ELECTRICITY & MAGNETISM (Fall 2011) LECTURE # 10 BY MOEEN GHIYAS.

ELECTRICITY & MAGNETISM (Fall 2011)

LECTURE # 10

BY

MOEEN GHIYAS

TODAY’S LESSON

(Ohm’s Law, Power & Energy – Chapter 4)

(Series Circuit – Chapter 5)

Introductory Circuit Analysis by Boylested (10th Edition)

Today’s Lesson Contents

Chapter 4

• Energy

• Circuit Breakers, GFCIs and Fuses

• Applications

• Solutions to Problems

Chapter 5

• Introduction

• Series Circuit

• Voltage Sources in Series

Energy

• Power is the rate of doing work

• A motor may have the horsepower to run a heavy load,

but unless the motor is used over a period of time,

there will be no energy conversion

• The energy (W) lost or gained by any system is

therefore determined by

• The unit of energy is the wattsecond or joule

Energy

• The wattsecond, however, is too small a quantity

for practical purposes, so watthour (Wh) and

kilowatthour (kWh) were defined, as follows:

• To develop some sense, consider 1 kWh energy

is dissipated by a 100-W bulb in 10 h.

Energy – Kilowatthour Meter

• The kilowatthour meter is an instrument for measuring

the energy supplied to the residential or commercial

user of electricity

• It is normally connected directly to the lines at a point

just prior to entering the power distribution panel of

the building


• Example – How long can a 205-W television set be on

before using more than 4 kWh of energy?

• Solution:


• Example – What is the cost of using a 5-hp motor for

2 h if the rate is 9¢ per kilowatthour?

• Solution:


• Example – What is the total cost of using all of the

following at 9¢ per kilowatthour?

– A 1200-W toaster for 30 min

– Six 50-W bulbs for 4 h

– A 400-W washing machine for 45 min

– A 4800-W electric clothes dryer for 20 min

• Solution:

Circuit Breakers, GFCIs and Fuses

• To save the costly equipments from short circuit and

thus limit the current level, fuses or circuit breakers are

installed in panels where the power enters the house or

installation from outside feeder lines

• The fuses have an internal metallic conductor through

which the current will pass; a fuse will begin to melt if

the current through the system exceeds the rated value


• In homes built in recent years, fuses have been

replaced by circuit breakers

• When the current exceeds rated conditions, an

electromagnet in the device will have sufficient strength

to draw the connecting metallic link in the breaker out

of the circuit and open the current path

• When conditions have been corrected, the breaker can

be reset manually


• Note the load on each breaker should not exceed 80%

of its rating,

• 15 A circuit breaker to 12 A

• 20 A circuit breaker to 16-A

• 30 A circuit breaker to 24-A


• GCFIs (ground fault current interrupt) are designed to

trip more quickly than the standard circuit breaker.

The commercial unit trips in 5 ns and are generally

employed in bathroom and other sensitive areas.

• The GFCI is able to react as quickly by sensing the

difference between input and output currents to outlet

• An errant path such as through an individual

establishes a difference in the two current levels and

causes breaker to trip and disconnect power source

Applications – Microwave Oven

• Most microwaves are rated at 500 W to 1200 W at a

frequency of 2.45 GHz (2.45 billion cycles per second)

• The heating occurs because water molecules in food are

vibrated at such a high frequency that the friction with

neighbouring molecules causes the heating effect



• The microwave uses a 120 / 220 V ac supply which is

then converted through a high voltage transformer to

one having peak values approaching 5000 V (at

substantial current levels)

• Through the rectifying process, a high dc voltage of a

few thousand volts will be generated that will appear

across a magnetron.

• The magnetron (initially invented for WWII radar units),

will generate the required 2.45-GHz signal for oven


• If the microwave is plugged into a 120-V outlet, the

current drawn is

I = P/V = 1200 W/120 V = 10.0 A

• The amount of power dedicated solely to the cooking

process is determined by efficiency level. That is,

Po = ηPi = (0.55)(1200 W) = 660 W

• The energy transferred to the food over a period of 5

min can then be determined from

W = Pt = (660 W)(5 min)(60 s/1 min) = 198 kJ


• The number of kilowatthours drawn by the unit is

determined from

W = Pt/1000 = (1200 W)(5/60 h)/1000 = 0.1 kWh

• At a rate of 10 Rs / kWh we find that we can cook the

food for 1 Rs — relatively speaking, pretty cheap

• However, the cooking time with a microwave oven is

related to the amount of food in the oven, also it is not

a linear relationship so it won’t take twice as long —

perhaps 75% to 90% longer time thus raising cost.

Applications – Household Wiring

• One specification that defines household wiring is the

maximum current that can be drawn from the power

lines since the voltage is fixed at 110 V or 220 V.

• For most homes a 100-A service is the normal. Today,

with all the electronic systems becoming commonplace

in home, many people are opting for 200-A service.

• A 100-A service specifies that the maximum current

that can be drawn through power lines into your home

is 100 A.

Applications – Household Wiring

• Using the line-to-line rated voltage and the full-service

current we can determine the maximum power that can

be delivered using the basic power equation:

P = EI = (208 V)(100 A) = 20,800 W = 20.8 kW

• This rating reveals that total rating of all units turned on

in the home cannot exceed 20.8 kW at any one time.

• If it did, we could expect the main breaker at the top of

the power panel to open or even cause damage to

transformers.

Solution to Problems

• # 13

a) If an electric heater draws 9.5 A when connected to a

120-V supply, what is internal resistance of heater?

b) How much energy is converted in 1 h?

• Solution:

Solution to Problems

• # 33 - A calculator with an internal 3-V battery draws

0.4 mW when fully functional.

a) What is the current demand from the supply?

b) If the calculator is rated to operate 500 h on the same

battery, what is the ampere-hour rating of the battery?

• Solution:

Chapter 5 - Introduction

• Two types of currents

• Direct current (dc), in which ideally the flow of charge

(current) does not change in magnitude (or direction)

with time.

• Sinusoidal alternating current (ac), in which the flow of

current is continually changing in magnitude (and

direction) with time.

• In this semester we will restrict to circuit analysis

purely from a dc approach.

Chapter 5 - Introduction

• we generally consider the wire to be an ideal

conductor (that is, having no opposition to flow)

• The current is limited only by the resistor R

• The higher the resistance, the less the current, and

conversely, as determined by Ohm’s law.

Series Circuit

• A circuit consists of any number of elements joined at

terminal points, providing at least one closed path

through which charge can flow.

• If all the elements are in series, the network is called a

series circuit

• A branch of a circuit is any portion of the circuit that

has one or more elements in series

Series Circuit

Two elements in a circuit are in series if

1) They have only one terminal in common (i.e., one lead

of one is connected to only one lead of the other).

2) The common point between the two elements is not

connected to another current-carrying element.

Series Circuit

• Note that the current is the same through series

elements.

• The total resistance of a series circuit is the sum of the

resistance levels.

Series Circuit

• Note that the total resistance is actually the

resistance “seen” by the battery as it “looks”

into the series combination of elements

• The only resistance the source “sees” is the

total resistance.

• It is totally unaware of how the elements

are connected to establish RT

• RT of N resistors of same value in series is

Series Circuit

• Once RT is known, the current drawn from

the source can be determined using Ohm’s

law,

• Since E is fixed, the magnitude of the

source current will be totally dependent on

the magnitude of RT.

• A larger RT will result in a relatively small

value of Is, while lesser values of RT will

result in increased current levels

Series Circuit

• The fact that the current is the same through each

element permits a direct calculation of the voltage

across each resistor using Ohm’s law; that is,

Series Circuit

• The power delivered to each resistor can be calculated,

e.g for R1

• The power delivered by the source is

• The total power delivered to a resistive circuit is

equal to the total power dissipated by the resistive

elements.

Series Circuit - Example

a) Find the total resistance for the series circuit of fig.

b) Calculate the source current Is.

c) Determine the voltages V1, V2, and V3.

d) Calculate the power dissipated by R1, R2, and R3.

e) Determine the power delivered by the source, and

compare it to the total power dissipated.

Series Circuit – Solution to Example

a) Find the total resistance for the series circuit of fig.

b) Calculate the source current Is.

c) Determine the voltages V1, V2, and V3.

Series Circuit – Solution to Example

d) Calculate the power dissipated by R1, R2, and R3.

e) Determine the power delivered by the source, and

compare it to the total power dissipated.

Series Circuit - Example

• Example – Given RT and I, calculate R1 and E for the

circuit of fig.

• Solution:

Voltage Sources in Series

• Voltage sources can be connected in series to increase

or decrease the total voltage applied to a system

• The net voltage is determined simply by summing the

sources with the same polarity and subtracting the total

of the sources with the opposite “pressure.”

• The net polarity is the polarity of the larger sum.

Summary / Conclusion

Chapter 4

• Energy

• Circuit Breakers, GFCIs and Fuses

• Applications

• Solutions to Problems

Chapter 5

• Introduction

• Series Circuit

• Voltage Sources in Series

ELECTRICITY & MAGNETISM (Fall 2011) LECTURE # 10 BY MOEEN GHIYAS.

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