An inductor of inductance L, a capacitor of capacitance C and a resistor of resistance ‘R’ are connected in series to an ac source of

Answer: (4) 5 Ω

In a series LCR circuit, the voltages across L and C are opposite in phase, and the voltage across R is at 90° to both. So the source voltage is

\[V = \sqrt{V_R^{2} + (V_L - V_C)^{2}}\] \[= \sqrt{40^{2} + (40-10)^{2}} = 50\ \mathrm{V}\]

These voltages are rms values. The current amplitude is \(I_0 = 10\sqrt{2}\) A, so the rms current is

\[I_{rms} = \frac{I_0}{\sqrt{2}} = 10\ \mathrm{A}\]

Impedance:

\[Z = \frac{V_{rms}}{I_{rms}} = \frac{50}{10} = 5\ \Omega\]

Check: \(R = 40/10 = 4\ \Omega\), \(X_L - X_C = 30/10 = 3\ \Omega\), and \(\sqrt{4^{2}+3^{2}} = 5\ \Omega\). Option (2) \(5\sqrt{2}\ \Omega\) is what you get by mixing values, i.e. dividing the peak voltage \(50\sqrt{2}\) V by the rms current 10 A.