91Ó°ÊÓ

1. Emf equation,$\mathrm{Î\mu }={\mathrm{Î\mu }}_m\sin \left({\mathrm{Ó\neg }}_{d}t-\frac{\mathrm{Ï€}}{4}\right)$
2. Amplitude of emf,${\mathrm{Î\mu }}_m=30\text{V}$
3. Angular frequency,${\mathrm{Ó\neg }}_d=350\text{rad/s}$
4. Current equation, $i\left(t\right)=I\sin \left({\mathrm{Ó\neg }}_{d}t-\frac{3\mathrm{Ï€}}{4}\right)$
5. Amplitude of current,$I=620\text{mA}$

The obstruction offered in the path of alternating current by the inductor is called inductive reactance. Using the value of emf and current in the given problem. These values are maximum when the sine term has a value of unity. Using this, we find the time when the generator emf first reaches its maximum and the current first reaches maximum. By identifying the phase angle of current and emf, we can find the element of the circuit other than the generator. Using the relation of current amplitude and voltage amplitude, we can find the inductance of the inductor.

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

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

The inductive reactance of the inductor,

${\mathit{X}}_{\mathit{L}}\mathbf{=}{\mathit{Ó\neg }}_{\mathit{d}}\mathit{L}$ (ii)

Here,$\mathit{L}$ is the inductance,${\mathit{I}}_{\mathit{L}}$ is the current flowing through the inductor,${\mathit{Ó\neg }}_{\mathit{d}}$is the driving angular frequency.

The generator emf is given as follows:

$\mathrm{Î\mu }={\mathrm{Î\mu }}_m\sin \left({\mathrm{Ó\neg }}_{d}t-\left(\frac{\mathrm{Ï€}}{4}\text{rad}\right)\right)$

The emf is maximum when

$$\begin{gathered} \sin \left({\mathrm{Ó\neg }}_{d}t-\left(\frac{\mathrm{Ï€}}{4}\text{rad}\right)\right)=1 \\ \left({\mathrm{Ó\neg }}_{d}t-\left(\frac{\mathrm{Ï€}}{4}\text{rad}\right)\right)=\left(\frac{\mathrm{Ï€}}{2}Â\pm 2n\mathrm{Ï€}\right)\text{rad} \end{gathered}$$

This will be maximum when $n=0$for the first time. Thus, the above value becomes

$$\begin{gathered} \left({\mathrm{Ó\neg }}_{d}t-\left(\frac{\mathrm{Ï€}}{4}\text{rad}\right)\right)=\left(\frac{\mathrm{Ï€}}{2}\text{rad}\right)Â\pm 2\left(0\right)\mathrm{Ï€} \\ \left({\mathrm{Ó\neg }}_{d}t-\left(\frac{\mathrm{Ï€}}{4}\text{rad}\right)\right)=\left(\frac{\mathrm{Ï€}}{2}\text{rad}\right) \\ {\mathrm{Ó\neg }}_{d}t=\left(\frac{3\mathrm{Ï€}}{4}\text{rad}\right) \end{gathered}$$

The above equation can also be written as-

$$\begin{gathered} t=\left(\frac{3\mathrm{Ï€}}{4}\text{rad}\right)×\frac{1}{{\mathrm{Ó\neg }}_d} \\ =\left(\frac{3\mathrm{Ï€}}{4}\text{rad}\right)×\frac{1}{\left(350\text{rad/s}\right)} \\ =6.73×{10}^{-3}\text{s} \end{gathered}$$

Hence, the emf is maximum when time is $6.73×{10}^{-3}\text{s}$.

The alternating current is given as:

$i\left(t\right)=I\sin \left({\mathrm{Ó\neg }}_{d}t-\left(\frac{3\mathrm{Ï€}}{4}\text{rad}\right)\right)$

The current is maximum when

$$\begin{gathered} \sin \left({\mathrm{Ó\neg }}_{d}t-\left(\frac{3\mathrm{Ï€}}{4}\text{rad}\right)\right)=1 \\ \left({\mathrm{Ó\neg }}_{d}t-\left(\frac{3\mathrm{Ï€}}{4}\text{rad}\right)\right)=\left(\frac{\mathrm{Ï€}}{2}Â\pm 2n\mathrm{Ï€}\right)\text{rad} \end{gathered}$$

This will be maximum when$n=0$for the first time. Thus, the above value becomes

localid="1664186765125" $$\begin{gathered} \left({\mathrm{Ó\neg }}_{d}t-\left(\frac{3\mathrm{Ï€}}{4}\text{rad}\right)\right)=\left(\frac{\mathrm{Ï€}}{2}\text{rad}\right)Â\pm 2\left(0\right)\mathrm{Ï€} \\ \left({\mathrm{Ó\neg }}_{d}t-\left(\frac{3\mathrm{Ï€}}{4}\text{rad}\right)\right)=\left(\frac{\mathrm{Ï€}}{2}\text{rad}\right) \\ {\mathrm{Ó\neg }}_{d}t=\left(\frac{5\mathrm{Ï€}}{4}\text{rad}\right) \end{gathered}$$

The above equation can also be written as-

localid="1664186770953" $$\begin{gathered} t=\left(\frac{5\mathrm{Ï€}}{4}\text{rad}\right)×\frac{1}{{\mathrm{Ó\neg }}_d} \\ =\left(\frac{5\mathrm{Ï€}}{4}\text{rad}\right)×\frac{1}{\left(350\text{rad/s}\right)} \\ =1.12×{10}^{-2}\text{s} \end{gathered}$$

Hence, the value of the time is localid="1664186725589" $1.12×{10}^{-2}\text{s}$.

According to the given information, it is clear that current is lagging behind emf by a phase angle $\frac{\mathrm{Ï€}}{2}\text{rad}$. Thus, the circuit is purely inductive and the component connected is a inductor.

For the pure inductive circuit,$V_L={\mathrm{Î\mu }}_m$

Thus, the value of the inductance is given using equation (ii) in equation (i) as follws:

$$\begin{gathered} I_L=\frac{{\mathrm{Î\mu }}_m}{{\mathrm{Ó\neg }}_{d}L} \\ L=\frac{{\mathrm{Î\mu }}_m}{{\mathrm{Ó\neg }}_d{I}_L} \\ =\frac{30\text{V}}{\left(620×{10}^{-3}\text{A}\right)\left(350\frac{\text{rad}}{\text{s}}\right)} \\ =0.138 \text{H} \end{gathered}$$

Hence, the value of the inductance is $0.138\text{H}$.