Question

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

• A. 2.3 µF
• B. 0.32 µF
• C. 63 µF
• D. 0.62 µF

Hint:

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.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This question describes a series RLC circuit. The current in such a circuit is maximum when the circuit is in a state of resonance. Resonance occurs when the inductive reactance (\(X_L\)) becomes equal to the capacitive reactance (\(X_C\)), causing the total impedance of the circuit to be at its minimum value (equal to the resistance \(R\)).

Step 2: Key Formula or Approach:
The condition for resonance in a series RLC circuit is: \[ X_L = X_C \] where \(X_L = 2\pi f L\) is the inductive reactance and \(X_C = \frac{1}{2\pi f C}\) is the capacitive reactance. By equating these, we get the formula for the resonant frequency, or we can solve for the capacitance \(C\) needed for resonance at a given 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: Detailed Explanation:
Given data:
Inductance, \(L = 500 \, \text{mH} = 500 \times 10^{-3} \, \text{H} = 0.5 \, \text{H}\).
Frequency, \(f = 0.4 \, \text{kHz} = 0.4 \times 10^3 \, \text{Hz} = 400 \, \text{Hz}\).
Calculation:
Substitute the values into the formula for capacitance at resonance: \[ C = \frac{1}{4\pi^2 f^2 L} \] \[ C = \frac{1}{4\pi^2 (400)^2 (0.5)} \] \[ C = \frac{1}{4\pi^2 (160000) (0.5)} \] \[ C = \frac{1}{2\pi^2 (160000)} = \frac{1}{320000 \pi^2} \, \text{F} \] Using the approximation \(\pi^2 \approx 9.87\): \[ C \approx \frac{1}{320000 \times 9.87} = \frac{1}{3158400} \, \text{F} \] \[ C \approx 3.166 \times 10^{-7} \, \text{F} \] To express this value in microfarads (µF), we 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: Final Answer:
The required capacitance to achieve maximum current (resonance) is approximately 0.32 µF.