In Figure 5 E 150 kV C 3800 pF R1 R2 R3 680 M At time t

In Figure 5, E = 15.0 kV, C = 3800 pF, R₁ = R₂ = R₃ = 6.80 M?.     At time t = 0 (i.e., immediately after the circuit is connected) the current through R₁ = ___________. (Give your answer in the form of \"a.bc x 10^(x)\" mA)

At t = 0 (i.e., immediately after the circuit is connected) the current in R₂ = _______. (Give your answer in the form of \"a.bc x 10^(x)\" mA.)

At t = 0 (i.e., immediately after the circuit is connected) the current in R₃ = _______. (Give your answer in the form of \"a.bc x 10^(x)\" mA.)

What is the current in R₃ at t = ? ? (Give your answer in the form of \"a.bc x 10^(x) A\".)

What is the potential difference across R₁ at t = ? ?. (Give your answer in the form of \"a.bc x 10^(x) V\".)

What is the potential difference across R₂ a very long time after the circuit is connected? Give your answer in the form \"a.bc x 10^(4)\" V.

Solution

at time t=0 , there would be no charge in capacitor so it would act as short circuit.

R2 and R3 are in parallel.

so R23 = R2.R3/R2+R3 =3.4 ohm

Now R1 and R23 are in series so R123=3.4 +6.8= 10.2 Mohm

So current through battery which is equal to current through R1 = V/R123= 15 *10^3/10.2*10^6 = 1.47 * 10^-3 A.

Now current through R2 and R3 are divided which is inversely propotional to R2 and R3 ratio.

So current through R2= Current Through R3= 1.47 *10^-3/2 = 7.35 * 10^-4 A.

at t= capacitor would act like open and there would be no current through R2. so R1 and R3 would be in series and equivalnet resistance would 13.6 . so current through R3= 15*10^3/13.6 * 10^6 = 1.1 * 10^-3 A

Potetnial difference across R1= 1.1 * 10^-3 * 6.8 * 10^6 =7.5 * 10^3 V.

there wouldn\'t be any potential difference across R2 as there will not be any current flowing.