2 In the circuit of Figure with C completely uncharged switc

2. In the circuit of Figure, with C completely uncharged, switch S is suddenly closed. Opens at 0s 60 1012 2.0 (a) Find the surrent through each resistor and the chazze Q on the capacitor immediately after the switch is closed;

Solution

a) when switch is closed charge on capacitor would be 0. it would act as short circuit.

so resistor 40 and 10 ohm would be in parallel and 60ohm would be in series

so eqv resistance = 60+(40*10/50) = 68 ohm

so current through 60 ohm= current through eqv = 100/68 =1.47 ampere

this current would be divided in ration 4:1 so current through 40 ohm would be 1.47/5 =.294 A

and current through 10 ohm would be 4*1.47/5 = 1.176 A

b)after long time there would be no current through capacitor .it would acts as open switch.

current through 10 ohm would be 0

now current through 40 ohm and 60 ohm would be 100/(60+40) = 1 ampere

charge on capacitor = V/C = 100/2*10^-6 = 50 * 10^6 C

c) when switch is opened 60 ohm resistor wouldn\'t be part of circuit.

the capacitor will dischargde through 40 and 10 ohm resistor with time constant (40+10) *2 * 10^-6 = 10^-4 s

Q\'=Qe^(-t/10^-4)

Q\'=Q/5

so 1/5 = e^(-t/10^-4)

t = 10^-4 * ln 5 = 1.6 * 10^-4 s