# Volt drop in a reactance / resistance circuit

Discussion in 'General Electronics Discussion' started by Chickwolf, Jan 10, 2014.

1. ### Chickwolf

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0
Jan 3, 2014
Hello all, working through a circuit with both reactive and resistive components. I've worked out the Total resistance and the total current but im having trouble working out the volt drops over the components. I understand how to work out the volt drop over the resistor but im not sure what the method is for working it out over reactive components (the capacitor and inductor). Thumbnail attached shows my workings and the circuit.
Thanks, Chick.

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5,164
1,081
Dec 18, 2013
No dc apart from leakage current of cap because cap blocks dc. I think they are trying to trick you. The power supply is a battery right?

3. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,418
2,788
Jan 21, 2010
Yes Chickwolf, what Arouse1973 says!

Unless they tell you about the initial charge on the capacitor and the initial inductor current, then the answer is simple.

If they give both of the above (and there are other ways to put it) then the result is a function of time that (in this case) will be a series of ever reducing swings in voltage and current. The answer will be complex (in that it involves the square root of -1) and will require calculus to solve.

If any of what I've said in the preceding paragraph sounds weird, bizzare, or way more than you've been taught, then it's pretty certain that the simple answer is what they're after (essentially that is the final result of the tricky stuff above)

Remember that to a steady DC voltage, inductors are a short (0 ohms) in the long term, and capacitors are open (infinite ohms) in the long term.

4. ### Chickwolf

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Jan 3, 2014
Ok, so if the don't give you the starting charge on the capacitor and the starting current on the inductor then it is easier to work out than if they do give it to you? Would it be possible for you to show me a method that can be used when they don't give you the initial values?
Yes, the power supply is a battery, and I drew this circuit so I'm sorry if it is incorrect.
Thanks a lot for the help.
Chick.

5. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
The thing you need to ask yourself is "How much current flows through a capacitor when it has a constant DC voltage across it?"

One you know the answer to that...

6. ### Chickwolf

36
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Jan 3, 2014
Ok, I think I have got this now.
If I work out the volt drops over each component:
R = 392x10^-3x 25 = 9.8v
Xc = 392x10^-3x 15 = 5.88v
Xl = 392x10^-3x 20 = 7.84v
Then subtract Xc from Xl, which equals 1.96v
And lastly use Pythagoras's theorem to work out 1.96^2 +9.8^2 which equals 9.99v
That is now correct is it not?
Thanks, Chick.

Last edited: Jan 13, 2014
7. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,418
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Jan 21, 2010
Ahhhh... No.

First tell us what the current through the capacitor is. This is a very simple question assuming you're not supposed to give an answer which is expressed as a function over time.

For the simple case, the current in all components is *exactly* the same.

The big question is, are you supposed to use calculus or not? What is the level at which you are studying?

8. ### Laplace

1,252
184
Apr 4, 2010
Capacitors and inductors do not have a reactance value unless in the steady-state AC domain. Nor are phasor diagrams used unless in the steady-state AC domain. So was the original problem given with a battery symbol for the AC power source? Or was it the OP's idea to represent the AC voltage source with a DC battery symbol? How was the reactance value calculated without knowing the AC frequency? What was the exact wording of the original problem statement, and where did the battery symbol come from? Is this a steady-state AC phasor problem or is it a transient time domain problem?

9. ### Chickwolf

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Jan 3, 2014
Ok, the circuit is powered by an AC power source but i have mistakenly copied up the circuit with a DC battery.The question is a steady-state AC phasor problem and the reactance was not calculated but given at the start of the problem.
The circuit was not given by a teacher but it is one I came up with so that I can practice my method for working out AC phasor problems.
Given that this is a phasor problem with a steady AC supply, are my workings correct?

10. ### Laplace

1,252
184
Apr 4, 2010
What I got for the current is I= 5/13 -j(1/13), |I|= sqrt(2/13) = 0.39223227

I don't see any problem with your calculations.

11. ### Chickwolf

36
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Jan 3, 2014
Ok thanks everyone for helping me out with that, im sorry for the circuit being unclear!