Please compute the initial currents through each of the resi

Please compute the initial currents through each of the resistors. At this time a dielectric (k=4) is suddenly placed in between the plates. Conceptually explain how the currents in the resistors: will be affected at that moment how they evolve afterwards.

Solution

a) R2 and R3 in series, their equivalent, R\' = R2 + R3

R \' = 2 + 6 = 8 ohm

now R\' and R1 are in parallel,

1/R\" = 1/R\' + 1/R1 = 1/8 + /14

R\" = 5.09 ohm

now R\" and R4 are in series.

Req = R\" + R4 = 5.09 + 4 = 9.09 ohm

initially Voltage across capacitor will be zero.

hence current through battery, I = e / Req = 18 / 9.09 = 1.98 A

I4 = 1.98 A

PD across R\" = 18 - (4 x 1.98) = 10.08 Volt

I1 = 10.08 / 14 = 0.72 A

I2 = I3 = 10.08 / (2 + 6) = 1.26 A

B) Capacitance of capacitor, C = e0 A / d

    Area is not visible properly .

I am taking A = 1.5 x 10^-3 m^2 (if value is different then put accordingly)

C = (8.854 x 10^-12 x 1.5 x 10^-3) / (2 x 10^-6) = 6.64 x 10^-9 F

Q = C V = 6.64 x 10^-9 x 18 = 1.195 x 10^-7 C

c) now capacitance will change,

C\' = k C = 4 x 6.64 x 10^-9 = 26.56 x 10^-9 F

initially current will be same.

now time constant will increase.

so current will decrease slowly then brfore.

Capacitor will take more time to fully charge.

and final charge on capacitor will be , Q = C\' V = 26.56 x 10^-9 x 18 = 4.78 x 10^-7 C