Asked by AAA
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?
Answers
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.
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.
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