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4. Consider the Circuit below. Let C = 12.0 muF, R1 = 240 ohm, R2 = 180 ohm, R3 = 65.0 ohm, and e = 20.0 V. a. Calculate the current supplied by the battery immediately after the switch is closed b. Calculate the current supplied by the battery a very long time after the switch has been closed. c. Calculate the energy stored in the capacitor a very long time after the switch has been closed. d. The switch remains closed for a long time. The switch is then opened. How long until the capacitor discharges to 25% of the maximum charge?

Explanation / Answer

Immidiately after closing switch capacitor acts as a conductor so

Req= R1||R2+R3 =720/7 +65 ohms

=> current through the battery = 20/(720/7+65)= 0.11915 Amps

b) After a long time capacitor acts as an insulator, So current passes through battery = 20/(65+240) =0.0656 Amps

c) Voltage across capacitor = 20*240/(240+65) = 15.7377 V

So energy stored = .5 CV2 =1486.05 J

d)After switch is opened charge on capacitor q=q0exp(-t/RC) ,q0=188.8524microC , R=240+180=420ohms

let at t= t, q=q0/4 =>t=RCln4

t=6986.92358microsec or t= 6.987msec

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