Sajjad
Mar8-06, 05:48 AM
The charge flowing through a circuit is
q(t)=[3e^(-t) - 5e^(-2t)]--------(1)
find the value of t and then current i.
as i=dq/dt.
i am doing it like this
i=-3e^-t + 10e^-2t..................taking derivative
let e^-t=u...........for eaze
i=10[u^2-3u/10]
let 1=0 then
10[u^2-3u/10]=0
10[u^2 - 2(u)(3/20) + (3/20)^2 - (3/20)^2]=0 .....using formula
10[u-3/20]^2 - 10[3/20]^2=0
10[u-3/20]^2=10[3/20]^2
10[u-3/20]^2=9/40
[u-3/20]^2=9/400
u-3/20=sqrt[9/400]
u=sqrt[9/400] + 3/20
e^-t = 3/20 + 3/20.......as u=e^-t
Taking log on both side
ln[e^-t]= ln[3/10]
-t= -1.204
-----------------|
t= 1.204 seconds |------ am i doing ok till here?
-----------------|
by putting this in equation 1 we will get the vale for charge,q.
q(t)=[3e^(-t) - 5e^(-2t)]--------(1)
find the value of t and then current i.
as i=dq/dt.
i am doing it like this
i=-3e^-t + 10e^-2t..................taking derivative
let e^-t=u...........for eaze
i=10[u^2-3u/10]
let 1=0 then
10[u^2-3u/10]=0
10[u^2 - 2(u)(3/20) + (3/20)^2 - (3/20)^2]=0 .....using formula
10[u-3/20]^2 - 10[3/20]^2=0
10[u-3/20]^2=10[3/20]^2
10[u-3/20]^2=9/40
[u-3/20]^2=9/400
u-3/20=sqrt[9/400]
u=sqrt[9/400] + 3/20
e^-t = 3/20 + 3/20.......as u=e^-t
Taking log on both side
ln[e^-t]= ln[3/10]
-t= -1.204
-----------------|
t= 1.204 seconds |------ am i doing ok till here?
-----------------|
by putting this in equation 1 we will get the vale for charge,q.