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schematic

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?

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  • \$\begingroup\$ Is it just me, or are your units off here? You appear to be measuring charge in farads there... \$\endgroup\$ Jan 18 '15 at 14:20
  • \$\begingroup\$ 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. \$\endgroup\$
    – Nedd
    Jan 18 '15 at 16:57
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An ideal capacitor has no resistance and therefore no heat will be dissipated by the capacitors in your circuit. The only place in that circuit (assuming all ideal parts) that electrical energy will be converted to heat is the resistor, so what you need to find is the power dissipated by the resistor, which involves the charges stored in the capacitors as well as the voltage sources E1 and E2.

Naturally, real parts have what is called an ESR (equivalent series resistance) but based on the way this question is phrased, it seems we are looking at a theoretical rather than actual circuit.

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}\$.

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  • \$\begingroup\$ We've spend quite some time learning about energy in capacitors, so I don't what are telling me? \$\endgroup\$
    – Desperado
    Jan 18 '15 at 14:32
  • \$\begingroup\$ Energy is stored in a capacitor, but only converted to heat by the resistor. \$\endgroup\$
    – Edward
    Jan 18 '15 at 14:37
  • \$\begingroup\$ @Desperado: I've added some detail to the answer, maybe it will help. \$\endgroup\$
    – Edward
    Jan 18 '15 at 15:53

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