac{3 - 1 + i_{1}}{R} = 0 \Rightarrow i_{1} = \frac{- (R + 4)}{3 R + 2} \\ A g a i n f r o m E q s . (i) a n d (i i), w e g e t \end{array}$

$\begin{array}{l} \frac{2 i_{2} - 1}{1} + i_{2} + \frac{1 - 2 i_{2}}{R} = 0 \Rightarrow i_{2} = \frac{R - 1}{3 R + 2} \\ N o w i_{3} = - i_{1} - i_{2} = \frac{5}{3 R + 2} \end{array}$
i. Current i3 cannot be zero
ii. Current i1 cannot be zero
iii. $i_{2} = 0 \Rightarrow R = 1 Ω$
iv. $i_{3} = \frac{5}{8} \Rightarrow \frac{5}{3 R + 2} = \frac{5}{8} \Rightarrow R = 2 Ω$

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Column-I	Column-II
(A) The value of R such that current through 3 V battery is zero.	(p) $1 Ω$
(B) The value of R such that current through 1 V battery is zero.	(q) $2 Ω$
(C) The value of R such that current current through 2 V battery is zero	(r) $3 Ω$
(D) The value of R for which current through 3 V battery is 5/8 A	(s) Not possible for any value of R