ELECTRICITY & MAGNETISM (Fall 2011) LECTURE # 12 BY MOEEN GHIYAS.

ELECTRICITY & MAGNETISM (Fall 2011)

LECTURE # 12

BY

MOEEN GHIYAS

TODAY’S LESSON

(Series Circuit – Chapter 5)

Introductory Circuit Analysis by Boylested (10th Edition)

Today’s Lesson Contents

• Notation

• Internal Resistance of Voltage Sources

• Voltage Regulation

• Measurement Techniques

• Applications

• Solution to Problems

Notation – Voltage Sources and Ground

• Three ways to sketch the same series dc circuit

• If two grounds exist in a circuit and no connection

is shown between them, even then such a

connection exists for the continuous flow of charge.


• On large schematics

where space is at a

premium and clarity is

important, voltage

sources may be

indicated as in fig (a)

rather than as

illustrated in fig (b)


• On large schematics where space is at a premium and

clarity is important, voltage sources may be indicated

as fig (a) rather than as illustrated in fig (b)


• In schematics, the potential

levels may also be indicated

to permit a rapid check of the

potential levels at various

points in a network with

respect to ground to ensure

that the system is operating

properly

Notation – Double-Subscript Notation

• Voltage is an across variable and exists between two

points resulting in a double-subscript notation

• In fig, since a is the first subscript for Vab, point a must

have a higher potential than point b if Vab = +ve value.

• If point b is at a higher potential than point a, then

Vab = -ve value.

Notation – Single-Subscript Notation

• The single-subscript notation Va specifies the voltage at

point a with respect to ground (zero volts). Thus for

voltage at point b w.r.t to ground, we have Vb

• If the voltage is less than zero volts, a negative sign

must be associated with the magnitude of Va

Notation – General Comments

• Also the voltage Vab can be determined using

Eq. Vab = Va – Vb

• For fig below:

Notation

• Example – Find the voltage Vab for the conditions of fig

• Solution:

• Note the negative sign to reflect the fact that point b is

at a higher potential than point a.

Notation

• Example – Find voltage Va for the configuration of Fig

• Solution:

Notation

• Example – Find the voltage Vab for the configuration.

• Solution:

Notation & Voltage Divider Rule

• Example – Using the voltage divider rule, determine

the voltages V1 and V2 of fig.

• Solution: Circuit Redrawn,

• . From

• . voltage divider rule,


• Example – For the network of fig

a) Calculate Vab.

b) Determine Vb.

c) Calculate Vc.


a) Calculate Vab.

Solution:

c) Determine Vc.

Solution:


b) Determine Vb.

• Solution:

• or

Internal Resistance of Voltage Sources

• Every voltage source, whether a generator, battery, or

laboratory supply (fig (a)) has some internal resistance.

• The equivalent circuit of any voltage source appears as

shown in fig (b).


• The effect of the internal resistance on the output

voltage is important to study in order to understand

unexpected changes in terminal characteristics of

voltage source.


• The ideal voltage source has no internal resistance and

an output voltage of E volts with no load or full load as

shown in fig (a).


• In the practical case [fig(b)], where we consider the

effects of the internal resistance, the output voltage will

be E volts only when no-load (IL = 0) conditions exist.


• When a load is connected [fig (c)], the output voltage of

the voltage source will decrease due to the voltage drop

across the internal resistance.


• By applying Kirchhoff’s voltage law around the indicated

loop of fig (c), we obtain

• Since

• We have

• And we get an important relation

• If Rint is not known, it can be found


• A direct consequence of the loss in output voltage is a

loss in power delivered to the load.

• Multiplying both sides of eq.

by the current IL in the circuit, we obtain


• For a dc generator, a plot of the output voltage versus

current appears in fig


• Note that any increase in load demand causes a drop

in terminal voltage due to the increasing loss in

potential across the internal resistance.


• At maximum current IFL, the voltage across the internal

resistance is Vint = IFLRint = (10 A)(2 ) = 20 V, and the

terminal voltage has dropped to 100 V—a significant

difference (from 120V) even if you stay below the listed

full-load current.


• Eventually, if the load current were permitted to

increase without limit, the voltage across the internal

resistance would equal the supply voltage, and the

terminal voltage would be zero.


• The larger the internal resistance, the steeper is the

slope of the characteristics of fig

• In fact, for any chosen interval of voltage or current, the

magnitude of the internal resistance is given by


• Example – Before a load is applied, the terminal voltage

of the power supply of is set to 40V. When a load of

500Ω is attached, the terminal voltage drops to 38.5 V.

What happened to the remainder of the no-load voltage,

and what is the internal resistance of the source?


• Solution:

• The difference of 40 V – 38.5 V = 1.5 V now appears

across the internal resistance of the source.

• The load current IL = 38.5 V/0.5 kΩ = 77 mA.

• . Applying Eq.

38.5V38.5V


• Example – The battery of fig has an internal resistance

of 2Ω . Find the voltage VL and the power lost to the

internal resistance if the applied load is a 13Ω resistor.

• Solution:

Voltage Regulation

• If a supply is set for 12 V, it is desirable that it maintain

this terminal voltage, even though the current demand

on the supply may vary.

• A measure of how close a supply will come to ideal

conditions is given by the voltage regulation

characteristic.

Voltage Regulation

• By definition, the voltage regulation (VR) of a supply

between the limits of full-load and no-load conditions

(Fig. 5.56) is given by the following:

Voltage Regulation

• We see for ideal conditions, VFL = VNL and VR% = 0.

• Therefore, the smaller the voltage regulation, the less

the variation in terminal voltage with change in load.

• It can be shown with a short derivation that the voltage

regulation is also given by

• The smaller the internal resistance for the same load,

the smaller the regulation and more ideal the output.

Voltage Regulation

• Example - Calculate the voltage regulation of a supply

having the characteristics of Fig. 5.53.

• Solution:

Voltage Regulation

• Example - Determine the voltage regulation of the

supply of fig with internal resistance of 19.48Ω and

load resistance as 500Ω.

• Solution:

38.5V38.5V

Measurement Techniques - Ammeters

• Ammeters are placed in series with the branch in which

the current is to be measured

• For minimal impact on the network behaviour,

ammeter’s resistance should be very small (ideally zero

ohms) compared to the other series elements of the

branch


• If the meter resistance approaches or exceeds 10% of

branch resistance R, it will have a significant impact

on the current level it is measuring.

• It is also noteworthy that the resistances of the

separate current scales of the same meter are usually

not the same.

• In fact, the meter resistance normally increases with

decreasing current levels.


• For an up-scale (analog meter)

or positive (digital meter)

reading, an ammeter must be

connected with current entering

the positive terminal (Red) of

the meter and leaving the

negative terminal (Black), as

shown in fig.

Measurement Techniques - Voltmeters

• Voltmeters are always hooked up across

the element for which the voltage is to

be determined.

• An up-scale or positive reading on a

voltmeter is obtained by connecting

positive terminal (red lead) to the point

of higher potential and the negative

terminal (black lead) is connected to the

lower potential


• The internal resistance of a supply cannot be

measured with an ohmmeter due to the voltage

present.


• However, the no-load voltage can be measured by

simply hooking up the voltmeter as shown.

• The internal resistance of the voltmeter is usually

sufficiently high to ensure that the resulting current is

so small that it can be ignored.

Measurement Techniques

• It seems we can find by Ohm’s law: Rint = ENL /ISC.

• However, Rint of the voltage supply may be very low,

resulting in high current levels which could damage the

ammeter and supply.

• A better approach would be to apply a RL resistive load

and then measure current and use following eq. To

calculate Rint.

Applications – Holiday Lights

• If one wire enters and leaves the bulb casing, they are

in series.

• If two wires enter and leave, they are probably in

parallel.


• When bulbs are connected in series, if one burns out

(the filament breaks and circuit opens), and all the

bulbs should go out


• However, the holiday bulbs are specially designed to

permit current to continue to flow to the other bulbs

when the filament burns out.

• Note that only one flasher unit is required per 50 bulb

panel


• The bulbs of fig are rated 2.5 V at 0.2 A or 200 mA.

• Since there are 50 bulbs in series, the total voltage will

be 50 x 2.5 V = 125 V which matches voltage available

• Since the bulbs are in series, the current through each

bulb will be 200 mA.

• The power rating of each bulb is therefore P = VI (2.5

V)(0.2 A) = 0.5 W with a total wattage demand of

50 x 0.5 W = 25 W.


• When each set is connected together, they will actually

be in parallel

• Note that the top line is the hot line to all the connected

sets, and the bottom line is the return, neutral, or

ground line for all the sets.

Applications - Microwave

Solution to Problems

• #24a – Determine the voltages Va, Vb, and Vab for the

network

• Solution:

Va = 12 - 8 = 4V

Vb = -8V

Vab = 12V

Solution to Problems

• #32b – Find the voltage VL and the power loss in the

internal resistance for the configuration of fig

• Solution:

IL = 12 / (0.05+3.3)

= 12 / 3.35 = 3.58 A

VL = E – IL Rint = 12 – 3.58 x 0.05

= 12 – 0.179 = 11.82 V

Summary / Conclusion

• Notation

• Internal Resistance of Voltage Sources

• Voltage Regulation

• Measurement Techniques

• Applications

• Solution to Problems

ELECTRICITY & MAGNETISM (Fall 2011) LECTURE # 12 BY MOEEN GHIYAS.

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