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.
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 .
1 answer
0 answers