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:
- Since there's no current in both states, we can disregard the resistor.
- I calculate voltage of C1, when switch is open $$ U = Q/C = 2V $$
- Switch closes, voltage is $$ E1-E2=8V $$ it divides on capacitors 4/3V on C1 and 20/3 on C2.
- 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?