Asked by Ti
Calculate the time it takes for the voltage across the resistor to reach 11.0V after the switch is closed.
the total resistance is 10.0kΩ , and the battery's emf is 27.0V . If the time constant is measured to be 34.0μs .
the total resistance is 10.0kΩ , and the battery's emf is 27.0V . If the time constant is measured to be 34.0μs .
Answers
Answered by
Henry
I'm assuming R and C are in series.
R = 10k ohms
RC = 34us = 34*10^-6 s.
Vr = E/e^(t/RC) = 11 Volts.
27/e^(t/RC) = 11
e^(t/RC) = 27/11 = 2.45455
Take Ln 0f both sides:
(t/RC)*Ln e = Ln 2.45455
(t/RC)*1 = 0.89794
Multiply both sides by RC:
t = 0.89794RC
t=0.89794*(34*10^-6)=30.5*10^-6 s =
30.5 Microseconds(uS).
NOTE: At the instant the switch is closed(t=0), all of the voltage is across the resistor. It decreases to
11 volts after 30.5 seconds.
R = 10k ohms
RC = 34us = 34*10^-6 s.
Vr = E/e^(t/RC) = 11 Volts.
27/e^(t/RC) = 11
e^(t/RC) = 27/11 = 2.45455
Take Ln 0f both sides:
(t/RC)*Ln e = Ln 2.45455
(t/RC)*1 = 0.89794
Multiply both sides by RC:
t = 0.89794RC
t=0.89794*(34*10^-6)=30.5*10^-6 s =
30.5 Microseconds(uS).
NOTE: At the instant the switch is closed(t=0), all of the voltage is across the resistor. It decreases to
11 volts after 30.5 seconds.
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