Question

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

Trial V (volts) I (mA)
1 1.00 7.2
2 2.10 14.0
3 3.10 20.7
4 4.00 27.2
5 4.90 32.2
Determine the slope and y-intercept of the graph, and report these values below. (
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 measurements to find the equivalent resistance of the arrangement.
V (volts) I (mA)
2.0 5.5
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

Kindly check explanation

Step-by-step explanation:

Given the data:

Trial V (volts) I (mA)

1 1.00 7.2

2 2.10 14.0

3 3.10 20.7

4 4.00 27.2

5 4.90 32.2

Slope = Rise / Run

Rise = y2 - y1 = 32.2 - 7.2 = 25

Run = x2 - x1 = 4.9 - 1.0 = 3.9

Slope = 25 / 3.9 = 6.410

y = mx + c

The intercept, c

Take the point ( 1; 7.2)

Put x = 0

7.2 = 6.410(1) + C

7.2 - 6.410 = C

C = 0.79