Question

An $\mathbf{50}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{mH}$inductor is connected, as in Figure to an ac generator with ${\mathbf{\in }}_{\mathbf{m}}\mathbf{=}\mathbf{30}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{V}$.(a) What is the amplitude of the resulting alternating current if the frequency of the emf is $\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{kHz}$? and (b) What is the amplitude of the resulting alternating current if the frequency of the emf is $\mathbf{8}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{kHz}$?

An inductor is connected across an alternating-current generator.

Short answer

1. The amplitude of the resulting alternating current if the frequency of the emf has a value$f=1.00\mathrm{kHz}$is $95.5\mathrm{mA}$.
2. The amplitude of the resulting alternating current if the frequency of the emf has a value $f=8.00\mathrm{kHz}$is $11.9\mathrm{mA}$.

Step-by-step solution

The given data

1. Emf of the ac generator,${\in }_m=30.0\mathrm{V}$
2. Series inductance,$L=5.0\mathrm{mH}=5\times {10}^{-3}\mathrm{H}$

Understanding the concept of capacitive reactance and Ohm’s law

An inductor's obstruction in the current flow is known as inductive reactance. By determining the reactance due to the oscillation frequency within an inductor, we can get the current value by substituting this value in the voltage equation of Ohm's law.

The inductive reactance of a capacitor,

$X_L=2{\mathrm{\pi f}}_{\mathrm{d}}\mathrm{L}$...... (i)

The voltage equation from Ohm’s law,

$i_L=\frac{{}_{{\in }_m}}{X_L}$........ (ii)

Here, $f_d$is the driving frequency of oscillations, $L$is the inductance of the inductor, and ${\in }_m$is the emf applied across the circuit.

a) Calculation of the resulting current for frequency 1.00 kHz

Since the circuit contains only the inductor and sinusoidal generator, the potential difference across the inductor and sinusoidal generator are the same.

$V_L={\in }_m$

${\in }_m$is the amplitude of the sinusoidal generator.

Thus, substituting equation (ii) in equation (i), we can get the resulting current through the inductor for frequency$f=1.00\mathrm{kHz}$ as follows:

$$\begin{gathered} I_L=\frac{{\in }_m}{2{\mathrm{\pi f}}_{\mathrm{d}}\mathrm{L}} \\ {I}_L=\frac{30\mathrm{V}}{2\mathrm{\pi }\left(1\times {10}^3\mathrm{Hz}\right)\left(50\times {10}^{-3}\mathrm{H}\right)} \\ {I}_L=0.0955\mathrm{A} \\ {I}_L=95.5\mathrm{mA} \end{gathered}$$

Hence, the value of current is $95.5\mathrm{mA}$.

b) Calculation of the resulting current for frequency 8.00 kHz

The amplitude of alternating current if the frequency of emf is$8\mathrm{kHz}$.

In this case, the frequency is eight times larger than part a). Thus, the value of the resulting current is given as:

$$\begin{gathered} I_L=\frac{1}{8}\left(\frac{{\in }_m}{2{\mathrm{\pi f}}_{\mathrm{d}}\mathrm{L}}\right) \\ {I}_L=\frac{95.5\mathrm{mA}}{8} \\ {I}_L=11.9\mathrm{mA} \end{gathered}$$

Hence, the value of the current is $11.9\mathrm{mA}$.