Question

A $0.30\mu F$capacitor is connected across an AC generator that produces a peak voltage of $10V$. What is the peak current to and from the capacitor if the emf frequency is

(a) $100Hz$?

(b)$100kHz$?

Short answer

$\left(\mathrm{a}\right){\mathrm{I}}_o=1.9\times {10}^{-3}\mathrm{A}$

role="math" localid="1649956592903" $\left(\mathrm{b}\right)\hspace{0.17em}{\mathrm{I}}_o=1.9\mathrm{A}$

Step-by-step solution

Part (a) Step 1: Given Information:

We have been given that

Capacitance ($C$)= $0.30\mu F$

Peak voltage ($E_o$)= $10V$

Emf frequency (n)= localid="1649958487243" $100Hz$

Part (a) Step 2: Solving:

Peak current I_ois given by :${\mathbf{I}}_o={\mathbf{E}}_o\times \omega \times \mathbf{C}$

Putting the value of$\left(\omega \right)\text{as}2\pi n$in the above equation:${\mathrm{I}}_o={\mathrm{E}}_o\times 2\pi \mathrm{n}\times \mathbf{C}$

Using the given values in the updated equation we get:${\mathrm{I}}_o=\left(10\right)\times 2\pi \left(100\right)\times \left(0.30\times {10}^{-6}\right)$

There the peak value of current for a frequency of $100Hz$is${\mathrm{I}}_o=1.9\times {10}^{-3}\mathrm{A}$

Part (b) Step 1: Given Information:

We have been given that

Capacitance ($C$)= $0.30uF$

Peak voltage ($E_o$)= $10V$

Emf frequency (n)=$100kHz$

Part (b) Step 2: Solving:

Using the expression of I_o that we earlier used:${\mathrm{I}}_o={\mathrm{E}}_o\times 2\pi \mathrm{n}\times \mathbf{C}$

Putting the known values:${\mathrm{I}}_o=\left(10\right)\times 2\pi \left(100\times {10}^3\right)\times \left(0.30\times {10}^{-6}\right)$

${\mathrm{I}}_o=1.9\mathrm{A}$

Therefore the peak current value for an emf frequency of$100kHz$is ${\mathrm{I}}_o=1.9\mathrm{A}$