  Question:
Published on: 22 June, 2022

A two element series circuit consumes 700 V of power and has power factor of 0.707 leading when energized by a voltage source of waveform $$v=141\sin\left(314t+30^{\circ}\right)$$. Find out the circuit element.

The power dissipated by the coil is $$P_C=I_{rms}V_{rms}\ cos\ \phi$$

For this problem $$V_{rms}=\frac{V_{peak}}{\sqrt2}=\frac{141}{\sqrt2}=100\ V$$

Given that

$$I_{rms}V_{rms}\ cos\ \phi=700$$

$$I_{rms}=\frac{700}{V_{rms}\ cos\ \phi}=\frac{700}{100\times0.707}=7\sqrt2\ Amp$$.

Impedance of the coil $$Z_c=\frac{V_{rms}}{I_{rms}}=\frac{100}{7\sqrt2}=10.04\ \mathrm{\Omega}$$

The power factor of the coil is 0.707 leading which means that the coil has a resistor and a capacitor.

$$\phi=\cos^{-1}{0.707}={45}^0$$

Resistance $$R=10.04\ cos\ {45}^0=10.04\times0.707=7.1\ \mathrm{\Omega}$$

$$X_C=\frac{1}{\omega C}=10.04\times sin\ {45}^0=10.04\times0.707=7.1\ \mathrm{\Omega}$$

$$C=\frac{1}{\omega X_C}=\frac{1}{2\pi f\times7.1}=0.45\ mF$$

So the circuit elements are a 7.1 \(\mathrm{\Omega}) resistor and 0.45 mF capacitor.

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