Potentiometer to measure cell voltage

[dba]

Homework Statement

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We did an experiment to measure E mf or the cell voltage of bateries. A thin wire (slide-wire potentiometer) is connected into the circuit. From point a to c the resistance along the wire is 36 Ohm. Depending on where the slider is placed (point b) the resistance along the wire changes in the relation
Lb/Lc = Rb/Rc.
Lb: length from a to b
Lc: length from a to c
Rb: resistance on length a-b
Rc: resistance on length a-c

We set the potential difference Vac or the voltage drop for the wire portion to 4V (connected from point a to point c) and slided the contact so that the current I_2 was zero.
With the length of the wire and the voltage drop we could calculate the cell voltage of the batery. the voltage of the power supply was V0=20V. I understood this so far.

Then we decreased V0 to 5V and tried to find the point with the slider where the current was zero. We could not find this point.

The question now is, why we could not find the null current for this setting. We had a batery with 1.5V (1D cell).

My second question is about the given equation:
Vac = [Rc/(R+Rc)] where R is the 150 Ohm resistor.

Where does the Rc/(R+Rc) comes from?

Homework Equations

for I_2 = 0
I_3=I_2

Vac=I_1*Rc

Lb/Lc = [(R+Rc)/Rc] * (E/V0) where E is the cell voltage E= Vac

The Attempt at a Solution

For the first part I was trying to look at the entire circuit. I have two voltage supplies. V0=5V and the batery V=1.5V. I have also resistor R=150 Ohm and the wire with adjustable Rb. The resistors ar in parallel which means that one equivalent resistor would be Req= (R*Rb)/(R+rb). The current I_2 for the lower circuit should be zero. I_3 needs to be zero also, since it is part of the same "loop." So I would have I_1.

But I am stuck how I can explain the question:
For a given V_) it is possible that the unknown E mf may be such that a null current may not be found. How should V0 be changed so as to allow a null current reading?


Thanks for any help.

 

[gneill]

The condition for I2 to be zero is that the slider must contact a point where the potential Vab on the wire is equal to the potential of the cell under test. When these potentials are equal, there is no potential difference to drive current I2.

Consider for a moment just the portion of the circuit that includes the power supply, the 150 Ohm resistor, and the 36 Ohm length of wire. These components form a voltage divider. What is the potential difference, Vac, across the wire when the supply is 20V? How about when it is 4V?

 

[dba]

I am not sure about this but Vac = I_1 * RC with Rc = 36 Ohm but how do I include the 20V and the other 150 Ohm? I think my problem is to understand how two voltage supplies in a circuit are related.

 

[gneill]

Can you calculate I1 given the total resistance in the loop?

 

[dba]

V=I*R
I_1 = V/R

V=20V
Resistors R=150 Ohm and Rc=36 Ohm are parallel, so
R_equivalent = R*Rc/R+Rc = (150*36)/(150+36) = 900/31 Ohm

I_1 = 20V/ (900/31)Ohm = 0.68a

I am still not sure about the second voltage supply of the battery. So, i calculated it now as it wasn't there.

 

[gneill]

The 150 Ohm resistor and the 36 Ohm resistance are NOT in parallel, they are in series.

 

[dba]

Ok, if they are in series than R_equivalent = R+Rc = 150 + 36 = 186 Ohm
And I_1 = 20V/186 Ohm = 0.1 A

 

[gneill]

Ok, if they are in series than R_equivalent = R+Rc = 150 + 36 = 186 Ohm
And I_1 = 20V/186 Ohm = 0.1 A

Okay, that's the current in the series circuit consisting of the voltage source and the two resistances alone (we're still ignoring the other cell and its resistance at the moment), when the voltage source is 20V. Given that value for the current, what is the voltage across the wire resistor from end to end?

Now repeat the calculation for when the voltage source is reduced to 4V.

 

[dba]

For V0=20V circuit
Rc=36 Ohm, I_1=0.1A
Vab = 0.1A * 36 Ohm = 3.6 Volt

For V0=4V
I_1 = 4V/186 Ohm = 0.02 A
Vab = 0.02 A * 36 Ohm = 0.72 Volt

 

[gneill]

For V0=20V circuit
Rc=36 Ohm, I_1=0.1A
Vab = 0.1A * 36 Ohm = 3.6 Volt

So, the slider when moving from point a to point c can "pick off" any potential from 0 to 3.6V when the voltage source is 20V.

What is the range when the voltage source is 4V?

 

[dba]

Any potential from 0 to 0.72 Volt which is smaller than the voltage of the battery of 1.5 Volt. And Vab has to be equal to the battery voltage to allow the current to be zero. Is that correct? is that the reason why I could not find the null current?

 

[gneill]

Any potential from 0 to 0.72 Volt which is smaller than the voltage of the battery of 1.5 Volt. And Vab has to be equal to the battery voltage to allow the current to be zero. Is that correct? is that the reason why I could not find the null current?

Yes! You've got it.

 

[dba]

Thank you so much for your help!