For the RC circuit of Figure P8.1. R = 2.5 k Ohm and C = 50 mu F. If vC(0) = 10
ID: 1803018 • Letter: F
Question
For the RC circuit of Figure P8.1. R = 2.5 k Ohm and C = 50 mu F. If vC(0) = 10 V, find vC(t). Plot your answer for 0 t 5 tau, where tau is the circuit time constant. If vC(0) = 10 V, find iR(t) = -iC(t) without differentiating your answer to part (a). Plot your answer for 0 t 5 tau, where tau is the circuit time constant. At what time is the energy stored in the capacitor about 1% of it initial value? Compute iR(t) for vC(0) = 5V and vC(0) = 20V without doing any further calculations, i.e., by using the principle of linearity.Explanation / Answer
Given:-
R=2.5K;C=50F;V0=10V;
a).=RC; Vc(t)=V0e-t/RC=V0e-t/;
For t=0; Vc(t)=V0e-0=V0=10
For t=; Vc(t)=V0e-1=0.368(V0)=3.68;
t=2=V0e-2=0.1353(V0)=1.353;
t=4=V0e-4=0.0183(V0)=0.183;
t=5=V0e-5=0.006(V0)=0.006;
b)Applying KVL we get
-Ri(t)-Vc(t)=0;
i(t)=-Vc(t)/R;
For t=0;i(t)=-Vc(t)/R=10/2.5K=4mA
For t=; i(t)=-Vc(t)/R=3.68/2.5K=1.4mA
t=2=i(t)=-Vc(t)/R=1.353/2.5K=0.54mA
t=4=i(t)=-Vc(t)/R=0.183/2.5K=0.07mA
t=5=i(t)=-Vc(t)/R=0.006/2.5K=2.4A
c) Using principle of linearity;
The currents would be half the upper (i-e "b" part)values for v(0)=5V;And double the upper values for v(0)=20V;
For Example;
For v(0)=5V
at t=0;i(t)=2mA;
For v(0)=20V;
at t=0;i(t)=8mA;
HOPE THAT HELPS:)