A resistor with R = 300 Ω and an inductor are connected in series across an ac source that has voltage amplitude 500 V. The rate at which electrical energy is dissipated in the resistor is 286 W. What is (a) the impedance Z of the circuit; (b) the amplitude of the voltage across the inductor; (c) the power factor?

An inductor is a passive two-terminal device that stores energy in a magnetic field when current passes through it. When an inductor is attached to an AC supply, the resistance produced by it is called inductive reactance (X_L).

Resistance is a measure of opposition to the flow of current in a closed electrical circuit. It is measured in Ohm (Ω).

Impedance is defined as the effective resistance of an electric circuit to the flow of current due to the combined effect of resistance (offered by a resistor) and reactance (offered by a capacitor and inductor).

The power factor of an AC circuit is defined as the ratio of real power absorbed to the apparent power flowing in the AC circuit. It is a dimensionless quantity and its value lies in the interval [-1,1].

Resistance of the resistor, R = 300 Ω

The inductance of the coil, L = 0.400 H

The amplitude of voltage source, V = 500 W

Rate of Energy dissipation, P_av = 280 W

Current in the circuit can be determined using the relation:

$$\begin{gathered} P_{av}=\frac{1}{2}{I}^{2}R \\ I=\sqrt{\frac{2P_{av}}{300\Omega }} \\ =\sqrt{\frac{2\left(286W\right)}{300\Omega }} \\ =1.38A \end{gathered}$$

The impedance of the circuit can be found using ohm’s law:

$$\begin{gathered} V=I.Z \\ I=\sqrt{\frac{2P_{av}}{R}} \\ =\sqrt{\frac{2\left(286W\right)}{300\Omega }} \\ =1.38A \end{gathered}$$

Therefore, the impedance of the electric circuit is 362.32 Ω

By Ohm’s Law, Voltage drop across resistor is given by

$$\begin{gathered} V_R=I.R \\ =\left(1.38A\right)\left(300\Omega \right) \\ =414V \end{gathered}$$

Impedance of a R-L circuit is given by

$$\begin{gathered} \Rightarrow Z=\sqrt{R^2+X_L^2} \\ \Rightarrow X_L^2=Z^2-R^2 \\ \Rightarrow {\left(I{V}_L\right)}^2={\left(IV\right)}^2-{\left(I{V}_R\right)}^2 \\ \Rightarrow V_L=\sqrt{V^2-V_R^2} \end{gathered}$$

Thus, Voltage drop across inductor can be determined as

$$\begin{gathered} V_L=\sqrt{V^2-V_R^2} \\ =\sqrt{{500}^2-{414}^2} \\ =280V \end{gathered}$$

Therefore, the amplitude of the voltage across inductor is 280 V

Power factor of the circuit is given by

$$\begin{gathered} \mathrm{P}.\mathrm{F}=\mathrm{cos\varphi }=\frac{\mathrm{R}}{\mathrm{Z}}=\frac{300\mathrm{\Omega }}{362.32\mathrm{\Omega }}=0.828 \\ Also,theangleis\varphi =+34.11° \end{gathered}$$

Therefore, the Power factor for the circuit is 0.828 and the phase angle is +34.11°.