Homework Help: Potentiometer question

1. Jun 16, 2010

daysrunaway

1. The problem statement, all variables and given/known data
The attached graphic shows a potentiometer circuit. The uniform wire AB is 100 cm long and has a resistance of 0.038 ohms/cm. The standard cell has an emf of 1.35 V. The galvanometer registers zero when the sliding contact is at L = 48.0 cm. If the sliding contact were moved to L = 56.0 cm, in what direction would electrons flow through the standard cell and why?

2. Relevant equations
V = IR
emf = V + Ir = Ve + Vi = L(Vo/Lo)

3. The attempt at a solution
I'm really confused by this question. The only thing I've been able to figure out so far is that, using I = V/R, I equals 0.74 A when the potentiometer is in balance, and equals 0.63 A when the contact is at 56.0 cm. I'm not sure what this means though. Doesn't the potentiometer mean that no current flows when it is in balance?

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Last edited: Jun 16, 2010
2. Jun 16, 2010

Mindscrape

Implicit in the problem is that on that left hand side there is a signal ground (or earth ground is fine to think about too if signal ground confuses you). All of your voltages are defined with reference to it. This info isn't super useful for the problem, but might help orient you a little more.

By changing where the galvanometer is you are changing the resistance in parallel with it. Current likes to take the path of least resistance, so by changing the resistance in parallel with the galvanometer, you are changing the path options for the current to travel. At 48 cm, just as much current wants to flow to the right as flow to the left (wrt the galvanometer) because the paths are equal. (In a word of perfect batteries and galvanometers we would of course expect the balance point to be 50, but it's not a perfect world.)

The thinking part comes in when you put the connection a little bit to the right. I'll let it marinate in you for a bit. Also, it might help you think of the wire segments as actual resistors.

Last edited: Jun 16, 2010
3. Jun 16, 2010

collinsmark

Yes, that's what I would assume, but by 'potentiometer' I think you might mean 'galvanometer' (The potentiometer is the 100 cm resistor thing with the sliding contact; the galvanometer is the thing that senses the current flow, labeled G in the figure). But be careful what zero current means. There are multiple parts to this circuit, and the galvanometer is attached to only one of these parts. Rephrasing, it means,

When the galvanometer registers zero, it means that there is no current passing through the galvanometer. (Again though, 'galvanometer' is not to be confused with 'potentiometer'. They are different components in the circuit.)

Zero current flowing through the galvanometer doesn't necessarily mean (and in this case certainly does not mean) that current is zero everywhere else too.

4. Jun 16, 2010

daysrunaway

Now I'm even more confused. Mindscrape, do you mean that the resistance of the standard cell is equal to the resistance of the length of wire (48.0 cm x 0.038 ohms/cm) when the potentiometer is balanced? If this were the case, I (= V/R) would be greater in the circuit with less resistance, which would be the part of the circuit with the standard cell. But collinsmark's comment says that there is no current when the potentiometer is balanced, and I don't understand how the fact that the resistances are the same means that no current flows through one part of the circuit.

5. Jun 16, 2010

Mindscrape

Sorry about the confusion. I mean to say if you treated the galvanometer as your reference point, if you thought about it as a battery, then the current coming out of the galvanometer would neither care about going left or right as the resistance is the same when it is balanced. You can think about the galvanometer as a voltage pushing around your electrons, and the battery at the top as simply a reservoir of electrons (they have come from somewhere). Of course the galvanometer is based on the premise of magnetic deflection, so if you have no current going into it then it doesn't deflect any.

Basically what the galvanometer is doing is using a cell to bias the voltage so you are now looking at that biased voltage as a reference point for your current. So when the galvanometer is balanced, the voltage bias (the galvanometer battery cell if you will) wants to put at just as much current as will go in, so the currents cancel out. You could do a KVL on the two loops and see the mathematical relation.

Last edited: Jun 16, 2010
6. Jun 16, 2010

daysrunaway

There's a lot of vocabulary in there that I don't understand (this is a physics 12 course) and I'm still really confused. I'll try to start again at the beginning and ask a slightly different question and see if that gets me anywhere.

So if the potentiometer is not hooked up to the simple circuit of battery and resistor, conventional current will flow through that circuit clockwise.

Considering the potentiometer circuit (with some resistance wire) only, if it isn't hooked up to the rest of the wire and the battery, its current will flow counterclockwise.

When the two are connected and the length of the wire is set to 48.0 cm, no current flows through the galvanometer, but current does flow through the whole resistance wire (including the section 'L'). I don't understand why this is? It can't be because the electric potential energy due to the standard cell equals the electric potential energy due to the terminal voltage of the battery, because these are both constant.

7. Jun 17, 2010

Mindscrape

Mathematically we see that the current going through the galvanometer is (see a tutorial on node voltage analysis if you want to get to this quickly, or to go through it slowly use KVL and KCL):

$$I_G = \frac{V_G}{R_G} - \frac{V_B-V_G}{R_1}$$

Where R_G is the resistance in parallel with the galvanometer, and R1 is the resistance from the galvanometer node to part B. At the particular length of 48cm, I_G goes to zero because the currents cancel.

Now, if you make R1 a little bit smaller... what will happen?

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