Question

The peak current to and from a capacitor is $10mA$. What is the peak current if

a. The emf frequency is doubled?

b. The emf peak voltage is doubled (at the original frequency)

Short answer

Result is:

$\text{a.)}{I}_o^{\text{'}}=20\mathrm{mA}$

$\text{b.)}{I}_o^{\text{'}}=20\mathrm{mA}$

Step-by-step solution

Part (a) Step 1: Given information

We have been given that$I_o=10\mathrm{mA}$

we will use the formula to sovle is:

$I_o=\left(2\pi f\right)V_{o}C\text{and}{I}_o\alpha f$

Part(a) Step 2: Simplify

if $f$is the frequency doubled:

$I_o^{\text{'}}=2·I_o$

As,$I_o^{}=10mA$

$I_o^{\text{'}}=2·10mA$

$I_o^{\text{'}}=20mA$

Part (b) Step 1: Given information

We have been given that$I_0=10\mathrm{mA}$

According to the formula:

$I_o=\left(2\pi f\right)V_{o}C\text{and}{I}_o\alpha f$

Part (b) Step 2: Simplify

$I_o\alpha V_o\Rightarrow V_o\left(\text{doubled}\right)\Rightarrow I_o^{\text{'}}=2I_o$

$I_o^{\text{'}}=2·10\mathrm{mA}$

$I_o^{\text{'}}=20mA$