Find both the internal resistance of a battery and EMF

[Shinwasha]

Homework Statement

When a nonideal battery is connected to a 3.0-Ω resistor in a circuit, the current in the circuit is 2.0A . When the same battery is connected to a 1.0-Ω resistor in a circuit, the current in the circuit is 3.0 A.

What is the internal resistance of the battery?

What is the emf of the battery?

Homework Equations

I=ε/(R_battery+R)
I=ε/R

The Attempt at a Solution

I've looked at this to solve the emf for each individual situation, which gives me for the first scenario
2.0=ε/3.0 ε=6.0
2nd scenario
3.0=ε/1.0 ε=3.0

I realized that plugging this in wouldn't work because this requires me to have two different EMFs when it only wants one.

However using it I get for first scenario

2.0=6.0/(rbattery + 3.0), = rbattery + 3.0 = 3.0, = r battery = 0

second scenario

3.0 = 3.0 /(rbattery + 1.0), = r battery + 1 = 1, = r battery = 0

Considering the equation has to unknowns that I need to find out figuring out how to nail one of the variables down would help solve this.

 

[Simon Bridge]

You have two unknowns, so you need two simultaneous equatons.
Write it out in symbols first:

Lets call the EMF just V for voltage - it is the open circuit voltage of the battery, and what you normally think of as the battery voltage, so why not?
The two load resistors are R1 and R2.
The internal resistance is r - lower case because the internal resistance is expected to be small.
The two currents are I1 and I2.

You have already done: V = I1(R1 + r)
That's one euation - you need another one. Can you see one that relates I2 and R2 to V and r?

 

[Shinwasha]

The only thing I can possibly see is to take the EMF (which is unknown). Looking through the book I'm not seeing anything, and using google every thing I seeing gives one of the two unknowns. It seems like there just isn't enough information and that the voltage using V=IR should be the same in both cases but it's not. I'm stumped on how to go about solving this as I can't see any way of going about the way you are describing.

 

[andrevdh]

The emf stays the same (unknown value) for both circuits.

 

[Shinwasha]

The emf stays the same (unknown value) for both circuits.

Which doesn't add up with the values I have. If in the second case the current was 6 amperes it would make sense. But since Current = EMF/Resistance it doesn't in this case.

That's one euation - you need another one. Can you see one that relates I2 and R2 to V and r?

Only thing I could think of here is the -IR is the same thing as V.

 

[Shinwasha]

I think I got it now.

Using Ohm's law

V=IR

For case 1
V = 2(3.0+R_i)
V= 6+2R_i

Case 2
V = 3(1.0+R_i)
V = 3.0+3R_i

6+2R_i= 3+3R_i

3 = R_i

Thus

Emf = 6.0+2*3

EMF = 12

Just wasn't seeing the R part as having to pieces.

 

[Simon Bridge]

Well done... the trick is to understand what the equations are telling you instead of just substituting.
In the relationship I=V/R, that V is the potential difference across the load R... so, when you used it, that was not the same as the emf.
That relationship, btw, is also Ohm's law.

Following the notation in post #2

V=I1(R1+r)
V=I2(R2+r)

Here you have two equations and two unknowns.
Notice how it is clearer when you use variables instead of numbers?

 

[Shinwasha]

Well done... the trick is to understand what the equations are telling you instead of just substituting.
In the relationship I=V/R, that V is the potential difference across the load R... so, when you used it, that was not the same as the emf.
That relationship, btw, is also Ohm's law.

Following the notation in post #2

V=I1(R1+r)
V=I2(R2+r)

Here you have two equations and two unknowns.
Notice how it is clearer when you use variables instead of numbers?

It is actually easier for me to use numbers than variables because I find it easier to relate values to something.

 

[Simon Bridge]

That is something that will change - and will have to if you are to advance.
Good luck.