Lab 1: DC Circuits. Student 1, Partner : Student 2,


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1 Lab Date Lab 1: DC Circuits Student 1, Partner : Student 2, I. Introduction The purpose of this lab is to allow the students to become comfortable with the use of lab equipment, through exercises involving simple DC circuits. This lab introduces us to the use of a power supply, multimeter, and voltmeter. We examine the properties of DC circuits containing resistors and diodes. We also become familiar with the use of an oscilloscope and a function generator. In this lab we will be connecting a DC power supply to circuits containing various resistive sources (resistors, light bulbs, diodes). We plot voltage (xaxis) vs current (yaxis) and compare the plots for the various setups. This will allow us to understand the following: how voltage and current are affected by devices that follow Ohm s law, devices that are not ideal resistors, and devices that do not follow Ohm s law. We also try to calculate and build a Thevenin equivalent circuit.
2 II. Results 11. Ohm s Law A. Experimental Setup We connect a power supply, voltmeter, multimeter, and resistor to the breadboard. Circuit Diagram: Breadboard Diagram:
3 B. Results B.1 Values of V and I for 20K and 10K resistor. 18 kω Resistor V (V) I (ma) kω Resistor V (V) I (ma) B.2 Plot the data points as I (y axis) vs V (x axis). 1 V vs. I Current (ma) k 12k Voltage (V) B.3 Compare to calculated from Ohm s Law The curves are clearly linear, as expected by Ohm s Law. The slopes are ohm and ohm, close to the resistance of the corresponding resistor used. B.4 How to fix circuit to get voltmeter to measure what we want B.5 Accuracy of current measurement B.6 Ideal voltmeter should do B.7 Ideal ammeter should do
4 In the circuit diagram above we can see that the voltmeter is measuring any voltage drop across the resistor or ammeter. If the ammeter were ideal, then its internal resistance would be zero and there would be no voltage drop across it. In reality, the ammeter will have nonzero resistance. We can try to fix the problem by moving the terminals of the voltmeter to cross only the resistor. However, this introduces a new problem. The voltmeter is now directly in parallel to the resistor, so that the measured current will now be the current passing through the resistor plus the current passing through the voltmeter. If the voltmeter were ideal, it would have infinite resistance and there would be no current passing through it. In that case, our measurements would not be affected. B.8 Measure R of ammeter B.9 Measure R of voltmeter It is possible to measure the internal resistance of the ammeter and voltmeter (configured for this experiment) by placing each device in series with a resistor comparable to the internal resistance of the device. In the case of the voltmeter, we placed it in series with a 20M resistor, and set the input voltage to 5.00 V. The voltage measured by the voltmeter was only 1.66 V. We can compare the internal resistance knowing the voltage divider equation: V_meas = [ R_int / (R_int + R_ext)] * V_in, or R_int = [Vmeas / (V_in V_meas)] * R_ext. R_int = [1.66 / 3.34] * 20 Mohms = 9.94 Mohms, ~10 Mohms In the case of the ammeter, we placed it in series with a 1kohm resistor and set the input voltage to be 0.20 V. We know I_meas * (R_ext + R_int) = V_in, or, after solving, R_int = (V_in / I_meas) R_ext, R_int = (0.20V / ma) (1.0 kohm) = 150 ohms. B.10 Quantitative view: how large is each error, given 20K resistor Knowledge of the internal resistance of our measuring devices allows us to estimate the error for the two possible positions our voltmeter may be in. If the voltmeter is across only the resistor, then it draws about R_ext / (R_ext + R_int) = 20k/10M = (or
5 0.2%) of the current away from the 20K resistor. If the voltmeter is across both resistor and ammeter, then the ammeter draws about R_int / (R_ext + R_int) = 150/20k = (or 0.8%) of the voltage away from the resistor. B.11 Which of two alternative hookups is preferable B. 12 Estimate error for 20 M resistor It is preferable by a small margin to place the voltmeter across the resistor. This conclusion would be reversed if we were trying to measure a 20M resistor since then placing the voltmeter directly across the resistor will have it draw out 1/3 of the current.
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