Question

A capacitor is connected to a $15kHz$oscillator. The peak current is $65mA$when the rms voltage is $6.0V$. What is the value of the capacitance $C$?

Short answer

The value of the capacitance is$C=0.081\times {10}^{-7}$.

Step-by-step solution

Given information

We have been given that a capacitor is connected to a$15kHz$oscillator and the peak current is$65mA$when the rms voltage is $6.0V$.

We need to find the value of the capacitance$C$.

Simplify

Given values:

A capacitor is connected to a $15kHz$oscillator i.e. frequency is $f=15\times {10}^{3}Hz$.

Therefore,$\omega =2\pi f=30\pi \times {10}^{3}rad/s$.

The peak current is $I_c=65mA$.

The rms voltage is $V_{rms}=6.0V$.

The peak voltage is $V_c=\sqrt{2}{V}_{rms}=6\left(\sqrt{2}\right)V$.

As we know the formula :

$I_c=\omega V_{c}C$ (Let this equation be ($1$))

Divide both sides of equation ($1$) by $\omega V_c$,

$\frac{I_c}{\omega V_c}=\frac{\omega V_{c}C}{\omega V_c}$

$C=\frac{I_c}{\omega V_c}$ (Let this equation be (localid="1649930919468" $2$))

Substituting the values of $I_c,\omega ,V_c$in equation (localid="1649930928779" $2$), we get

localid="1649930937509" $C=\frac{65}{30\pi \times 6\left(\sqrt{2}\right)}\times {10}^{-6}$

On simplifying,

localid="1649930945170" $C=0.081\times {10}^{-7}F$