- #1

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## Homework Statement

If I have an impedance of: 10+j4 ohms and was told to add a capacitance of -j15 in parallel to this, would this make the impedance 10-j11 ohms??

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- Thread starter bengaltiger14
- Start date

- #1

- 138

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If I have an impedance of: 10+j4 ohms and was told to add a capacitance of -j15 in parallel to this, would this make the impedance 10-j11 ohms??

- #2

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- #3

berkeman

Mentor

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## Homework Statement

If I have an impedance of: 10+j4 ohms and was told to add a capacitance of -j15 in parallel to this, would this make the impedance 10-j11 ohms??

Expecting less than a 2 hour turnaround on a homework question is not reasonable. Do not bump like this again.

But since you asked.... Show us your work. Show us the math where you calculated the parallel impedance based on what relevant equations....?

- #4

- 1,763

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Can you think of what you have to do now?

- #5

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0+j4 in calculations?

- #6

- 1,763

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To convert series to parallel assuming real and imaginary do the following.

Rp = (Rs^2 + Is^2)/Rs

Ip = (Rs^2 + Is^2)/Is

Now add the capacitor

Ip(total) = 1/(1/Ip + 1/Ic)

Convert back to series

Rs = Rp/(Rp^2 + Ip^2)

Is = Ip(total)/((Rp^2 + Ip(total)^2)

where Rs = series resistance and Rp = parallel resistance.

Let me know how they tell you to work the problem. I derived these formulas myself. I don't know the official method. When I was doing bench level RF design I used to work these in my head for practice.

- #7

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After I added the bottom, I would change it to polar and finish the division. I just do not understand what I am suppose to use for the real part of -j15. I can do the math just unsure about that.

- #8

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- #9

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Since the -j15 is the series value, how much is the series resistance... zero.

- #10

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Converting to polar and back again is another good way to solve it.

- #11

- 138

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SQRT(0^2 + 15^2)

tan-1 (15)

= 15 angle(86.19)

Rectangular = 15cos(86.19) = .996

= 15sin(86.19) = j14.96

- #12

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Convert 0-j15 to polar... Mag 15, Ang -90 or I prefer 15@-90

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