In the circuit of the figure below, b = 1.6 kV, C = 7.0 mu F, R_1 = R_2 = R_3 =

Question

In the circuit of the figure below, b = 1.6 kV, C = 7.0 mu F, R_1 = R_2 = R_3 = 0.76 M ohm. With C completely uncharged, switch S is suddenly closed (at t = 0). At t = 0, what Is current i_1 in resistor 1? A At t = 0, what Is current i_2 in resistor 2? A At t = 0, what is current i_3 in resistor 3? A Repeat for t = infinity (that is, after many time constants.) What is current i_1? A What is current i_2? A What is current i_3? A What is the potential difference V_2 across resistor 2 at t = 0? v What s V_2 at t = infinity? v Sketch v_2 versus t between these two extreme times. (Do this on paper. Your instructor may ask you to turn in this sketch.)

Explanation / Answer

a), b), and c)

At t = 0, when the capacitor is completely discharged (q = 0), there is no voltage across C and it is effectively a “short.” The circuit can thus be drawn as a simple array of resistors, giving

Reff = R + [1/R + 1/R]1 = R + 0.5R = 1.5*R initially.

The net current (which flows through R1) is then

I1 = E/Reff = 1600/(1.5*0.76*10^6) = 1.4 mA

At junction b, I1 splits into two equal parts I2(0) = I3(0) = 0.5*I1(0) = 0.7 mA

d), e) and f)

As t the capacitor gets fully charged and no more current flow onto it: I3() = 0 . Then we can just ignore that part of the circuit and calculate the current through the other two resistors in series as

I1 = I2 = E/2R = 1600/(2*0.76*10^6) = 1.05 mA

g) V2(t = 0) = 0.76*10^6*0.7*10^-3 = 0.532 kV

h) V2(t = ) = 0.76*10^6*1.05*10^-3 = 0.798 kV

i) We anticipate an initial voltage drop of V2(0) = R*I2(0) heading monotonically toward the final value of V2() = R*I2() and the initial change should be more rapid — this is a qualitative description of exponential decay of the difference between V2 and its equilibrium value of V2().

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