The current in a resistive series circuit is 7.0 A. On adding a resistance of 1.6 Ω in the same circuit in series, the current drops to 5.0 A. The original resistance of the circuit is

4.0 Ω

The correct answer is Option (2) → 4.0 Ω

Given:

Initial current = $I_1 = 7.0 \text{ A}$

New current = $I_2 = 5.0 \text{ A}$

Added resistance = $R' = 1.6 \, \Omega$

Let the original resistance be $R$ and supply voltage be $V$.

From Ohm’s law, $V = I_1 R = I_2 (R + R')$

Substitute values:

$7R = 5(R + 1.6)$

$7R = 5R + 8$

$2R = 8$

$R = 4 \, \Omega$

The original resistance of the circuit is $4 \, \Omega$.