# Help with capacitor circuit calculations

Hopefully somebody will be able to help me better understand how to calculate capacitor change and everything else. I just started studying them. The problem and my attempt to solve it is below.

How would I go about to get E (energy) and Q for each capacitor? thanks!!!

GIVEN:

C1 = 20 uF

C4 = 2 uF

C6 = 70 uF

C8 = 10 uF

V(total) = 3 kV

Find V voltage, E energy, Q electrical charge for each capacitor.

simulate this circuit – Schematic created using CircuitLab

First I simplify the circuit and calculate C46:

simulate this circuit

$$C_{46}=C4 + C6 = 2uF + 70uF = 72uF$$

$$C_{total} = \frac{1}{\frac{1}{\text{C1}}+\frac{1}{\text{C8}}+\frac{1}{\text{C46}}}=\frac{1}{\frac{1}{\text{20}}+\frac{1}{\text{10}}+\frac{1}{\text{72}}}=6.1017 uF=6.1017*10^{-6} F$$

$$Q = Q1 = Q8 = Q46 = 6.1017*10^{-6}*3000 = 1.83051*10^{-2} C$$

$$V1 = \frac{\text{Q1}}{\text{C1}} = \frac{1.83051*10^{-2}}{2*10^{-5}} = 9.15255*10^{2} V$$

$$V8 = \frac{\text{Q8}}{\text{C8}} = \frac{1.83051*10^{-2}}{1*10^{-5}} = 1.83051 * 10^{3} V$$

$$V46 = \frac{\text{Q46}}{\text{C46}} = \frac{1.83051*10^{-2}}{7.2*10^{-5}} = 2.5424 * 10^{2} V$$

$$E = \frac{\text{V*Q}}{\text{2}} = \frac{3000*1.83051*10^{-2}}{2} = 2.7458*10^{-2} J$$

$$E1 = \frac{\text{V1*Q1}}{\text{2}} = \frac{9.15255*10^{2}*1.83051*10^{-2}}{2} = 8.3769 J$$

$$E46 = \frac{\text{V46*Q46}}{\text{2}} = \frac{2.5424*10^{2}*1.83051*10^{-2}}{2} = 2.3269 J$$

$$E8 = \frac{\text{V8*Q8}}{\text{2}} = \frac{1.83051*10^{3}*1.83051*10^{-2}}{2} = 1.6754 *10^{1} J$$

$$Q4 = C4*V4 = 2*10^{-6} * 2.3225*10^{3} = 4.6450*10^{-3} V$$ $$Q6 = C6*V6 = 7*10^{-5} * 2.3225*10^{3} = 1.6258*10^{-1} V$$

$$E4 = \frac{\text{V4*Q4}}{\text{2}} = \frac{2.3225*10^{3}*4.6450*10^{3}}{2} = 5.3940*10^{6} J$$

$$E6 = \frac{\text{V6*Q6}}{\text{2}} = \frac{2.3225*10^{3}*1.6258*10^{5}}{2} = 1.88796*10^{8} J$$

• Check your math/thoughts for C46.... – Trevor_G Sep 14 '17 at 18:23
• This can't be solved without knowing the history of the circuit. Were all capacitors initially fully discharged and then the power source was connected? – The Photon Sep 14 '17 at 18:23
• @Trevor indeed, I can move it down and draw in parallel. Thats true, I didn't see that :) So I should calculate it in parallel and go from there... :) Thanks – gointern Sep 14 '17 at 18:48
• There is a substitution error when you calculate V46. Instead of putting 72uF you are putting Ctotal on C46. – next-hack Sep 14 '17 at 19:46
• We can only assume $Q=Q1=Q8=Q46$ if all capacitors are perfect (no leakage) and they all started with the same charge: For example they all started fully discharged and so each had a charge of 0 coulombs. Otherwise this can't be solved. – Warren Hill Sep 14 '17 at 19:54

If we consider two capacitors in series

simulate this circuit – Schematic created using CircuitLab

Then we know $V_{C1} + V_{C2} = V_{tot}$ but we have no way of determining how this is shared because lets say $C1 = C2 = 1 \mu F$ and $V_{tot} = 300 \text{V}$ for example then there is nothing to say the voltage has to be shared equally because one capacitor may have had some initial charge which makes this claim invalid. Also since no current is flowing at DC the voltages across the capacitors will be dominated by the leakage resistance of the capacitors and not the capacitor values.

If we are assuming a purely theoretical answer then we can ignore leakage resistance and if we assume all capacitors were initially fully discharged we can assume that $Q_{C1} = Q_{C2}$ since $Q = \int i \text{ dt}$ and since all capacitors initially had zero charge and have had identical current flowing in them at all times since.

With that caveat in place the approach to this type of this problem is:

1. Calculate total capacitance (you have done this already).
2. Calculate total charge $Q = C{tot} \cdot V{tot}$.
3. Calculate the charge in each series element, they all have the same $Q = Q_{C1} = Q_{C2}$.
4. Calculate the voltage across each capacitor $Q = C \cdot V \Rightarrow V = \frac{Q}{C}$
5. Calculate the energy in each capacitor $E = \frac{1}{2} \cdot C \cdot V^2$
• Sorry @Trevor you are correct, edited – Warren Hill Sep 14 '17 at 21:21
• :) No worries.. it always gets me too. The truth seems counter-intuitive. – Trevor_G Sep 14 '17 at 21:33