Engineering How Do You Calculate Instantaneous Voltage Across a Capacitor in an AC Circuit?

AI Thread Summary
To calculate the instantaneous voltage across a capacitor in an AC circuit, the correct approach involves understanding the phase relationship between voltage and current. The phase shift for a capacitor is typically 90 degrees, meaning the voltage lags the current. The user initially struggled with incorporating the phase shift into their calculations but later realized that using the formula Vc = Vmsin(2πft - π/2) simplifies the process. By calculating the reactance and using phasor diagrams, the user found the correct voltage across the capacitor. Ultimately, recognizing the role of phase shift clarified the solution, leading to accurate results.
Flupps
Messages
3
Reaction score
0

Homework Statement



Series_RC_Circuit.jpg


For the circuit in question1above: R=(3.74x10^2) (Ω), C=(3.8700x10^-7) (F), f=(4.610x10^3) (Hz) and the peak amplitude of the voltage of the source (ES)=(6.185x10^1) (V). Calculate the instantaneous value of the voltage across the capacitor at t=(4.110x10^-5) (s) after the positive zero crossing of the current waveform.

Homework Equations



v=Vmsin(2*pi*f *t + θ)
Xc = 1/(2*pi*f*C)
θ = tan^-1(Xc/R)

The Attempt at a Solution



The part that's confusing me is where it says "after the positive zero crossing of the current waveform". I've tried using the phase shift angle and subtracting 90 degrees to put into equation:
v=Vmsin(2*pi*t + θ). I've also tried using just the phase shift angle, using only 90 degrees, and using 90 degrees + phase shift angle, and none have gotten me an answer any where near what it's supposed to be.

What I've done so far is calculated Xc = 89.2 ohms
then calculated phase shift = 13.41 degrees
then tried a number of different combinations of angles using the formula v=Vmsin(2*pi*t + θ)

Could anyone tell me where I'm going wrong? This question is driving me nuts :(

after looking at the question some more, perhaps this is what I am supposed to do?

vr = vmsin(2*pi*f*t - 13.41 degrees (0.216368996rad) = 51.16253852 V
vs = vmsin(2*pi*f*t) = 57.43068007 V

vc = vs - vr = 57.43068007 V - 51.16253852 V = 6.268 V
 
Last edited:
Physics news on Phys.org
You have the voltage waveform - what is the corresponding current waveform?
(i.e. what is the relationship between voltage and current?)
 
Isn't the capacitor voltage lagging the current by 90 deg? I'm not sure how to incorporate the phase shift though, and I've tried this question a few more times and still end up with an answer that's way off.
 
The phase shift comes out automatically if you use complex impedances ... if you are just using reactances, you should use a phasor diagram: the loop law voltages add head-to-tail.
 
Ah thanks! With the same problem with different numbers I had R = 350, C = 380nF, f = 3.91kHz, (Es)m = 29.86v

Got an answer of -3.12v which was correct.

It became much simpler now that I know the phase shift isn't used... just used formula (Vc)msin(2*pi*f*t-(pi/2)) after calculating the other stuff and it gave me the right answer

Thanks for the help!
 

Similar threads

Replies
5
Views
2K
Replies
26
Views
3K
Replies
2
Views
2K
Replies
2
Views
2K
Replies
2
Views
1K
Replies
5
Views
3K
Back
Top