Electrical Engineering - RC Circuits - Discussion
Discussion Forum : RC Circuits - General Questions (Q.No. 18)
18.
When the frequency of the voltage applied to a series RC circuit is increased, the phase angle
Discussion:
5 comments Page 1 of 1.
Aleksandr said:
10 years ago
Phase angle = arctg ((XL-XC)/R), XL = 0, phase angle = arctg(-XC/R),
So tg(alpha) = -XC/R, when XC decrease tg (alpha) from minus go to 0 to be increased.
So alpha also from minus go to 0. So the question is how you consider angle increase or decrease. So if alpha from -40 goes to 0 it's decrease or increase?
So tg(alpha) = -XC/R, when XC decrease tg (alpha) from minus go to 0 to be increased.
So alpha also from minus go to 0. So the question is how you consider angle increase or decrease. So if alpha from -40 goes to 0 it's decrease or increase?
Yeji said:
4 years ago
- If you want to see it visually you can draw a phasor diagram relationship between Xc and R.
- Decreasing the frequency means your Xc will also decrease.
- Xc now is < than R.
- The angle formed between them will also decrease.
- Decreasing the frequency means your Xc will also decrease.
- Xc now is < than R.
- The angle formed between them will also decrease.
Amit Kumar Saha said:
1 decade ago
We know, Xc=1/2*pi*f*c ------ i
Hence, from eqn. i, if f(frequency) increased, Xc gradually decreased.
Also we know, Phase angel= tan^-1(Xc/R)
So if Xc decreased, also phase angle gradually decreased.
Hence, from eqn. i, if f(frequency) increased, Xc gradually decreased.
Also we know, Phase angel= tan^-1(Xc/R)
So if Xc decreased, also phase angle gradually decreased.
Niru said:
8 years ago
Alpha=tan-1(xc/r).
xc=k(1/f).
Thus xc dec.
Alpha dec.
xc=k(1/f).
Thus xc dec.
Alpha dec.
Tapash said:
1 decade ago
Any wide description about it.
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