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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 ___________ $\mu\text{F}$. 

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

The maximum current \(I_{\text{max}}\) is given as \(50 \, \text{mA} = 0.05 \, \text{A}\).

The voltage across the circuit is \(V = 5 \sin(100t)\), so the peak voltage \(V_0 = 5 \, \text{V}\).

The angular frequency \(\omega\) is \(100 \, \text{rad/s}\).

For maximum current, the circuit is at resonance: \(Z = \frac{V_0}{I_{\text{max}}}\), where \(Z\) is the impedance. So, \(Z = \frac{5}{0.05} = 100 \, \Omega\).

At resonance in an LCR circuit, \(Z = R\), meaning \(R = 100 \, \Omega\).

The resonance condition is \(\omega L = \frac{1}{\omega C}\), where \(L = 2 \, \text{H}\).

Thus, \(\omega^2 = \frac{1}{LC}\). Substituting the known values, \(100^2 = \frac{1}{2C}\).

Solve for \(C\):

\(C = \frac{1}{2 \times 10000} = \frac{1}{20000} = 50 \times 10^{-6} \, \text{F} = 50 \, \mu\text{F}\).

The calculated capacitance \(C = 50 \, \mu\text{F}\), which falls within the expected range of [50, 50].

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Important Questions from Electricity and Magnetism

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