logarithm to solve current time

[lena49]

the current flowing in a circuit is given by

i = 2e^(-2t) + 5e^-t

amperes, where t is the time in seconds. Determine the value of t when i = 3 amperes

 

[Bob Brown MSEE]

~0.7 sec can be obtained using Newton's Method or graping calculator. Can you get more digits? What have you tried?

 

[Deleted member 4993]

the current flowing in a circuit is given by

i = 2e^(-2t) + 5e^-t

amperes, where t is the time in seconds. Determine the value of t when i = 3 amperes

Substitute

u = e^-t

then your equation becomes:

3 = 2u² + 5u

2u² + 5u - 3 = 0

Solve the above quadratic equation using your favorite method.

 

[lena49]

Substitute

u = e^-t

then your equation becomes:

3 = 2u² + 5u

2u² + 5u - 3 = 0

Solve the above quadratic equation using your favorite method.

Thank you
ans:
t = ln 2

 

[soroban]

Hello, lena49!

This approach is equivalent to Subhotosh's solution.

\(\displaystyle \text{The current flowing in a circuit is given by: }\:i \:=\:2e^{-2t} + 5e^{-t}\text{ amperes,}\)
\(\displaystyle \text{where }t\text{ is the time in seconds.}\) .\(\displaystyle \text{Determine the value of }t \text{ when }i=3.\)

We are given: .\(\displaystyle 3 \:=\:2e^{-2t} + 5e^{-t}\)

Multiply by \(\displaystyle e^{2t}\!:\;\;3e^{2t} \:=\:2 + 5e^t \quad\Rightarrow\quad 3e^{2t} - 5e^t - 2 \:=\:0\)

Factor: .\(\displaystyle (3e^t + 1)(e^t-2) \:=\:0\)


And we have two equations to solve.
. . One of them has no real root.

 

[lena49]

Then is this correct?

i = 2e^(-2t) + 5e^(-t)
if i = 3
3 = 2e^(-2t) + 5e^(-t)
2(e^(-t))^2 + 5e^(-t) = 3
let e^(-t)=x
2x^2 + 5x = 3
x^2 + 5x/2 = 3/2
x^2 + 5x/2 + 25/16 = 49/16
(x + 5/4)^2 = 49/16
x + 5/4 = 7/4 or x + 5/4 = - 7/4
x = 1/2 or x = -3 rejected
e^(-t) = 1/2
t = ln (2)
t = 0.693

 

[Deleted member 4993]

Then is this correct?

i = 2e^(-2t) + 5e^(-t)
if i = 3
3 = 2e^(-2t) + 5e^(-t)
2(e^(-t))^2 + 5e^(-t) = 3
let e^(-t)=x
2x^2 + 5x = 3
x^2 + 5x/2 = 3/2
x^2 + 5x/2 + 25/16 = 49/16
(x + 5/4)^2 = 49/16
x + 5/4 = 7/4 or x + 5/4 = - 7/4
x = 1/2 or x = -3 rejected
e^(-t) = 1/2
t = ln (2)
t = 0.693

Correct ..... good work.

 

[HallsofIvy]

Then is this correct?

i = 2e^(-2t) + 5e^(-t)
if i = 3
3 = 2e^(-2t) + 5e^(-t)
2(e^(-t))^2 + 5e^(-t) = 3
let e^(-t)=x
2x^2 + 5x = 3
x^2 + 5x/2 = 3/2
x^2 + 5x/2 + 25/16 = 49/16
(x + 5/4)^2 = 49/16
x + 5/4 = 7/4 or x + 5/4 = - 7/4
x = 1/2 or x = -3 rejected
e^(-t) = 1/2
t = ln (2)
t = 0.693

Yes, though I would not have written the final answer as an approximation but left it as t= ln(2).

 

[lena49]

ok I should leave it to t = ln (2), thanks everyone

 

[lookagain]

ok I should leave it to t = ln (2), thanks everyone

No, it depends on whether the instructor wants you to state the amount of seconds to a rounded decimal. Either the instructions for the problem should state to give the answer to the nearest tenth of a second, for example, or it is put out ahead of time what number of decimal places of approximation is desired. If a student reports at the end that it is t = ln(2) s, then that isn't pragmatic. I want to know that the student know approximately how large ln(2) is by reporting a rounded value.