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

[science.girl]

Homework Statement

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 V_xy. The quantities I and V_xy 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

Homework Equations

\epsilon = IR + Ir
\DeltaV = IR

The Attempt at a Solution

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

Help! What values do I use?

 

[nickjer]

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.

 

[science.girl]

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.

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?

 

[nickjer]

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

 

[science.girl]

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

Wonderful -- it works!

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

 

[nickjer]

Both are correct.

 

[science.girl]

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)?

 

[science.girl]

Both are correct.

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

 

[rl.bhat]

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?

 

[Redbelly98]

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

1. Use the equation

ε = V_XY + I r​

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

2. Look at the V_XY-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.)

 

[nickjer]

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.

 

[Redbelly98]

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