Fundamentals of Electrical Circuits 1 WHAT IS CURRENT? Electrons are charge carriers Unit of charge is the Coulomb (C) Current is the rate of flow of charge

Fundamentals of Electrical Circuits

1

WHAT IS CURRENT?

• Electrons are charge carriers

• Unit of charge is the Coulomb (C)

• Current is the rate of flow of charge

• 1C of negative charge = total charge carried by 6.242×1018 electrons

• Charge of 1 electron = 1/ 6.242×1018 = 1.6×10-19C

• Charge can either be positive or negative

• Electric current exists when there is a net transfer of charge in a material

For example: If you inject electrons into a copper wire, they travel through the wire and emerge at the other end → current in the wire

• Current = rate at which charge is transferred

• Unit of Current = Ampere (Amp)

• 1A = rate of flow of charge of 1C in 1 second i.e.

• Current (I) = Charge (Q) / Time (s)


2

WHAT IS VOLTAGE?

• To establish a flow of charge through a conductor, we need to exert a force on the electrons that carry the charge

• This is called electromotive force (emf)• To sustain flow, electrons need a destination such as the positive and

negative terminals of a battery• Unit of EMF is the volt• Named after Alessandra Volta• The greater the voltage of a source of EMF, the greater the current it can

produce• EMF is also called electric potential, which is the same as talking about the

ability (potential) of a voltage source to produce current• We say E volts across the voltage source or component• Symbol for voltage source and its terminals

+ _


3

WHAT IS RESISTANCE?

• It is the measure to the extent to which a material interferes with, or resists, the flow of current through it

• A conductor has small resistance, an insulator has high resistance

• Unit of resistance is the Ohm (George Ohm) - R• The symbol of the Ohm is Ω• For a perfect conductor, its resistance is 0 Ω• A perfect insulator has a resistance of ∞ Ω

• Conventional current needs a complete path to flow• There must be a destination that will accept electrons, and

there must also be a source of electrons

E+

-R

I


4

OHM’S LAW

• The greater the voltage at a source, the greater the current it can produce

• Current produced in a resistor is directly proportional to the voltage of the source

• Resistance reduces the flow of current• Current is inversely proportional to resistance, i.e. the

greater the resistance, the less the current• For a fixed resistance R, the current I increases with an

increase in the voltage V at the source• This is Ohm’s Law, which is a linear relationship• V = IR; I = V/R; R = V/I

V

I

V /I = R2 4V+

- 8 I

Question:How much voltage is necessary to create a flow of 0.24C in 0.8s through a resistance of 500 Ω?


5

THE VOLTMETER

• Used to measure voltage across a component

• Always connected in parallel to a component

• The voltmeter in the first three circuits will always read 6V

• The voltmeter in the forth circuit reads -6V

• Voltmeters have black (negative or common) and red (positive) terminals

6 V +

-R V

+

-

6 V+

-

+

-

V

6 V+

-

+

--

RV

6 V +

-R V

+

-


6

THE AMMETER

• Used to measure current

• To measure current flowing in a resistance, you must disconnect the resistance and insert the ammeter in such a way so that all the current flowing in the resistance also flows through the ammeter, i.e. in series with the resistance

• The ammeter in these circuits will read 3A

2 4V+

- 83 A

A

2 4 V+

-

+

-

8

A2 4 V

+

-

+

-

8


7

WORK, ENERGY AND POWER

• Work – energy spent to overcome a restraint to achieve a physical change

• Energy is the ability to do work

• Energy and work have the same SI units – Joules (J)

• Power – rate at which energy is expended (Joules/sec)

• Unit of power = watt = Joules per second = Js-1

• Power (P) = Work (W)/Time (t); W = Pt

• When electrical current flows through a resistance, electrical energy is converted to heat energy at a rate that depends on the voltage across the resistance and the value of current through it, i.e.

• Power (P) = Voltage (V) × Current (I) (watts)

• P = V2/R = I2R (watts)

• Power ‘delivered’ to a resistance = power dissipated in resistance

• A kilowatt-hour (kWh) is the total energy delivered or consumed in one hour, and is used in industry

• kWh = (P in kW) × (t in hours)


8

WORK, ENERGY AND POWEREXAMPLES

• A 12V DC power source is connected to a 500Ω resistor that has a tolerance of ±5%. What is the maximum and minimum power that can be dissipated in the resistance?

• Electrical energy costs £0.12/kWh. For how long could a 900W oven be operated without costing more than 36p?


9

POWER RATING OF RESISTORS

• If the power rating of a resistor is too small for a particular application, then the resistor will not be able to dissipate heat at a rate rapid enough to prevent destructive temperature build up

• Resistors that are physically large in size have a greater surface area, so they dissipate heat faster

• Question: A 200Ω resistor has a 2W power rating. What is the maximum current that can flow in the resistor without exceeding the power rating?


10

RESISTOR COLOUR CODES

• 0 Black

• 1 Brown

• 2 Red

• 3 Orange

• 4 Yellow

• 5 Green

• 6 Blue

• 7 Violet

• 8 Grey

• 9 White

• 5% Gold

• 10% Silver

• 20% None

• Yellow; violet; orange; silver = 47×103Ω ±10%

• Brown,black,red?

• Blue,grey,black,gold?

1st

2nd

3rd

Tolerance


11

CONDUCTANCE

• Conductance is the reciprocal of resistance

• Conductance is the ability of a material to pass electrons

• The higher the resistance, the lower the conductance

• Symbol of Conductance is G

• Unit of conductance is the Siemen (S)

• G = 1/R Siemens

• What are the resistance and conductance ranges of a 75kΩ ± 10% resistor?


12

REAL AND IDEAL SOURCES

• Ideal voltage source – one that maintains a constant terminal voltage, no matter how much current is drawn from it

• E.g. a 12V source should theoretically maintain 12V across 1MΩ resistor (I = 12μA), across 1k Ω (I = 12mA) and across 1 Ω (I = 12A)

• But all real voltage sources have an internal resistance that causes the terminal voltage to drop if the current is made large, i.e. if a small value of resistance is connected across the terminals

• However, for ease in practical analysis, it is convenient to assume that a voltage source is ideal, i.e. we can neglect its internal resistance


13

REAL AND IDEAL SOURCES

• An ideal current source (or constant current source) supplies the same current to any resistor connected across its terminals.

• Symbol

• The voltage across the terminals of the source will change if the resistance R changes.

• A 2A current source with a 10Ω resistor, E = 20V

• A 2A Current source with a 100 Ω, E = 200V

• However an ideal current source cannot be constructed because it will always have an internal resistance that causes a current drop if the voltage becomes very large (i.e. for a large R)

• Current source is really a voltage source with certain characteristics (we’ll see this later)

• Active components – voltage and current sources as they ‘furnish’ electrical energy to a circuit

• Passive components – Such as resistors

18mA


14

READ AND IDEAL SOURCESEXAMPLE

• A constant current source develops a terminal voltage of 9V when a 500Ω resistor is connected across its terminals. What is the terminal voltage when the 500Ω resistor is replaced by a 1.5kΩ resistor?

• Solve the question with the aid of circuit diagrams.


15

LINEARITY

• An electrical device is linear if its V v I graph is a straight line

• Many circuit analysis techniques can only be applied to circuits composed of linear devices, such as resistors.

• Graph of a linear component such as a resistor, where the gradient ΔV / ΔI = R

• The voltage across a device is directly proportional to the current through it

V

I

ΔV / ΔI = R


16

ELECTRIC CIRCUITS

• An electric circuit is a configuration of interconnected resistors, active sources or other electrical components through which current flows

• A complete circuit is a circuit where there is a path from the source back to the source

• A circuit diagram is also known as a schematic

• Circuit analysis is the process of determining current flows and voltages that exist is various parts of a circuit.

• There are two basic kinds of connections that we will encounter when we analyse circuits

• 1) Series

• 2) Parallel


17

SERIES CIRCUITS (1)

• Two components have a common terminal when there is a path of zero resistance from the terminal of one to the terminal of the other

• Two components are connected in series if they have exactly 1 common terminal and if no other component has a terminal that shares that common connection

• Important property – current is the same in every series connected component

• Current in one component = current in a component in series with it

• For resistors connected in series RT = R1 + R2 +…+ Rn

• For a voltage source in series → Ohm’s Law IT = V/RT


18

SERIES CIRCUITS EXAMPLES

• It is necessary to limit the current in a certain light emitting diode (LED) to 50mA. The resistance of the LED is 250Ω and it is connected in series with a 5V source. How much resistance should be inserted in series with the LED?

• Find the current in and voltage across each resistor of the circuit below.

E = 24V

R1 = 12Ω R2 = 6Ω

R3 = 10ΩR4 = 20Ω

IT↑

+V1-

+V2

-+V3

-

-V4+


19

KIRCHHOFF’S VOLTAGE LAW (1)

• Kirchhoff’s Voltage Law: The sum of the voltage drops around any closed loop equals the sum of the voltage rises around the loop

• Example Consider the series circuit below containing a voltage source and 3 resistors

• What is the total series resistance, and the current flow?• What are the voltage drops across each resistor?• How can we verify Kirchhoff’s voltage law?

2 0 V+

-

+++ ---3 5 2


20

KIRCHHOFF’S VOLTAGE LAW (2)

• A circuit is a closed loop circuit (see previous slide)

• When the loop passes through a component from positive (+) to negative (-), then the voltage across that component is a drop

• When the loop passes from – to +, it is a voltage rise

• Voltage rises occur from source voltages

• Voltage rise – voltage drop = 0

• Σ(voltage drops) = Σ(voltage rises)

• The direction of the loop is merely for analytical purposes, and does not have any affect on the result of Kirchhoff's voltage law

• The direction of the loop does not have to be the same as the direction of the current flow

• We measure voltage across source terminals, which we label. Here Vab = -Vba

a

b

+

-Vab

a

b

-

+

Vba


21

KIRCHHOFF’S VOLTAGE LAWEXAMPLES

• By drawing a loop to represent current flow, find E and hence the total current in the circuit above.

• Use Kirchhoff’s Voltage Law to find Vab in the circuit above. You may find it useful to identify the number of loops there are in this circuit.

E

3.9kΩ2.2kΩ

1.8kΩ

1.1kΩ

-13.2V+ -23.4V+

+6.6V-

-10.8V+

24V

R1 = 2kΩ R2 = 1kΩ

R3 = 3kΩa+

b-


22

OPEN CIRCUITS

• An open circuit is a gap, break or interruption in a circuit path.

• Open circuits result in no current flow in a series circuit where there is a break.

• If the current I = 0, then from Ohm’s Law R = V/I, which leads to V/0 = ∞ = R

• Therefore at an open circuit, there is an infinite resistance

• We may think of an open circuit as a fault, but it can be useful for circuit analysis

• Loads are often measured in terms of its resistance, but we may need to study a circuit when its load is removed, i.e. when RL = ∞

• Important V ≠ 0 in an open circuit.

• From the above, I = 0 regardless of the value of V


23

SERIES CONNECTED SOURCES

• Series connected sources are when two or more voltage sources are connected in series

• To the right, we have a series aiding configuration, where the net effect of the sources are equivalent to that of a single source, whose voltage equals the sum of the two sources

• Here to the right, we have a series opposing configuration. Current is produced in opposite directions, and the net effect on the circuit is the same as a single voltage source whose magnitude equals the difference in the sources. Here E1>E2.

E 1+ E 2E 1+ E 2+

-

E 2

E 1E 1

E 2

+

-

E 1-E 2+

-

E 2

E 1

E 1-E 2

E 1

E 2

+

-


24

VOLTAGE DIVIDER RULE

• Consider the above circuit, where the total current flow is I = VIN / RT = VIN / (R1+R2)

• The sum of the voltage drops across each resistor is equal to the source voltage VIN = IR1 + IR2 = V1 + V2

• V2 is the same as VOUT

• Substituting, we get:• VIN = (VIN R1)/ (R1+R2) + (VIN R2)/ (R1+R2) • VIN = V1 + V2; the voltage drops have been calculated

without knowledge of I• This is the voltage divider rule Vx = VIN × Rx / RT

• We can use potentiometers to adjust resistance. Also known as a variable resistor and will affect the voltage drop

R 1

R 2

V IN

+

-

V O U T


25

VOLTAGE DIVIDER RULE EXAMPLES

• Use the voltage divider rule to find voltages Vab and Vac in the circuit below.

• A certain electronic device is activated when a voltage level of 5V ±10% is applied to it. In one application where it is used, the only DC power available is a 24V source.

a) Design a voltage divider that will provide the required activation voltage across a resistor, which must not draw more than 10mA from the source.

b) Assuming that only standard value resistors having 5% tolerance can be used, draw a schematic of the final design, and verify that the design criteria are met.

36V

100Ω 200Ω

150Ω

a b

50Ω c


26

PARALLEL CIRCUITS

• When two components are connected in parallel, they have two common terminals

• Every parallel – connected component has the same voltage across it

• To calculate resistances in parallel

• Conductance of resistors in parallel:

• GT = G1+…+Gn

• GT = 1/RT and is measured in Siemens

nT RRRRR

11111

321


27

PARALLEL CIRCUITS EXAMPLE

• In the circuit below, find the current in each resistor. What is the circuit’s total resistance? What is its total conductance? What is the total current flow in the circuit?

E = 32V

R1=1.6kΩ

R2=320Ω

R3=100kΩ

R4=4kΩ

I1 ↓ I2 ↓ I3 ↓ I4 ↓


28

KIRCHHOFF’S CURRENT LAW

• The sum of all currents entering a junction, or any portion of a circuit, equals the sum of currents leaving the same

• Kirchhoff’s current law is a useful technique for problem solving, and it is often used to find unknown current values, as we shall see.

I5

I1

I4I3

I2

12345 IIIII


29

KIRCHHOFF’S CURRENT LAW EXAMPLES

• Find the current in the 150Ω resistor.

• Find the currents I1, I2, I3, and I4.

270Ω

150Ω

330Ω100Ω

← 0.2A

0.8A ↓

0.1A →

48V 12kΩ 8kΩ 9.6kΩ

I1 →

I2 →

I3 →

←I4


30

THE CURRENT DIVIDER RULE

V+

-

IR

I 1 I 2

R 1 R 2

• The current I entering the junction of two parallel resistors divides into two paths. The smaller the resistance of a path, the greater its share of the total current

• Because of the parallel nature of the circuit

• Using the above, we get

• Generally, current divider rule when current I enters a junction of an arbitrary number of parallel resistors is Ix=IRT / Rx

21

21

21

21

RR

RRIIRV

RR

RRR TT

KCL states that I = I1+I2

Parallel: V = I1R1 = I2R2

So I1=V/R1 and I2=V/R2

21122121 ; RRIRIRRIRI


31

THE CURRENT DIVIDER RULE EXAMPLES

• Find the current in the 470Ω resistor using the current divider rule. Find the current in the 330Ω resistor using the current divider rule, and verify the result using Kirchhoff’s current law.

• What should the value of R be in the circuit below if the current in it must be 0.1A?

R1 = 470Ω

R2 = 330Ω

160mA→

→I1

I2→

0.4A 30Ω R60Ω

↓ 0.1A


32

SHORT CIRCUITS

• A short circuit is a path of zero resistance• R = 0• When a component is short circuited, all current is diverted

through the path that shorts it• This follows on from the fact that on encountering a junction,

most current will flow through the path with the least resistance• Short circuiting techniques may be useful when analysing

circuits. We shall see this later on

I I

I SS = I

R

I R = 0

Documents

Fundamentals of Electrical Circuits 1 WHAT IS CURRENT? Electrons are charge carriers Unit of charge is the Coulomb (C) Current is the rate of flow of charge