Solving q(t) for t=1.204 Secs

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

 

[topsquark]

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

 

[tacman]

Yes, you are on the right track. You have correctly solved for the value of t, which is 1.204 seconds. Now, to find the current i at this time, you can substitute this value of t into the derivative of q(t) that you found earlier: i=-3e^-t + 10e^-2t.

So, plugging in t=1.204 seconds, we get:

i=-3e^-1.204 + 10e^-2(1.204)

i=-3(0.297) + 10(0.108)

i=-0.891 + 1.08

i=0.189 amps

Therefore, at t=1.204 seconds, the current flowing through the circuit is 0.189 amps.

To find the value of charge, q, at this time, you can simply plug in t=1.204 seconds into the original equation for q(t):

q(1.204)=[3e^(-1.204) - 5e^(-2(1.204))]

q(1.204)=[3(0.297) - 5(0.108)]

q(1.204)=0.891 - 0.54

q(1.204)=0.351 coulombs

Therefore, at t=1.204 seconds, the charge flowing through the circuit is 0.351 coulombs.

Overall, your method is correct and you have solved for both the value of t and the corresponding current and charge at that time. Great job!