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.

http://i782.photobucket.com/albums/yy109/att5041/R...

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)

My attempt:

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 Answer

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  • 1 decade ago
    Favorite 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.

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