Question

Use the exact values you enter to make later calculations.

Part A

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

Trial	ΔV (volts)	I (mA)
1	1.00	5.3
2	1.90	9.5
3	3.00	16.3
4	4.00	20.6
5	4.90	25.3

where ΔV 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 Ω.)

slope =

y-int =

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.

ΔV (volts)	I (mA)
3.75	8.9

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

Based on your value of the equivalent resistance, what must the arrangement be? (series or 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.

Answer 1

15 Series1 -Linear (Series1) Vertical (Value) Axis Minor Gridl 3 4 4.9

From the graph plotted above, the slope is 5.11 kOhm and the y-intercept is 0.07 mA.

PART-B:-

Now the value of Req = 2.37 kOhm,So since the eqivalent resistance has decreased so it will be a parallel connenction.

PART-C:-

Let the unknown resistance be R.

$\frac{1}{R}+\frac{1}{5.11}=\frac{1}{2.37}$

Solving we get R = 4.42 kOhm.