91Ó°ÊÓ

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$C$.

Given values:

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

Therefore,$\mathrm{Ó\neg }=2\mathrm{Ï€}f=30\mathrm{Ï€}×{10}^{3}rad/s$.

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 :

$I_c=\mathrm{Ó\neg }{V}_{c}C$ (Let this equation be ($1$))

Divide both sides of equation ($1$) by $\mathrm{Ó\neg }{V}_c$,

$\frac{I_c}{\mathrm{Ó\neg }{V}_c}=\frac{\mathrm{Ó\neg }{V}_{c}C}{\mathrm{Ó\neg }{V}_c}$

$C=\frac{I_c}{\mathrm{Ó\neg }{V}_c}$ (Let this equation be (localid="1649930919468" $2$))

Substituting the values of $I_c,\mathrm{Ó\neg },V_c$in equation (localid="1649930928779" $2$), we get

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

On simplifying,

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