ECE2001W
Section 3
Lab 2
1.
This step involved finding the voltage in two resistors in parallel. The first resistor
was 1.1 kiloohms and the second resistor was 1.6 kiloohms. A voltage of 0.995
volts was supplied and gave 0.406 volts in the 1.1 kiloohm resistor and 0.589
volts in the 1.6 kiloohm resistor. The theoretical voltages came out within a ten
thousandth of a volt which means there was not much error in the system.
V=IR
2.
This step involved finding the current in two resistors used in step one except in
parallel. The current through the 1.1 kiloohm resistor was 0.86 mA and the
current through the second resistor was 0.59 mA. The theoretical currents came
out within a couple hundredths of a milliamp which is a very small amount of
error.
V=IR
3.
This step involved finding the voltage and current in a circuit with a power supply
and a resistor, then plotting the data, and finding the resistance by using the slope
of the graph. The slope of the graph came out to be 1.68 kiloohms. The resistor
used was 1.6 kiloohms, the difference could be because of the tolerance of the
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 Spring '08
 Ayers
 Volt, Resistor, power supply

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