The electric current in a discharging R-C circuit is given by `i=i_0e^(-t/(RC))` where `i_0` `R `and `C `are constant parameters and `t `is time. Let `i_0=2.00A`, `R=6.00xx10^5 ohm` and `C=0.500 muF`.
a. Find the current at t=0.3 s.
b. Find the rate of change of current at t=0.3 s.
Find approximately the current at t=0.31 s.

Answer

(a)2/e (b)-20/3e (c)5.8/3e

Explanation

Equation `i =i_0.e^(-t/RC)`, where
i_0=2A, R=6x10^5 ohm
C=0.0500xx10^(-6)F
=5x10^(-7)F
=5xx10^(-7)F
i=2.0 e^)-t/0.3)
a. i=2xxe^(-1)= 2/e amp.
b.di/dt=(-i_0)/RC.e^(-t/RC)
when t=0.3 sec, then di/dt=2/(0.3) e^)(-0.3)/(0.3))(-20)/(3e) amp/sec
c. At t= 0.31 st
i= 2e((-0.3)/0.3), 5.8/(3e) amp (approx.)