The current through a resistor in a circuit is 0.6 A. On putting a resistance of 2 Ω in parallel with the resistor, the current drops to 0.9 A. The resistance of the resistor is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 1.0 Ω
$I_1 = 0.6 \ \text{A}, \quad I_2 = 0.9 \ \text{A}$
$R = ?$
$\text{Let emf } = V$
$V = I_1 R = 0.6R$
$R_{\text{parallel}} = \frac{2R}{R + 2}$
$V = I_2 \cdot R_{\text{parallel}}$
$0.6R = 0.9 \cdot \frac{2R}{R+2}$
$0.6R(R+2) = 1.8R$
$0.6R^2 + 1.2R = 1.8R$
$0.6R^2 - 0.6R = 0$
$0.6R(R - 1) = 0$
$R = 1 \ \Omega$
The resistance is $1 \ \Omega$.
Note: There is a minor error in the question as given by NTA. The question says: “the current drops to 0.9 A”. But the current changes from 0.6 A to 0.9 A, which is actually an increase, not a drop.