Question

The current through 1$\Omega$ resistance in the following circuit is

Answer 1

1.5 A

Answer 2

Since the bridge is balanced, no current flows through the galvanometer between Q and S. Thus, the circuit reduces to two parallel branches:

1. Branch P → Q → R: Series resistances = 5Ω + 1Ω = 6Ω
2. Branch P → S → R: Series resistances = 12.5Ω + 2.5Ω = 15Ω

Let I₁ = current through branch P–Q–R, and I₂ = current through branch P–S–R with I₁ + I₂ = 2.1 A.

Using current division:

$$I_1 = \frac{1/6}{1/6 + 1/15} \times 2.1 = \frac{(1/6)}{(5/30 + 2/30)} \times 2.1 = \frac{1/6}{7/30} \times 2.1 = \frac{30}{42} \times 2.1 = \frac{5}{7} \times 2.1 = 1.5\,\text{A}$$

Thus, the current through the 1Ω resistor is 1.5 A.