Problem

A coil of pure inductance $318 \, \mathrm{mH}$ is connected in series with a pure resistance of $75 \, \Omega$. The voltage across the resistor is $150 \, \mathrm{V}$, and the frequency of the power supply is $50 \, \mathrm{Hz}$. Calculate the voltage of the power supply and the phase angle.

Data

• Inductance: $L = 318 \, \mathrm{mH} = 318 \times 10^{-3} \, \mathrm{H}$
• Resistance: $R = 75 \, \Omega$
• AC frequency: $f = 50 \, \mathrm{Hz}$
• Voltage across the resistor: $V_R = 150 \, \mathrm{V}$

Prerequisite Concepts

1. Inductive Reactance:

$$X_L = 2 \pi f L$$

2. Impedance in an R-L Series Circuit:

$$Z = \sqrt{R^2 + X_L^2}$$

3. Current through the Resistor:

$$I = \frac{V_R}{R}$$

4. Supply Voltage:

$$V = I Z$$

5. Phase Angle:

$$\theta = \tan^{-1} \left(\frac{X_L}{R}\right)$$

Solution

Step 1: Calculate the Inductive Reactance

$$X_L = 2 \pi f L$$ $$X_L = 2 \times 3.14 \times 50 \, \mathrm{Hz} \times 318 \times 10^{-3} \, \mathrm{H}$$ $$X_L = 99.852 \, \Omega$$

Step 2: Calculate the Current through the Resistor

$$I = \frac{V_R}{R}$$ $$I = \frac{150 \, \mathrm{V}}{75 \, \Omega}$$ $$I = 2.00 \, \mathrm{A}$$

Step 3: Calculate the Total Impedance

$$Z = \sqrt{R^2 + X_L^2}$$ $$Z = \sqrt{(75 \, \Omega)^2 + (99.852 \, \Omega)^2}$$ $$Z = \sqrt{5625 + 9970.63}$$ $$Z = \sqrt{15595.63} = 124.88 \, \Omega$$

Step 4: Calculate the Supply Voltage

$$V = I Z$$ $$V = 2.00 \, \mathrm{A} \times 124.88 \, \Omega$$ $$V = 249.76 \, \mathrm{V} \approx 250 \, \mathrm{V}$$

Step 5: Calculate the Phase Angle

$$\theta = \tan^{-1} \left(\frac{X_L}{R}\right)$$ $$\theta = \tan^{-1} \left(\frac{99.852 \, \Omega}{75 \, \Omega}\right)$$ $$\theta = 53.06^\circ$$

Answer

• Supply Voltage: $V = 250 \, \mathrm{V}$
• Phase Angle: $\theta = 53.06^\circ$

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