Part A A group of students performed the same \"Ohm\'s Law\" experiment that you

Question

Part A
A group of students performed the same "Ohm's Law" experiment that you did in class. They obtained the following results:

where V is the voltage difference across the resistor and I is the current traveling through the resistor at the same time.

(a) Analyze the data. (You will not submit this spreadsheet. However, the results will be needed later in this problem.)

(i) Enter the above data into an Excel spreadsheet.
(ii) Make a plot of the voltage difference vs. current.
(iii) Use the trendline option in Excel to fit the data of voltage difference versus current to get the slope and intercept.

(b) Determine the slope and y-intercept of your graph, and report these values below. (Use ohm for .)

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Part B
Your mischievous lab partner takes the resistor that you just experimented with and assembles it in a network with one other resistor and places them inside a black box. He challenges you to tell him the configuration of the resistors inside the box. Being an industrious physics student you connect the leads of the black box to your power source, voltmeter (in parallel), and ammeter (in series) and take the following simultaneous measurements.

Use the above measurements to find the equivalent resistance of the arrangement. (Use ohm for .)
Req =  

Based on your value of the equivalent resistance, what must the arrangement be?

in series or in parallel?

Part C
Now that you've answered his challenge, your lab partner asks you to give the resistance of the resistor that he added to the one you experimented with. Using the information you obtained in parts A and B, predict this value of the resistance of the second resistor.

Trial V (volts) I (mA) 1 1.00 20.5 2 2.10 42.0 3 2.90 57.5 4 3.90 78.0 5 4.90 98.0

Explanation / Answer

Hi,

Part A. In this part we use Excel to graph the data, using the voltage as the dependent variable and the current as independent variable. Once we do that, we get a straight line with a correlation of 0.9999 which is both very high and expected (because the data must follow the Ohm's Law)

Its slope is equal to: m = 0.0503

Its y-intercept is equal to: b = -0,015

I = 0.0503*V - 0.015   (mA)

However, something is wrong with this expression. According to Ohm's Law, the relation between the voltage and the current is:

V = R I ; where V is the voltage, I is the current and R is the resistance.

As the Omh's Law predicts a straight line that crosses the origin, the y-intercep should be cero. Once that is specified in Excel, the new line has a correlation of 0.9999 (which is the same as before) and we use this one instead of the other, so the new equation should be:

V = 0.05 I ; here we can see that the new value of the slope is the resistence which would have a value of:

R = 0.05 k

Part B. For this part you already know that you must find a new line. This new straight line must contain that new point you have (32.1 mA, 4.18 V), but also must cross the oringin (0 mA, 0 V). With this facts you can create the new straight line:

Req = ( 4.18 V / 32.1 mA) = 0.130 k

As an extra, the voltimeter must be connected in parallel in order to have the same voltage than the arrangement of resistors; something similar happends with the ammeter, which must be connected in series in order to have the same current than the arrangement of resistors.

As this new resistance is bigger than the orignial one (0.05 k), the two resistors must be in series.

Part C. Once we know the arrengement we can easyly find the value of the other resistance:

Req = R1 + R2 ::::::: R2 = Req - R1 = (0.130 - 0.05) k = 0.08 k

I hope it helps.

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