Question

For the circuit shown in figure, match the two columns.

Column I	Column II
(a) Current in wire ae	(p) 1 A
(b) Current in wire be	(q) 2 A
(c) Current in wire ce	(r) 0.5Ω
(d) Current in wire cle	(s) None of these

(A) a – p, b – s, c – q, d – r 

(B) a – s, b – r, c – q, d – p 

(C) a – q, b – s, c – q, d – s 

(D) a – s, b – q, c – p, d – r

Answer 1

The correct option (C) a – q, b – s, c – q, d – s

Explanation:

Let voltage at point e be Ve.

at node e, applying Kirchoff's current

law, I₁ + I₂ + I₃ + I₄ = 0

∴ {(Ve – 2) / 1} + {(Ve – 6) / 1} + {(Ve – 4) / 2} + {(Ve – 4) / 2} = 0

2Ve – 8 + Ve – 4 = 0

3Ve = 12

Ve = 4V

(a) current in ae = I₁ = {(Ve – 2)/1} = 4 – 2 = 2A

current in be = I₄ = {(Ve – 4)/2} = 0A

current in ce = I₃ = {(Ve – 6)/1} = – 2A

current in de = I₂ = {(Ve – 4)/2} = 0A