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

Question

The figure below shows how a bleeder resistor (R = 266 kohm) is used to discharge a capacitor (C = 53.7 mu F) after an electronic device is shut off, allowing a person to work on the electronics with less risk of shock. What is the time constant? How long will it take to reduce the voltage on the capacitor to 0.100% of its full value once discharge begins? If the capacitor is charged to a voltage VQ through a 173 ohm resistance, calculate the time it takes to rise to 0.865/o (this is about two time constants).

Explanation / Answer

while discharging the capacitor loses the potential as v = Vo*e^-t/T

T = time constant = RC

(a)

time constant T = RC = 266*10^3*53.7*10^-6 = 14.3 s

(b)

for v = 0.001Vo

0.001vo = vo*e^(-t/14.3)

t = 98.8 s

(c)

while charging

v = vo(1-e^-t/T)

T = 173*53.7*10^-6 = 0.0093 s

0.865 = 1-e^-(t/0.0093)

t = 0.0186 s

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