Solve Capacitor Circuits Homework: 20nF, 5V Peak Voltage, 50mA Peak Current

[Sarah88]

Homework Statement

A 20 nF capacitor is connected across an AC generator that produces a peak voltage of 5.0 V.
- At what frequency f is the peak current 50 mA
- What is the instantaneous value of the emf at the instant when ic= Ic?

Homework Equations

Ic= C(2*pi*f*Vc)
Ic= Vc/Xc
Xc= 1/(2*pi*f*C)
possibly E= Eocos(2*pi*f*t)

The Attempt at a Solution

I understood the first part of the question and found the frequency to be 79365 Hz. I tried using this value with the equation Ic=C(2*pi*f*Vc) and solving for Vc (assuming Vc is emf) so : 50*10^-3 A= (20*10^-9 F) (2*pi* 79365 Hz*Vc) therefore Vc= 5 V. However, the answer to the problem is 0 and I'm not sure how that answer was obtained. Any hints, or useful equations would be great, thank you!

 

[ideasrule]

I think you're misunderstanding your first equation. Ic= C(2*pi*f*Vc) gives the peak current, Ic, in terms of the peak voltage, Vc. Alternatively, it gives you the RMS current in terms of the RMS voltage, since peak and RMS voltages/currents differ by only a factor of sqrt(2).

However, peak current and peak voltage do not occur at the same time because the capacitor has impedance. What is the phase angle between current and voltage for a purely capacitive circuit?

 

[Sarah88]

Oh, ok, that explanation of the Ic equation definitely makes a lot more sense. Logically, the phase angle between current and voltage for a purely capacitive circuit seems to be 90 degrees, and then it appears the Emf of an AC voltage source equation can be utilized (and cosine of 90 is 0, which correlates to the book's answer). Thank you for your help!