The figure below shows how a bleeder resistor (R = 266 k) is used to discharge a

Question

The figure below shows how a bleeder resistor (R = 266 k) is used to discharge a capacitor (C = 59.7 F) after an electronic device is shut off, allowing a person to work on the electronics with less risk of shock. Electronic circuit (a) What is the time constant? (b) How long will it take to reduce the voltage on the capacitor to 0.100% of its full value once discharge begins? (c) If the capacitor is charged to a voltage Vo through a 163 resistance, calculate the time it takes to rise to 0.86 5V0 (this is about two time constants)

Explanation / Answer

a) Time constant of the circuit

T = RC = 266*10^3*59.7*10^-6 = 15.88 sec

b) Discharging equation of the capacitor

V(t) = Vo*e^(-t/T)

=> (0.1/100)*Vo = Vo*e^(-t/15.88)

taking logarithm both sides

=> ln(0.001) = -t/15.88

=> -6.907 = -t/15.88

=> time taken , t = 109.68 sec

c) new time constant T = RC = 163*59.7*10^-6 = 9.73*10^-3 sec

charging equation of the RC circuit

V(t) = Vo*( 1 - e^(-t/T))

=> 0.865*Vo = Vo*( 1 - e^(-t/T))

=> e^(-t/T) = 0.135

taking logarithm both sides

=> -t/T = ln(0.135) = -2.0

=> t = 2.0*9.73*10^-3 = 0.0194 sec  

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