Question

In the circuit show in figure, match the following two columns:-

Column I	Column II
(a) Potential difference across battery A	(p) Zero
(b) Potential difference across battery B	(q) 1
(c) Net power supplied/consumed by A	(r) 2
(d) Net power supplied/consumed by B	(s) 3

(A) a – p, b – q, c – r, d – s 

(B) a – s, b – q, c – r, d – p 

(C) a – s, b – r, c – s, d – r 

(D) a – q, b – r, c – s, d – s

Answer 1

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

Explanation:

applying Kirchhoff’s law 

– I(1) – I(1) – I(1) – 4 + 1 = 0 

∴ 3I = – 3 

I = – 1A 

Potential across A = – 4 – I(1) = – 4 + 1 = – 3V (direction of current is opposite of we assumed) 

Potential across B = 1 – I(1) = 1 + 1 = 2V 

Power by A = V ∙ I = (+ 3) × (+ 1) = 3W ----- 

Power by B = V ∙ I = (2) (+ 1) = + 2W ---- 

considering magnitude only 

V_A = 3V 

V_B = 2V 

P_A = 3W 

P_B = 2W