# Theory in approaching Faraday law/circuit question

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1. Oct 3, 2016

### RoboNerd

1. The problem statement, all variables and given/known data
I have a square conducting loop of size L that contains two identical lightbulbs 1 and 2.

The magnetic field that goes into the page varies B(t) = a*t + b.

Lightbulbs have resistance of R0

Here is a question and the sample response. Disregard the part of the question that starts with the words "Now" and ends with "removed."

My question is why the resistances in the two loops are considered to be halved.
If anyone could explain why that to me, that would be great!
2. Relevant equations

R = resistivity * length /cross sectional area

3. The attempt at a solution

OK.

by the attached formula above, all I do is look at the length of each loop.

Initially I have a square loop with no wire in the center. Thus the total loop length is 4L

Then, I have a square loop and I add a wire in the center. Considering each half-loop, I have two sides of length L + two sides of length L/2, so I have total length of 3L not 2L.

Thus, the resistance is not cut in half by the formula, or so it seems.

Any input is appreciated, and thanks in advance.

2. Oct 3, 2016

### TSny

The wires are considered to have negligible resistance. Only the bulbs have resistance.

3. Oct 4, 2016

### andrevdh

Without the extra wire the two lightbulbs are in series, that is you have a circuit with two lightbulbs in it.

4. Oct 6, 2016

### RoboNerd

And then adding the extra wire makes the two bulbs in parallel, thus halving the total resistance?

5. Oct 6, 2016

### TSny

I don't believe this is the way to look at it. The solution given in post #1 is comparing the blue loop in the left figure to the green loop in the right figure.

How do the emf's in these two loops compare? How do the resistances of these two loops compare? How do the induced currents in these loops compare?

6. Oct 10, 2016

### RoboNerd

EMF's for left loop is 2 times as big as for the right loop due to twice as large of an area.

The resistances: for the left loop is twice as large as the right one.

Current: for the left loop is twice as less as the right one.

7. Oct 10, 2016

### TSny

Yes.

Can you give a reason for this answer?

8. Oct 14, 2016

### RoboNerd

Yes,

in the left loop there is one bulb and in the right loop there are two bulbs in series. Each bulb has a resistance "r" so, the resistance in the left loop is less than the one in the right so the current in the left loop is larger than that in the right loop.

I mistyped this:

9. Oct 14, 2016

### TSny

Yes
Not necessarily. The current depends not only on the resistance but also on something else. See Ohm's law.

10. Oct 21, 2016

### RoboNerd

In the left current, we have the resistance being 2R and in the right current we have the total resistance being R

EMF for left loop is twice as big as for the right one. But a 1 : 1 ratio between Voltage and resistance gives the same current. Right?

11. Oct 21, 2016

### TSny

I'm not sure I understand the 1 : 1 ratio statement. But, yes, the current will be the same. Ohm's law states $I = \varepsilon/R$. If $\varepsilon$ and $R$ are both doubled, then you can see that $I$ will remain unchanged.

12. Oct 21, 2016

### RoboNerd

That is exactly what I meant. Thanks for the help. I really appreciate it.