Question:medium

An inductor of 500 mH is in series with a resistance and a variable capacitor connected to a source of frequency 0.4 kHz. The value of capacitance of the capacitor to get a maximum current will be

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The condition for maximum current in a series RLC circuit is resonance. The key formula to remember is \(X_L = X_C\), which leads to the resonant frequency \(f_0 = \frac{1}{2\pi\sqrt{LC}}\). You can rearrange this formula to solve for L or C if the frequency is given.
Updated On: Mar 27, 2026
  • 2.3 µF
  • 0.32 µF
  • 63 µF
  • 0.62 µF
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The Correct Option is B

Solution and Explanation


Step 1: Concept Introduction: This question pertains to a series RLC circuit. Maximum current in such a circuit occurs at resonance. Resonance is achieved when inductive reactance (\(X_L\)) equals capacitive reactance (\(X_C\)), minimizing the circuit's total impedance to the value of resistance (\(R\)).

Step 2: Governing Equation: The condition for resonance in a series RLC circuit is: \[ X_L = X_C \] where \(X_L = 2\pi f L\) (inductive reactance) and \(X_C = \frac{1}{2\pi f C}\) (capacitive reactance). Equating these yields the resonant frequency formula or allows for calculating the capacitance (\(C\)) required for resonance at a specified frequency (\(f\)). \[ 2\pi f L = \frac{1}{2\pi f C} \implies C = \frac{1}{(2\pi f)^2 L} = \frac{1}{4\pi^2 f^2 L} \]

Step 3: Calculation Details: Provided Values: Inductance, \(L = 500 \, \text{mH} = 0.5 \, \text{H}\). Frequency, \(f = 0.4 \, \text{kHz} = 400 \, \text{Hz}\). Computation: Utilize the resonance capacitance formula with the given values: \[ C = \frac{1}{4\pi^2 f^2 L} \] \[ C = \frac{1}{4\pi^2 (400)^2 (0.5)} \] \[ C = \frac{1}{4\pi^2 (160000) (0.5)} \] \[ C = \frac{1}{320000 \pi^2} \, \text{F} \] Approximating \(\pi^2 \approx 9.87\): \[ C \approx \frac{1}{320000 \times 9.87} = \frac{1}{3158400} \, \text{F} \] \[ C \approx 3.166 \times 10^{-7} \, \text{F} \] To express in microfarads (µF), multiply by \(10^6\): \[ C \approx 3.166 \times 10^{-7} \times 10^6 \, \mu\text{F} = 0.3166 \, \mu\text{F} \] This value is approximately 0.32 µF.

Step 4: Conclusion: The capacitance necessary to achieve resonance (maximum current) is approximately 0.32 µF.

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