Solution:
Removing C gives \(X_L = R\) and removing L gives \(X_C = R\). Hence, \(X_L = X_C\), indicating resonance. Impedance at resonance is \(Z = R = 100\ \Omega\). Current \(I = V/Z = 100/100 = 1\text{ A}\).
An L-C-R series circuit with \(100\ \Omega\) resistance is connected to an ac source of \(100\text{ V}\) and angular frequency \(300\text{ rad/s}\). When the capacitance is removed, the current lags behind the voltage by \(45^\circ\). When the inductance is removed, the current leads the voltage by \(45^\circ\). The current flowing in the original circuit will be:
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