AC Circuit Quiz #1 [Archive]

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Ether

20-02-2011, 17:08

Manoel

20-02-2011, 17:14

Are mentors elligible for the challenge? :P

Ether

20-02-2011, 17:19

Are mentors elligible for the challenge? :P

Sure. Interested students can learn by studying your answer.

Manoel

20-02-2011, 17:53

Suppose you have an air-core coil with inductance 200uH and resistance 0.09 ohms, connected in series with a 10 ohm resistor.

You apply a 15KHz 12 volt peak-to-peak sine wave across this series circuit.

What is the power dissipated in the coil?

What is the rms voltage measured across the coil?

Let's work exclusively with rms quantities: 12 Volts peak-to-peak is 6 Volts amplitude, and dividing by the square root of 2 (~ 1.41), we get an rms voltage of 4.24 V from the source.

At 15 kHz, the coil's reactance is given by:

XL = omega*L = 2*pi*f*L = 2*pi*15*10^3*200*10^-6 = 18.85 ohms

The coil's impedance, then

Zcoil = 0.09 + j*18.85

The circuit total impedance, considering the 10 ohm series resistor:

Ztotal = 10.09 + j*18.85

We can now determine the circuit's total current, in magnitude:

|I| = |V|/|Z| = |4.24|/|10.09+j*18.85| = 4.24/21.38 = 0.198 A

The power dissipated in the coil, then, is due to its small resistance:

P = i^2 * R = 0.198^2 * 0.09 = 0.0035 = 3.5 mW

We can determine the voltage drop in the coil using a voltage divider:

Vcoil = Vsource * Zcoil/Ztotal = 4.24*(0.09+j*18.85)/(10.09+j*18.85)
Vcoil = 3.3062+j*1.7495

The rms voltage that you would measure with a voltmeter is the magnitude of Vcoil:

|Vcoil| = |3.3062+j*1.7495|
|Vcoil| = 3.74 V

Ether

20-02-2011, 18:03

...

Nicely presented.

Students interested in AC circuit theory should carefully study your answer and ask questions if they need help understanding.

Matt Krass

20-02-2011, 18:51

XL = omega*L = 2*pi*f*L = 2*pi*15*10^-3*200*10^-6 = 18.85 ohms

I believe you meant 2*pi*15*10^3.... not 10^-3. The math checks out otherwise.

Or am I just tired?

Matt

Manoel

20-02-2011, 19:21

I believe you meant 2*pi*15*10^3.... not 10^-3. The math checks out otherwise.

Or am I just tired?

Matt

Yes, absolutely! The calculations are correct, I just typed it out wrong. :rolleyes:

Matt Krass

20-02-2011, 19:34

Yes, absolutely! The calculations are correct, I just typed it out wrong. :rolleyes:

Yep, I did the math out before I said anything, I get the same impedance as you, just a minor typo.

This was a fun challenge, any particular reason for it? It almost sounds like a motor controller to motor, if the wave was square instead of sine.

Ether

20-02-2011, 20:04

This was a fun challenge, any particular reason for it?

Just to raise awareness among interested students here on CD how AC calculations differ from the simple V=IR DC stuff they learned, and maybe motivate some to get into studying it.

It almost sounds like a motor controller to motor, if the wave was square instead of sine.

Yeah, it was inspired by some recent discussions here involving motor inductance and how that affects power calculations.

Matt Krass

20-02-2011, 20:14

Just to raise awareness among interested students here on CD how AC calculations differ from the simple V=IR DC stuff they learned, and maybe motivate some to get into studying it.

AC power stuff is almost a form of black magic. :)

Seriously though, it is pretty interesting, especially once you get in to reactance, systems can do weird things. I find students tend to get very interested in inductive backlash (which is what I call the tendency of inductive loads to cause interesting voltage fluctuations in an attempt to maintain current flow)

JDL

21-02-2011, 02:50

Eww math. I did too much of that today already, load calculations and voltage drops, motor conductor sizing, I slacked off on some bids and paperwork this week :rolleyes:

Students, while math is boring and you find it stupid (not that I blame you, no one really gives a darn where X is, it's more fun when you can apply it and make money off it) it is important if you want a technical career.