RC-circuit

[TsAmE]

In the circuit shown the capacitor is initially uncharged. (The battery may be regarded as ideal.)

1. After some time the circuit reaches equilibrium. Determine the charge on the capacitor.

2. Switch S is now opened once again. How long does it take for the charge on
the capacitor to drop by a factor of e–1 = 0.369?

(Referring to attached answer):

1. Why is the voltage at the top of the capacitor (Vax) = voltage drop across the 20Ω? Shouldnt it equal 12 - voltage drop across the 20Ω resistor? (same reasoning for Vay but with 10Ω instead).

2. How can the charge drop if the circuit is broken and hence there is no loop for the capacitor to discharge through?

 

[nsklavos]

When the switch is closed, you have 12v across each 30 ohm leg of the circuit. Vax = 12v/30 ohm*10 ohm = 4v. Vay = 12v/30 ohm*20 ohm = 8v. Therefore you have 4v across the capacitor. When the switch opens you have 15 ohms (the two 30 ohm legs in parallel) across the capacitor to discharge it.

 

[TsAmE]

I also got Vax = 4V and Vby = 8V and then I said the Vax - Vay (since Vax is at the top and Vay at the bottom) = 4 - 8 = -4V , but how can this be right if the voltage across the capacitor is negative?

 

[nsklavos]

It's all relative to how you reference it. It's just a difference in potential betwenn Vax and Vay. If you went Vay-Vax then the voltage would be positive.

 

[TsAmE]

To get the voltage across a component, dont you always have to subtract the voltage of the bottom part of the circuit from the top? Since the top should normally be at a higher potential?

 

[crutschow]

To get the voltage across a component, dont you always have to subtract the voltage of the bottom part of the circuit from the top? Since the top should normally be at a higher potential?

That's common but not mandatory. If the circuit is not grounded (as is the case here) then it's completely arbitrary.