FIGURE 8.61 For the circuit depicted in Fig. 8.62, (a) compute the circuit time
ID: 1800142 • Letter: F
Question
FIGURE 8.61 For the circuit depicted in Fig. 8.62, (a) compute the circuit time constant (b) determine v in the instant just before the switch is closed; (c) obtain an expression for v(t) valid for t > 0; (d) calculate v (3 ms). FIGURE 8.62 The switch drawn in Fig. 8.62 has been open ponderously long time. (a)Determine the value of the current labeled i just prior to the switch being closed. (b) Obtain the value of i just after the switch - closed, (c) Compute the power dissipated in each resistor over the range of 0Explanation / Answer
a) before closing (the capacitor is open)
i = 2 mA
b) just after the swith closes the capacitor is like short circuit so,
i = 1 mA
c) power
for the capacitor part
current through capacitor = I = 2-i for t > 0
Q = CV
I = dQ/dt = C dV/dt
V = i*10 - (2-i)*10 = 20i-20
I = dQ/dt = C dV/dt = C*d(20i -20)/dt
2 - i = C*d(20i -20)/dt
2-i = C*20*di/dt
di/(2-i) = dt/(C*20)
integrate
-ln(2-i) = t/C*20 + K
at t = 0 , i= 1mA
-ln(2-1) = +K
k = 0
-ln(2-i) = t/C*20
2-i = e(-t/C*20)
i = 2-e(-t/C*20)
now power dissipation through 10Kohm resistor not in series with capacitor = i210 = ( 2-e(-t/C*20))210 mW
power dissipation through 10 10Kohm resistor in series with capacitor = I210 =( e(-t/C*20))210 mW
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