Module 2: Power, Heat, Light - Electrical Power and Resistance, Lab Reports of Electrical Engineering

Power is the rate of doing work per unit of time. It is a measure of work done within a specific length of time. In electricity, the unit of power measurement is the watt (W), named for James Watt who perfected the steam engine. One watt of power is the work done in one second by one volt of electrical pressure in moving one coulomb of charge. Since one coulomb per second is one ampere, power in watts is the product of volts times amperes

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46
MODULE 2: POWER, HEAT, LIGHT
At the end of this experiment, the student should be able to:
TLO 1: Demonstrate that electrical power is a function of voltage and current by
calculating and measuring the power dissipated in a resistance as the
voltage is increased.
TLO 2: Determine that the resistance of a lamp filament varies with temperature.
INTRODUCTION:
Power is the rate of doing work per unit of time. It is a measure of work done within
a specific length of time. In electricity, the unit of power measurement is the watt (W),
named for James Watt who perfected the steam engine. One watt of power is the work
done in one second by one volt of electrical pressure in moving one coulomb of charge.
Since one coulomb per second is one ampere, power in watts is the product of volts
times amperes.
๐‘ƒ๐‘œ๐‘ค๐‘’๐‘Ÿ (๐‘Š๐‘Ž๐‘ก๐‘ก๐‘ )= ๐‘‰๐‘œ๐‘™๐‘ก๐‘  ๐‘ฅ ๐ด๐‘š๐‘๐‘’๐‘Ÿ๐‘’๐‘ 
๐‘œ๐‘Ÿ
๐‘ƒ = ๐ธ ๐‘ฅ ๐ผ
When current flows through a resistance, heat is produced. This release of energy
(work) in the form of heat indicates that power is being generated by the power source
and dissipated (given off) by the resistance of the circuit. Because power is dissipated by
the circuit resistance, power is often calculated in terms of resistance. Two equations for
calculating power when resistance is known are derived from the basic equation
(๐‘ƒ = ๐ธ ร— ๐ผ) as follows:
(1) Substitute the Ohmโ€™s Law equivalent for current (๐ธ ๐‘…
โ„) for ๐ผ in the basic
equation. This results in
๐‘ƒ = ๐ธ ร— (๐ธ ๐‘…
โ„) ๐‘กโ„Ž๐‘ข๐‘  ๐‘ƒ = ๐ธ2
๐‘…
โ„
(2) Substitute the Ohmโ€™s Law equivalent for voltage (๐ผ ร— ๐‘…) for ๐ธ in the basic
equation. This results in
๐‘ƒ = (๐ผ ร— ๐‘…) ร— ๐ผ ๐‘กโ„Ž๐‘ข๐‘  ๐‘ƒ = ๐ผ2 ร— ๐‘…
The second equation reflects the power loss resulting from current flow through the
resistance of a circuit. Such power losses are frequently referred to as ๐ผ2 ๐‘… losses.
ENGAGE
1. How does voltage and current affect the power output of a light bulb or electric
appliance?
pf3
pf4
pf5

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MODULE 2 : POWER, HEAT, LIGHT

At the end of this experiment, the student should be able to:

TLO 1 : Demonstrate that electrical power is a function of voltage and current by

calculating and measuring the power dissipated in a resistance as the

voltage is increased.

TLO 2 : Determine that the resistance of a lamp filament varies with temperature.

INTRODUCTION:

Power is the rate of doing work per unit of time. It is a measure of work done within

a specific length of time. In electricity, the unit of power measurement is the watt (W),

named for James Watt who perfected the steam engine. One watt of power is the work

done in one second by one volt of electrical pressure in moving one coulomb of charge.

Since one coulomb per second is one ampere, power in watts is the product of volts

times amperes.

When current flows through a resistance, heat is produced. This release of energy

(work) in the form of heat indicates that power is being generated by the power source

and dissipated (given off) by the resistance of the circuit. Because power is dissipated by

the circuit resistance, power is often calculated in terms of resistance. Two equations for

calculating power when resistance is known are derived from the basic equation

(๐‘ƒ = ๐ธ ร— ๐ผ) as follows:

(1) Substitute the Ohmโ€™s Law equivalent for current (๐ธ โ„๐‘… ) for ๐ผ in the basic

equation. This results in

๐‘ƒ = ๐ธ ร—

2

(2) Substitute the Ohmโ€™s Law equivalent for voltage (๐ผ ร— ๐‘…) for ๐ธ in the basic

equation. This results in

๐‘ƒ = (๐ผ ร— ๐‘…) ร— ๐ผ ๐‘กโ„Ž๐‘ข๐‘  ๐‘ƒ = ๐ผ

2

ร— ๐‘…

The second equation reflects the power loss resulting from current flow through the

resistance of a circuit. Such power losses are frequently referred to as ๐ผ

2

๐‘… losses.

ENGAGE

  1. How does voltage and current affect the power output of a light bulb or electric

appliance?

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  1. What is the importance of determining power ratings of the electrical appliances

and devices we use every day?

EXPLORE

  1. Refer to this module and read in advance to have a better understanding of the

activities to be performed.

  1. Read about Ohmโ€™s Law and Power using the references stated at the end of this

module or references taken online.

EXPLAIN & ELABORATE

EQUIPMENT/ MATERIALS NEEDED:

Power Source - 0 โ€“ 30 Vdc, 25 mA

Milliammeter - 0 โ€“ 0.1/ 1 Adc

Electronic VOM

Practical Electronics Trainer

R1 - 1 kฮฉ, 1W

R2 - 1.5 kฮฉ, 1W

R3 - 3.3 kฮฉ

R4 - 10 kฮฉ, 1W

R5, R6 - 100 ฮฉ, I W

DS1 - Miniature Lamp

Universal Experiment Board K

PROCEDURES:

Note: In the absence of an actual laboratory set-up, you may answer the module based

on the concepts learned from the lecture subjects. As a verification, perform a simulation

with your simulator of choice. For every circuit, attach a screenshot of the set-up in your

report.

Examples of simulators: LTspice (Free), Electronic Workbench, Circuit Construction Kit: DC

at phet.colorado.edu

TLO 1 : Demonstrate that electrical power is a function of voltage and current by

calculating and measuring the power dissipated in a resistance as the voltage is

increased.

  1. Connect resistors R1 and R2 in a parallel, then connect them through the

milliammeter to the power source as shown in Fig. 2 โ€“ 1.

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  1. Set the milliammeter to the 100mAdc range.
  2. Adjust the power source to 2Vdc as indicated by the VOM.
  3. Record the current reading from the milliammeter in Table 2 โ€“ 2.
  4. Adjust the power supply to each of the remaining voltages listed in Table 2 โ€“ 2 and

record the current for each voltage. After the 4 โ€“ volt reading is recorded, reset

the VOM to the 15Vdc range and the milliammeter to the 1Adc range for the

other voltages.

  1. Return the voltage to zero.
  2. Now calculate and record the power for each of the voltages and currents listed

in Table 3 โ€“ 3.

VOLTAGE

E

CURRENT

I

POWER

W

Table 2 โ€“ 2

  1. Compare the calculated values of Table 2 โ€“ 1 with the measured values of Table

2 โ€“ 2. What factors would account for any differences in the two sets of values?

_________________________________________________

_________________________________________________

_________________________________________________

Resistor tolerances, accuracy of the meters, and measurement error

(reading the meter at an angle) can cause your measured values to differ from

your calculated values. However, this difference should be no greater than 5 to

10 percent.

TLO 2 : Determine that the resistance of a lamp filament varies with temperature.

  1. Set the electronic VOM to the LPฮฉ function and measure the cold resistance of

lamp DS

Rcold = _______________ ฮฉ

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You should measure approximately six ohms.

  1. Connect lamp DS1 through the milliammeter to the power source as shown in Fig.

A

V

0 โ€“ 5Vdc

0 โ€“ 1Adc

DS

Fig. 2 โ€“ 2

  1. Set the Electronic VOM to +DCV, 5V range and connect it across the lamp, as

shown.

  1. Set the milliammeter to the 1Adc range.
  2. Adjust the power source, in turn, to each of the voltages listed in Table 2 โ€“ 3.

Measure and record the current flow for each voltage.

  1. Return the voltage to zero.
  2. Calculate and record the power dissipated by the filament for each set voltage

and current values recorded in Table 2 โ€“ 3.

  1. Calculate and record the filament resistance of the lamp for each voltage and

current values recorded in Table 2 โ€“ 3.

  1. Examine your recorded measurements in Table 2 โ€“ 3. As the voltage and current

increase, does the resistance of the filament increase? ____

E

Vdc

I

mAdc

P

W

R ฮฉ 1 2 3 4