Calculating Heat Generated in a Coil due to Changing Current and Resistance

[AdityaDev]

Homework Statement

What amount of heat will be generated in a coil of resistance R due to a charge q passing through it if the current in the coil decreases down to zero uniformly during a time interval $\Delta t$

Homework Equations

$dH=i^2Rdt$
$i=\frac{dq}{dt}$

The Attempt at a Solution

$dH=i^2Rdt$
$dH={\frac{dq}{dt}}^2Rdt$
$dH=\frac{{dq}^2R}{dt}$
How do I solve this multi variable differential equation?

 

[haruspex]

Homework Statement

What amount of heat will be generated in a coil of resistance R due to a charge q passing through it if the current in the coil decreases down to zero uniformly during a time interval $\Delta t$

Homework Equations

$dH=i^2Rdt$
$i=\frac{dq}{dt}$

The Attempt at a Solution

$dH=i^2Rdt$
$dH={\frac{dq}{dt}}^2Rdt$
$dH=\frac{{dq}^2R}{dt}$
How do I solve this multi variable differential equation?

Your are told the current decreases uniformly with time. Write that as an equation.

 

[AdityaDev]

Your are told the current decreases uniformly with time. Write that as an equation.

$i=-at$
$di=-adt$
$dH=iR^2dt$
$dH=-aR^2tdt$
$H=-aR^2\frac{t^2}{2}$

 

[BvU]

I see an unknown a appearing, whereas the known (given) q has disappeared.
And the math leaves to be desired:

• $i = -at$ can't be right (try $t=0$ and $t = \Delta t$)
• Any reason for $i^2R$ changing into $iR^2$ ?

With $t$ in the answer I suppose you mean $\Delta t$ ?
Any chance to present the answer in terms of the givens ?

 

[AdityaDev]

• Any reason for $i^2R$ changing into $iR^2$ ?

Big typing mistake sorry
$di/dt = -a$
This shows that i decreases with time, just like dN/dt=-N for nuclear decay.
Hence you will again get I=-at.

 

[haruspex]

Big typing mistake sorry
$di/dt = -a$
This shows that i decreases with time, just like dN/dt=-N for nuclear decay.
Hence you will again get I=-at.

You're forgetting the constant of integration. The current starts positive and finishes at zero. And you need to relate a to q.
And it is not like nuclear decay which follows a negative exponential with time, not linear.

 

[AdityaDev]

You're forgetting the constant of integration. The current starts positive and finishes at zero. And you need to relate a to q.
And it is not like nuclear decay which follows a negative exponential with time, not linear.

$i = i_0 - at$
$dH= (i_0 - at)^2Rdt$
$H=i_0^2Rt - aRt^2/2$

 

[BvU]

Still this annoying a ...

And a strange result for the integral...

 

[AdityaDev]

Still this annoying a ...

$i=i_0-bt$
Please help me solve this

 

[AdityaDev]

You can't exclude a because the equation will become dimensionally incorrect

 

[BvU]

• $i = -at$ can't be right (try $t=0$ and $t = \Delta t$)

Now it fits for t=0. Next: make it fit for $t = \Delta t$.
And (to keep in the back of your mind): $q = \int_0^{\Delta t}\; i \;dt$ to eliminate $I_0$

[edit] from zero to Delta t

 

[AdityaDev]

Now it fits for t=0. Next: make it fit for $t = \Delta t$.

$i=\frac{q}{\Delta t} - at$

 

[AdityaDev]

No that doesn't seem right.

 

[AdityaDev]

$i=i_0 - \frac{i_0t}{\Delta t}$

 

[BvU]

$i=i_0-at$ not good ? try $i=i_0-bt$ or $i=i_0-ct$ :)

Reminded me of this joke where you see Einstein in front of a blackboard with $E=ma^2$ and $E=mb^2$ stricken through, and then $E=mc^2$ ! (with an exclamation mark). Sorry.

What I am pushing towards: $i=i_0-at$ is correct, but not good enough. t = 0 works fine, now fill in $\Delta t$ when you know $i(\Delta t) = 0$ you get a.
Bit now you have $i_0$ as unknown. However, ... (see post #11)

[edit] crossed post #14. Bingo !

 

[AdityaDev]

What I am pushing towards: $i=i_0-at$ is correct, but not good enough. t = 0 works fine, now fill in $\Delta t$ when you know $i(\Delta t) = 0$ you get a.
Bit now you have $i_0$ as unknown. However, ... (see post #11)

Ok...here comes my updated equation
$i=\frac{q}{\Delta t} - \frac{q}{(\Delta t)^2}t$

 

[BvU]

Almost :)

 

[AdityaDev]

Is it correct?
Now can I use i2Rt?

 

[BvU]

Almost means: not there yet: $$q \; {?\atop =}\ \int_0^{\Delta t}\; i \;dt = {q\over \Delta t} \int_0^{\Delta t} \;(1-{t\over \Delta t}) \; dt = {q} \int_0^{\Delta t} \;(1-{t\over \Delta t}) \; d({t\over \Delta t}) = q\int_0^1 (1-u) du = {q\over 2}$$so there is a 2 missing. Obvious if you sketch i(t), a triangle.

 

[AdityaDev]

At t=0,you get $q=i\Delta t$.
Here you get Q=q/2
What is the difference between the two 'q's?
(I got the answer after using post 19. Thank you very much)

 

[BvU]

No, at t=0 you get $i = i_0 = {2q\Delta t}$

Q = q/2 ? There should come q = q !

And the picture refuses to be uploaded, sorry. Can't copy and paste a picture either. Must be some issue thanks to the security mafia.
1000 words:

a triangle from (0, $I_0$) to ($\Delta t$, 0).

Area ${1\over 2} I_0\Delta t \equiv q \Rightarrow I_0 = ...$

 

[haruspex]

Still this annoying a ...

And a strange result for the integral...

I don't know why you had this objection to introducing a. It should then have been a simple matter to find a in terms of q and $\Delta t$ by integrating.

 

[BvU]

I don't know why you had this objection to introducing a. It should then have been a simple matter to find a in terms of q and $\Delta t$ by integrating.

Yes, we're working on that. Nearly done (see post #16). Or completely -- I don't know what I can deduce from post #20.

It was more the duo $i_0$ and a , of which I picked a to eliminate. But you are referring to post #8; we are at over 20 now ...