Current prob

  • Thread starter Sajjad
  • Start date
5
0
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.
 
335
4
Sajjad said:
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.
How many places are you going to post this question? It's been answered.

-Dan
 

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