Magnitude of voltage across impedance


by KK25
Tags: impedance, magnitude, voltage
KK25
KK25 is offline
#1
Oct4-10, 11:10 PM
P: 7
1. The problem statement, all variables and given/known data

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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SirAskalot
SirAskalot is offline
#2
Oct5-10, 05:03 AM
P: 136
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.
KK25
KK25 is offline
#3
Oct5-10, 05:35 AM
P: 7
Ok. I see that in this diagram:



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

SirAskalot
SirAskalot is offline
#4
Oct5-10, 04:08 PM
P: 136

Magnitude of voltage across impedance


Nope. The total impedance is Z=27-j2.8

You cant 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
KK25
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#5
Oct6-10, 07:20 AM
P: 7
I've come up with a figure of 3.69 A

How does that sound?

Thanks for posting back by the way :)
SirAskalot
SirAskalot is offline
#6
Oct6-10, 04:16 PM
P: 136
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.
KK25
KK25 is offline
#7
Oct7-10, 07:01 AM
P: 7
Gah... Back to the drawing board.

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

Thanks
KK25
KK25 is offline
#8
Oct10-10, 07:52 AM
P: 7
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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