Physics: RC Circuits?
In the circuit below, the switch S is initially open, and has been open for a long time. The switch is closed at t = 0.
Immediately after the switch is closed AND after the switch has been closed for a long time:
A. What is the charge and voltage on the capacitor?
B. What's the current through R3? (the resistor in parallel with the capacitor)
I'm not sure how to do it immediately after the switch is closed.
For after the switch has been closed for a long time:
Q = C(emf) = .012 C
The voltage across the capacitor = voltage across R3
V = iR, and the i for R3 = 6A
This is also where I get confused..isn't the current through the capacitor 0? Does this affect R3?
Can someone please show/explain how to do this problem? Is any of my work correct? Thanks.
- 1 decade agoFavorite Answer
Just after t=0, The voltage drop across R3 is V/3 (because the 3 resistors are equal in magnitude). At this point capacitor has not had enough time to build a charge. The current through R3 is the same current that is going through the other two resistors (the capacitor is an open for DC).
When t--->infinity, the capacitor has built up enough charge to match the emf across it (across R3), so the current through R3 is I=(V)/4*R3, because there are 4 equal drops in voltage that sum to V. The definition of capacitance is C=Q/V(across) ---> Q=C*V/4.