Magnitude of voltage across impedance

In summary, the three impedances are connected in series across a 100-V, 20-kHz supply. The current through the circuit is 1.54A. The magnitude of the voltage across the coil is 40.8V, the magnitude of the voltage across the both 15 Ohm resistor and the capacitor is 75.9V.
  • #1
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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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  • #2
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.
 
  • #3
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:
 
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  • #4
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
 
  • #5
I've come up with a figure of 3.69 A

How does that sound?

Thanks for posting back by the way :)
 
  • #6
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.
 
  • #7
Gah... Back to the drawing board.

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

Thanks
 
  • #8
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
 

Related to Magnitude of voltage across impedance

1. What is impedance and how does it relate to voltage?

Impedance is a measure of the total opposition to the flow of alternating current in a circuit. It is made up of two components: resistance and reactance. The magnitude of voltage across impedance is affected by the value of impedance in the circuit, as well as the frequency of the alternating current.

2. How is the magnitude of voltage across impedance calculated?

The magnitude of voltage across impedance can be calculated using Ohm's law, which states that voltage equals current multiplied by impedance. It can also be calculated using the formula V = IZ, where V is voltage, I is current, and Z is impedance.

3. What factors can affect the magnitude of voltage across impedance?

The magnitude of voltage across impedance is affected by the value of impedance in the circuit, as well as the frequency of the alternating current. Additionally, the type of circuit and the components used can also impact the magnitude of voltage across impedance.

4. How is the magnitude of voltage across impedance measured?

The magnitude of voltage across impedance can be measured using a voltmeter, which is a device that measures the voltage difference between two points in a circuit. The voltmeter is connected in parallel with the impedance in order to measure the voltage across it.

5. Why is the magnitude of voltage across impedance important in circuit analysis?

The magnitude of voltage across impedance is important in circuit analysis because it allows us to understand the behavior of the circuit and how it responds to the flow of current. It can also help us determine the power and efficiency of the circuit, as well as identify any potential issues or problems in the circuit.

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