Q:
Two capacitors are connected in parallel through a switch.
C1= 1uF, C2=0.25uF.
Initially the switch is open, C1 is charged
to 10V. What happens if we close the switch?
No losses in the wires and capacitors.
A:
Hint from Hitequest
The equation for charge is Q=CU
After closing the switch the charge will be distributed between the two caps.
Since there are no losses in the circuit the amount of charge remains the same:
U1C1 + U2C2 = U3(C1+C2)
Both caps will share the same voltage potential U3
U3 = (U1C1+U2C2)/(C1+C2) = (10*1 + 0*0.25)/1+0.25 = 8
U3= 8V
Anonymous
What about that voltage on capacitor can not change
immediately? It will be the spark and system will lose
some energy.
RK, MSEE
Voltage on the cap C2 in this scenario builds up almost
instantly. Why, because Tau=R*C2 and since R is assumed as
zero, Tau =0s. Thus, there appears to be a sharp voltage
build-up on C2.
As for the spark, some energy will be lost and the voltage
on both caps be slightly less that 8V.
