anonymous
  • anonymous
please help. medal and fan
Mathematics
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
  • anonymous
please help. medal and fan
Mathematics
katieb
  • katieb
See more answers at brainly.com
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 this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • 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
  • anonymous
I'm having trouble differentiating it
anonymous
  • 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
  • anonymous
with the numbers plugged in Ie^(10t) = integral: (V(t)e^(10t))/5
anonymous
  • anonymous
That's as far as I got
anonymous
  • anonymous
maybe treat V(t) as a constant?
anonymous
  • anonymous
anonymous
  • anonymous
I also plugged in 10 for V(t) and got -1/5 for C when I(0)=0.
sparrow2
  • sparrow2
i(r/L) this is constant,isn't it?
anonymous
  • anonymous
What do you mean?
sparrow2
  • sparrow2
because r/L is constant becasue r and L are constants
sparrow2
  • sparrow2
or i'm missing smth
anonymous
  • anonymous
ya when i plugged those values in I got e^10T
sparrow2
  • sparrow2
v(t) is also constant it's 10
anonymous
  • anonymous
yep, but the question is asking me to find an equation for V(t)
sparrow2
  • sparrow2
v(t)=10 this is equation
anonymous
  • anonymous
I wish it was, I already tried that lol
sparrow2
  • sparrow2
when it is writte above v(t)=10v v here is volt not a variable i guess
anonymous
  • anonymous
i think it is a constant but i'm not sure where i need to apply it
anonymous
  • anonymous
The only other part of the question is V(t) =
anonymous
  • anonymous
triciaal
  • 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
  • triciaal
|dw:1433057995387:dw|
amoodarya
  • 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
  • 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
  • amoodarya
v(t) is voltage of ? resistant or self ?
anonymous
  • anonymous
yesh veryish goosh sir

Looking for something else?

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