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

To find the capacitance \(C\) of the capacitor used in the LCR circuit, follow these steps:

Given the maximum current \(I_{\text{max}} = 50 \text{ mA} = 0.05 \text{ A}\) and voltage \(V = 5 \sin(100t)\), we can see the maximum voltage \(V_{\text{max}} = 5\) V. 

The angular frequency \(\omega = 100 \, \text{rad/s}\) as given in the voltage function.

The impedance \(Z\) of the LCR circuit is given by:

\(Z = \sqrt{R^2 + (X_L - X_C)^2}\)

\(X_L = \omega L = 100 \times 2 = 200 \, \Omega\)

\(X_C = \frac{1}{\omega C}\)

Using Ohm's law \(V_{\text{max}} = I_{\text{max}} \cdot Z\), we have:

\(5 = 0.05 \times \sqrt{100^2 + (200 - X_C)^2}\)

\(\Rightarrow 100 = \sqrt{100^2 + (200 - X_C)^2}\)

Square both sides:

\(10000 = 10000 + (200 - X_C)^2\)

\((200 - X_C)^2 = 0\)

Calculate \(C\) from \(X_C\):

\(X_C = \frac{1}{\omega C} = 200\)

\(C = \frac{1}{200 \times 100} = 5 \times 10^{-5} \, \text{F} = 50 \, \mu\text{F}\)

Thus, the capacitance \(C\) is 50 µF, which fits within the given range.

This implies \(X_C = 200 \, \Omega\).