Determine the value of the resistance R that will produce a current

1. May 15, 2009

science.girl

1. The problem statement, all variables and given/known data
A battery with emf $$\epsilon$$ and internal resistance r is connected to a variable resistance R at points X and Y. Varying R changes both the current I and the terminal voltage Vxy. The quantities I and Vxy are measured for several values of R and the data are plotted in a graph, as shown above on the right.
a. Determine the emf $$\epsilon$$ of the battery.
b. Determine the internal resistance r of the battery.
c. Determine the value of the resistance R that will produce a current I of 3 amperes.
d. Determine the maximum current that the battery can produce.
e. The current and voltage measurements were made with an ammeter and a voltmeter. On the diagram below, show a proper circuit for performing these measurements. Use A to represent the ammeter and V to represent the voltmeter.

**Diagram is #4 from this link: http://mrmaloney.com/mr_maloney/ap/past-AP-tests/BII_91.pdf

2. Relevant equations
$$\epsilon$$ = IR + Ir
$$\Delta$$V = IR

3. The attempt at a solution

a) I know a battery is a constant source of emf.
$$\epsilon$$ = IR + Ir
$$\epsilon$$ = $$\Delta$$V + Ir
From the diagram:
$$\epsilon$$ = (4-2) + Ir

Help! What values do I use?

2. May 15, 2009

nickjer

Re: emf

The $$\Delta V$$ is the voltage drop across the resistor R, and not the total change of voltage when changing the resistace.

You know...
$$\epsilon = V_{XY}+Ir$$

You also know $$V_{XY}$$ and I at multiple values. From those multiple values you can solve for $$\epsilon$$ and r.

3. May 15, 2009

science.girl

Re: emf

So, solve a system of equations?:

$$\epsilon$$ = V_{XY}+Ir[/tex]
\epsilon = 4 + r
\epsilon = 3 + 3r

Therefore, r = 1/2 and \epsilon = 4.5?

4. May 15, 2009

nickjer

Re: emf

Yes. You can even check your answers with the 3rd point on the graph.

5. May 15, 2009

science.girl

Re: emf

Wonderful -- it works!

Now, for (c), the data on the graph shows that:
I (in Amps) of 3 corresponds to Vxy of 3 (Volts).

Would V = IR apply? Therefore, 3 = 3R; R = 1?

OR

I = ($$\epsilon$$)/(R + r); 3 = (4.5)/(R + .5) ... Oh! R = 1.

Is this method correct?

6. May 15, 2009

nickjer

Re: emf

Both are correct.

7. May 15, 2009

science.girl

Re: emf

Hmm... now, maximum current...
It seems as though there's already quite a few pieces of information that we have to work with. However, is there a more direct approach to finding maximum current (i.e., an equation specifically for max current)?

8. May 16, 2009

science.girl

Re: emf

Thank you so much for your help, Nickjer. I appreciate it.

9. May 16, 2009

rl.bhat

Re: emf

d. Determine the maximum current that the battery can produce.
I think the above statement is not correct. With the variable resistance in the external circuit, the power delivered by the battery changes. SO you can ask what is the maximum power delivered by the battery and when it is possible?

10. May 16, 2009

Redbelly98

Staff Emeritus
Re: emf

There is a maximum current. There are 2 ways I can think of to look at this.

1. Use the equation

ε = VXY + I r

Solve for I, and think about what value the variable VXY should have to give the maximum I.

2. Look at the VXY-vs-I graph. Theoretically, how far does the line extend? Hint: the answer is not "to ∞ in both directions".

(The graph method may help understand what is happening, but you'll still need to use the equation to get the answer.)

11. May 16, 2009

nickjer

Re: emf

What Redbelly98 said is correct. Or you can use the equation you listed above:

$$I = \frac{\epsilon}{R+r}$$

Since $$\epsilon$$ and $$r$$ are fixed. What value of R will give the largest I. Since R is on the bottom of a fraction you want to find the limit that makes 'I' largest.

12. May 16, 2009

Redbelly98

Staff Emeritus
Re: emf

I like nickjer's suggestion, his equation (in post #11) is better to use than what I had said.