In the circuit of the figure 430 kV C 720 muF R1 R2 R3

In the circuit of the figure = 4.30 kV, C = 7.20 muF, R_1 = R_2 = R_3 = 0.940 Mohm. With C completely uncharged, switch S is suddenly closed (at t = 0). At t = 0, what are (a) current l_1 in resistor 1, (b) current l_2 in resistor 2, and (c) current l_3 in resistor 3? At t = infinity (that is, after many time constants), what are (d)l_1, (e)l_2, and (f)l_3? What is the potential difference V_2 across resistor 2 at (g)t = 0 and (h)t = infinity?

Solution

Immediately after switch is closed at t=0 ,the capacitor will act as short circuit,so equivalent resistance

R_eq =R₁+(R₂||R₃) =R₁+R₂R₃/(R₂+R₃)

R_eq = 0.94 + (0.94*0.94)/(0.94+0.94) =1.41 Mohms

Total Current flowing in the circuit is

I=E/R_eq =(4.3*10³)/(1.41*10⁶) =3.05*10^-3 A or 3.05 mA

So Current through each resistor is at t=0 is

a)

I₁=I = 3.05 mA or 3.05*10^-3A

b)

By Current divider rule

I₂=I*(R₃/R₂+R₃) =3.05*(0.94/0.94+0.94)=1.525*10^-3 A or 1.525 mA

c)

By Current divider rule

I₃=I*(R₂/R₂+R₃) =3.05*(0.94/0.94+0.94)=1.525*10^-3 A or 1.525 mA

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After long time i.e t=infinity ,capacitor will act as open circuit ,so equivalent resistance

R_eq=R₁+R₂ =1.88 Mohms

Total Current flowing in the circuit is

I=E/R_eq =4.3*10³/(1.88*10⁶)=2.29*10^-3 A or 2.29 mA

So Current flowing through each resistor is

d)

I₁=I=2.29*10^-3 A or 2.29 mA

e)

I₂=I=2.29*10^-3 A or 2.29 mA

f)

Since the capacitor acts as open circuit ,no current flows through R₃ so

I₃=0 A

g)

Voltage across R2 at t=0 is

V₂=I₂R₂ =(1.525*10^-3)(0.94*10⁶)

V₂=1433.5 Volts

h)

Voltage across R2 at t=infinity is

V₂=I₂R₂ =(2.29*10^-3)(0.94*10⁶)

V₂=2150 Volts