The value of $$\psi$$ is ___.

The circuit contains a pure resistor (the lamp) in series with a capacitor, connected to a 200 V, 50 Hz supply. All data are r.m.s. values.

Step 1 : Current in the circuit
For a purely resistive element, $$P = VI$$. Hence for the lamp

$$I = \frac{P}{V_{\text{lamp}}}= \frac{500\;\text{W}}{100\;\text{V}} = 5\;\text{A}$$

This 5 A is the r.m.s. current everywhere in the series circuit.

Step 2 : Voltage across the capacitor
Take the current $$I$$ as the reference phasor along the positive x-axis. The lamp voltage $$V_R$$ (100 V) is in phase with the current, while the capacitor voltage $$V_C$$ lags the current by $$90^{\circ}$$ (points along the negative y-axis). The supply voltage $$V_s$$ is the vector sum:

$$V_s^2 = V_R^2 + V_C^2$$

Substituting $$V_s = 200\;\text{V}$$ and $$V_R = 100\;\text{V}$$:

$$200^2 = 100^2 + V_C^2 \quad\Longrightarrow\quad V_C^2 = 200^2 - 100^2 = 40000 - 10000 = 30000$$
$$\Rightarrow\; V_C = 100\sqrt{3}\;\text{V} \approx 173.2\;\text{V}$$

Step 3 : Capacitive reactance

$$X_C = \frac{V_C}{I} = \frac{100\sqrt{3}}{5} = 20\sqrt{3}\;\Omega \approx 34.64\;\Omega$$

Step 4 : Resistance of the lamp

$$R = \frac{V_R}{I} = \frac{100}{5} = 20\;\Omega$$

Step 5 : Phase angle between current and supply voltage
For a series $$R$$-$$C$$ circuit the phase angle $$\psi$$ (magnitude) satisfies

$$\tan\psi = \frac{X_C}{R}$$

Hence

$$\tan\psi = \frac{20\sqrt{3}}{20} = \sqrt{3}$$

Using $$\tan60^{\circ} = \sqrt{3}$$ (and the given hint $$\pi\sqrt{3}\approx 5$$ which implies $$\sqrt{3}\approx1.732$$),

$$\psi \approx 60^{\circ}$$

Answer
The magnitude of the phase angle is approximately $$60^{\circ}$$, which lies in the range 55.00 - 62.00.