Question

PLEASE PLEASE PLEASE URGENT! You connect an unknown resistor to a battery that can provide the electric potential up to 10 V to form a closed loop circuit m8.5ade of the battery and the resistor. Using an ammeter to measure the current coming out of the battery and using a voltmeter to measure the electric potential difference across the resistor. In this experiment you vary the electric potential V provided by the battery from 0 V to 10 V in a step of 2 V. At each step you measure the current I out of the battery. When you plot V (V) vs. I (A) on a graph, the data points spread evenly around a straight line whose slope is 8.5 V/A. Find the measured esistance of the resistor in Ohm Keep at least two significant figures. No units are needed for the answer, and you input just a numerical value for the specified units.part 2) if the resistor is rated at 10 Ohm by the manufacturer with a possible 5% error on the resistance,calculate the %difference using your measurement and the rating in %. Keep at least two significant figures. No units are needed for the answer, and just input a numerical value for the specified units.

Answer 1

The measured resistance of the resistor is 8.5 ohms using Ohm's law. The percentage difference between the measured value and the manufacturer's rating is 15%.

Step-by-step explanation:

To find the measured resistance of the resistor, we can use Ohm's law, which states that V = IR. Given that the slope of the V versus I graph is 8.5 V/A, this value is effectively the resistance of the resistor because slope = ΔV/ΔI = R. Therefore, the measured resistance is 8.5 ohms.

For the second part of the question, we calculate the percentage difference between the measured resistance and the manufacturer's rating. The manufacturer rates the resistor at 10 ohms with a ±5% error. The percentage difference is calculated as |measured value - true value| / true value × 100%, which equals |8.5 - 10| / 10 × 100% = 15%.