a) Find the current through each resistor and the charge Q on the capacitor at t

Question

a) Find the current through each resistor and the charge Q on the capacitor at t=0.

c) The switch is now open at t=0. Find the time interval required for the charge on a capacitor to fall to two-fifths its initial value.

A circuit is constructed with four resistors, one capacitor, one battery and a switch as shown. The values for the resistors are: R1= R2= 65 Theta R 3 = 3 45 Omega R4 = 120 Omega Capacitance is C= 86 Mu F Voltage is V = 24 V a) Find the current through each resistor and the charge Q on the capacitor at t=0. b) Find the current through each resistor and the charge Q on the capacitor at t= infinity c) The switch is now open at t=0. Find the time interval required for the charge on a capacitor to fall to two-fifths its initial value.

Explanation / Answer

a) at t = 0, capacitor acts as short ckt.so, there is no effect of R2 and R3.

Rnet = R1 + R4

= 65 + 120

= 185 ohms
so, Current through R1 and R4,
I = V/Rnet

= 24/185

= 0.13 A <<<<<<<<<-----------------Answer

Current through R2 and R3 is zero. <<<<<<<<<-----------------Answer

charge on the capacitor, Q = 0 <<<<<<<<<-----------------Answer

b) at t = 0, capacitor acts as open ckt.
so,

Rnet = R1 + R2 + R3 + R4

= 65+65+45+120

= 295 ohms

so, Current through each resistor(R1,R2,R3 and R4),
I = V/Rnet

= 24/295

= 0.08 A <<<<<<<<<-----------------Answer

Potential differnce across is equal to potenatil difference across (R2+R3)

so, V_C = I*(R2+R3)

= 0.08*(65+45)

= 8.8 volts

charge on the capacitor, Q = V_C*C

= 8.8*86*10^-6

= 7.57*10^-4 C or 757 micro C <<<<<<<<<-----------------Answer

c) when switch is open, Time constant T = (R2+R3)*C

= (65+45)*86*10^-6

= 9.46*10^-3 s

In discharging,

Q = Qmax*e^(-t/T)

at time t, Q = Qmax/5

so,

Qmax/5 = Qmax*e^(-t/T)

e^(t/T) = 5

t = T*ln(5)

= 9.46*10^-3*ln(5)

= 1.52*10^-2 s or 152 m s <<<<<<<<<-----------------Answer

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