# Amount of heat developed in capacitors

simulate this circuit – Schematic created using CircuitLab

Q10 is equal to 10μC. No charge on C2. I need to find the amount of electric work that is converted into heat, from the moment the sw closes until circuit goes into stationary state.

Here's how I go:

1. Since there's no current in both states, we can disregard the resistor.
2. I calculate voltage of C1, when switch is open $$U = Q/C = 2V$$
3. Switch closes, voltage is $$E1-E2=8V$$ it divides on capacitors 4/3V on C1 and 20/3 on C2.
4. Use $$We= 1/2*C*(ΔU)^2$$ Use it on both capacitors, sum them and get the wrong result, 70/3 instead of 15μJ.

What did I do wrong?

• Is it just me, or are your units off here? You appear to be measuring charge in farads there... Jan 18 '15 at 14:20
• Just because energy is transferred it does not mean that heat is generated. Heat is created only if the transfer dissipates the energy. Heat can be considered a form of energy, but it is not the only form. Eg: If I were to hand you a charged battery you might say I transferred energy, but I did not generate "Heat". If you want to calculate heat generated from switch closure to stationary you must calculate it from the brief current flow through the resistor. Your calculation above is trying to show the energy transferred in Joules.
– Nedd
Jan 18 '15 at 16:57

The work done is the change in energy, $W = E_f - E_0$. The initial energy is stored in $C_1$ and is $\displaystyle E_0 = \frac{1}{2}\frac{Q^2}{C_1}$. The final energy is $\displaystyle E_f = \frac{1}{2}C_tV^2$ where $\displaystyle C_t = \frac{1}{\frac{1}{C_1}+\frac{1}{C_2}}$. Putting it all together we get $$W = \frac{1}{2}\left(\frac{1}{\frac{1}{C_1}+\frac{1}{C_2}}V^2 - \frac{Q^2}{C_1}\right)$$ If you plug in the numbers with $V=8\text{V}$ and $Q=10\mu\text{J}$, you should get $16.67\mu \text{J}$.