Question

A 20 \(\Omega\) resistance is connected across a supply of 400 V. If a resistance \(R\) is connected in parallel with the above 20 \(\Omega\) resistance, the current drawn from the supply gets doubled. The value of the unknown resistance \(R\) is:

• A. 80 \(\Omega\)
• B. 40 \(\Omega\)
• C. 20 \(\Omega\)
• D. 10 \(\Omega\)

Hint:

When two resistors are in parallel, their equivalent resistance is always less than the smallest resistance.

Answer 1

The Correct Option is C

Let the current drawn by the 20 \(\Omega\) resistance be \(I_1\). The current when the resistance \(R\) is added in parallel becomes \(I_2 = 2I_1\). From Ohm's law: \[ I_1 = \frac{V}{R_1} = \frac{400}{20} = 20 \, \text{A} \] Now, the equivalent resistance of the parallel combination is: \[ R_{\text{eq}} = \frac{R_1 R}{R_1 + R} = \frac{20R}{20 + R} \] The current doubles, so: \[ I_2 = \frac{400}{R_{\text{eq}}} = 2I_1 = 40 \, \text{A} \] Solving for \(R\), we get \(R = 20 \, \Omega\).