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I'm having some problems solving the voltages across the capacitors in the following circuit:

Circuit

Because it is a parallel circuit, we know the voltage across C3 must be 6V.

But how can you figure out the voltage across C1 and C2? Does C1 store all the energy and leave C2 with no voltage? Or is it proportional?

I tried solving it in CircuitLab, and it calculates 6V across C1 and 0V across C2...

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The charge on C1 and C2 must be equal by conservation of charge because the node between them is isolated. The voltage of C1 and C2 must sum to 6V. Use q=CV and solve for the voltages.


Reworked by RM:

Take 3:

The same current flows in C1 & C2.
As charge is, by definition, proportional to current (Q = I x t) then
the charge on C1 and C2 must be equal.

But, also by definition Charge = capacitance x Voltage (Q = C x V).
Or, rearranging, V = Q/C.
So, for equal charges in each, capacitor voltage will be inversely proportional to capacitance.

The voltage of C1 and C2 must sum to 6V. Use q=CV and solve for the voltages.

Wrap


Take 2:

The relationship between charge Q, voltage V and capacitance C is given by the expression
Q = C x V.
Rearranging, V = Q/C.

Charge Q is defined as the summation of current with time ie Q = i x t
As an identical current must flow C1 and C2, they both experience the same current x time profile so their charges are equal.

But from above, Vcap = Q/C.
ie for equal charges, cap voltages will be inversely proportional to capacitor sizes.

So, in this example, the voltages on C1 & C2 will be inversely proportional to capacitor sizes so VC1 = 2 x VC2.
By inspection, VC2 = 2, VC1 = 4.

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  • \$\begingroup\$ Ah, so 6(V1) = 12(V2). So C1 voltage is 4V and C2 voltage is 2V. Got it! \$\endgroup\$ – mr_schlomo Jun 11 '12 at 2:43
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    \$\begingroup\$ @mr_schlomo - a tip with simulation, try setting "start supply voltages at 0V" (or an equivalent, e.g step voltage from 0 to 6V) and it should simulate correctly (at least it does in LTSpice) \$\endgroup\$ – Oli Glaser Jun 11 '12 at 4:20
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    \$\begingroup\$ Why must they be equal? \$\endgroup\$ – stevenvh Jun 11 '12 at 4:25
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    \$\begingroup\$ This answer is poorer than @stevenh 's and should not be the accepted answer. No offense madrivereric, but steven shows much more effort. \$\endgroup\$ – abdullah kahraman Jun 11 '12 at 6:16
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    \$\begingroup\$ @abdullah - thanks! But OP accepted the answer before I had posted mine, and he hasn't been here since. Personally I also think he needs to explain his first sentence. Note that the accepted answer often doesn't change anymore, even if the total of votes is negative. \$\endgroup\$ – stevenvh Jun 11 '12 at 6:23
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"The charge on C1 and C2 must be equal" doesn't explain it. Why must they be equal?

Just think of it as a voltage divider, but while for resistors \$ V_{R2} = \dfrac{R2}{R1 + R2} V1 \$, for capacitors: \$ V_{C2} = \dfrac{C1}{C1 + C2} V1 \$. Notice the different indices in the numerator.


Proof

WARNING. Ugly equations ahead. Don't worry if they look too complicated; all you need in practice is the equation above.

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\$ V_{C2} = \dfrac{Z_{C2}}{Z_{C1} + Z_{C2}}V1 = \dfrac{\dfrac{1}{j\omega C2}}{\dfrac{1}{j\omega C1} + \dfrac{1}{j\omega C2}} V1 = \dfrac{\dfrac{C1}{j\omega C1 C2}}{\dfrac{C2}{j\omega C1 C2} + \dfrac{C1}{j\omega C1 C2}} V1 = \dfrac{C1}{C1 + C2} V1 \$

The last step is only allowed for \$\omega \ne 0\$. For DC we have to take the limit:

\$ V_{C2} = \displaystyle \lim_{\omega \to 0} \dfrac{\dfrac{C1}{j\omega C1 C2}}{\dfrac{C2}{j\omega C1 C2} + \dfrac{C1}{j\omega C1 C2}} V1 = \dfrac{C1}{C1 + C2} V1 \$

You don't need to do this derivation each time, just remember that it's the inverse of a resistor divider.

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    \$\begingroup\$ +1 but I'd like more an explanation involving charge conservation (should come from the fact that in the central plates there is the same charge as well). I think that it would be a stronger explanation \$\endgroup\$ – clabacchio Jun 11 '12 at 7:24
  • \$\begingroup\$ @clabacchio - Russell commented to the other answer "S's answer is OK but beginner's eyes are going to glaze over at the j.omegas." It might look scary to a beginner who just wants to know what voltage on each cap. \$\endgroup\$ – stevenvh Jun 11 '12 at 7:35
  • \$\begingroup\$ @stevenvh: well, the trick is to use Laplace's s :) \$\endgroup\$ – clabacchio Jun 11 '12 at 7:37
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    \$\begingroup\$ @clabacchio - That's right, the equations look simpler, until I have to explain the transform :-/ \$\endgroup\$ – stevenvh Jun 11 '12 at 7:40
  • \$\begingroup\$ @clabacchio - sorry for the trouble you took, but I rolled back. I could have kept the subscripts. But you can't just eliminate \$j\omega C1 C2\$ if \$\omega\$ is zero. That's the DC from the question. You can't eliminate \$\infty/\infty\$. That's where the limit comes in. I checked with the guys at math.stackexchange and they agreed. \$\endgroup\$ – stevenvh Jun 11 '12 at 7:58

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