# Heat generated in coil

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1. Feb 12, 2015

1. The problem statement, all variables and given/known data

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 $\Delta t$
2. Relevant equations

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

3. 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?

2. Feb 12, 2015

### haruspex

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

3. Feb 13, 2015

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

4. Feb 13, 2015

### 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 ?

5. Feb 13, 2015

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.

6. Feb 13, 2015

### haruspex

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.

7. Feb 13, 2015

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

8. Feb 13, 2015

### BvU

Still this annoying a ...

And a strange result for the integral...

9. Feb 13, 2015

$i=i_0-bt$

10. Feb 13, 2015

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

11. Feb 13, 2015

### BvU

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 $I_0$

 from zero to Delta t

12. Feb 13, 2015

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

13. Feb 13, 2015

No that doesn't seem right.

14. Feb 13, 2015

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

15. Feb 13, 2015

### 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)

 crossed post #14. Bingo !

16. Feb 13, 2015

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

17. Feb 13, 2015

### BvU

Almost :)

18. Feb 13, 2015

Is it correct?
Now can I use i2Rt?

19. Feb 13, 2015

### 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.

20. Feb 13, 2015

At t=0,you get $q=i\Delta t$.