- #1

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The problems is as follows:

Find the voltage V1 across the first capacitor.

Express your answer in terms of V.

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- Thread starter lylos
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- #1

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The problems is as follows:

Find the voltage V1 across the first capacitor.

Express your answer in terms of V.

- #2

ranger

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What would be your first step?

- #3

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What would be your first step?

The first step is finding that they all have the same charge. Then, I think, I would find the equivalent capacitance which is equal to 11/6 C. From there I don't know where I would get the voltage for each one.

- #4

ranger

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Do you know also know of a formula that relates capacitance, voltage, and charge?

- #5

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Do you know also know of a formula that relates capacitance, voltage, and charge?

ofcourse, c=q/v... So Ceq=Q/Veq...

- #6

ranger

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ofcourse, c=q/v... So Ceq=Q/Veq...

Good. But instead of Ceq=Q/Veq (not sure what you mean by Veq). It should be Ceq=Q/V, where V is the total source voltage.

We know what Ceq and V is simply V, whats left to be determined? Remember that this quantity would be the same for all caps.

- #7

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Good. But instead of Ceq=Q/Veq (not sure what you mean by Veq). It should be Ceq=Q/V, where V is the total source voltage.

We know what Ceq and V is simply V, whats left to be determined? Remember that this quantity would be the same for all caps.

Ok by the Veq I meant the total voltage... So what we have is that the total voltage is equal to V. The C equivalence is equal to 11/6 C. And that the charge on each capacitor is Q. Now from that information, I am still at a loss as to how to determine the voltage across the first capacitor.

The answer is 6/11 V according to the system my homework is in...

- #8

ranger

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- #9

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Ok, I see that I made the mistake there... It is 6/11 C. So we would then set that 6/11 C = Q(for each capacitor) / (Vtot).

So Q = 6 (Vtot) (C) / 11 for each capacitor.

Now that we have that, we could then find the voltage by knowing the q=v/c equation.

C = (6/11 C Vtot / 11)/V1

C V1 = 6/11 C Vtot

V1 = 6/11 Vtot

And that's the answer!!! Thank you for your help!!!

The problem that I had was knowing to get the charge for each capacitor and then work backwards...

- #10

ranger

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You're very much welcome. And well done on your part

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