a) what value of resistance must be used to cause a capacitorC=100x10E-6F to dis

Question

a) what value of resistance must be used to cause a capacitorC=100x10E-6F to discharage one half of its initial charge in 10seconds? b) Physically, why does a higher value of resistance cause thecapacitor to discharge more slowly?
c) Assume the capacitor int he RC circuit shown were initially charged toa voltage of 10 volts and the time constant measured as in this experiment. If the initial voltage was now doubled to 20 volts, what would be the value of the measured time constant?
c) Starting with a capacitor charged to 30V, how long will ittake to completely discharge it? Explain. a) what value of resistance must be used to cause a capacitorC=100x10E-6F to discharage one half of its initial charge in 10seconds? b) Physically, why does a higher value of resistance cause thecapacitor to discharge more slowly?
c) Assume the capacitor int he RC circuit shown were initially charged toa voltage of 10 volts and the time constant measured as in this experiment. If the initial voltage was now doubled to 20 volts, what would be the value of the measured time constant?
c) Starting with a capacitor charged to 30V, how long will ittake to completely discharge it? Explain.

Explanation / Answer

a) t = RC ln2

10 = R * 10^(-4) * 0.693

R = 144300ohm

b) because a higher resistance tries to decrease the current, thus reducing the rate of flow of charge leading to higher discharge time

c)the time constant would remin unchanged, stil equal to 144300* 10^(-4) = 14.43s

d) since the charge decays exponentially, it decreases by a certain fractin ina given time. So ideally it will take infinite time for it to decay completely

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