Solving q(1.204): Charge Flow in Circuits

[Sajjad]

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
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t= 1.204 seconds |------ am i doing ok till here?
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by putting this in equation 1 we will get the vale for charge,q.

 

[berkeman]

What do you mean find the value of t? The value of t when what happens?

 

[Sajjad]

i think!

everyone knows that i= q/t in general but to find current i at some instant we have i= di/dt. In the given equation above if we can find the value of t we can put this value of t back in the equation and can find the charge q. by having charge q and time t, current i can be found by using formula i=q/t and the value of t can only be determined by taking derivative of the equation. at least i see only this possibility.

 

[Hootenanny]

So basically you have
$$q(t) = 3e^{-t} - 5e^{-2t}$$
$$q '(t) = 10e^{-2t} - 3e^{-t}$$
You now can't solve any further without having more information. As berkeman said, when do you need to know t? Do you want to find t when q is maximum / minimum etc.