When a solenoid is connected to a 48.0-V dc battery that has negligible internal resistance, the current in the solenoid is 5.50 A. When this solenoid is connected to an ac source that has voltage amplitude 48.0 V and angular frequency 20.0 rad/s, the current in the solenoid is 3.60 A. What is the inductance of this solenoid?

By Ohm’s law, Resistance is given by

$\mathbf{R}\mathbf{=}\frac{\mathbf{V}}{\mathbf{I}}$, where V is the dc voltage supply and I is the current through the circuit

Z is defined as the impedance of the circuit which is the effective resistance of an electric circuit or component to alternating current, arising from the combined effects of ohmic resistance and reactance.

$\mathbf{Z}\mathbf{=}\frac{\mathbf{V}}{{\mathbf{I}}_{\mathbf{2}}}$, where V is the ac voltage supply and${\mathrm{I}}_2$ is the current through the circuit by the ac voltage supply.

The inductive reactance in absence of a capacitor is given by,

${\mathbf{X}}_{\mathbf{L}}\mathbf{=}\sqrt{{\mathbf{Z}}^{\mathbf{2}}\mathbf{-}{\mathbf{R}}^{\mathbf{2}}}$, where is the circuit resistance

Also it depends on the inductance of the coil ${\mathbf{X}}_{\mathbf{L}}\mathbf{=}\mathbf{\omega L}$

Equate the last 2 equations to get$\mathbf{L}\mathbf{=}\frac{\sqrt{{\mathbf{Z}}^{\mathbf{2}}\mathbf{-}{\mathbf{R}}^{\mathbf{2}}}}{\mathbf{\omega }}$

Calculate R using dc voltage and current

$$\begin{gathered} \mathbf{R}=\frac{48\mathrm{v}}{5.5\mathrm{A}} \\ =8.72\mathrm{\Omega } \end{gathered}$$

Obtain impedance using ac voltage and current

$$\begin{gathered} \mathbf{Z}=\frac{48\mathrm{V}}{3.6\mathrm{A}} \\ =13.33\mathrm{\Omega } \end{gathered}$$

Plug in values of R,Z and$\omega$ to get L

$$\begin{gathered} \mathrm{L}=\frac{\sqrt{{\left(13.33\mathrm{\Omega }\right)}^2-{\left(8.72\mathrm{\Omega }\right)}^2}}{20\mathrm{rads}/\mathrm{s}} \\ =0.504\mathrm{H} \\ =504\mathrm{mH} \end{gathered}$$

Therefore inductance of the solenoid L = 504mH