The current through a resistor in a circuit is 0.6 A. On putting a resistance of \( 2\,\Omega \) in parallel with the resistor, the current rises to 0.9 A. The resistance of the resistor is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 1.0\,\Omega \)
Take the potential difference across the resistor as fixed (an ideal source), so 0.9 A is the total current drawn after the \( 2\,\Omega \) resistor is added in parallel. The current in the resistor itself stays 0.6 A, and the extra current flows through the \( 2\,\Omega \) branch.
\( \displaystyle V = 0.6\,R \)
\( \displaystyle I_{2\Omega} = 0.9 - 0.6 = 0.3\,\mathrm{A} \)
\( \displaystyle V = 0.3\,\mathrm{A} \times 2\,\Omega = 0.6\,\mathrm{V} \)
\( \displaystyle R = \frac{V}{0.6\,\mathrm{A}} = \frac{0.6\,\mathrm{V}}{0.6\,\mathrm{A}} = 1.0\,\Omega \)