A friend asked me to help him answer a question for his past paper revision notes, and apparantly i got it wrong. Now i look silly!. The question was:
What value of capacitor must be added in series with a 1k ohm resistor, and signal source (1v peak @ 1khz) to complete the potential divider and reduce the peak voltage to half.
It should be 92nf right???
His Teacher marked it wrong and said the answer was 159uf - with a little note on the side "Reactance of 90nf does NOT equal 1k ohm at 1khz".
Vo/Vin=(1/RC)/(jw+1/RC), where w=radian frequency=2*pi*F
|Vo/Vin|=(1/RC)/sqrt(w^2+(1/RC)^2)=0.5, by problem definition
F=1000, R=1000
solving for C,
C=sqrt(3)/(w*R)
C=275.66nF
The teacher's solution (probably 159nF, not uF) does indeed give Xc=1k@1kHz, but due to the associated 45 degree phase shift, the resulting amplitude is 0.707V.
Tell your friend to take this back to the teacher.
I agree with Megamox. If you assume the output voltage is across the resistor then.
Vo/Vi = R/(R + 1/SC) = 0.5 Let SC = jwC
Vo/Vi = R/(R - j/wC) = 0.5
Thus wRC/(wRC -j) = 0.5
Take the modulus wRC/sqrt{(wRC)^2 + 1} = 0.5
Square both sides and rearrange 4(wRC)^2 = (wRC)^2 + 1
3(wRC)^2 = 1 gives C = 1/{sqrt(3) wR}
C = 92 nF
Edit: After posting this, I noticed Ron's latest post. He has the output across the capacitor. Whereas, I assumed it was across the resistor. So I'll re-do the maths to see if it agrees with Ron's simulation.
If R = 1000 ohm, w = 2 pi * 1000
I would appear that the difference is whther you take the output across the resistor or the cap. I'll do the maths and see if I get the same answer as Ron if the output is across the cap.
I transposed the R and C and re-did the maths. This way, the answer is 275 nF as Ron said.
So we are all right! (except your teacher)
I had a feeling that the Xc of the cap would not = 1 k for the output to be half.
The relationship between the supply voltage and the voltages across the R and the C is not linear. They are related by Pythagorus (I'm not sure about the spelling)
Len, you're right. Megamox, please accept my apologies. As Len said, I assumed the lowpass configuration. I think the original wording implied the highpass config, which I didn't even consider. I guess I was only half wrong.
Anyhow, the teacher needs some re-education.
Yeah, I know. I just felt bad for chiding Megamox while not considering that there were two solutions.
God, I hope Megamox has a way to get this back to the teacher. We have enough guys giving bad advice in this forum. We don't need teachers creating more of them! :roll:
The teacher probably meant that the capacitor with a reactance equal to the resistor will produce a half-power output from across the capacitor.
The voltage is at 0.707, therefore the output is 3dB down which is half-power.
Len, I see you've been on this forum for 2 years. You need an avatar so we can more easily spot your posts, and also to see what sort of foolish picture you select for it. :lol:
The teacher probably meant that the capacitor with a reactance equal to the resistor will produce a half-power output from across the capacitor.
The voltage is at 0.707, therefore the output is 3dB down which is half-power.