Theory in approaching Faraday law/circuit question

[RoboNerd]

Homework Statement

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!

Homework Equations

R = resistivity * length /cross sectional area

The Attempt at a Solution

[/B]
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.

 

[TSny]

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

 

[andrevdh]

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

 

[RoboNerd]

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

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

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

 

[TSny]

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

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?

 

[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.

 

[TSny]

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.

Yes.

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

Can you give a reason for this answer?

 

[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:

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

 

[TSny]

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

Yes

so the current in the left loop is larger than that in the right loop.

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

 

[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?

 

[TSny]

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?

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.

 

[RoboNerd]

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