Question

An ac generator has emf $\mathit{\epsilon }\mathbf{=}{\mathit{\epsilon }}_{\mathbf{m}}\mathit{s}\mathit{i}\mathit{n}{\mathit{\omega }}_{\mathbf{d}}\mathit{t}$, with ${\mathit{\epsilon }}_{\mathbf{m}}\mathbf{=}\mathbf{25}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{V}$and ${\mathit{\omega }}_{\mathbf{d}}\mathbf{=}\mathbf{377}\mathbf{}\raisebox{1ex}{$\mathbf{rad}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{s}$}\right.$. It is connected to an $\mathit{L}\mathbf{=}\mathbf{12}\mathbf{.}\mathbf{7}\mathbf{}\mathbf{H}$inductor. (a) What is the maximum value of the current? (b) When the current is a maximum, what is the emf of the generator? (c) When the emf of the generator isrole="math" localid="1663220843017" $\mathbf{-}\mathbf{12}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{V}$ and increasing in magnitude, what is the current?

Short answer

1. The maximum value of the current is $5.22\times {10}^{-3}\mathrm{A}$.
2. Emf of the generator when current is maximum is $0$.
3. When the emf of the generator is$-12.5\mathrm{V}$ and increasing in magnitude, the current is $4.51\times {10}^{-3}\mathrm{A}$.

Step-by-step solution

The given data

1. Amplitude of emf ${\epsilon }_m=25.0\mathrm{V}$
2. Angular frequency ${\omega }_d=377\raisebox{1ex}{$\mathrm{rad}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.$
3. Inductance $L=12.7\mathrm{H}$
4. Amplitude of emf${\epsilon }_m=-12.5\mathrm{V}$

Understanding the concept of a purely inductive circuit

An electrical circuit made by connecting an inductor only is called a purely inductive circuit. We use the condition of the purely inductive circuit, and substituting the value of inductive reactance, we get the maximum current value. When the current is maximum, its first derivative is zero. Using this idea, we can find the emf of the generator.

The voltage equation of an inductor due to Ohm’s law,

${\mathit{V}}_{\mathbf{L}}\mathbf{=}{\mathit{l}}_{\mathbf{L}}{\mathit{X}}_{\mathbf{L}}$.......(i)

The emf equation due to the current change in the inductor,

${\mathit{\epsilon }}_{\mathbf{L}}\mathbf{=}\mathbf{-}\mathit{L}\frac{\mathbf{di}}{\mathbf{dt}}$.........(ii)

The current equation for sinusoidal varying current,

${\mathit{i}}_{\mathbf{L}}\left(t\right)\mathbf{=}{\mathit{l}}_{\mathbf{L}}\mathit{s}\mathit{i}\mathit{n}\left({\omega }_{d}t-\varphi \right)$........(iii)

The inductive reactance of the inductor,

${\mathit{X}}_{\mathbf{L}}\mathbf{=}{\mathit{\omega }}_{\mathbf{d}}\mathit{L}$..........(iv)

Here, $\mathit{L}$is the inductance, ${\mathit{l}}_{\mathbf{L}}$is the current flowing through the inductor, ${\mathit{\omega }}_{\mathbf{d}}$is the driving angular frequency, and$\mathit{\varphi }$is the phase constant of the current.

a) Calculation of the maximum current

For a pure inductive circuit, the potential difference $V_L$across the inductor is always equal to the emf.

${\epsilon }_m=V_L$.....(a)

Thus, using equation (a) and equation (iv) in equation (i), we get the maximum current value as:

role="math" localid="1663224542349" $$\begin{gathered} l_L=\frac{{\epsilon }_m}{{\omega }_{D}L} \\ {l}_L=\frac{25.0\mathrm{V}}{\left(377{\displaystyle \raisebox{1ex}{$\mathrm{rad}$}\!\left/ \!\raisebox{-1ex}{$s$}\right.}\right)\left(12.7\mathrm{H}\right)} \\ {l}_L=5.22\times {10}^{-3}\mathrm{A} \end{gathered}$$

Hence, the value of the current is $5.22\times {10}^{-3}\mathrm{A}$.

b) Calculation of the emf of the generator when current is maximum

$i_L\left(t\right)$If the current in a coil changes with time, an emf is induced in the coil. When the current is maximum, its first derivative is zero.

$\frac{di}{dt}=0$

Using equation (ii), this implies ${\epsilon }_L=0$

In another way, in an ac circuit with pure inductive load, the alternating current $i_L\left(t\right)$in the inductance lags the alternating potential difference $\epsilon \left(t\right)$by $90°$.

Thus, using this value in equation (iii), we get the equation of current and emf as:

role="math" localid="1663224779750" $$\begin{gathered} i_L\left(t\right)=l_L\sin \left({\omega }_{d}t-\frac{\mathrm{\pi }}{2}\right)...........\left(a\right) \\ {\epsilon }_L\left(t\right)={\epsilon }_m\sin {\omega }_{d}t \end{gathered}$$

From the above two relations, we can say that when the current is at its maximum, the alternating emf is zero.

c) Calculation of the current when emf of the generator is -12.5 V

The Emf value, in this case, is exactly half the previous value and negative.

$\epsilon =-\frac{{\epsilon }_m}{2}$

And it is increasing in magnitude. Since it already has a negative sign, thus increase in its magnitude results in a more negative value.

${\epsilon }_L\left(t\right)={\epsilon }_m\sin {\omega }_{d}t$

Taking the first derivative oftheabove relation, we get the emf equation as:

$$\begin{gathered} \frac{d}{dt}{\epsilon }_L\left(t\right)=\frac{d}{dt}\left({\epsilon }_m\sin {\omega }_{d}t\right) \\ \frac{d}{dt}{\epsilon }_L\left(t\right)={\omega }_d{\epsilon }_m\cos \left({\omega }_{d}t\right) \end{gathered}$$

And this result should be negative. To get the negative value, $\cos \left({\omega }_{d}t\right)$must be negative. That implies:

$$\begin{gathered} {\omega }_{d}t=2n\mathrm{\pi }-\frac{5\mathrm{\pi }}{6} \\ {\mathrm{\omega }}_{\mathrm{d}}\mathrm{t}<0 \end{gathered}$$

Substituting this value in the alternating current of equation (iii), we get the current equation as:

$$\begin{gathered} i_L\left(t\right)=l_L\sin \left({\omega }_{d}t-\frac{\mathrm{\pi }}{2}\right) \\ {i}_L\left(t\right)=l_L\sin \left(2n\mathrm{\pi }-\frac{5\mathrm{\pi }}{6}-\frac{\mathrm{\pi }}{2}\right) \\ {i}_L\left(t\right)=l_L\sin \left(2n\mathrm{\pi }-\frac{4\mathrm{\pi }}{3}\right)...................\left(\mathrm{v}\right) \end{gathered}$$

$\sin \left(2n\mathrm{\pi }-\frac{4\mathrm{\pi }}{3}\right)=\sin 2n\mathrm{\pi cos}\frac{4\mathrm{\pi }}{3}-\cos 2\mathrm{n\pi sin}\frac{4\mathrm{\pi }}{3}$

As, $\sin 2n\mathrm{\pi }=0$and $\cos 2n\mathrm{\pi }=1$, we get

$$\begin{gathered} \sin \left(2n\mathrm{\pi }-\frac{4\mathrm{\pi }}{3}\right)=\sin \frac{4\mathrm{\pi }}{3} \\ \sin \left(2n\mathrm{\pi }-\frac{4\mathrm{\pi }}{3}\right)=-\raisebox{1ex}{$\sqrt{3}$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \end{gathered}$$

Thus, using these above values, equations (v) become:

$$\begin{gathered} i_L\left(t\right)=l_L\left(\frac{\sqrt{3}}{2}\right) \\ {i}_L\left(t\right)=\left(5.22\times {10}^{-3}\mathrm{A}\right)\left(\frac{\sqrt{3}}{2}\right) \\ {i}_L\left(t\right)=4.51\times {10}^{-3}\mathrm{A} \end{gathered}$$

Hence, the value of the current is $4.51\times {10}^{-3}\mathrm{A}$.