Using a variable frequency a.c. 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.

($$Z$$) of a series LCR circuit

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

($$X_L = \omega L$$) and ($$X_C = \frac{1}{\omega C}$$):

$$Z = \sqrt{R^2 + \left(\omega L - \frac{1}{\omega C}\right)^2}$$

$$I_0 = \frac{V_0}{Z}$$

$$I_0 = \frac{V_0}{\sqrt{R^2 + \left(\omega L - \frac{1}{\omega C}\right)^2}}$$

For the current $$I_0$$ to be maximum, the denominator (the impedance $$Z$$) must be at its least possible value.

Since $$R^2$$ is constant and positive, $$Z$$ is minimum when the squared term is zero:

$$\left(\omega L - \frac{1}{\omega C}\right)^2 = 0$$

$$\omega L = \frac{1}{\omega C}$$

$$100 \times 2 = \frac{1}{100 \times C}$$

$$C = \frac{1}{200 \times 100}$$

$$C = 5 \times 10^{-5} \text{ F}$$

$$C = 50 \ \mu\text{F}$$