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Lab 5: BASIC CIRCUITS
( Resistors, Voltage, and Current with MATLAB adapted from P-178 DC Circuit Labs )
Introduction:
Electric circuits can be defined as closed or continuous paths in which electric currents are confined and around which electric currents can be caused to flow. Electrical circuits are an essential part of daily living, and may be found in heavy and light industry, commercial installations and operations, and residential applications. Modern life and its many conveniences seem inconceivable without the use of electric circuits.
The total resistance of a circuit is the sum of the individual resistances of the power source, the wiring, and the load. The load resistance is generally much higher than either the resistance of the power source or the wiring. The resistances of the wiring are usually neglected in classroom laboratory experiments. Very rarely is circuit wiring significant in experimental work. In these cases we consider the loads resistances to be the only resistance. Wiring resistance may be considerable in the case of transmission cables, as well as telephone lines, which are many miles long, and we have a lab which investigates and calculates the resistance in such cables and the lost power and energy due to these lengths.
If an arbitrary load of relatively low resistance were connected to an existing power supply or voltage source, an excessive current might flow to the load, causing burn up or other malfunctions with the load and wiring. The current can be reduced
by reducing the source voltage, but this is not always feasible and is frequently impossible. The resistances of the voltage source or the load could be increased, but these are usually built right into the source or load. Resistances of connecting wires are so low that miles would be needed to increase the circuit resistance by more than a few dozen ohms. A selection of materials for connecting wires might be useful, but a better method would be to creation of a device that is specifically a resistor that can be included with the circuit to give the net or total resistance needed to provide the desired current for the voltage source involved.
In any DC circuit, the total current is equal to the power source voltage divided by the total or equivalent resistance. For a Series Circuit, this is the only current. This means that if the current in some portion of the circuit is known, the total current and the current through every part of the circuit is known. The sum of the voltage drops across the resistors in series is equal to the power supply voltage.
In Parallel Circuits, the total current from the power source divides into different paths as in approaches the parallel branches. The voltage drop across parallel branches is the same for all the branches. If the voltage drop for one branch is known, the voltage drop for all the parallel branches is known. The sum of the currents in the various branches is equal to the current from the power supply.
Circuits with combinations of Series and Parallel portions or sections are more complex. The current through different sections is not the same and the voltage drop across various branches is not the same either.
Ohm’s law is the relationship between voltage, current, and resistance, and is used to calculate current, voltages, or resistances in relatively simple circuits that may be reduced to a simple circuit consisting on one voltage source and on resistance. For more complex circuits, including series and parallel portions and branches that might not be reducible or readily reducible to a single resistor, the application of Kirchhoff’s Rules is used to solve for various branch currents, branch voltages, total current and power.
Kirchhoff’s Rules are two, the Junction Rule and the Loop Rule. The Junction Rule says that the sum of currents entering a node must equal the sum of currents leaving the node. The Loop Rule says that the sum of the voltage sources around a loop must equal the sum of the voltage drops or i*R drops around the same loop. Algebraically these would be:
Σ i = 0.0 (Node Rule)
and ΣV = Σ i*R or Σ V
– Σ i*R = 0.0 (Loop Rule)
Various sections are from Total Circuit Resistance: Mileaf, 7th ed pages 2-20, 2-22, 2-106
Equipment for each Group:
PASCO Electronics Kit.
Large Power Supply.
Multimeter.
Additional connecting wiring as needed.
Procedure:
From the PASCO Electronics Kit select 5 resistors with values approximately equal to those tabulated here:
|
Number of Resistor |
1 |
2 |
3 |
4 |
5 |
|
Resistance (ohms) |
98. |
330. |
550. |
995. |
33. |
Task 1
.
a. (5) Measure and Tabulate the values of the Resistors in the Data Sheet provided.
b. Place the Resistors in the arrangement shown in the Lecture Notes of Lect_Circuits.
Note there are no resistors in Series or in Parallel. The circuit is irreducible.
c. Connect the Power Supply and set it to 15.0 volts.
d. ( 5) Measure and record the Voltages across each resistor in the Data Sheet provided.
e. (10) Calculate and record the Currents through each Resistor using Ohm’s Law.
Task 2
.
During today’s Lab Session complete these steps f – i :
f. ( 6) Write the Kirchhoff’s Node Rule for nodes b and c;
g. (12) Write the Kirchhoff’s Loop Rule for loops I, II, and III.
h. ( 5) Substitute the values of the Power Supply and the five Resistors into the equations.
i. ( 5) Arrange the five equations in Matrix Format as preparation for entry into MATLAB.
Task 3
.
When you have access to MATLAB complete the following steps j – l :
j. (15) Solve the five simultaneous equations for the five currents using MATLAB and tabulate the values of current in the
Data Sheet below the measured values. Print out the MATLAB script file or workspace and results.
k. ( 5) Calculate and tabulate the Percent Differences between the measured values of current and the ‘Theoretical’ values
of current from MATLAB; use the MATLAB values as the denominator.
l. ( 5) Calculate and record the Equivalent Resistance of the circuit using the measurements and again using the
MATLAB results.
m. (1) Think about how you would calculate and fill in the various ‘Total’ entries in the Data Sheet, and fill them in.
Deliverables and Typed Memo Write-Up from Each Student:
1. (5) Brief Introduction discussing the importance of Circuits and Ohm’s law. Do not simply copy the material I sent.
2. Neatly completed Data sheet and a paper copy of the MATLAB script file used to calculate the currents.
Equations for Kirchhoff’s Node Rule for nodes b and c;
Equations for Kirchhoff’s Loop Rule for loops I, II, and III.
The equations with values of Resistors and Power Supply substituted into the equations.
The five equations written in Matrix Format as preparation for entry into MATLAB.
3. Discussion Questions, Conclusions,
a. (2) Which set of values of current do you believe to be the more accurate?
b. (2) Which measurements contributed to the uncertainty of obtaining the currents from Task 1?
c. (2) Which measurements contributed to the uncertainty of obtaining the currents from Task 3?
d. R1 and R2 appear to be in parallel; are they? How do you know:
i. ) (3) Experimentally
ii ) (3) Theoretically
e. R3 and R5 appear to be in series; are they? How do you know:
i. ) (3) Experimentally
ii ) (3) Theoretically
4. Conclusions
a. (3) What did you learn from this lab?
b. Suggested recommendations for improvements, if any.
I will post the data sheet when the deal is done
Lecture Circuits: September 17, 2012 Page 1
Circuits Lab 01, Voltage and Resistors
Thomas J. Crawford, PhD, PE
July 12, 2013
ENGR1181_Lect_Circuits.ppt
Circuits Lab 1:
Ohm’s Law : V = I * R Power: P = V * I = I2 * R = V2 / R
Example: One resistor
Given: V = 12.0 volts, R = 48.0 ohms (Ω )
Find: current I, power P.
Solution: I = V / R = 12.0 / 48.0 = 0.25 amps = 250 milliamps
P = V * I = 12.0 * 0.25 = 3.0 Watts
Find: R if I = 0.0625 (1 / 16 ) amps.
Solution: R = V / I = 12.0 / 0.0625 = 192 ohms ( Ω )
Example: Series Resistors
Given: R1 = 50.0 ohms, R2 =200.0 ohms (Ω ) in series; V = 12.0 volts
Find: current I, ΔV1, ΔV2, P1, P2.
Solution: I = V / Req = 12.0 / (50. + 200.) = 0.048 amps = 48 milliamps
ΔV1 = I * R1 = 0.048 * 50 = 2.4 volts,
ΔV2 = I * R2 = 0.048 * 200 = 9.6 volts.
Or by the Voltage Divider Rule
ΔV1 = ΔVT * R1 / Req = 12.0 * 50.0 / 250.0 = 2.4 volts,
ΔV2 = ΔVT * R2 / Req = 12.0 * 200.0 / 250.0 = 9.6 volts.
P1 = I2 * R1 = (0.048)2 * 50.0 = 0.1152 watts
P2 = I2 * R2 = (0.048)2 * 200.0 = 0.4608 watts
Example: Parallel Resistors is on next page
Thomas J. Crawford, PhD, PE
July 12, 2013
ENGR1181_Lect_Circuits.ppt
Circuits Lab 1:
Ohm’s Law : V = I * R Power: P = V * I = I2 * R = V2 / R
Example: Parallel Resistors
Given: R1 = 21.0 ohms, R2 =42.0 ohms (Ω ) in parallel; V = 10.5 volts
Find: Req, currents I, I1, I2
Solution: 1 / Req = 1 / R1 + 1 / R2 = 1 / 21 + 1 / 42 Req = 14.0 ohms
I = IT = V / Req = 10.5 / 14.0 = 0.75 amps
I1 = ΔV1 / R1 = 10.5 / 21 = 0.50 amps
I2 = ΔV2 / R2 = 10.5 / 42 = 0.25 amps
Or by the Current Divider Rule
I1 = IT * Req / R1= 0.75 * 14.0 / 21 = 0.50 amps
I2 = IT * Req / R2= 0.75 * 14.0 / 42 = 0.25 amps Also next Page
Lecture Circuits: September 17, 2012 Page 2
Circuits Lab 01, Voltage and Resistors
Thomas J. Crawford, PhD, PE
July 12, 2013
ENGR1181_Lect_Circuits.ppt
Circuits Lab 1:
Example: Parallel Resistors
Given: R1 = 21.0 ohms, R2 =42.0 ohms (Ω ) in parallel; V = 10.5 volts
Find: Req, currents I, I1, I2
Alternative Solution: Re-Writing the Req
1 / Req = 1 / R1 + 1 / R2 Req = R1 * R2 / ( R1 + R2 )
Req = 21.0 * 42.0 / ( 21.0 + 42.0 ) = 21.0 * 42.0 / 63.0 Req = 14.0 ohms
Everything else remains the same:
I = IT = V / Req = 10.5 / 14.0 = 0.75 amps
I1 = ΔV1 / R1 = 10.5 / 21 = 0.50 amps and
I2 = ΔV2 / R2 = 10.5 / 42 = 0.25 amps or
Or
I1 = IT * Req / R1= 0.75 * 14.0 / 21 = 0.50 amps
I2 = IT * Req / R2= 0.75 * 14.0 / 42 = 0.25 amps
Note: This Alternative Solution gets complicated with 3 or 4 resistors:
3 resistors: 1 / Req = 1 / R1 + 1 / R2 + 1 / R3
Req = R1 * R2 * R3 / ( R1* R2 + R1*R3 + R2*R3 )
4 resistors: 1 / Req = 1 / R1 + 1 / R2 + 1 / R3 + 1 / R4
Req = R1 * R2 * R3 * R4 / ( you fill in the Rest !! )
Lecture Circuits: September 17, 2012 Page 3
Circuits Lab 01, Voltage and Resistors
Thomas J. Crawford, PhD, PE
August 31, 2013
ENGR 1181_Lect_Circuits.ppt
Illustrations:
How are these resistors connected ?
Illustrations:
How are these resistors connected ?
Illustrations:
How are these resistors connected ?
Illustrations:
Relative to the discussion of Circuits, What do these symbols represent?
Lecture Circuits: September 17, 2012 Page 4
Circuits Lab 01, Voltage and Resistors
R1 R2 R3
R2
R1 R3
Thomas J. Crawford, PhD, PE
August 31, 2013
ENGR 1181_Lect_Circuits.ppt
Illustrations:
How are these resistors connected ?
Illustrations:
How are these resistors connected ?
A
B
R2
R3
R1
A
B
Lecture Circuits: September 17, 2012 Page 5
Circuits Lab 01, Voltage and Resistors
R3
R2
R1
Thomas J. Crawford, PhD, PE
August 31, 2013
ENGR1181_Lect_Circuits.ppt
Circuits Lab 1:
Example:
Given: The Circuit shown below; R3 = 80.0, R1 = ? R2 = ? ohms, V = 24.00 Volts
Find: Current through R3 when the Switch is Open, and when the Switch is Closed.
Solution: Apply Ohm’s Law, note the Parallel configuration.
Switch Open: I3 = 0.00 amps
Switch Closed: I3 = ΔV3 / R3 = 24.00 / 80.00 I3 = 0.300 amps
Solution: We get the same result if we know the values of R1 & R2.
Let R1 = R2 = 10.00 ohms Req = 16.0 IT = 24 / 16 = 1.50 I3 = 0.30 amp
Let R1 = R2 = 40.00 ohms Req = 40.0 IT = 24 / 40 = 00.60 I3 = 0.30 a
Let R1 = R2 = 360.00 ohms Req = 72.0 IT = 24 / 72 = 1/3 I3 = 0.30 amp
V =
24.0
R3 = 80
R2 = ?
Lecture Circuits: September 17, 2012 Page 6
Circuits Lab 01, Voltage and Resistors
R1 = ?
Switch
Thomas J. Crawford, PhD, PE
August 31, 2013
ENGR1181_Lect_Circuits.ppt
Circuits Lab 1:
Example:
Given: The Circuit shown below; R1 = 80.0, R2 = 80.0 ohms, V = 24.00 Volts
Find: Current through R1, R2 when the Switch is Open.
Solution: Apply Ohm’s Law, note the position of the LED.
I1 = I2 = 0.00 amps
Find: Current through R1, R2 when the Switch is Closed.
Solution: Apply Ohm’s Law, etc
No current flows through R1, I1 0.00 amps
I2 = V / R2 = 24.0 / 80.0 I2 = 0.30 amps
V =
24.0
R2 = 80
R2
Lecture Circuits: September 17, 2012 Page 7
Circuits Lab 01, Voltage and Resistors
R1 = 80
Switch
Thomas J. Crawford, PhD, PE
July 26, 2013
ENGR 1182_Lect_RC_Circuits.ppt
Examples:
Given: R1 = 100 ohms, ΔVR1 = 2.80 volts ΔVLED = 2.45 volts, Switch is Open.
Find: Voltage of the Power Supply or Battery.
Solution: Ohm’s Law. V = ΔVR1 + ΔVLED = 2.80 + 2.45 = 5.25 volts
R1
RLED
V
V
R1
RLED
V
Find: Current through R1.
Solution: Ohm’s Law. I = ΔVR1 / R1 = 2.80 / 100. = 0.0280 amps
Find: Resistance of the LED; RLED
Solution: I is common to R1 and the LED; I = ΔV / R
Hence ΔVR1 / R1 = ΔVLED / RLED RLED = R1 * ΔVLED / ΔVR1 = 100 * 2.45 / 2.80
RLED = 87.50 ohms
EXTRA DRAWING
S
Lecture Circuits: September 17, 2012 Page 8
Circuits Lab 01, Voltage and Resistors
R1
RLED
V
Lecture Circuits: September 17, 2012 Page 9
Circuits Lab 02, Battery Resistance and Transmission Cables
Thomas J. Crawford, PhD, PE
February 16, 2013
ENGR1181_Lect_Circuits.ppt
Circuits Lab 2:
Internal Resistance of Dry Cell Battery
Example:
Given: Open Circuit Voltage E = 1.624, Loaded Circuit Voltage VT = 1.584
External Resistor R = 79.2 ohms
Find: Internal Resistance r, current I, and power P.
Solution: r = R * ( E / VT – 1 ) = 79.2 * ( 1.624 / 1.584 – 1 ) = 2.00 ohms
i = VT / R = 1.584 / 79.2 = 0.020 amps OR
i = ( E – VT ) / r = ( 1.624 – 1.584 ) / 2.00 = 0.020 amps
Various Powers
Resistor: P = i2 * R = (0.02)2 * 79.2 = 0.03168 Watts
P = VT2 / R = (1.584)2 / 79.2 = 0.03168 Watts
Internal: P = i2 * r = (0.02)2 * 2.00 = 0.0008 Watts
P = ( E – VT )2 / r = (1.624 – 1.584)2 / 2.00 = 0.0008 Watts
Total: P = i2 * (R + r) = (0.02)2 * ( 79.2 + 2.0) = 0.03248 Watts
P = E2 / (R + r) = (1.624)2 / ( 79.2 + 2.0 ) = 0.03248 Watts
Lecture Circuits: September 17, 2012 Page 10
Circuits Lab 02, Battery Resistance and Transmission Cables
Thomas J. Crawford, PhD, PE
June 17, 2013
ENGR1181_Lect_Circuits.ppt
Circuits Lab 2:
Transmission Cable:
Example:
Given: Single copper wire diameter d = 3.20 mm, L = 2400 m
Find: Resistance of the single wire.
Solution: R = ρ * L / A = 4 * ρ * L / ( π d2 ) =
4 * 1.72E-08 * 2400 / π ( 0.0032 )2 = 5.133 ohms
Example: Seven – strand Aluminum Transmission Cable
Given: Wire diameter d = 2.80 mm, L = 1800 m
Find: Resistance of the Transmission Cable.
Solution: R = ρ * L / A = 4 * ρ * L / ( π d2 ) Get Area first
Area A = N * π * d2 / 4 = 7 * π ( 2.8 )2 / 4 = 43.10 mm2
R = 2.75E-08 * 1800 * 1.00E+06 / 43.10 = 1.148 ohms.
Find: Max current the Cable may carry (Ampacity) if J = 2.20 amps / mm2
Solution: i = J * A = 2.2 * 43.10 = 94.8 amps
Given: The Transmission Cable Data Sheets
Find or Show: the Percent Line Loss of I2 * R = Percent Voltage Drop of ΔV Solution: Write out what each calculation is:
Percent Line Loss = I2 * R / Power = I2 * R / ( V * I ) = I *R / V
Percent Voltage Drop = ΔV / V = I * R / V, hence each percent is I * R / V
Similarly, Percents for Copper and Aluminum are essentially equal because the Max Current Densities J are proportional to Resistivity ( ρ ).
Lecture Circuits: April 06, 2013 Page 11
Circuits Lab 03, DC Circuits with MATLAB.
Thomas J. Crawford, PhD, PE
July 12, 2013
ENGR1181_Lect_Circuits.ppt
Circuits Lab 3:
Kirchhoff’s Rules and MATLAB:
Example:
Given: The Simple Circuit shown below; R1 = 80.0, R2 = 160.0 ohms. V = 12.00 V
Find: Req, Current I, and illustrate Kirchhoff’s Laws or Rules.
Solution: Apply Ohm’s Law and Kirchhoff’s Rules
Node Rule: Summation of currents equals zero, or Currents in = currents out;
Loop Rule: Summation of P.S. Voltages = summation of Voltage Drops.
Req = R1 + R2 = 80.0 + 160.0 Req = 240.0 ohms
I = V / Req = 12.0 / 240.0 I = 0.050 amps = 50. mAmps
Kirchhoff’s Loop Rule:
V – I1*R1 – I2*R2 = 0.0 12.0 – 0.05 * 80.0 – 0.05*160.0 = 0.0 ? Yes
12.0 – 4.0 – 8.0 = 0.0 Yes, it Does.
V
R1 = 80
R2 = 160
ΔV1 = I*R1 = 0.05 * 80 = 4.0 volts
ΔV2 = I*R2 = 0.05 * 160 = 8.0 volts
Lecture Circuits: April 06, 2013 Page 12
Circuits Lab 03, DC Circuits with MATLAB
Thomas J. Crawford, PhD, PE
March 30, 2013
ENGR1181_Lect_Circuits.ppt
Circuits Lab 3:
Five Resistors and Five Currents from P178.
R1
R2
R5
R3
R4
V
a
b
c
d
i1
i2
i3
i5
i4
I
II
III
Lecture Circuits: April 06, 2013 Page 13
Circuits Lab 03, DC Circuits with MATLAB.
Thomas J. Crawford, PhD, PE
May 13, 2013
ENGR1181_Lect_Circuits.ppt
Circuits Lab 3:
Kirchhoff’s Rules and MATLAB:
Example:
Given: The Circuit from P118, P-178 shown on the previous page.
Find: The equations for the five currents in the circuit.
Solution: Apply Kirchhoff’s Rules
Node Rule: Summation of currents equals zero, or Currents in = currents out;
Loop Rule: Summation of P.S. Voltages = summation of Voltage Drops.
SEND THIS PAGE 13 TO STUDENTS.
NEXT PAGE 14 HAS TOO MANY DETAILS AND RESULTS. MAY 13, 2013.
