please help. medal and fan

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions.

- anonymous

please help. medal and fan

- katieb

See more answers at brainly.com

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

Consider a series circuit consisting of a resistor of R ohms, an inductor of L henries, and variable voltage source of V(t) volts (time t in seconds). The current through the circuit I(t) (in amperes) satisfies the differential equation:
dI/dt + (R/L)I = (1/L)V(t)
Find the solution to this equation with the initial condition I(0)=0, assuming that R=50Ω, L=5 H, and V(t) is constant with V(t)=10 V.

- anonymous

I'm having trouble differentiating it

- anonymous

I(e^(RT/L)) = integral: (V(t)e^(RT/L))/L

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

with the numbers plugged in
Ie^(10t) = integral: (V(t)e^(10t))/5

- anonymous

That's as far as I got

- anonymous

maybe treat V(t) as a constant?

- anonymous

- anonymous

I also plugged in 10 for V(t) and got -1/5 for C when I(0)=0.

- sparrow2

i(r/L) this is constant,isn't it?

- anonymous

What do you mean?

- sparrow2

because r/L is constant becasue r and L are constants

- sparrow2

or i'm missing smth

- anonymous

ya when i plugged those values in I got e^10T

- sparrow2

v(t) is also constant it's 10

- anonymous

yep, but the question is asking me to find an equation for V(t)

- sparrow2

v(t)=10 this is equation

- anonymous

I wish it was, I already tried that lol

- sparrow2

when it is writte above v(t)=10v v here is volt not a variable i guess

- anonymous

i think it is a constant but i'm not sure where i need to apply it

- anonymous

The only other part of the question is V(t) =

- anonymous

- triciaal

don't have the solution,
my input:
dl/dt = dl/dv*dv/dt
for some reason I would be separating the variables and integrating
when you integrate you add the constant
here the constant is given as 10V

- triciaal

|dw:1433057995387:dw|

- amoodarya

\[\frac{ di }{dt }+\frac{ r }{ l }i=\frac{ 1 }{ l}v(t)\\r=50\\l=5\\v(t)=10\\\frac{ di }{dt }+\frac{ 50 }{ 5 }i=\frac{ 1 }{ 5}10\\\] and now solve this differential equation
\[\frac{ di }{dt }+10i=2\\ \rightarrow * e^{10t} \\e^{10t} (\frac{di}{dt}+10i)=2e^{10t}\\e^{10t} di +e^{10t} 10i dt =2e^{10t}dt\\now\\ \space \\\ d(e^{10t}i)=2e^{10t}dt \] now apply integral \[\int\limits_{0}^{t}d(e^{10t}i)=\int\limits_{0}^{t}2e^{10s}ds \\e^{10t}i(t)-e^0i(0)=2e^{10t}-2\]
and now apply I(0)=0
so you have
\[e^{10t}i(t)-0=2(e^{10t}-1)\\ \rightarrow * e^{-10t} \\i(t)=2(1-e^{-10t})\]

- anonymous

@amoodarya
I tried that but it was wrong.
The question is asking for V(t) and i'm not sure what they want

- amoodarya

v(t) is voltage of ? resistant or self ?

- anonymous

yesh veryish goosh sir

Looking for something else?

Not the answer you are looking for? Search for more explanations.