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

Step 1: Understand the condition of parallel resonance.
In a parallel RLC circuit at resonance, the inductive susceptance and capacitive susceptance are equal in magnitude and opposite in sign. Hence,
\[ I_L = I_C \quad \text{(equal magnitude, opposite phase)} \]
Therefore, the net reactive current drawn from the source is zero.
Step 2: Identify the source current.
Since the reactive currents cancel each other, the source current flows only through the resistive branch.
Thus,
\[ I_{\text{source}} = I_R = 1 \text{ mA} \]
Step 3: Analyze the branch currents.
Although the source current is only $1\,\text{mA}$, the individual branch currents $I_L$ and $I_C$ can be much larger due to resonance. This is known as current magnification in parallel resonance.
Step 4: Evaluate the options.
(A) $\lvert I_R \rvert<1\,\text{mA}$ is incorrect because $I_R = 1\,\text{mA}$.
(B) $\lvert I_R + I_L \rvert>1\,\text{mA}$ is correct because $I_L$ is large and adds vectorially with $I_R$.
(C) $\lvert I_R + I_C \rvert<1\,\text{mA}$ is incorrect because $I_C$ is also large.
(D) $\lvert I_L + I_C \rvert>1\,\text{mA}$ is incorrect since $I_L$ and $I_C$ cancel each other at resonance.
Step 5: Conclusion.
Hence, the correct statement for a parallel RLC circuit at resonance is
\[ \boxed{\lvert I_R + I_L \rvert>1 \text{ mA}} \]

Answer 2

At parallel resonance the source sees only the resistive branch, since the inductor and capacitor branch currents are equal in magnitude and opposite in phase and therefore cancel at the input; but that cancellation is between \(I_L\) and \(I_C\) only, not between \(I_R\) and either of them. Picking a representative case where the circuit's quality factor makes the reactive branch currents several times larger than \(I_R\), say \(I_R = 1\,\text{mA}\angle 0^\circ\) and \(I_L = 5\,\text{mA}\angle -90^\circ\) (with \(I_C = 5\,\text{mA}\angle 90^\circ\) canceling it), the combinations in each option can be tested directly.

1. \(|I_R| < 1\,\text{mA}\): Since the source current supplies only the resistive branch at resonance, \(I_R\) equals the full 1 mA source current, not less than it.
2. \(|I_R+I_L| > 1\,\text{mA}\): Adding \(1\angle 0^\circ\) and \(5\angle -90^\circ\) mA gives a resultant of magnitude \(\sqrt{1^2+5^2}=5.1\,\text{mA}\), which is indeed greater than 1 mA, because the large circulating current in the inductive branch dominates the sum.
3. \(|I_R+I_C| < 1\,\text{mA}\): Adding \(1\angle 0^\circ\) and \(5\angle 90^\circ\) mA gives magnitude \(5.1\,\text{mA}\) as well, which is larger than, not smaller than, 1 mA.
4. \(|I_L+I_C| > 1\,\text{mA}\): Since \(I_L\) and \(I_C\) are equal and opposite at resonance, their sum is zero, not greater than 1 mA.

Whatever the exact quality factor of the circuit, adding the resistive branch current to either large circulating reactive current always produces a magnitude at least as large as \(I_R\) itself and typically much larger, since the two contributions are at right angles rather than canceling.

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