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Can anyone please point out the steps needed (not the final answers!) in finding the following things from this circuit diagram?

It's difficult to find resources regarding complex circuit problems online.

enter image description here

1) Overall capacitance of the capacitors in series?

I'm aware that in series charge is less in total, and would be found like this: 1/C tot = (1/C1 + 1/C2 + etc). The issue here is what on earth does the resistor do to affect this value?

2) Overall capacitance of the circuit?

So here the total series capacitance is just added to the capacitance of the parallel capacitor. But what does the resistor do?

3) The max charge in the circuit?

Charge = Current X time. But what does the resistor do? How do I factor this in?

4) The max energy in the smallest most capacitor?

Now this bit I'm pretty lost on.

Thanks!

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closed as off-topic by Voltage Spike, ThreePhaseEel, Wesley Lee, uint128_t, Andrew Feb 20 '17 at 14:47

  • This question does not appear to be about electronics design within the scope defined in the help center.
If this question can be reworded to fit the rules in the help center, please edit the question.

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    \$\begingroup\$ show some effort \$\endgroup\$ – dirac16 Jan 31 '17 at 16:36
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    \$\begingroup\$ I'm voting to close this question as off-topic because its HW without an attempt at a solution \$\endgroup\$ – Voltage Spike Jan 31 '17 at 16:45
  • \$\begingroup\$ Nope. It's not homework. Actually revision. \$\endgroup\$ – Dolorem Ipsum Jan 31 '17 at 17:15
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    \$\begingroup\$ The resistor has NO effect on the total capacitance or the total charge, it only slows the charging process by acting as a current limiter. Because all the capacitors are connected together with direct connections, you can just treat them as one big capacitor, i.e. (60F in series with 70F) in parallel with 80F is...? I'll leave that part to you. Capacitance is only dependent on capacitors and charge is only dependent on the voltage across a capacitor and it's capacitance. \$\endgroup\$ – Sam Jan 31 '17 at 22:38
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Edit: The resistor limits the charge current, it does not prevent the capacitors from reaching 12 V, it only makes charging slower. The resistor is not relevant in this question at all.

Original:

  1. Series connected capacitors store the same charge (Q), but different voltage (U), C = Q*U.

Notice how these are similar to how you calculate total resistance, but the other way around.

Series: 1/C_Total = 1/C1 + 1/C2 + 1/C3 ...

Parallel: C_Total = C1 + C2 + C3 ...

  1. This one is easy now when you know the total capacitance of the series connected capacitors. Imagine that you replace the 60 and 70 pF capacitors with a single ?? pF capacitor. The voltage over the ?? pF capacitor must be the same as over the 80 pF capacitor.

  2. C = Q*U It could however be the power source and not the capacitors but I don't think that's a valid answer. :)

  3. If you know the capacitance of the ?? pF capacitor, you can calculate its charge, remember C = Q*U. Then you know that the 60 pF capacitor must contain the same charge. The magic formula for energy stored in a capacitor is E = (1/2)QU. (U in volts, Q in Coloumbs (Amperes*seconds), E in Joules)

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  • \$\begingroup\$ Oh okay. So the resistor doesn't affect charge even if it draws voltage. Thanks. \$\endgroup\$ – Dolorem Ipsum Jan 31 '17 at 17:33
  • \$\begingroup\$ The resistor does not drop any voltage when the capacitors have been charged. \$\endgroup\$ – Oskar Skog Jan 31 '17 at 18:01
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the capacitor 60f and 70f are in series use the formula below to calculate the capacitance in series

capacitance in series

This computed capacitor is in parallel to 80F so add the computed capacitor with 80F, the final circuit is a reduced to simple RC circuit

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