Question

The parallel RLC circuit shown in the figure is in resonance. In this circuit, 

• A. $\lvert I_R \rvert<1 \text{ mA}$
• B. $\lvert I_R + I_L \rvert>1 \text{ mA}$
• C. $\lvert I_R + I_C \rvert<1 \text{ mA}$
• D. $\lvert I_L + I_C \rvert>1 \text{ mA}$

Hint:

In parallel resonance, source current is minimum, but branch currents can be very large due to current magnification.

Answer 1

The Correct Option is B

At parallel resonance, \(I_L\) and \(I_C\) are equal in magnitude and opposite in phase, so they cancel each other, leaving the source to supply only \(I_R = 1\,\text{mA}\).

1. \(|I_R| < 1\,\text{mA}\): Wrong, since \(I_R\) is exactly the 1 mA source current at resonance.
2. \(|I_R+I_L| > 1\,\text{mA}\): Correct, because \(I_L\) is generally much larger than \(I_R\) at resonance (current magnification) and sits at 90 degrees to it, so their combined magnitude exceeds 1 mA.
3. \(|I_R+I_C| < 1\,\text{mA}\): Wrong, since combining a large \(I_C\) with \(I_R\) at 90 degrees can only increase the magnitude beyond \(I_R\), not shrink it.
4. \(|I_L+I_C| > 1\,\text{mA}\): Wrong, since \(I_L\) and \(I_C\) cancel at resonance, giving zero, not something greater than 1 mA.

So the correct statement is that \(|I_R+I_L|\) exceeds 1 mA.

Answer 2

The branch currents at parallel resonance can also be understood through the circuit's quality factor \(Q\), defined as the ratio of the reactive branch current to the resistive branch current, \(Q = I_L/I_R = I_C/I_R\). For any real circuit with even a modest \(Q\) (greater than zero), \(I_L\) and \(I_C\) individually exceed \(I_R\), and since \(I_L\) sits 90 degrees out of phase with \(I_R\), their vector sum has magnitude \(I_R\sqrt{1+Q^2}\), which is always larger than \(I_R\) alone whenever \(Q>0\).

1. \(|I_R| < 1\,\text{mA}\): Incorrect, since at resonance the entire 1 mA source current flows through the resistive branch alone.
2. \(|I_R+I_L| > 1\,\text{mA}\): Correct, since \(I_R\sqrt{1+Q^2} > I_R = 1\,\text{mA}\) for any nonzero \(Q\), which is the case in a practical resonant circuit.
3. \(|I_R+I_C| < 1\,\text{mA}\): Incorrect for the same reason; the combination can only grow in magnitude relative to \(I_R\) alone, not shrink.
4. \(|I_L+I_C| > 1\,\text{mA}\): Incorrect, since by definition of resonance \(I_L\) and \(I_C\) are equal and opposite, summing to zero.

Therefore, the correct answer is \(|I_R+I_L| > 1\,\text{mA}\).