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
Answer & explanation
Correct answer: option 2
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 $4 \, \Omega$.