91Ó°ÊÓ

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

The angular frequency ${\mathrm{Ó\neg }}_d=350\text{rad/sec}$,

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$\sin \left({\mathrm{Ó\neg }}_{d}t-\frac{\mathrm{Ï€}}{4}\right)=1$.

For maximum current$\sin \left({\mathrm{Ó\neg }}_{d}t+\frac{\mathrm{Ï€}}{4}\right)=1$.

The capacitive reactance $X_c=\frac{1}{\mathrm{Ó\neg }C}$,

Here, $\mathrm{Ó\neg }$is the angular frequency and $C$is the capacitance.

The current $I=X_c\mathrm{Ó\neg }$,

Maximum emf is produced when $\sin \left({\mathrm{Ó\neg }}_{d}t-\frac{\mathrm{Ï€}}{4}\right)=1$

It means, the angle is,

$\begin{array}{rcl}\left({\mathrm{Ó\neg }}_{d}t-\frac{\mathrm{Ï€}}{4}\right)& =& {\sin }^{-1}\left(1\right)\\ & =& \frac{\mathrm{Ï€}}{2}\end{array}$

$\begin{array}{rcl}t& =& \frac{3\mathrm{Ï€}}{4×{\mathrm{Ó\neg }}_d}\\ & =& \frac{3×3.14}{4×350}\\ & =& 6.73×{10}^{-3}\sec \end{array}$

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

Time at which current reaches its maximum value

Maximum emf is produced when

$\begin{array}{rcl}sin\left({\mathrm{Ó\neg }}_{d}t-\frac{\mathrm{Ï€}}{4}\right)& =& 1\\ {\mathrm{Ó\neg }}_{d}t-\frac{\mathrm{Ï€}}{4}& =& {\sin }^{-1}\left(1\right)\\ \left({\mathrm{Ó\neg }}_{d}t-\frac{\mathrm{Ï€}}{4}\right)& =& \frac{\mathrm{Ï€}}{2}\end{array}$

$\begin{array}{rcl}t& =& \frac{\mathrm{Ï€}}{4×{\mathrm{Ó\neg }}_d}\\ & =& \frac{3.14}{4×350}\\ & =& 2.24×{10}^{-3}\sec \end{array}$

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

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

$X_c=\frac{1}{\mathrm{Ó\neg }C}$

$I=X_c\mathrm{Ó\neg }$

Therefore, the capacitance is,

$\begin{array}{rcl}C& =& \frac{I}{{\mathrm{Î\mu }}_m\mathrm{Ó\neg }}\\ & =& \frac{6.2×{10}^{-3}}{30×350}\\ & =& 5.9×{10}^{-5}F\end{array}$

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