Lab 215
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School
New Jersey Institute Of Technology *
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Course
111A
Subject
Electrical Engineering
Date
Dec 6, 2023
Type
Pages
6
Uploaded by kmontemayor16
Physics Laboratory Report
Lab 215: Electric Field by Point Charges
Objectives:
-
To learn to measure current with an ammeter, voltage with a voltmeter, and resistance
with an ohmmeter.
-
To study the properties of Ohm;s Law and circuits of series, parallel and combination of
series and parallel
-
To study the characteristics of non-Ohmic devices (LED) by measuring the voltage and
corresponding current.
Equipment List:
-
Electronic connection board
-
DC power supply
-
Electronic board connectors
-
Banana cables
-
Resistors (1K, 2.2K, 5.6K, 10K) Ohm
-
LED
-
Digital Multimeter (x2, yellow one for accurate voltage measurement and black one for
accurate current measurement)
Experiment Procedure:
Part 1: Ohm’s Law and Measuring Current Through an Electric Circuit
-
In this part of the experiment, we had an electric current passing through a resistor,
-
We use the equation V = IR, with V being voltage, I being current, and R being resistance
-
We use two digital meters for this experiment, one for voltage measurements and one for
current measurements.
-
We connected a 56000 Ohm resistor to the voltage port of the DC power supply,
aswe set the voltage across the resistor for 4,6 ,8 and 10 volts this is the results we
got
-
V = 4, I = 0.7*10^-3
-
V = 6, I = 1.0*10^-3
-
V = 8, I = 1.4*10^-3
-
V = 10, I = 1.7*10^-3
-
As you can see the table is a straight line with the slope being very consistent among the
increments
-
The digital meter can also be used to measure resistance-
-
The color coded value of 5600 Ohms is the same result that we got from the graph of the
slope of the graph in voltage vs current plot.
Part II: Series Circuits
-
Procedure:
-
Connect three resistors together in a series circuit, digital multimeters and cables
-
Measure the exact resistance of each resistor used in your experiment by using the
digital multimeter and record the values
-
We used two 1K resistors, and 1 10k resistor
-
1K = 1.01
-
1K = 1.01
-
10k = 10.02
-
Total Voltage Drop:
-
Vtotal = 12.02
-
V1 = 1.2
-
V2 = .989
-
v3 = 10.02
-
As you can see from the data here, by adding V1,V2,and V3, we get that the sum of the
voltage drops add up to 12V.
-
We use Ohm’s law to determine the current which you would expect to be flowing
through each resistor, I = V/R
-
This calculation gives the same current flowing through each resistor
-
For example:
-
I_1 = V1/R1
-
1.2/1.01 = 1.18
-
Total equivalent Resistance Req = R1 + R2 + R3 = 12.02
-
12/12.02 is approximately equal to 1,
Part III Parallel Circuits:
-
Procedure:
-
Connect the same three resistors used in the series circuit together into a parallel
circuit
-
Set the voltage input across the three resistors connected in parallel in 12 V. Use
the digital meter to measure the total current coming from the power supply.
-
Our total voltage is equal to 11.53 amps.
-
Current is equal to 24.1 mA
-
To solve for the current, we had each individual current through each resistor
- Through V1, = 11.39
-
Through V2, = 11.5
-
Through V3 = 1.17
-
= 24
-
Use Ohm’s Law to determine the current you would expect to be flowing through each
resistor, Ix = V/Rx
R1 = 12/11.39 = 1.053
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Question 6
Feedback
Vin
Incorrect.
+
R1
Vb
+
Given a diode circuit with Vin = 1.5 V, R1 =1.8 kOhms, Vb 5.9 V. Determine the current along the diode. Consider the diode as ideal.
ID=_____ MA
Round your answer to 2 decimal places.
Your Answer: -2.44
D1
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