Question

An ac generator has emf $\mathit{\epsilon }\mathbf{=}{\mathit{\epsilon }}_{\mathbf{m}}\mathit{s}\mathit{i}\mathit{n}\mathbf{(}{\mathbf{\omega }}_{\mathbf{d}}\mathbf{t}\mathbf{-}\frac{\mathbf{\pi }}{\mathbf{4}}\mathbf{)}$, where${\mathit{\epsilon }}_{\mathbf{m}}\mathbf{=}\mathbf{30}\mathbf{\text{V}}$ and${\mathit{\omega }}_{\mathbf{d}}\mathbf{=}\mathbf{350}\mathbf{}\mathbf{\text{rad/s}}$. The current produced in a connected circuit is$\mathit{i}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathit{I}\mathit{s}\mathit{i}\mathit{n}\mathbf{(}{\mathbf{\omega }}_{\mathbf{d}}\mathbf{t}\mathbf{-}\frac{\mathbf{3}\mathbf{\pi }}{\mathbf{4}}\mathbf{)}$, where $\mathit{I}\mathbf{=}\mathbf{620}\mathbf{}\mathbf{\text{mA}}$. At what time after $\mathit{t}\mathbf{}\mathbf{=}\mathbf{0}$does (a) the generator emf first reach a maximum and (b) the current first reach a maximum? (c) The circuit contains a single element other than the generator. Is it a capacitor, an inductor, or a resistor? Justify your answer. (d) What is the value of the capacitance, inductance, or resistance, as the case may be?

Short answer

1. The time when the generator emf first reaches a maximum is $6.73\times {10}^{-3}\text{s}$.
2. The time when the current first reaches a maximum is$1.12\times {10}^{-2}\text{s}$.
3. The single element other than the generator that the circuit contains is an inductor.
4. The value of the inductor is $0.138\text{H}$.

Step-by-step solution

The given data

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

Understanding the concept of inductive reactance and Ohm’s law

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,

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

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

a) Calculation of the time when the emf reaches maximum

The generator emf is given as follows:

$\epsilon ={\epsilon }_m\sin \left({\omega }_{d}t-\left(\frac{\pi }{4}\text{rad}\right)\right)$

The emf is maximum when

$$\begin{gathered} \sin \left({\omega }_{d}t-\left(\frac{\pi }{4}\text{rad}\right)\right)=1 \\ \left({\omega }_{d}t-\left(\frac{\pi }{4}\text{rad}\right)\right)=\left(\frac{\pi }{2}\pm 2n\pi \right)\text{rad} \end{gathered}$$

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

$$\begin{gathered} \left({\omega }_{d}t-\left(\frac{\pi }{4}\text{rad}\right)\right)=\left(\frac{\pi }{2}\text{rad}\right)\pm 2\left(0\right)\pi \\ \left({\omega }_{d}t-\left(\frac{\pi }{4}\text{rad}\right)\right)=\left(\frac{\pi }{2}\text{rad}\right) \\ {\omega }_{d}t=\left(\frac{3\pi }{4}\text{rad}\right) \end{gathered}$$

The above equation can also be written as-

$$\begin{gathered} t=\left(\frac{3\pi }{4}\text{rad}\right)\times \frac{1}{{\omega }_d} \\ =\left(\frac{3\pi }{4}\text{rad}\right)\times \frac{1}{\left(350\text{rad/s}\right)} \\ =6.73\times {10}^{-3}\text{s} \end{gathered}$$

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

b) Calculation of the time when the current first reaches a maximum

The alternating current is given as:

$i\left(t\right)=I\sin \left({\omega }_{d}t-\left(\frac{3\pi }{4}\text{rad}\right)\right)$

The current is maximum when

$$\begin{gathered} \sin \left({\omega }_{d}t-\left(\frac{3\pi }{4}\text{rad}\right)\right)=1 \\ \left({\omega }_{d}t-\left(\frac{3\pi }{4}\text{rad}\right)\right)=\left(\frac{\pi }{2}\pm 2n\pi \right)\text{rad} \end{gathered}$$

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

localid="1664186765125" $$\begin{gathered} \left({\omega }_{d}t-\left(\frac{3\pi }{4}\text{rad}\right)\right)=\left(\frac{\pi }{2}\text{rad}\right)\pm 2\left(0\right)\pi \\ \left({\omega }_{d}t-\left(\frac{3\pi }{4}\text{rad}\right)\right)=\left(\frac{\pi }{2}\text{rad}\right) \\ {\omega }_{d}t=\left(\frac{5\pi }{4}\text{rad}\right) \end{gathered}$$

The above equation can also be written as-

localid="1664186770953" $$\begin{gathered} t=\left(\frac{5\pi }{4}\text{rad}\right)\times \frac{1}{{\omega }_d} \\ =\left(\frac{5\pi }{4}\text{rad}\right)\times \frac{1}{\left(350\text{rad/s}\right)} \\ =1.12\times {10}^{-2}\text{s} \end{gathered}$$

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

c) Calculation of the element contained in the circuit

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

d) Calculation of the value of inductance

For the pure inductive circuit,$V_L={\epsilon }_m$

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

$$\begin{gathered} I_L=\frac{{\epsilon }_m}{{\omega }_{d}L} \\ L=\frac{{\epsilon }_m}{{\omega }_d{I}_L} \\ =\frac{30\text{V}}{\left(620\times {10}^{-3}\text{A}\right)\left(350\frac{\text{rad}}{\text{s}}\right)} \\ =0.138 \text{H} \end{gathered}$$

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