Question

In an open RC circut, the total resistance is 15Kohms and the battery emf is 24 and constant is measured to be 35 microsecond calculate (a) the total capacitance of the circut and the time take the voltage across the 16.0 V after the switch is closed.

Answer 1

a)Time constant

T=RC

=>C =T/R =35*10^-6/15*10^3

C=2.333*10^-9 F or 2.333 nF

b)

V=Vo[1-e^(-t/T)]

16=24[1-e^(-t/35*10^-6)]

ln(0.3333) =-t/35*10^-6

t=38.45 us

Answer 2

a. timeconstant T = RC

C = 35*10^-6/15000

C = 2.33nF

b. V = vo e^-t/RC

e^-t/RC = 16/24

-t/RC = ln (0.6667)

-t/RC =   --0.405

t = 0.405 * 35*10^-6

t = 14.175 us   or 1.4175 *10^-5 secs

Answer 3

Answer 4

R=15 k ohms

E=24 v

R*C=35*10^-6 s

C=35*10^-6/15*10^3

C=2.333*10^-9 F

C=2.33 nF

Vc=V(1-e^(-t/RC))

16=24(1-e^-(t/35*10^-6))

0.333=e^-(t/35*10^-6)

-0.477=-(t/35*10^-6)

t=16.7 *10^-6 s=16.7 us