An ac generator produces emf $\mathit{\epsilon }\mathbf{=}{\mathit{\epsilon }}_{\mathbf{m}}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\left({\omega }_{d}t-\frac{\pi }{4}\right)\mathbf{}$ , where ${\mathit{\epsilon }}_{\mathbf{m}}\mathbf{=}\mathbf{30}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{V}$and ${\mathit{\omega }}_{\mathbf{d}}\mathbf{=}\mathbf{350}\mathbf{}\mathbf{rad}\mathbf{/}\mathbf{sec}$. The current in the circuit attached to the generator is $\mathit{i}\left(t\right)\mathbf{=}\mathit{I}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\left({\omega }_{d}t-\frac{\pi }{4}\right)$, where $\mathit{I}\mathbf{}\mathbf{=}\mathbf{}\mathbf{620}\mathbf{}\mathbf{mA}$ . (a) At what time after$\mathit{t}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}$does the generator emf first reach a maximum? (b) At what time after$\mathit{t}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}$does 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?

The emf ${\epsilon }_m=30.0\mathrm{V}$,

The angular frequency ${\omega }_d=350\text{rad/sec}$,

The current $I=650\mathrm{mA}$,

To find time at which emf and current is maximum, use. To find capacitance, use the basic formula for capacitance.

Formula:

For maximum emf$\sin \left({\omega }_{d}t-\frac{\pi }{4}\right)=1$.

For maximum current$\sin \left({\omega }_{d}t+\frac{\pi }{4}\right)=1$.

The capacitive reactance $X_c=\frac{1}{\omega C}$,

Here, $\omega$is the angular frequency and $C$is the capacitance.

The current $I=X_c\omega$,

Maximum emf is produced when $\sin \left({\omega }_{d}t-\frac{\pi }{4}\right)=1$

It means, the angle is,

$\begin{array}{rcl}\left({\omega }_{d}t-\frac{\pi }{4}\right)& =& {\sin }^{-1}\left(1\right)\\ & =& \frac{\pi }{2}\end{array}$

$\begin{array}{rcl}t& =& \frac{3\pi }{4\times {\omega }_d}\\ & =& \frac{3\times 3.14}{4\times 350}\\ & =& 6.73\times {10}^{-3}\sec \end{array}$

Hence, the time at which emf reaches maximum is$6.73\times {10}^{-3}\sec$

Time at which current reaches its maximum value

Maximum emf is produced when

$\begin{array}{rcl}sin\left({\omega }_{d}t-\frac{\pi }{4}\right)& =& 1\\ {\omega }_{d}t-\frac{\pi }{4}& =& {\sin }^{-1}\left(1\right)\\ \left({\omega }_{d}t-\frac{\pi }{4}\right)& =& \frac{\pi }{2}\end{array}$

$\begin{array}{rcl}t& =& \frac{\pi }{4\times {\omega }_d}\\ & =& \frac{3.14}{4\times 350}\\ & =& 2.24\times {10}^{-3}\sec \end{array}$

Hence, time at which current reaches maximum is$2.24\times {10}^{-3}\sec$ .

By finding the phase difference between the current and emf, we confirm that the current leads the emf by $\frac{\pi }{2}$so element must be a capacitor.

$X_c=\frac{1}{\omega C}$

$I=X_c\omega$

Therefore, the capacitance is,

$\begin{array}{rcl}C& =& \frac{I}{{\epsilon }_m\omega }\\ & =& \frac{6.2\times {10}^{-3}}{30\times 350}\\ & =& 5.9\times {10}^{-5}F\end{array}$

Hence, the value of capacitance is $5.9\times {10}^{-5}\mathrm{F}$