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A 35 μF capacitor is in a series with a 270 KΩ resistor. Initialy, the voltage across the capacitor is 300 V. A multimeter whose resistance is equal to twice the resistor resistance is placed in parallel with the resistor. How long does it take for the voltage to drop to 100 V?
5 years ago

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Answered by henry2,
300/e^x = 100.
e^x = 300/100 = 3
x*Lne = Ln3
X = 1.1 = T/RC.

R = (R1*R2)/(R1+R2) = (270*540)/(270+540) = 180k ohms.
RC = 180k * 35uF = 6300 ms = 6.3 s.
x = T/RC = 1.1
T/6.3 = 1.1
T = 6.9 s.
5 years ago

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