Phase Angle Between Current and Voltage in AC Circuit

To find the magnitude of the phase-angle $\varphi$ (in degrees) between the supply voltage and the current, we analyze the given series $R\text{-}C$ AC circuit step-by-step.

Step 1: Analyze the Lamp (Resistor $R$)

The lamp behaves as a purely resistive load $R$. We are given:

• Power consumed by the lamp, $P = 500\text{ W}$
• Voltage drop across the lamp (rms), $V_R = 100\text{ V}$

The power consumed by a resistor is given by: $P = V_R \cdot I$

where $I$ is the rms current flowing through the circuit. Substituting the given values: $500 = 100 \cdot I \implies I = 5\text{ A}$

Step 2: Determine the Voltage across the Capacitor

In a series $R\text{-}C$ circuit, the rms supply voltage $V$ is related to the voltage across the resistor $V_R$ and the voltage across the capacitor $V_C$ by the phasor relation: $V^2 = V_R^2 + V_C^2$

Given that the total supply voltage $V = 200\text{ V}$ and $V_R = 100\text{ V}$: $200^2 = 100^2 + V_C^2$ $40000 = 10000 + V_C^2$ $V_C^2 = 30000 \implies V_C = 100\sqrt{3}\text{ V}$

Step 3: Calculate the Phase Angle $\varphi$

The phase angle $\varphi$ between the supply voltage and the current is given by: $\cos\varphi = \frac{V_R}{V}$

Substituting the known values: $\cos\varphi = \frac{100}{200} = \frac{1}{2}$

Thus, the magnitude of the phase angle in degrees is: $\varphi = \arccos\left(\frac{1}{2}\right) = 60^\circ$

Alternatively, using the tangent relation: $\tan\varphi = \frac{V_C}{V_R} = \frac{100\sqrt{3}}{100} = \sqrt{3} \implies \varphi = 60^\circ$

The value of $\varphi$ is 60.