# Calculating Energy Stored in Capacitors in a Series Circuit

• Nimmy
In summary, the problem involves a resistance, two capacitors in series, and a voltage source (Vo). The switch is closed at t=0 and the question asks for the energy stored in C1 20.0 ms after the switch is closed. To solve this, we need to find the total capacitance (Ceq) and use it to find the voltage drop across both capacitors. Then, using Q=CV, we can find the total charge on the capacitors and use that value to calculate the energy using the equation U=1/2CV^2. Both the voltage drop and the charge are needed to solve the problem accurately.
Nimmy
Ok try to review for the finals but I need help in understanding this problem.

So I can say that there is a resistance, 2 capacitors in series, and voltage source (Vo).

Vo = 20.0 V
R = 5.00 kilo ohm
C1 = 12.0 micro F
C2 = 4.0 micro F
Switch is closed at t = 0.

How much energy will be store in C1 20.0 ms after the switched closed.

U = Q^2/2C = 1/2CV^2 = 1/2 QV

V(t) =Vo(1-e^(-taw/RC)
V(t) = 20V(1-e^-(20.0E-3 s/(5.00E3 ohms)(is this C the Ceq for the Capacitors because I only wrote C1 and I got it wrong)
V(t) = ______ volts

U = 1/2 _____volts x ______Faradays = _______ J

You have the incorrect initial voltage, because the voltage drop across the single capacitor won't be V_0. Instead, the voltage drop across both capacitors is V_0.

First solve for the total capacitance and use that equation to get the voltage drop across both capacitors.

Then find the total charge across this new total capacitor. You will remember that capacitors in series have the same charge. That means the total charge will be the same charge on the capacitor in question and you can use Q=CV to get the voltage across it.

nickjer said:
You have the incorrect initial voltage, because the voltage drop across the single capacitor won't be V_0. Instead, the voltage drop across both capacitors is V_0.

First solve for the total capacitance and use that equation to get the voltage drop across both capacitors.

Then find the total charge across this new total capacitor. You will remember that capacitors in series have the same charge. That means the total charge will be the same charge on the capacitor in question and you can use Q=CV to get the voltage across it.

Ok so you are saying this:

Ceq = 1/C1 + 1/C2 = 1/12 E -6 + 1/4 E -6 but its the inverse because we are trying to find Ceq from 1/Ceq. So I got 3 E-6 F.

I don't know what you mean to find the total charge?

You still need to find the voltage across the total capacitor. Then you know the equation Q=CV to get the total charge.

nickjer said:
You still need to find the voltage across the total capacitor. Then you know the equation Q=CV to get the total charge.

OHHHH...I think I understand now

You use OHM's Law

V = IR = IoR = Vo/R e (-t/taw) so that value x 5.00 E 3 ohms = ______ V

V(t) = _____ V (from last line) (1-e(-t/RCeq) = ______ V

Q = 3 E-6 F x ____ V (that I solved previously) = _________ C

So I plug that into Q of the energy equation to solve the problem

U = 1/2 (_________ C) (V(t))^2 = ________ J ?

Last edited:
The battery is 20V but the capacitor isn't fully charged.

nickjer said:
The battery is 20V but the capacitor isn't fully charged.

Huh? I revised my post...

You have a resistor and two capacitors in series. You will have a voltage drop across the resistor, another voltage drop across C1, and a last voltage drop across C2. All three drops must add up to 20V. So it is not possible that the voltage across C1 is 20V otherwise everything else would have 0 voltage drops.

So you need to solve for the voltage drop across both capacitors combined using your voltage equation after 20ms. This will give you the charge Q on this total capacitor. Then you know that capacitors in series have the same charge as if they were one giant capacitor. That means the charge on C1 is Q.

You also know the energy equation is U1 = Q^2/(2*C1). So you solved it.

nickjer said:
You have a resistor and two capacitors in series. You will have a voltage drop across the resistor, another voltage drop across C1, and a last voltage drop across C2. All three drops must add up to 20V. So it is not possible that the voltage across C1 is 20V otherwise everything else would have 0 voltage drops.

So you need to solve for the voltage drop across both capacitors combined using your voltage equation after 20ms. This will give you the charge Q on this total capacitor. Then you know that capacitors in series have the same charge as if they were one giant capacitor. That means the charge on C1 is Q.

You also know the energy equation is U1 = Q^2/(2*C1). So you solved it.

Just making sure...the answer I posted was correct right?

I don't see an actual answer, just blanks. And you can't use the voltage drop you solved for to get the stored energy. Because that voltage drop is the drop across both capacitors and not just C1. That is why you need to solve for the charge, Q.

q = Ceq Vo (1-e^(-t/RC)

To find the charge Q?

Yes, if by Ceq you mean C_total, that is what I have been saying before :)

Just use the C_total in the exponential as well.

nickjer said:
Yes, if by Ceq you mean C_total, that is what I have been saying before :)

Just use the C_total in the exponential as well.

Thank you! Sorry for being hard head! lol.

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