Question

In a Wheatstone’ s bridge, the four resistances arms of the bridge are AB = 1 Ω, BC = 2 Ω, CD = 3 Ω and DA = 2 Ω. A cell of emf 2 V and negligible internal resistance is connected across AC and a galvanometer of resistance 4 Ω is connected between B and D. Find the current through the galvanometer.

Answer 1

There are other methods also, like applying KVL on each loop, delta to star method. I find this method easier.

Just mark the potentials at different junctions and apply KCL.

Hope, this helps...