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?
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?