EXPERIMENT 7
POWER IN DC CIRCUITS
PART 2
I. OBJECTIVES:
1. To determine the power dissipated in a resistive DC circuit.
2. To show that the power can be found by using any one of three methods.
II. THEORETICAL CONSIDERATIONS:
Electrical power in a DC circuit can be found by using the equation:
P = E x I
where:
--------------------------------- (Equation 1)
P: power, in watts
I: current, in amperes
E: voltage, in volts
Since the voltage E and the current I are related by the
resistance R
(by Ohms Law) two new equations for the power may be derived from Equation
1. By substituting IR for E, equation (1) becomes:
P = IR x I
P = I2 x R
--------------------------------- (Equation 2)
And since:
I = E/R, you can also substitute E/R for I in Equation 1
P = E/R x E
P = E2/R
--------------------------------- (Equation 3)
Thus, you can now calculate power for any DC circuit using the term
either E or I (it is not necessary to know the values of both E and I).
NOTE: The device dissipating the calculated power must be a pure
resistance only and not have any impendence. (Impendence is another form of
resistance that is used for AC circuits).
The law of conservation of energy requires that the power dissipated
by any number of resistance elements must equal the power supplied by the
source. This face will be verified experimentally in this laboratory
experiment. When electric energy is supplied to a resistor it is immediately
converted into heat, with the result that the resistor warms up.
LBYEC11 Experiment Manual
Experiment 7: Power in DC Circuit, Part 2
Page 1 of 9
The greater the electric power supplied, the higher the temperature
will become, until a point is reached where either the resistor or nearby
components will burn out. In order to maintain an acceptable temperature,
resistors that have to dissipate large amounts of power are made physically
large, while those that dissipate little energy can be made smaller.
It is therefore, clear that the physical size of a resistor depends
not on its resistance value but almost exclusively upon the power which it
has to dissipate. This is why 100-watt lamps are physically larger than 60watt lamp. The increased size offers better cooling both by convection and
by radiation.
III. MATERIALS AND EQUIPMENT:
1
---------------- DC Power Supply
---------------- EMS Resistance Module A and B
---------------- Digital Voltmeter (0-300VDC)
---------------- Digital Ammeter (0-10ADC)
10 ---------------- Connecting Wires
IV. PROCEDURE:
1. Connect the circuit shown in Figure 7.1, being careful to observe in
correct meter polarities.
2. T
u
r
n
ON the switch of the DC power supply and adjust the control knob until the
voltmeter across R1 indicates 30 VDC. Measure the current through R1.
3. Calculate and record the power dissipated by R1. Turn OFF the switch of
the DC power supply.
4. Calculate the BTU dissipated per hour by R1.
5. Change the value of R1 to 50 ohms. Repeat Steps 1 and 2, return the
voltage to zero and Turn OFF the switch of the DC power supply
LBYEC11 Experiment Manual
Experiment 7: Power in DC Circuit, Part 2
Page 2 of 9
6. Calculate the power dissipated by R1 (50 ohms) using each of the three
equations given in the theory section.
7. Reconnect the circuit as shown in Figure 7.2. Note that the three load
resistors are connected in series. (The same voltmeter will be used to measure
the voltage across each of the three resistors).
8. Turn ON the switch of the DC power supply and adjust VT = 30 VDC as indicated
by the DC Voltmeter.
a. Measure and record the current I and the voltage E across R1.
b. Remove the voltmeter leads from R1 and place them across R2. Turn ON the
switch of the DC power supply and adjust for V T = 30 VDC. Measure the voltage
across R2.
c. Repeat (b), this time measuring the voltage across R3. Return the voltage to
zero and Turn OFF the switch of the DC power supply.
9. Calculate the power dissipated
Calculate the power supplied.
by
summing
the
three
dissipated
powers.
10. You will now determine the dissipated powers without knowing the voltage
drops across the resistors. Connect the circuit shown in Figure 7.3. (This
circuit is the same one shown in Figure 7.2 but with the voltmeters removed.)
LBYEC11 Experiment Manual
Experiment 7: Power in DC Circuit, Part 2
Page 3 of 9
11. Turn ON the switch of the DC power supply and adjust for 30 VDC, and
measure and record the current. Return the voltage and Turn OFF the DC power
supply. Calculate the power dissipated in each resistor. Find the sum of the
total power dissipated and then calculate the total power supplied.
12. Connect the circuit as shown in Figure 7.4. With an assumed input of 30
VDC, calculate the power dissipated in each resistor and the total power
dissipated.
13. Knowing that the DC power supply must furnish the total power PT and that
supply voltage is 30 VDC, calculate the value of the supply current IT.
14. Insert the ammeter in the circuit as shown in Figure 7.4 and turn ON the
switch of the DC power supply and measure the total circuit current.
LBYEC11 Experiment Manual
Experiment 7: Power in DC Circuit, Part 2
Page 4 of 9
FINAL REPORT
Date Performed: 10/07/2014
Experiment Number: 7
Experiment Title: POWER IN DC CIRCUIT, Part 2
Group Number:
Signature:
Group Leader:
Group Members Present:
Ryback Po
_______________
Rohit Nihalani_
Luigi Ocampo
Hannah Pituk
________________
_______________
_______________
_______________
_______________
_______________
V. DATA AND RESULTS:
Step
2. IR1 =
0.303
Step
3. PR1 = IR1 x ER1 =
A
0.303 A
30.02 V
9.096
Step
4. PR1 x 3.43 =
Step
5. I =
Step
6. PR1 =
Step
Step
0.604
9.096 W
3.43 =
30.02 V
I2
0.365 A
PR1 =
E2
901.20 V
0.604 A
50
x
0.132
b. ER1 =
6.53
c. ER3 =
9.92
9. a. PR1 =
ER1
IR1 =
13.2 V
b. PR2 =
ER2
IR2
c. PR3 =
ER3 x
IR3 =
18.132
50
18.24
=
18.024
13.2
0.132 A
1.74
6.53 V
0.132 A
0.862
9.92 V
0.132 A
1.31
Power
Dissipated
e. Ps
Es
Is
ER1 =
d. Total
=
BTU/hr.
PR1 =
8. a. I
31.20
= _ 3.912 _ W
29.65 V
x _0.132 A
3.91
LBYEC11 Experiment Manual
Experiment 7: Power in DC Circuit, Part 2
Page 5 of 9
Step
Step
11.
I =
0.133
a. PR1 =
I2 x
R1
0.018 A
100 =
1.8
b. PR2 =
I2
R2
0.018 A
50
0.9
c. PR3 =
I2
R3
0.018 A
75
1.35_ W
d. Total
Power
Dissipated
e. Ps
Es
Is
4.05
30.04 V
0.133 A
12.PR1 =
901.2 V
100
9.012
PR2 =
901.2 V
50
18.024
PT =
=
PR1 +
Step
13.IT
PT /
Step
14.IT(measured) =
PR2 =
E
9.012 W
27.036 W
0.894
W
3.99
W
W
18.024 W
/
30.02 V
=
=
27.036
0.9_ W
Instructors Signature: ____________
Grade: ___________
LBYEC11 Experiment Manual
Experiment 7: Power in DC Circuit, Part 2
Page 6 of 9
VI. COMPUTATIONS/ GRAPHS:
VII. ANALYSIS AND CONCLUSIONS:
In concise point form, draw appropriate conclusions from the results
of the experiment.
LBYEC11 Experiment Manual
Experiment 7: Power in DC Circuit, Part 2
Page 7 of 9
VIII. QUESTIONS TO BE ANSWERED:
1. Do all the results of Step 6, do you agree?
Yes.
Step 6) PR1 = 18.132 W
PR1 = 18.24 W
PR1 = 18.024 W
2. Is there an agreement between (d) and (e) of Step 9?
Yes.
d) Total Power Dissipated = 3.912 W
e) Ps = 3.91 W
3. Is there an agreement between (d) and (e) of Step 11?
Yes.
d) Total Power Dissipated = 4.05 W
e) Ps = 3.99 W
4. Is there an agreement between the calculated (Step 13) and
measured values (Step 14) of IT?
Yes.
Step 13)
IT =
0.9 A |
e)
IT(measured)
0.894 A
5. Calculate the power dissipated in each resistance as well as the
total power for each of the circuits shown in Figure 7.5.
6. Round a 12 gauge of copper wire and has a resistance of 1.6 ohms
per thousand feet.
a. Calculate the power loss in a 12 gauge of copper wire
conductor of 200 feet long carrying a current of 10 amps.
LBYEC11 Experiment Manual
Experiment 7: Power in DC Circuit, Part 2
Page 8 of 9
b. What is the voltage between the two ends of the conductor in
(a)?
R 1
V
7. The shunt field winding of a DC motor has a resistance of 240
ohms, Calculate the power loss when the voltage across it is 12
VDC?
8. A 1-ampere fuse has a resistance of 0.2 ohm. It will melt or blow
when the instantaneous current through it is large enough to
create power loss of 5 watts. What is the value of the melting
current?
V I
9. An earth ground at the base of transmission line tower has a
resistance of 2 ohms.
a. If a lightning stroke of 20,000 amperes and strikes the tower,
what would be the power dissipated in the ground?
b. What would be the voltage drop across the ground during (a)?
10. A DC motor draws a current of 50 A at 230 V. If 1,200 W is
dissipated in the form of heat in the motor, how much power is
available for mechanical work?
LBYEC11 Experiment Manual
Experiment 7: Power in DC Circuit, Part 2
Page 9 of 9
