Solve emf of battery with resistors and voltmeter

[saw176]

Homework Statement

Two 9.4 k-ohm resistors are placed in series and connected to a battery. A voltmeter of sensitivity 1000 ohm/V is on the 3-V scale and reads 2 V when placed across either resistor. What is the emf of the battery? (Ignore it's internal resistance)

Homework Equations

V=IR
Emf = IR

The Attempt at a Solution

First of all, I'm kind of unsure what "on the 3-V scale" means. My guess is that it means it takes 3 volts for a full-scale deflection of the voltmeter? Because it has a sensitivity of 1000 ohm/V and is reading 2 volts:
I = V/R
I = 2 V / 2000 ohm
I = 1 x 10^-3A

Then I tried to find the total resistance of the circuit, first with the voltmeter and one resistor in parrallel and the other resistor in series:
1/R = 1/9.4 k-ohm + 1/2000 ohm = 1.65 k-ohm
R = 1.65 k-ohm + 9.4 k-ohm = 11.1 k-ohm

Finally, I went Emf = IR
Emf= 1 x 10^-3 (11.1 k-ohm)
Emf = 11.1 V

The correct answer is 10 V, so my answer is close, but I am unsure if that's just by chance, because I never actually used the fact that the voltmeter is "on the 3.0 V scale". I really have no idea if my thought process is correct for this problem, am I on the right track?

Thanks!

 

[gneill]

The 3V scale means that if 3V is presented to the test leads, the meter pointer will have maximum deflection -- it will swing to the full scale position.

The Ohms-per-volt rating of a voltmeter is the resistance of the voltmeter divided by the full-scale reading. So in this case, if R is the resistance of the meter and it is set on the 3V scale, then R/3V = 1000 Ohms/Volt. You should be able to determine the actual resistance of the meter from that fact.

 

[saw176]

The 3V scale means that if 3V is presented to the test leads, the meter pointer will have maximum deflection -- it will swing to the full scale position.

The Ohms-per-volt rating of a voltmeter is the resistance of the voltmeter divided by the full-scale reading. So in this case, if R is the resistance of the meter and it is set on the 3V scale, then R/3V = 1000 Ohms/Volt. You should be able to determine the actual resistance of the meter from that fact.

Thanks, that makes sense. So the actual resistance of the meter would be 3000 ohm then, if I use R/3V = 1000 Ohms/Volt. I took this resistance and repeated my process:
I = 2 V/ 3000 ohm = 6.67 x 10^-4A

Total Circuit resistance = 1/9.4 kohm + 1/3000 ohm = 2.27 k-ohm
2.27 kohm + 9.4 kohm = 11.67 k-ohm

Emf = 11.67 k-ohm (6.67 x 10^-4A)
E = 7.78 Volts

So my process must still be wrong somewhere then to be getting an answer of 7.78? Thanks!

 

[gneill]

The problem is with your assumption that the current is flowing only through the meter. It flows through the parallel combination of the meter and one of the resistors. So think again about how you calculate the current.

 

[saw176]

Ah! Got it I think!

I = 2 / (2.27 k-ohm)
I = 8.81 x 10^-4

Total resistance is still 11.67 k-ohm, so:

Emf= 8.81 x 10^-4 (11.67 k-ohm)
Emf= 10.3 = 10 V

Thanks for your help!