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High-pass filter output voltage calculation

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dans37buick

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Hello,

I have a question that is on a study guide for a final exam.

In a high-pass filter, Xc = 500 Ω and R=2.2kΩ. What is the output voltage (Vout) when the input voltage (Vin)=10 V.

The given answer is 9.75 /_ 12.8 degrees. I cannot get this answer from my calculations below.

I have calculated it two different ways and come up with two different answers that don't match the given one.

Z= √(500^2+2200^2)=2256.1Ω

10V/2256.1Ω= 4.4 mA

V[SUB][/SUB]r =.0044*2200=9.68 V
V[SUB][/SUB]Xc=.0044*500= 2.2 V

Tan^-1 (2.2/9.68)=12.80

When I plug the numbers into my TI-89 to get the polar coordinates I get 9.92/_12.8 degrees

Second try

10V/2700Ω= 3.7 mA

Vr=.0037*2200=8.14 V
VXc=.0037*500= 1.85 V

Tan^-1 (1.85/8.14)=12.80

When I plug the numbers into my TI-89 to get the polar coordinates I get 8.34/_12.8 degrees

I have been pulling my hair out for the last couple days trying to figure this one out.

Thanks for the help
 
Lets do the calculation at 1Hz.
 
Your problem starts here:
10V/2256.1Ω= 4.4 mA

Rounding to 4.4, when you would be better by keeping a few more decomal places:
10/2256.1 = 4.4324

Then 2200 x 4.4324 = 9.7513, which is more like the expected answer.

JimB
 
Guess I should have seen that one coming! :D

Thanks for the help, I thought I was missing a huge step that I could not find in my text.
 
Mike I haven't used Spice before we use Multisim at school. Where did you get it, a quick search of the internet didn't reveal much besides lots of files that other people have made.
 
Hi,

instead "pulling your hair out for the last couple days" you simply should have checked if your results are exact enough.
Everything OK - except your rounding approach.

The exact current is 4.432 mA and with this value you arrive at the correct voltages Vout=Vr=9.7513 V and |Vxc|=2.216 V.
The corresponding angle is 12.8 deg.


Remark: Right after placing my reply I have noticed: JimB was a bit faster! Congratulation.
 
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