Magnitude of voltage across impedance

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SUMMARY

The discussion focuses on calculating the magnitude of voltage across impedances in an AC circuit consisting of a 12-Ω resistor, a coil with 100 μH inductance and 5-Ω resistance, and a 390-nF capacitor in series with a 15-Ω resistor, powered by a 100-V, 20-kHz supply. The total impedance (Z) was determined to be 32 - j7.8 Ω, leading to a calculated current of 3 A at a phase angle of -13.7 degrees. The magnitudes of the voltages across the coil and the combined capacitor and resistor were found to be 40.8 V and 75.9 V, respectively.

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Homework Statement



The following three impedances are connected in series across a 100-V, 20-kHz supply:
(1) a 12-Ω resistor, (2) a coil of 100 μH inductance and 5-Ω resistance and (3) a 390-nF
capacitor in series with a 15 Ω resistor. Sketch the circuit diagram, impedance diagram and the phasor diagram; take the supply voltage V as the reference phasor.

(a) Calculate the magnitude of the voltage V2 across the second impedance.

(b) Calculate the magnitude of the voltage V3 across the third impedance.2. The attempt at a solution

Ok, I'm pretty lost here, but this is what I have done so far...

I'm trying to get the current amount flowing in the circuit, so thought I should get a total impedance Z for the circuit.

First resistor = 12 Ohms
For the coil = 5 + 2 * Pi * (20*10^3) * (100*10^-6) = 17.6 Ohms
For the Capacitor = 1 / 2 * Pi * (20*10^3) * (390*10^-9) = 20.4 Ohms
Second resistor = 15 Ohms

Total Z for the circuit = 12+17.6+20.4+15 = 65 Ohms.

That is as far as I have gone so far. Is that the correct way to get total Z?

Next I will find the current by using I = V/Z ?

I guess that would be 1.54 A ?
 
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Keep in mind that capacitors and inductors are complex values.

X_l=j17.6 ohm
X_c=-j20.4 ohm

Total impedance is then Z=R+j(X_l-X_c)

Sketch up the impedance diagram(complex plane) and you see why.
 
Ok. I see that in this diagram:

[PLAIN]http://macao.communications.museum/images/exhibits/small/2_4_4_1_eng.png

Does that mean my total Z is in fact 24.2 Ohms?

I came to that conclusion by using 27 + (-2.8)

Sorry if I'm not getting it, I'm finding it hard to get my head around these AC basics :blushing:
 
Last edited by a moderator:
Nope. The total impedance is Z=27-j2.8

You can't sum real and imaginary numbers. Just like vectors, the impedance consist of an angle and magnitude.

Current is then: I=U/R=100+j0 / 27-j2.8=...A
 
I've come up with a figure of 3.69 A

How does that sound?

Thanks for posting back by the way :)
 
Wrong. It's supposed to be a complex number with both real and imaginary parts. (i.e I=2+j3 A or 3A <(angle) 36*)
Using a scientific calculator does this job easy, calculating by hand requires using complex conjugate. Refer to your calculus textbook.

Said in other words, the current has a phase angle in reference to the voltage source (leading or lagging current, Power factor and so forth)

BTW: I have to correct my answer Z=32 - j2.8 ohm
Didnt read the text thoroughly Total resistance is: 12 + 5 + 15 =32 ohm
But still, your current must have a phase angle.
 
Gah... Back to the drawing board.

Was working on other parts of my assignment today, I'll give this another shot tomorrow.

Thanks
 
Ok... Had another attempt.

Total Z = 32 - j7.8 (rather than 2.8 as before)

I = 3 A /_ -13.7

Magnitude of Voltage across coil = 40.8V

Magnitude of Voltage across both 15 Ohm resistor & Capacitor = 75.9V
 

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