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To an AC power supply of $220\text{ V}$ at $50\text{ Hz}$, a resistor of $20\,\Omega$, a capacitor of reactance $25\,\Omega$ and an inductor of reactance $45\,\Omega$ are connected in series. The corresponding current in the circuit and the phase angle between the current and the voltage is, respectively:
A
7.8 A and $30^\circ$
B
7.8 A and $45^\circ$
C
15.6 A and $30^\circ$
D
15.6 A and $45^\circ$
Explanation
Impedance is $Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{20^2 + 20^2} = 20\sqrt{2}\,\Omega$. Current $I = 220 / (20\sqrt{2}) \approx 7.8\text{ A}$ and $\tan\phi = 20/20 = 1 \implies \phi = 45^\circ$.
Detailed Solution
Given $R = 20\,\Omega$, $X_L = 45\,\Omega$, $X_C = 25\,\Omega$. Net reactance $X = X_L - X_C = 45 - 25 = 20\,\Omega$. Total circuit impedance: $Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{20^2 + 20^2} = 20\sqrt{2} \approx 28.28\,\Omega$. The RMS current is $I_{rms} = \frac{V_{rms}}{Z} = \frac{220}{28.28} \approx 7.78\text{ A} \approx 7.8\text{ A}$. The phase angle $\phi$ is $\tan\phi = \frac{X_L - X_C}{R} = \frac{20}{20} = 1 \implies \phi = 45^\circ$.