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## anonymous one year ago please help. medal and fan

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1. 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.

2. anonymous

I'm having trouble differentiating it

3. anonymous

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

4. anonymous

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

5. anonymous

That's as far as I got

6. anonymous

maybe treat V(t) as a constant?

7. anonymous

@rational

8. anonymous

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

9. sparrow2

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

10. anonymous

What do you mean?

11. sparrow2

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

12. sparrow2

or i'm missing smth

13. anonymous

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

14. sparrow2

v(t) is also constant it's 10

15. anonymous

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

16. sparrow2

v(t)=10 this is equation

17. anonymous

I wish it was, I already tried that lol

18. sparrow2

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

19. anonymous

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

20. anonymous

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

21. anonymous

@triciaal

22. 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

23. triciaal

|dw:1433057995387:dw|

24. 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 $e^{10t}i(t)-0=2(e^{10t}-1)\\ \rightarrow * e^{-10t} \\i(t)=2(1-e^{-10t})$

25. anonymous

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

26. amoodarya

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

27. anonymous

yesh veryish goosh sir

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