# A capacitor is connected to a 15 kHz AC source. The peak current is 50 mA when the rms voltage is...

## Question:

A capacitor is connected to a 15 kHz AC source. The peak current is 50 mA when the rms voltage is 8.1 V. What is the peak voltage of the AC source? What is the angular frequency of the AC source? What is the value of the capacitance C? What is the voltage at t =15 {eq}\mu {/eq}s? What is the current through the capacitor at t =15 {eq}\mu {/eq}s?

• What is the peak voltage of the AC source?

We are given the rms value of the voltage, so the peak value equivalent is given by:

{eq}V_p = \sqrt{2} V_{rms} {/eq}

{eq}V_p = \sqrt{2} (8.1 \ V) {/eq}

{eq}V_p = 11.46 \ V {/eq}

• What is the angular frequency of the AC source?

From the given frequency, we have:

{eq}\omega = 2\pi f {/eq}

{eq}\omega = 2\pi (15 \ kHz) {/eq}

So the angular frequency is equal to:

{eq}\omega = 94,248 \ rad/s {/eq}

• What is the value of the capacitance C?

We can use the relationship between the peak values of the current and the voltage to solve for the capacitance.

{eq}I_p = \omega C V_p {/eq}

{eq}C = \frac{I_p }{\omega V_p} {/eq}

{eq}C = \frac{50 \ mA }{ (94,248 \ rad/s)(11.46 \ V)} = 4.63\times 10^{-8} \ F {/eq}

{eq}C = 46.3 \ nF {/eq}

• What is the voltage at t =15 {eq}\mu {/eq}s?

Now, let us use rms value of the voltage that varies with time.

{eq}V(t) = \frac{1}{\omega C} I_p \sin (\omega t +90) {/eq}

So;

{eq}V(15) = \frac{1}{(94,248 \ rad/s)(46.3 \ nF)} (50 \ mA) \sin ((94,248 \ rad/s)(15 \ \mu s)\frac{180}{2\pi} +90) {/eq}

{eq}V(15) = 8.71 \ V {/eq}

• What is the current through the capacitor at t =15 {eq}\mu {/eq}s?

Similarly, let use the rms current with respect to time:

{eq}I(t) = \omega C V(t) {/eq}

So using directly the answer in the previous part,

{eq}I(15) = \omega C V(15) {/eq}

{eq}I(15) = (94,248 \ rad/s)(46.3 \ nF)(8.71 \ V) = 0.038 \ A {/eq}