A large electromagnetic coil is connected to a 120-Hz ac source. The coil has resistance 400 Ω, and at this source frequency the coil has inductive reactance 250 Ω. (a) What is the inductance of the coil? (b) What must the rms voltage of the source be if the coil is to consume an average electrical power of 450 W?

Inductive reactance is the resistance offered by an inductor in a circuit. It is denoted as$X_L$ and is calculated using the formula, $X_L=\omega L$, where,$\omega$ is the angular frequency of the A/C signal, andL is the inductance.

Impedance of a circuit is the total resistance offered by the circuit.Denoted by Z, the formula for impedance of a LR circuit is $Z\sqrt{R^2+X_L^2}$.

Average power of an A/C circuit is given by the formula $P_{avg}=V_{rms}{I}_{rs}\cos \varphi$, where$P_{avg}$ is the average power,$V_{rms}$ is the rms value of the voltage,$I_{rms}$ is the rms value of current,$\varphi$ is the phase difference between voltage and current. It is known that $\cos \varphi =\frac{R}{Z}$, and $I_{rms}=\frac{V_{rms}}{Z}$. Using these two,$P_{avg}$ the formula for comes out to be-$P_{avg}=\frac{V_{rms}R}{Z^2}$ .

Given that-

$$\begin{gathered} \begin{array}{cc}{X}_L=250\Omega \\ \omega =2\mathrm{\pi }\times 120 \\ =753.98\mathrm{rad}/\mathrm{s}& V_{rms}=\sqrt{P_{avg}Z} \\ =\sqrt{450\times 471.7} \\ =145.69V\end{array} \end{gathered}$$

Now, from previous discussion it is known that, $X_L=\omega L$. Therefore,

$$\begin{gathered} L=\frac{X_L}{\omega } \\ =\frac{250}{753.98} \\ =0.33H \end{gathered}$$

Hence, the inductance of the coil is 0.33 H.

The impedance of the circuit is can be given by,$Z=\sqrt{R^2+X_L^2}$, therefore,

$$\begin{gathered} Z=\sqrt{{400}^2+{250}^2} \\ =471.7\Omega \end{gathered}$$

Now, average power is given by, $P_{avg}=\frac{V_{rms}^{2}R}{Z^2}$, therefore,

$$\begin{gathered} V_{rms}=Z\sqrt{\frac{P_{avg}}{R}} \\ =471.7\times \sqrt{\frac{450}{400}} \\ =500.31V \end{gathered}$$

Therefore, rms value of voltage is 500.31 V