Question

In the given circuit, current through branch de and the ammeter reading will be

(1) 0.A and 9/13A

(2) 0.A and 9/40 A

(3) 0.A and 1.A

(4) 0.5A and 0.5A

Answer 1

(3) 0.A and 1.A

The effective resistance between points a and f.

Hence current drawn from battery i = 9/9 = 1A

Therefore current through branch de = 0 (d and e being equal potential points, and ammeter reading 1A).