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A capacitor of capacitance C = 6.5 muF is initially uncharged. It is connected in series with a switch of negligible resistance, a resistor of resistance R = 9.5 K Omega, and a battery which provides a potential difference of VB = 65 V.

C = 6.5 ?F

R = 9.5 k?

VB = 65 V

Part (a) Calculate the time constant \tau for the circuit in seconds.

Part (b) After a very long time after the switch has been closed, what is the voltage drop VC across the capacitor in terms of VB?

Part (c) Calculate the charge Q on the capacitor a very long time after the switch has been closed in C.

Part (d) Calculate the current I a very long time after the switch has been closed in A.

Part (e) Calculate the time t after which the current through the resistor is one-third of its maximum value in s.

Part (f) Calculate the charge Q on the capacitor when the current in the resistor equals one third its maximum value in C.

Added Sun, 26 Jul '15

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Solutions

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1) time constant RC= 61.75 milli seconds

2) VC= VB

3)Q= C*V= ( 6.5 *10^-6)* 65V= 422.5 micro coulomb

4) I= 0 beacause circuit is opened

5) t= 67.839 milli seconds

6) Q= 281.66 micro columb

Added Sun, 26 Jul '15

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