An ∃mazing Physics thread (Unit 2)

[Guest]

For b.), if you look at the y-intercept of the graph, you get a current when there is no voltage across the output; that is, when all the voltage is going across the battery's internal resistance. Using , we get a value of for R.

Thanks for the swift reply. Appreciate you taking time to reply during your VCE. +1 for your explanation for part (a). I understand it wholly now 

But for part (b), I don't quite understand everything what you said. So essentially, do we have to find the y-intercept of the graph?
I don't see how it equals 5.
And could you also explain why the principals you outlined still apply to what seems to be a non-ohmic resistor?

[SocialRhubarb]

Firstly, we'll try to demonstrate an ohmic resistor can behave in the manner shown on the graph.

Imagine you have this circuit:


Let's say that at first, the resistance of R is . What's the voltage across and current through R?

It's a pretty simple calculation, I=0.5 A, V=5 V.

But now lets make the resistance of R . What's the voltage across and current through R now?

Same maths. I=0.1 A, V=9 V.

How can this happen? Clearly voltage has come up and current has gone down. Well, this has happened because:
1. Resistance has changed.
2. The resistor is in series with another resistor.

Both of these, though unstated, are true in your question.

In more of a direct answer to your question, for part b.), when the graph shows a voltage of 0V, what it really means is that there is a voltage of 0V across the load, but the voltage has to drop somewhere, since the sum of the voltage gains is equal to the sum of the voltage losses. If you connect 6V power source to circuit, your 6V is going to go somewhere, be it wires, resistors or even the air. In this case there is a voltage drop across your 'internal resistance' of your battery. Since the external voltage is 0V, ALL of the EMF must be going to your internal resistance, so the voltage across the internal resistance is 1.6V, and if you look at the graph, the current is 320 mA or 0.32 A when the external voltage is 0V. Using these we can calculate the value of the internal resistance, which is 1.6V/0.32A= 5 Ohms.

[lzxnl]

Alternatively...we can use algebra and make everything clearer (:

Let the actual voltage of the battery be V. Let the internal resistance be R and the resistance of the resistor be r. Similarly, let the voltage measured across the resistor be v.

What we want is a relationship between v and I. V=I(R+r) = IR + v.
Therefore v=V-IR
As the current increases, the measured voltage should drop linearly.
Which is what SocialRhubarb was getting at, but now in figures.

Just for fun:
Using voltage divider formulas, v/V=r/(R+r)
v=Vr/(R+r)=V(1-R/(R+r))
If you increase the resistance of the resistor, the fraction becomes smaller, and the measured voltage becomes larger. However, no matter how large your resistance, v<V. Always. That's the issue with real life batteries.