Question

Using a variable frequency ac voltage source the maximum current measured in the given LCR circuit is 50 mA for V = 5 sin (100t) The values of L and R are shown in the figure. The capacitance of the capacitor (C) used is_______ µF.

Hint:

When the peak current is exactly $V_{peak}/R$, it implies the circuit is in resonance, meaning the inductive and capacitive reactances cancel each other out.

Answer 1

Correct Answer: 50

Step 1: Understanding the Concept:
In an AC LCR circuit, the maximum current (\(I_{max}\)) is limited by the impedance (\(Z\)). The peak voltage \(V_{peak}\) is given by the amplitude of the sine function. We use Ohm's law for AC: \(V_{peak} = I_{max} Z\).
[Image of a series LCR circuit with an AC source]
Step 2: Key Formula or Approach:
1. Impedance \(Z = \sqrt{R^2 + (X_L - X_C)^2}\).
2. \(X_L = \omega L\) and \(X_C = \frac{1}{\omega C}\).
3. Peak current \(I_{peak} = \frac{V_{peak}}{Z}\).
Step 3: Detailed Explanation:
From \(V = 5 \sin(100t)\), we have \(V_{peak} = 5\) V and \(\omega = 100\) rad/s.
Given \(I_{max} = 50\) mA \(= 0.05\) A.
\[ Z = \frac{V_{peak}}{I_{max}} = \frac{5}{0.05} = 100 \Omega \] Assume the circuit values (based on standard problem versions) are \(R = 100 \Omega\). If \(Z = R\), the circuit is at resonance: \[ X_L = X_C \implies \omega L = \frac{1}{\omega C} \] \[ C = \frac{1}{\omega^2 L} \] If \(L = 10\) H (typical value for this problem): \[ C = \frac{1}{(100)^2 \times 10} = \frac{1}{10^5} = 10 \times 10^{-6} \text{ F} = 10 \mu\text{F} \]
Step 4: Final Answer:
The capacitance is 10 µF.