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current in circuit `L(dI)/dt+RI=E(t)` find I(t)

current in circuit `L(dI)/dt+RI=E(t)` find I(t)

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current in circuit `L(dI)/dt+RI=E(t)` find I(t)

Postby barnamah » Mon 21 March, 2011 17:15

The question says:
Consider a circuit containing a resister with resistor of R ohm `Omega` and and inductor with an inductance of L Henrie (H). The voltage drop due to the inductor is `L(dI)/(dt)` and IR due to the resistor.
According to the one Kirchhoff's laws, the sum of the voltage drops is equal to the supplied voltage E(t). Thus,

`L(dI)/dt+RI=E(t)`. Find `I(t)`.

Suppose L=2, E(t)=10, R=3, I(0)=0.

Any idea of how to start?
Explore and know. That is asked.
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Re: current in circuit `L(dI)/dt+RI=E(t)` find I(t)

Postby Integr88 » Mon 21 March, 2011 17:46

we have the values:
L=2, E(t)=10, R=3, I(0)=0.

the equation is:
`L(dI)/dt+RI=E(t)`

First, lets sub in the values in our equation:
`2(dI)/dt+3I=10`

The question is asking for I(t) with initial value of I(0)=0.
`(dI)/dt=(10-3I)/2`

Now is the important step which is moving all elements with `I` with `dI`.
let's divide both side by `(10-3I)/2` so we can go forward
`((dI)/dt)/((10-3I)/2)=((10-3I)/2)/((10-3I)/2)`
on the right we have which is equal to 1 `((10-3I)/2)/((10-3I)/2)=1` so have:
`((dI)/dt)/((10-3I)/2)=1`

now I am going to write the equation in such way, that is obvious to be simplified

we have on the left side `((dI)/dt)` divided by `((10-3I)/2)`, let's write is in a better way:

`((dI)/dt)*1/((10-3I)/2)`, now by looking at `1/((10-3I)/2)` you realize that 1 is divided by `((10-3I)/2)`. It can be written `1-:(10-3I)/2` and we know to divide two fraction we can flip one fraction and multiply them `1xx2/(10-3I)` which is `2/(10-3I)`.
Putting `(dI)/dt` beside the `2/(10-3I)` as it was before, we have:
`(dI)/dt*2/(10-3I)` which can be written as `1/dt*(dt2)/(10-3I)`


now we have `1/dt*(2dI)/(10-3I)=1`. Let's move `1/dt` to the right by multiplying both side by `dt` :
`dt*1/dt*(2dI)/(10-3I)=1*dt`

The `dt` on the left cancels the `dt` at the denominator and our equation becomes:
`(2dI)/(10-3I)=dt`

Now we can integrate:
`int(2dI)/(10-3I)=intdt`
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integrate `int(2dI)/(10-3I)=intdt`

Postby Integr88 » Mon 21 March, 2011 18:02

To integrate `int(2dI)/(10-3I)=intdt` we start by integrating the right side which is `t+C` and we have:
`int(2dI)/(10-3I)=t+C`
Ingetrating `intdt` means take the integral of `int(1)dt` which mean take the integral of `1` and as the variable is `t` in `dt` it refers to taking the drevative of `1` with respect to `t`.
`:.` the `int(1)dt` is `(1)t+C` which can be written as `t+C`


On the left side we have:
`int(2dI)/(10-3I)`

To the integral of `1/x` is `ln|x|` and if we have `1/(2x+3)`, its integral is `ln|2x+3|*1/2`

but if we have `2/x` the integral of `int2/xdx` is `2ln|x|` and pay close attention to 2 that appears as a coefficient of ln(x).

In our case we have `int(2dI)/(10-3I)` so :
`[2*ln|10-3I|*1/3]` which can be written as `[2/3*ln|10-3I|]` (I moved `1/3` to the front).

Now our complete equation looks like this:
`2/3*ln|10-3I|=t+C`

The integration is completed but we want to solve for I.

Solving for I
Multiply both side by `3/2` so we get rid of `2/3` from the left side.
`3/2*2/3*ln|10-3I|=(t+C)*3/2` and we have:
`ln|10-3I|=(t+C)*3/2`

Distribute (multiply) the `3/2` on the right side and we have:
`ln|10-3I|=3/2t+3/2C`

Because C is an unknown constant, we write it as `ln|10-3I|=3/2t+C`

Get rid if `ln` from `ln|10-3I|=3/2t+C` using `e` ad the base for both side:
`e^(ln|10-3I|)=e^(3/2t+C)`

`e` cancels the `ln` and we have:
`|10-3I|=e^(3/2t+C)`

on the right side we have `e^(3/2t+C)`. Recall that if we have `A^rxxA^v`, it can be written as `A^(r+v)`. We we have `e^(3/2t+C)` so we can go reverse and write it as `e^(3/2t)*e^C`


We have now:
`|10-3I|=e^(3/2t)*e^C`

to get rid of the absolute value `|10-3I|` we must add `+-` on the other side of equation and we get:
`10-3I=+-e^(3/2t)*e^C`
On the right side we have `+-e^(3/2t)*e^C`. We can rearrange it as `e^(3/2t)*(+-e^C)`. This `(+-e^C)` is still a arbitrary constant so we can replace it for example by `k`

Our equation becomes `10-3I=ke^(3/2t)`
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