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Not a clue where to begin on this one.

http://tinyurl.com/2j9vr4

lol 😕

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Have a look at http://en.wikipedia.org/wiki/Internal_resistance.

You can use Ohm's law to calculate the current which flows when the switch is closed and then just put the numbers into the formular.

To derive this equation consider a cell to be a voltage source wired in series with a resitor, of resistance 'r'. Connect this arrangment in series with a load resistance, or resistance 'R'. The cell terminals are effectively either side of the load resitor. Now use Kirchhoff's voltage law, that the total voltage around the loop is zero, and the required relationship comes out.

NB: the relationship you will get with this method will be:

VL = VS / (1+(r/R)), where VL is the voltage when the switch is closed and VS is the voltage when the switch is open. This is equivalent to the one from wikipedia.

I hope this is helpful. :-)

2 edits
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Not exactly a showstopper, I assume you know ohms law, I=E/R,
E=IR, R=E/I, I=current E=voltage, R=resistance.
The differance of 0.3 volts off and on means 1.2/4=0.3 amps flowing so the 0.3 volt divided by 0.3 amp gives one ohm as the internal resistance of the cell.

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Originally posted by Bad wolf
Not a clue where to begin on this one.

http://tinyurl.com/2j9vr4

lol 😕
Well, you know that the there is a 1.2 V potential difference accross the 4 ohm resistor. Therefore, by Ohm's law, you get the current to be .3 Amps.

Now, when the switch is open, you can ignore the lower part of the circuit. It tells you the potential difference accros the voltmeter is 1.5 V. Again, using Ohm's law, with the current you;ve found, you get the internal resistance to be 5 ohms.

I hope I'm right.

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Originally posted by abejnood
Well, you know that the there is a 1.2 V potential difference accross the 4 ohm resistor. Therefore, by Ohm's law, you get the current to be .3 Amps.

Now, when the switch is open, you can ignore the lower part of the circuit. It tells you the potential difference accros the voltmeter is 1.5 V. Again, using Ohm's law, with the current you;ve found, you get the internal resistance to be 5 ohms.

I hope I'm right.
Unfortunatly this method is incorrect becasue no current flows when the switch is open, so the 0.3ohm current (which is correct for when the switch is closed) cannot be used. The answer is 1ohm which can be found using either the method I outlined or from Sonhouse's method, which are very similar.

1 edit
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Originally posted by abejnood
Well, you know that the there is a 1.2 V potential difference accross the 4 ohm resistor. Therefore, by Ohm's law, you get the current to be .3 Amps.

Now, when the switch is open, you can ignore the lower part of the circuit. It tells you the potential difference accros the voltmeter is 1.5 V. Again, using Ohm's law, with the current you;ve found, you get the internal resistance to be 5 ohms.

I hope I'm right.
You can ignore the resistor only for the time the switch is open. That represents the battery with no load, no current flow, so you cannot measure the resistance that way. You have to use the voltage differance between the off and on position.

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Originally posted by sonhouse
Not exactly a showstopper, I assume you know ohms law, I=E/R,
E=IR, R=E/I, I=current E=voltage, R=resistance.
The differance of 0.3 volts off and on means 1.2/4=0.3 amps flowing so the 0.3 volt divided by 0.3 amp gives one ohm as the internal resistance of the cell.
Why is the difference between the voltages used? I don't remember seeing it worked like that before...

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Originally posted by Bad wolf
Why is the difference between the voltages used? I don't remember seeing it worked like that before...
The cell terminals are being treated like the branch in a potential devider between the internal resistance and the load resistance.

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Originally posted by MattP
The cell terminals are being treated like the branch in a potential devider between the internal resistance and the load resistance.
I see. 🙂

I wasn't thinking of it like that at all.

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Originally posted by Bad wolf
I see. 🙂

I wasn't thinking of it like that at all.
You also gotta factor in Gay-Lussac's Law of Combining Volumes and then remember that the volume of a gas is inversely proportional to its pressure. After that you can use the R constant and get the flaming symbalus.

5 edits
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Here's my solution.

http://tinyurl.com/3xetnn

I made a little note to show to myself why the internal resistance isn't something to worry about when working out current flow, when using the p.d. across the terminals.
🙂

Is 36 ohms the right answer?


edit: apologies for my crappy handwriting.
🙁

edit: lol, sorry for all the edits....

1 edit
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Originally posted by Bad wolf
Here's my solution.

http://tinyurl.com/3xetnn

I made a little note to show to myself why the internal resistance isn't something to worry about when working out current flow, when using the p.d. across the terminals.
🙂

Is 36 ohms the right answer?


edit: apologies for my crappy handwriting.
🙁

edit: lol, sorry for all the edits....
I have to take the AP Physics test in a week and a day.

I'm in so much trouble.

EDIT:MattP, want to take it for me? 🙂

1 edit
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Originally posted by Bad wolf
Here's my solution.

http://tinyurl.com/3xetnn

I made a little note to show to myself why the internal resistance isn't something to worry about when working out current flow, when using the p.d. across the terminals.
🙂

Is 36 ohms the right answer?


edit: apologies for my crappy handwriting.
🙁

edit: lol, sorry for all the edits....
No, the answer is 1ohm.

The working goes like this:

when the switch is closed you have the cell in series with a load resistance, R = 4ohms. Model the internal resistance as a resistance, r, in series with the cell.

Now you have a voltage source, Vs, wired in series with an internal resistance, r, and a load resistance, R.

Using Kirchhoff's voltage law (are you familiar with this law? by the way) the total voltage drops around the loop sum to zero as follows:

Vs - Ir - IR = 0 equation {1}

where Vs is 1.5v (the EMF of the cell), and I is the current.

However, IR is the load voltage (using ohm's law), i.e the voltage which is across the cell terminals when the switch is closed = 1.2V. Think of this as the EMF - lost volts - P.d =0 as you put it in your workings.

so Vs - Ir - p.d = 0

Now rearange to get: r = (Vs - p.d)/I

where p.d = load voltage = 1.2v. I is calculated using ohms law, I = 1.2/4 = 0.3A.

Now r = (1.5 - 1.2)/0.3 = 1ohm.

I hope this is helpful, sorry if it isnt very clear.

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Originally posted by MattP
No, the answer is 1ohm.

The working goes like this:

when the switch is closed you have the cell in series with a load resistance, R = 4ohms. Model the internal resistance as a resistance, r, in series with the cell.

Now you have a voltage source, Vs, wired in series with an internal resistance, r, and a load resistance, R.

Using Kirchhoff's voltag ...[text shortened]... 3A.

Now r = (1.5 - 1.2)/0.3 = 1ohm.

I hope this is helpful, sorry if it isnt very clear.
like I said....

1 edit
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Originally posted by MattP
Using Kirchhoff's voltage law (are you familiar with this law? by the way) the total voltage drops around the loop sum to zero as follows:

Vs - Ir - IR = 0 equation {1}
I just looked at my textbook, its in there alright, but I've not had anyone explain it to me, nor had any questions on it.
🙁