Question

A coil has a resistance of 10 Ω and an inductance of 0.4 henry. It is connected to an AC source of 6.5 V, $\frac{30}{\mathrm{\pi }}\mathrm{Hz}$. Find the average power consumed in the circuit.

Answer 1

Given:
Resistance of coil, R = 10 Ω
Inductance of coil, L = 0.4 Henry
Voltage of AC source, Erms = 6.5 V
Frequency of AC source, $f$ = $\frac{30}{\mathrm{\pi }}\mathrm{Hz}$
Reactance of resistance-inductance circuit $\left(Z\right)$ is given by,
Z = $\sqrt{R^2+{X_L}^2}$
Here, R = resistance of the circuit
X_L = Reactance of the pure inductive circuit
Z = $\sqrt{R^2+{\left(2\mathrm{\pi }fL\right)}^2}$
= $\sqrt{{\left(10\right)}^2+{\left(2\times \mathrm{\pi }\times \frac{30}{\mathrm{\pi }}\times 0.4\right)}^2}$
Average power consumed in the circuit (P) is given by,
P = E_rmsI_rmscos$\varphi$
cos$\varphi$ = $\frac{R}{Z}$, I_rms = $\frac{E_{rms}}{Z}$
$$\begin{gathered} \therefore P=6.5\times \frac{6.5}{Z}\times \frac{R}{Z} \\ \Rightarrow P=\frac{6.5\times 6.5\times 10}{{\left(\sqrt{R^2+{\left(\omega L\right)}^2}\right)}^2} \\ \Rightarrow P=\frac{6.5\times 6.5\times 10}{100+576} \\ \Rightarrow P=\frac{6.5\times 6.5\times 10}{676} \\ \Rightarrow P=0.625=\frac{5}{8}\mathrm{W} \end{gathered}$$