MHB Application to Electric Circuits

[ineedhelpnow]

I don't know under what forum to post this but since I am learning about it in calculus, i guess ill just put it here for now any maybe itll get moved to the right place. I'm given a simple circuit with resistance $12 \varOmega$ and the inductance is $4 H$ with a voltage of $40 V$. Find I(t) and I(0.1). Here's what I did:

$L=2$
$R=10$
$E(t)=40$

$L \frac{dI}{dt}+RI=E(t)$

$2 \frac{dI}{dt}+10I=40$

$\frac{dI}{dt}+5I=20$

$e^{\int \ 5 dt}=e^{5t}$

$e^{5t} \frac{dI}{dt}+5e^{5t}I=20e^{5t}$

$\frac{dI}{dt}(e^{5t}I)=20e^{5t}$

$e^{5t}I=\int \ 20e^{5t}dt=4e^{5t}+C$

$I(t)=4+Ce^{-5t}$

so to solve for C:

$I(0)=0$ so $4+C=0$ and $C=-4$

$I(t)=4(1-e^{-5t})$

solving for I(0.1):

$I(0.1)=4(1-e^{-5(0.1)})= 4(1-e^{-1/2})$

did i do it right?

 

[I like Serena]

I don't know under what forum to post this but since I am learning about it in calculus, i guess ill just put it here for now any maybe itll get moved to the right place. I'm given a simple circuit with resistance $12 \varOmega$ and the inductance is $4 H$ with a voltage of $40 V$. Find I(t) and I(0.1). Here's what I did:

$L=2$
$R=10$
$E(t)=40$

$L \frac{dI}{dt}+RI=E(t)$

$2 \frac{dI}{dt}+10I=40$

$\frac{dI}{dt}+5I=20$

$e^{\int \ 5 dt}=e^{5t}$

$e^{5t} \frac{dI}{dt}+5e^{5t}I=20e^{5t}$

$\frac{dI}{dt}(e^{5t}I)=20e^{5t}$

$e^{5t}I=\int \ 20e^{5t}dt=4e^{5t}+C$

$I(t)=4+Ce^{-5t}$

so to solve for C:

$I(0)=0$ so $4+C=0$ and $C=-4$

$I(t)=4(1-e^{-5t})$

solving for I(0.1):

$I(0.1)=4(1-e^{-5(0.1)})= 4(1-e^{-1/2})$

did i do it right?

Assuming that you mixed up 2 problems and that for the actual problem at hand you have $L=2\text{ H}$, $R=10\ \mathrm{\Omega}$, and $V=40 \text{ V}$, yep, you did it right!

 

[ineedhelpnow]

haha very funny. (Tongueout) believe it or not, that is the ORIGINAL problem this time :D