Question

In given AC circuit, reading of voltmeter across capacitor and resistor reads $V_0$ volt for source EMF of $\sqrt{3}V_0$ volts. Reading of voltmeter connected across $L-C$ is a $V_0$, then $10\alpha$ is _______ (use $\sqrt{2}=1.4$, $\sqrt{3}=1.7$)

Answer 1

450

Answer 2

From the given voltmeter readings $V_R = V_0$ and $V_C = V_0$, we deduce $R = X_C$. From $V_{LC} = |V_L - V_C| = V_0$ and $V_C = V_0$, we get $|V_L - V_0| = V_0$, leading to $V_L = 2V_0$ (assuming $V_L > 0$). This implies $X_L = 2X_C$. Using $R = X_C$, we get $X_L = 2R$. The phase angle $\alpha$ is given by $\tan \alpha = \frac{X_L - X_C}{R} = \frac{2R - R}{R} = 1$. Thus, $\alpha = 45^\circ$. The question asks for $10\alpha$, which is $10 \times 45 = 450$. The inconsistency with the source EMF is noted but does not prevent calculation of $\alpha$ from component voltage relations.