Please answer all parts of question number 2 1. What is ohm’s law? Look at your

Question

Please answer all parts of question number 2
1. What is ohm’s law? Look at your graph from Part A. What evidence, if any, does it show that the rheostat is an ohmic device? What does the slope of your straight-line graph represent? All parts of this question must be in bullet form or points will be deducted. 2. How does the value of resistance as determined by your graph compare to the value measured using your multimeter? Considering the uncertainty in the last digit displayed by the power supply, what is the minimum uncertainty in your current measurements? What percentage uncertainty does this result in for your lowest current measurement in Part A? What percentage uncertainty does this result in for your highest current measurement in Part A? In part B, are the discrepancies between the values you measured with the ohmmeter and the values you calculated random (some high, some low), or systematic (consistently high or low)? Discuss possible causes for these discrepancies. All parts of this question must be in bullet form or points will be deducted. 001 +99 Part A: Verifving Ohm's Law Total Resistance of Rheostat (from multimeter)OQ Voltage V IOo 1t 105.4 2. Current, I R from Ohms Law % Difference from multimeter 200 0.91 98 ,924 4.632 0132 103 5 -024 L8L3 Bi Investizating the Rheostat, exeunimenterekal Approx Voltage % Current R from Ohms R from multimeter Difference Law position 4.99 12 84. 03 42.5 5.02 8842 3.43 0.94 1 9471 40 115.02 | 0.23 42.94 15.0210.35. 15.01 0. 21.2-2 213 0.096 2 0.2 10 278 100008 0.83

Explanation / Answer

2.

a. The value of the resistance measured from graph using Ohm's Law. (Measured values are correct according to the table provided in part-A). When measuring the values of resistance using multimeter, resistance must be isolated. (Resistance must be removed from the circuit and measure the resistance).

b. lowest current uncertainity:

According to the value of resistance, given 106.1 ohms, using ohm's law, for V= 0.97, I= V/R = 0.97/106.1 = 0.0091 Amps.

but the current reading is shown as, I = 0.01 Amps

difference, = 0.01-0.0091 = 0.0009 amps

percentage uncertainity = 0.0009/0.01 = 0.9 %.

c) highest current uncertainity:

According to the value of resistance, given 106.1 ohms, using ohm's law, for V= 8.54, I= V/R = 8.54/106.1 = 0.081 Amps.

but the current reading is shown as, I = 0.08 Amps

difference, = 0.08-0.081 = 0.001 amps

percentage uncertainity = 0.001/0.08 = 1.25 %.

d) Yes, there are discrepancies between the values you measured with the ohmmeter and the values you calculated randomly.

* The value of the current measured might be reduced to two fractional digits.

* The value of the voltage measured might not be exactly equal in all cases.

* while measuring the resistance using ohmmeter or multimeter, the resistance must not be in the circuit.

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