Differential equations

Discussion in 'Homework Help' started by MrRockchip, May 25, 2010.

1. MrRockchip Thread Starter New Member

May 25, 2010
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Please, tell me, what are the Correct differential equations
for calculating the electric current on inductivity ?

http://myimgs.net/images/zynyf.jpg

I've created some equations, but at the end (after solving them)
I get the wrong result - so, they were wrong.

2. Papabravo Expert

Feb 24, 2006
10,340
1,850
Voltage, Vr, across a resistor R, according to Ohm's Law, is "I*R"
Voltage, Vl across an inductor L is L*(di/dt)

The result of writing KVL around a loop should be a first order equation. with a step function representing the closing or opening of the switch. Does that help?

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3. MrRockchip Thread Starter New Member

May 25, 2010
18
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For inductor:

i(-0)=E/R;

L(di/dt)=E / ((R*R)/R+R);
L(di/dt)=2*E/R;
(di/dt) = 2*E/(R*L)
Because E,R,L=const., after integrating :

i(t)=(2*E/(R*L))*t

Something is wrong here, because i(t) can't be infinite;
it's obvious that i(+inf)=E/R .

Last edited: May 26, 2010
4. t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
Here's a start ... note that R2 only determines the value of $I_{0}$

$L\frac{di}{dt}+R_{1}i=E$

$\frac{di}{(\frac{E}{R_{1}}-i)}=\frac{R_{1}}{L}dt$

$\int^{i}_{I_{0}}\frac{di}{(\frac{E}{R_{1}}-i)}=\int^{t}_{0}\frac{R_{1}}{L}dt$

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5. MrRockchip Thread Starter New Member

May 25, 2010
18
0
Thank you, Tnk!

Here's my solution:

We know that

So, the final answer is

It seems to be correct, because

Am i right ?

6. t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
Actually

$I_{0}=\frac{E}{R_{1}||R_{2}}$

You happened to have

$I_{0}=\frac{2E}{R_{1}}$

"correct" only because R1=R2 in the problem statement.

Otherwise you appear to have got a handle on the problem solution.

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