ECE 4501
Power Systems Laboratory Manual
Rev 1.0

1
–
1.1
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.
In a series circuit, the voltage across a group of resistances is equal to the sum of voltages across each
ii. The total current delivered to a parallel circuit is equal to the sum of the currents in each parallel branch
iii. The current is the same
in every resistance of a series circuit
iv. 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 curr
ent 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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Power Systems Laboratory Manual
Rev 1.0

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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
ECE 4501
Power Systems Laboratory Manual
Rev 1.0

3
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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 curren
ts in the circuit?
________________________________________________________________
________________________________________________________________
________________________________________________________________
If the polarity of the voltage in the fir
st circuit was reversed, what would happen to the other voltages and
polarities in the circuit?
________________________________________________________________
________________________________________________________________
_______________________________
_________________________________
1.2
DC CIRCUITS
–
MEASUREMENTS
1.2.1
OBJECTIVE
To verify experimentally the theoretical calcula
tions performed in Section 1.1
DC CIRCUITS
–
CALCULATIONS above.
1.2.2
DISCUSSION
In circuits there are junct
ion 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)
EMS 8821
Resistance Module
EMS 8311
DC Metering Module (200V, 500mA, 2.5A)
EMS 8412
Connection Leads
EMS 8941
DC Voltmeters and DC Ammeters

ECE 4501
Power Systems Laboratory Manual
Rev 1.0

4
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1.2.4
PROCEDURE
CAUTION!
–
High voltages are present in this Experiment. DO NO
T 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 c
ircuit, 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 experiment
al 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
Vs = 90 Volts
V1 = _______ Volts
V
2 = _______ Volts
Is = _______ Amps
I1 = _______ Amps
I2 = _______ Amps
Experimental Values
Vs = _______ Volts
V1 = _______ Volts
V2 = _______ Volts
Is = _______ Amps
I1 = _______ Amps
I2 = _______ Amps
REMARKS: ________
_______________________________________________
_______________________________________________________
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Power Systems Laboratory Manual
Rev 1.0

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B) 3 SERIES RESISTORS
* Adjust Vs until V2 reads 30 Volts
Calculated Values
V2 = 30 Volts
I2 = ______ Amps
I1 = ______ Amps
Is = ______ Amps
V1 = ______ Volts
V3 = ______ Volts
Vs = ______ Volts
Experimental Values
V2 = ______ Volts
I2 = ______ Amps
I1 = _NA__ Amps
Is = ______ Amps
V1 = ______ Volts
V3 = ______ Volts
Vs
= ______ Volts
REMARKS: _______________________________________________________
_______________________________________________________
C) 3 PARALLEL RESISTORS
Adjust Vs until I1 = 0.2 A
Calculated Values
I1 = 0.2 Amps
V1 = ______ Volts
V2 = ______ Volts
V3 = ______ Volts
Vs = ______ Volts
I2 = ______ Amps
I3 = ______ Amps
Is = ______ Amps
Experimental Values
I1 = _______ Amps
V1 = __NA___ Volts
V2 = __NA___ Volts
V3 =
_______Volts
Vs = _______Volts
I2 = ______ Amps
I3 = ______ Amps
Is = ______ Amps
REMARKS: _______________________________________________________
_______________________________________________________
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Power Systems Laboratory Manual
Rev 1.0

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D) COMPLEX CIRCUIT
Adjust Vs until I3 = 0.2 A
Calculated Values
I3 = 0.2 Amps
V3 = ______ Volts
V2 = ______ Volts
I2 = ______ Amps
Is = _______ Amps
V1 = ______ Volts
Vs = ______ Volts
Experimental Values
I3 =
______ Amps
V3 = ______ Volts
V2 =
_
NA
__ Volts
I2 = ______ Amps
Is = _______ Amps
V1 = ______ Volts
Vs = ______ Volts
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.
________
________________________________________________________________
________________________________________________________________________
________________________________________________________________________
What about a voltmeter? _________ Why
?
________________________________________________________________________
________________________________________________________________________
2) Could you measure the voltage of a flashlight cell using a DC Voltmeter with a scale of 0 to 150 Vo
lts?
_______________
Would such a measurement be useful?
________________________________________________________________________
________________________________________________________________________
ECE 4501
Power Systems Laboratory Manual
Rev 1.0

7
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1.3
DC POWER
–
CALCULATIONS
1.3.1
OBJEC
TIVE
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 lo
ad 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 produc
ing 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 = I
2
R ; P = V
2
/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 en
ergy. 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 pr
imary 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 circui
t 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.
ECE 4501
Power Systems Laboratory Manual
Rev 1.0

8
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A) 2 SERIES RESISTORS
Calculated Values
Vs = 90 Volts
V1 = _______ Volts
V2 = _______ Volts
Is = _______ Amps
I1 = _______
Amps
I2 = _______ Amps
Power Dissipated
P1 = _______ Watts
P2 = _______ Watts
Ptot = _______ Watts
Power Supplied
Ps = _______ Watts
REMARKS: _______________________________________________________
________________________
_______________________________
B) SERIES RESISTORS
Calculated Values
V2 = 30 Volts
I2 = ______ Amps
I1 = ______ Amps
Is = ______ Amps
V1 = ______ Volts
V3 = ______ Volts
Vs = ______ Volts
Power Dissi
pated
P1 = _______ Watts
P2 = _______ Watts
P3 = _______ Watts
Ptot = _______ Watts
Power Supplied
Ps = _______ Watts
REMARKS: _______________________________________________________
________________________________________
_______________
ECE 4501
Power Systems Laboratory Manual
Rev 1.0

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C) 3 PARALLEL RESISTORS
Calculated Values
I1 = 0.2 Amps
V1 = ______ Volts
V2 = ______ Volts
V3 = ______ Volts
Vs = ______ Volts
I2 = ______ Amps
I3 = ______ Amps
Is = ______ Amps
Pow
er Dissipated
P1 = _______ Watts
P2 = _______ Watts
P3 = _______ Watts
Ptot = _______ Watts
Power Supplied
Ps = _______ Watts
REMARKS: _______________________________________________________
________________________________
_______________________
D) COMPLEX CIRCUIT
Calculated Values
I3 = 0.2 Amps
V3 = ______ Volts
V2 = ______ Volts
I2 = ______ Amps
Is = _______ Amps
V1 = ______ Volts
Vs = ______ Volts
Po
wer Dissipated
P1 = _______ Watts
P2 = _______ Watts
P3 = _______ Watts
Ptot = _______ Watts
Power Supplied
Ps = _______ Watts
REMARKS: _______________________________________________________
________________________________
_______________________
ECE 4501
Power Systems Laboratory Manual
Rev 1.0

10
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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.
____________________________
___________________________________________________
_______________________________________________________________________________
2) The circuit in C) 3 PARALLEL RESISTORS has 300, 600and 1200 Ohm resistors in parallel. If all three
resistors were o
f the same physical size, which one would become hotter? Explain.
_______________________________________________________________________________
_______________________________________________________________________________
3) If both resistors in A
)

TWO SERIES RESISTORS were of the same physical size, which would become
hotter? Explain.
_______________________________________________________________________________
_______________________________________________________________________________
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.
________________________________________
_______________________________________
_______________________________________________________________________________
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 resis
tance is still ‘cold’)? What power does the
lamp dissipate at that instant?
_______________________________________________________________________________
_______________________________________________________________________________
________________
_______________________________________________________________
ECE 4501
Power Systems Laboratory Manual
Rev 1.0

11
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1.4
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 dissi
pated 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 exp
ression:
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 dissi
pated
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, elec
tric ovens, etc.
1.4.3
INSTRUMENTS AND COMPONENTS
Power Supply Module (0

120 V

DC)
EMS 8821
Resistance Module
EMS 8311
DC Metering Module (200V, 500mA, 2.5A)
EMS 8412
Connection Leads
EMS 8941
DC Voltmeters and DC Ammeters

1.4.4
PRO
CEDURE
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

Vol
t 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]
__________________________________________________________________
3) W
hich resistor can safely handle the most electric power? ___________________
ECE 4501
Power Systems Laboratory Manual
Rev 1.0

12
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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 p
olarities.
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, calc
ulate 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 designe
d 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 = I
2
R : ___
____ Amps
2
x _______ Ohms = _______ Watts
P = V
2
/R : ______ Volts
2
/ _______ Ohms = _______ Watts
Do your results agree? _________
Explain: ________________________________________________________________________
ECE 4501
Power Systems Laboratory Manual
Rev 1.0

13
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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 wit
h 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?
_____________________________________
____________________________________________
ECE 4501
Power Systems Laboratory Manual
Rev 1.0

14
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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
dete
rmine 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 Volts
P1 = (Vs)
2
/R1 = ______ W
P2 = (Vs)
2
/R2 =
______ W
Ptot = P1 + P2 = ______ W
Is = Ptot/Vs = ______ A
Measurements
Vs = 90 Volts
Is = _______ Amps
22) Turn on the power supply and adjust the voltage control until the supply voltage is 90 Volts, DC. Measure
and record the a
mmeter 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:
__________________________________________________________________________
__
________________________________________________________________________
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?
________________________________________________________________________
________________________________________________________________________
_________________
_______________________________________________________
ECE 4501
Power Systems Laboratory Manual
Rev 1.0

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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.
_________________________________________________
_______________________
________________________________________________________________________
________________________________________________________________________
3) A 1 amp fuse has a resistance of 0.2 Ohms. It will blow (melt) when the inst
antaneous value of current
passing through it dissipates 5 watts of power inside it. What is the amp value of that instantaneous current?
________________________________________________________________________
________________________________________
________________________________
________________________________________________________________________
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 contac
ts 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?
________________________________________________________________________
________________________
________________________________________________
________________________________________________________________________
5) Water Heater Example: One BTU is required to raise the temperature of one pound of water one degree
Fahrenheit. How long woul
d 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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