In the circuit of the figure = 2.60 kV, C = 7.10 mu F, R_1 = R_2 = R_3 = 0.790 M

Question

In the circuit of the figure = 2.60 kV, C = 7.10 mu F, R_1 = R_2 = R_3 = 0.790 M Ohm. With C completely uncharged, switch S is suddenly closed (at t = 0). At t = 0, what are current in resistor 1, current ¡_2 in resistor 2, and current ¡_3 in resistor 3? At t = infinity (that is, after many time constants), what are (d)i_l, (e)i_2, and (f)i_3? What is the potential difference V?2 across resistor 2 at (g)t = 0 and (h)t = infinity? Number Units Number Units Number Units Number Units Number Units Number Units

Explanation / Answer

a)

When switch is closed, the capacitor acts as a short circuit

So, current in resistor R1 = E/(R1 + R2 || R3)

= 2600/((0.79 + 0.79/2)*10^6)

= 2.19*10^-3 A = 2.19mA

b)

current in R2 = 2.19/2 = 1.1 mA

c)

current in R3 = 2.19/2 = 1.1 mA

d)

After a along time, the capactior charges fully and it acts as an open circuit

So, current in R1 =  2600/((0.79 + 0.79)*10^6) = 1.65 mA

e)

current in R2 = 1.65 mA

f)

current in R3 = 0 A

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