Download Experiment 1

ECE 4501
1.1
Power Systems Laboratory Manual
Rev 1.0
DC CIRCUITS – CALCULATIONS
1.1.1 OBJECTIVE
To calculate the voltages and currents in series and parallel DC circuits
1.1.2 DISCUSSION
Series and parallel DC circuits can be analyzed by applying Ohm’s Law, V = I*R, and the following rules:
i.
ii.
iii.
iv.
In a series circuit, the voltage across a group of resistances is equal to the sum of voltages across each
The total current delivered to a parallel circuit is equal to the sum of the currents in each parallel branch
The current is the same in every resistance of a series circuit
The voltage is the same across every resistance branch of a parallel circuit
1.1.3 INSTRUMENTS AND COMPONENTS
(None for this portion)
1.1.4 PROCEDURE
Using the above rules calculate the voltage and current values listed for each of the following circuits. Show
calculations as necessary.
A) 2 SERIES RESISTORS
Vs =
90 Volts
V1 = _______ Volts
V2 = _______ Volts
Is = _______ Amps
I1 = _______ Amps
I2 = _______ Amps
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ECE 4501
Power Systems Laboratory Manual
Rev 1.0
B) 3 SERIES RESISTORS
V2 = 30 Volts
V3 = ______ Volts
I2 = ______ Amps
Vs = ______ Volts
I1 = ______ Amps
Is = ______ Amps
V1 = ______ Volts
C) 3 PARALLEL RESISTORS
I1 =
0.2 Amps
I2 = ______ Amps
V1 = ______ Volts
I3 = ______ Amps
V2 = ______ Volts
Is = ______ Amps
V3 = ______ Volts
Vs = ______ Volts
D) COMPLEX CIRCUIT
I3 =
0.2 Amps
V1 = _____ Volts
V3 = ______ Volts
Vs = _____ Volts
V2 = ______ Volts
I2 = ______ Amps
Is = ______ Amps
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ECE 4501
Power Systems Laboratory Manual
Rev 1.0
1.1.5 CONCLUSIONS
1) If the power supply voltage in the first circuit (1.1.4 – A) were doubled, what would happen to the other
voltages and currents in the circuit?
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If the polarity of the voltage in the first circuit was reversed, what would happen to the other voltages and
polarities in the circuit?
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1.2
DC CIRCUITS – MEASUREMENTS
1.2.1 OBJECTIVE
To verify experimentally the theoretical calculations performed in Section 1.1 DC CIRCUITS –
CALCULATIONS above.
1.2.2 DISCUSSION
In circuits there are junction points (nodes) where wires meet and are joined together. According to Kirchoff’s
Current Law (KCL) the sum of all currents at the node equals zero. In other words, the sum of all currents
entering the node is equal to the sum of all currents exiting the node. The physical reason behind this is that
energy cannot be stored in the node, so all electrons arriving at the junction must quickly leave. The following
procedures attempt to verify KCL.
1.2.3 INSTRUMENTS AND COMPONENTS
Power Supply Module (0-120 V-DC)
Resistance Module
DC Metering Module (200V, 500mA, 2.5A)
Connection Leads
DC Voltmeters and DC Ammeters
EMS 8821
EMS 8311
EMS 8412
EMS 8941
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ECE 4501
Power Systems Laboratory Manual
Rev 1.0
1.2.4 PROCEDURE
CAUTION! – High voltages are present in this Experiment. DO NOT make any connections with the
power supply ON. Get in the habit of turning OFF the power supply after every measurement.
The circuits for the following procedures are identical to those used in Section 1.1 DC CIRCUITS –
CALCULATIONS above. For each circuit, perform each of the following:
1) Enter your CALCULATED values from section 1.1 in the spaces provided for each procedure.
2) Wire each circuit using the equipment listed in 1.2.3 above, being careful to observe the CORRECT
metering polarities. The built-in voltmeter in the EMS 8821 unit will be used to measure supply voltage.
Always make sure the supply switch is OFF and the output control knob is turned fully counterclockwise
(0 Volts) when making connections.
3) Turn on the power supply and slowly turn the voltage control clockwise until the voltmeter on the DC
power supply indicates the required voltage.
4) Take your measurements
5) Return the voltage to zero and turn off the supply.
6) Compare the calculated results with the experimental values. Indicate whether they agree or disagree. In
the case of disagreement, try to determine the cause (“Stuff Happens” is an insufficient explanation).
A) 2 SERIES RESISTORS
Calculated Values
Experimental Values
Vs =
Vs = _______ Volts
90 Volts
V1 = _______ Volts
V1 = _______ Volts
V2 = _______ Volts
V2 = _______ Volts
Is = _______ Amps
Is = _______ Amps
I1 = _______ Amps
I1 = _______ Amps
I2 = _______ Amps
I2 = _______ Amps
REMARKS: _______________________________________________________
_______________________________________________________
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ECE 4501
Power Systems Laboratory Manual
Rev 1.0
B) 3 SERIES RESISTORS
* Adjust Vs until V2 reads 30 Volts
Calculated Values
Experimental Values
V2 = 30 Volts
V2 = ______ Volts
I2 = ______ Amps
I2 = ______ Amps
I1 = ______ Amps
I1 = _NA__ Amps
Is = ______ Amps
Is = ______ Amps
V1 = ______ Volts
V1 = ______ Volts
V3 = ______ Volts
V3 = ______ Volts
Vs = ______ Volts
Vs = ______ Volts
REMARKS: _______________________________________________________
_______________________________________________________
C) 3 PARALLEL RESISTORS

Calculated Values
Experimental Values
I1 =
0.2 Amps
I1 = _______ Amps
V1 = ______ Volts
V1 = __NA___ Volts
V2 = ______ Volts
V2 = __NA___ Volts
V3 = ______ Volts
V3 = _______Volts
Vs = ______ Volts
Vs = _______Volts
I2 = ______ Amps
I2 = ______ Amps
I3 = ______ Amps
I3 = ______ Amps
Is = ______ Amps
Is = ______ Amps
Adjust Vs until I1 = 0.2 A
REMARKS: _______________________________________________________
_______________________________________________________
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ECE 4501
Power Systems Laboratory Manual
Rev 1.0
D) COMPLEX CIRCUIT

Calculated Values
Experimental Values
I3 =
0.2 Amps
I3 = ______ Amps
V3 = ______ Volts
V3 = ______ Volts
V2 = ______ Volts
V2 = _NA__ Volts
I2 = ______ Amps
I2 = ______ Amps
Is = _______ Amps
Is = _______ Amps
V1 = ______ Volts
V1 = ______ Volts
Vs = ______ Volts
Vs = ______ Volts
Adjust Vs until I3 = 0.2 A
REMARKS: _______________________________________________________
_______________________________________________________
1.2.5 CONCLUSIONS
1) Would an ammeter, such as those used in the experiment, burn out if connected to a circuit in such a manner
as to reverse the polarity of the meter? _______________ Explain.
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What about a voltmeter? _________ Why?
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2) Could you measure the voltage of a flashlight cell using a DC Voltmeter with a scale of 0 to 150 Volts?
_______________
Would such a measurement be useful?
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ECE 4501
1.3
Power Systems Laboratory Manual
Rev 1.0
DC POWER – CALCULATIONS
1.3.1 OBJECTIVE
To calculate the power dissipated in a direct current resistor and show that the power dissipated in a load is
equal to the power supplied by the source (discounting any losses).
1.3.2 DISCUSSION
The power source supplies electrical energy to a load where the energy is transformed into useful work. In the
realm of electricity, useful work is denoted by the movement of electrons (electric current) at the load. POWER
is the rate at which work is performed. An electromotive force of one volt producing one ampere of current
through a one ohm resistance produces one watt of electric power. This relationship between voltage, current,
resistance and power is summarized with the following equations:
P = VI ; P = I2R ; P = V2/R (Watts)
When electric energy is supplied to a resistor, that energy is immediately converted to heat energy, resulting in a
physical rise in temperature of the resistor. The greater the amount of power supplied to the resistor, the greater
the amount of heat generated in the resistor and thus the larger the rise in temperature. Since resistors are
manufactured to meet a specific operating temperature at rated power and voltage levels, they are constructed to
be physically large enough to dissipate the required heat energy. To avoid unacceptable temperatures, resistors
that are required to dissipate significant amounts of electric power must be made with a large surface area.
Increasing the physical size of a resistor improves both convection and radiation, the two primary means by
which heat is dissipated.
1.3.3 INSTRUMENTS AND COMPONENTS
(None)
1.3.4 PROCEDURE
The circuits in this procedure are identical to those analyzed in Section 1.1 DC CIRCUITS –
CALCULATIONS.
1) Enter the values calculated for each circuit in Section 1.1 in the spaces provided.
2) Use the power formulas given above to calculate power dissipation in each resistor in the circuit
(P1 = V1 x I1, etc.).
3) Calculate the power delivered to the circuit by the supply (Ps = Vs x Is) and record the result.
4) Compare the power dissipated to the power supplied and remark upon any discrepancies.
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ECE 4501
Power Systems Laboratory Manual
Rev 1.0
A) 2 SERIES RESISTORS
Calculated Values
Power Dissipated
Vs =
P1 = _______ Watts
90 Volts
V1 = _______ Volts
P2 = _______ Watts
V2 = _______ Volts
Ptot = _______ Watts
Is = _______ Amps
Power Supplied
I1 = _______ Amps
Ps = _______ Watts
I2 = _______ Amps
REMARKS: _______________________________________________________
_______________________________________________________
B) SERIES RESISTORS
Calculated Values
Power Dissipated
V2 = 30 Volts
P1 = _______ Watts
I2 = ______ Amps
P2 = _______ Watts
I1 = ______ Amps
P3 = _______ Watts
Is = ______ Amps
Ptot = _______ Watts
V1 = ______ Volts
V3 = ______ Volts
Vs = ______ Volts
Power Supplied
Ps = _______ Watts
REMARKS: _______________________________________________________
_______________________________________________________
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ECE 4501
Power Systems Laboratory Manual
Rev 1.0
C) 3 PARALLEL RESISTORS
Calculated Values
Power Dissipated
I1 =
0.2 Amps
P1 = _______ Watts
V1 = ______ Volts
P2 = _______ Watts
V2 = ______ Volts
P3 = _______ Watts
V3 = ______ Volts
Ptot = _______ Watts
Vs = ______ Volts
Power Supplied
I2 = ______ Amps
Ps = _______ Watts
I3 = ______ Amps
Is = ______ Amps
REMARKS: _______________________________________________________
_______________________________________________________
D) COMPLEX CIRCUIT
Calculated Values
Power Dissipated
I3 =
0.2 Amps
P1 = _______ Watts
V3 = ______ Volts
P2 = _______ Watts
V2 = ______ Volts
P3 = _______ Watts
I2 = ______ Amps
Ptot = _______ Watts
Is = _______ Amps
Power Supplied
V1 = ______ Volts
Ps = _______ Watts
Vs = ______ Volts
REMARKS: _______________________________________________________
_______________________________________________________
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ECE 4501
Power Systems Laboratory Manual
Rev 1.0
1.3.5 CONCLUSIONS
1) Knowing that one watt of electric power is equivalent to 3.43 BTU/Hr (British Thermal Unit per Hour),
calculate the BTU/Hr of heat dissipated by a hair dryer rated at 1500 Watts.
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2) The circuit in C) 3 PARALLEL RESISTORS has 300, 600and 1200 Ohm resistors in parallel. If all three
resistors were of the same physical size, which one would become hotter? Explain.
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3) If both resistors in A) - TWO SERIES RESISTORS were of the same physical size, which would become
hotter? Explain.
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4) If three resistors, a 200, a 300, and a 600 ohm, were all designed to operate at the same temperature at a
given voltage, which would be the largest physically? ______________ The smallest? __________ Explain.
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5) A 100 watt incandescent lamp has a resistance when cold (de-energized) that is only 1/12 of its hot
(energized) resistance value. What current does the lamp draw when energized by a 120 Vdc source (or 120
Vrms AC source)? What is the ‘hot’ resistance of the lamp? What is the ‘cold’ resistance? What current does
the lamp draw at the instant after energization (i.e. while the resistance is still ‘cold’)? What power does the
lamp dissipate at that instant?
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ECE 4501
1.4
Power Systems Laboratory Manual
Rev 1.0
DC POWER – MEASUREMENTS
1.4.1 OBJECTIVE
To measure the power dissipated by resistors in DC networks and verify that the law of conservation of energy
requires that the power dissipated by any number of resistive elements be equal to the power supplied by the
source (when losses are neglected).
1.4.2 DISCUSSION
As stated previously, power in a DC circuit is related to the applied voltage and resulting current by the
following expression:
Power, P = VI
(watts)
In a resistor, electric energy is converted to heat energy. The presence of heat energy creates a rise in the
ambient temperature of the resistor and its surroundings. The rate at which the heat energy can be dissipated
from the resistor to the surrounding environment is directly related to the physical size of the resistor.
Converting electric energy to heat can be useful for many things, including the resistive heating elements in
electric hot water heaters, electric ovens, etc.
1.4.3 INSTRUMENTS AND COMPONENTS
Power Supply Module (0-120 V-DC)
Resistance Module
DC Metering Module (200V, 500mA, 2.5A)
Connection Leads
DC Voltmeters and DC Ammeters
EMS 8821
EMS 8311
EMS 8412
EMS 8941
--
1.4.4 PROCEDURE
CAUTION! – High voltages are present in this Experiment. DO NOT make any connections with the
power supply ON. Get in the habit of turning OFF the power supply after every measurement.
1) Remove the Resistance Module EMS 8311 from the Lab-Volt Station examine the 300, 600 and 1200 Ohm
resistors inside
2) List the resistors in order of their heat dissipating capability (least to greatest): [Consider their physical size]
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3) Which resistor can safely handle the most electric power? ___________________
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Power Systems Laboratory Manual
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4) Connect the circuit shown below using the EMS Resistance, DC Metering and Power Supply Modules.
Make sure the power supply is OFF before wiring. Take care to observe meter polarities.
5) Turn on the power supply and advance the voltage output until the voltmeter across the resistor, R, indicates
120 Volts, DC. Measure the current indicated by the ammeter.
6) Let the circuit operate for three minutes. In the meantime, calculate the power dissipated in the resistor.
A) A SIMPLE DC CIRCUIT
CALCULATIONS AND MEASUREMENTS
Vr = 120 Volts
Ir = _______ Amps
Pr = Vr x Ir = ______ x ______
= _______ Watts
3.43 x Watts = __________ BTU/Hr
7) Return the voltage control to zero and turn OFF the power supply. Remove the resistance module from the
console. Place your hand near the 300 Ohm resistor inside the module, but DO NOT TOUCH! The
resistor should be quite warm since it is designed to operate at 350 C. Replace the module in the rack.
8) Calculate the BTU/Hr dissipated by the resistor: ____________________ BTU/Hr (1 watt = 3.43 BTU/Hr)
9) Change the value of the resistor to 600 Ohms and repeat steps 5)and 7) above.
For 600 Ohms, Ir = _______ Amps
10) Return the voltage control to zero and turn off the supply.
11) Calculate the power dissipated in the 600 Ohm resistor by three methods:
P = VI : _______ Volts x _______ Amps = ________ Watts
P = I2R : _______ Amps2 x _______ Ohms = _______ Watts
P = V2/R : ______ Volts2 / _______ Ohms = _______ Watts
Do your results agree? _________
Explain: ________________________________________________________________________
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ECE 4501
Power Systems Laboratory Manual
Rev 1.0
12) Connect the new circuit, B), shown below. Make sure the power supply is OFF.
13) Turn on the power supply and adjust the voltage control until the source voltage is 110 Volts, DC.
14) Measure and record the current and voltages.
15) Return the voltage control to zero and turn OFF the supply.
B) A SERIES CIRCUIT
Vs = 110
Volts
Is = _______ Amps
P1 = ______ Watts
V1 = ______ Volts
P2 = ______ Watts
V2 = ______ Volts
P3 = ______ Watts
V3 = ______ Volts
Ptot = P1+P2+P3
Ptot = ______ Watts
Ps = Vs x Is = ______ Watts
16) Calculate the power dissipated in each resistor using the equation, P = VI.
17) Add the three powers and compare with the power supplied by the source, Ps = VsIs.
Do the two values agree? ______________
18) Could P1, P2, and P3 be determined without using the three voltmeters across the resistors? In other words,
if the resistor values are known (and are accurate) and the source value is known, would the ammeter
provide sufficient information to determine the power dissipated by each resistor? ______________
What equations would be used to calculate P1, P2, P3?
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ECE 4501
Power Systems Laboratory Manual
Rev 1.0
19) Connect the circuit, C), shown below, but DO NOT turn on the power supply at this time.
20) Using the input voltage of 90 Volts, DC, calculate the power dissipated by each resistor. Add them to
determine the total power dissipated.
21) Knowing that the power supply must deliver all the dissipated power, determine the supply current, Is.
C) A PARALLEL CIRCUIT
Calculations
Vs =
90
Measurements
Volts
P1 = (Vs)2/R1 = ______ W
Vs =
90 Volts
Is = _______ Amps
P2 = (Vs)2/R2 = ______ W
Ptot = P1 + P2 = ______ W
Is = Ptot/Vs = ______ A
22) Turn on the power supply and adjust the voltage control until the supply voltage is 90 Volts, DC. Measure
and record the ammeter reading for Is.
23) Return the voltage control to zero and turn OFF the supply.
Does the measured value of Is agree with the calculated value? _____________
Explain:
__________________________________________________________________________
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1.4.5 CONCLUSIONS
1) Round copper wire, gauge 12 in size, has a resistance of 1.6 Ohms per thousand feet. Calculate the power
lost in 200 feet of 12 gauge copper wire carrying 10 amps of DC current. Also, what is the voltage drop
between the two ends of the wire?
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ECE 4501
Power Systems Laboratory Manual
Rev 1.0
2) The shunt field winding of a DC motor has a resistance of 240 Ohms. Calculate the power loss in the
winding when it is energized at 120 Vdc.
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3) A 1 amp fuse has a resistance of 0.2 Ohms. It will blow (melt) when the instantaneous value of current
passing through it dissipates 5 watts of power inside it. What is the amp value of that instantaneous current?
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4) A ground rod at the base of a transmission line structure has a grounding resistance of 2 Ohms. If a lightning
stroke of 20,000 amperes contacts the structure, what would be the power dissipated by the ground rod (and
the earth it contacts)? Also, what is the voltage drop across the ground rod?
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5) Water Heater Example: One BTU is required to raise the temperature of one pound of water one degree
Fahrenheit. How long would it take to heat 300 pounds of water (in a well insulated tank) from 60 F to
160 F using a 12 Ohm resistive element connected across 240 Volts? (one watt of electric power is
equivalent to 3.43 BTU/Hr)
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