- #1

zenterix

- 555

- 76

- Homework Statement
- I am having major issues getting around calculations of

- Relevant Equations
- $$\oint\vec{E}\cdot d\vec{l}=-L\dot{I}

Consider the circuit below.

Just from looking at this circuit, we would expect the positive current to flow in a clockwise direction.

Let's define the normal vector pointing out of the screen. Then the positive circulation direction is counterclockwise.

I think that means that we have

I am not sure about the following: when we define the positive circulation direction we decide the direction that we are setting the positive ##I## to.

Consider the integral

$$\mathcal{E}_L=\oint\vec{E}\cdot d\vec{l}\tag{1}$$

It seems that we have

$$\oint\vec{E}\cdot d\vec{l}=\mathcal{E} -IR=-L\dot{I}\tag{2}$$

$$\dot{I}-\frac{R}{L}I=-\frac{\mathcal{E}}{L}\tag{3}$$

This doesn't seem correct, however.

After all, it leads to the solution

$$I(t)=\frac{\mathcal{E}}{R}\left (1-e^{\frac{Rt}{L}}\right )\tag{4}$$

But then ##I## starts at zero and then grows in magnitude without bounds. The sign is negative so the current does flow clockwise as we would expect.

I think the issue is the sign on the rhs of (2), ie the sign on ##-L\dot{I}##. I'd like to be able to explain why it should not have the negative sign.

If I do the calculation for current in the other direction it comes out right.

$$-\mathcal{E}+IR=-L\dot{I}$$

$$\dot{I}+\frac{R}{L}I=\mathcal{E}$$

I am missing something in going from this calculation to the one before.