•Voltage is a measure of the energy carried by the charge. Strictly: voltage is the "energy per unit charge".

•The proper name for voltage is potential difference or p.d. for short, but this term is rarely used in electronics.

•Voltage is supplied by the battery (or power supply).

•Voltage is used up in components, but not in wires.

•We say voltage across a component.

•Voltage is measured in volts, V.

•Voltage is measured with a voltmeter, connected in parallel.

•The symbol V is used for voltage in equations.

How do we measure voltage?

Connecting a voltmeter in parallel

Current, I Current is the rate of flow of charge.

Current is not used up, what flows into a component must flow out.

We say current through a component.

Current is measured in amps (amperes), A.

Current is measured with an ammeter, connected in series. (To connect in series you must break the circuit and put the ammeter acoss the gap, as shown in the diagram)

The symbol I is used for current in equations.

How do we measure current?

Connecting an ammeter in series

Voltage is the Cause, Current is the Effect

Voltage attempts to make a current flow, and current will flow if the circuit is complete.

Voltage is sometimes described as the 'push' or 'force' of the electricity, it isn't really a force but this may help you to imagine what is happening.

It is possible to have voltage without current, but current cannot flow without voltage.

Voltage and CurrentThe switch is closed making a complete circuit

so current can flow.

Voltage but No CurrentThe switch is open so the circuit is broken

and current cannot flow.

No Voltage and No CurrentWithout the cell there is no source of voltage

so current cannot flow.

Voltage and current for components in series

Voltages add up for components connected in series. Currents are the same through all components connected in

series.

In this circuit the 4V across the resistor and the 2V across the LED add up to the battery voltage: 2V + 4V = 6V.

The current through all parts (battery, resistor and LED) is 20mA.

Voltage and Current for components in Parallel Voltages are the same across all components connected in

parallel. Currents add up for components connected in parallel.

In this circuit the battery, resistor and lamp all have 6V across them.

The 30mA current through the resistor and the 60mA current through the lamp add up to the 90mA current through the battery.

Resistance, RThe resistance of a component restricts the

flow of electric current.Energy is used up as the voltage across the

component drives the current through it and this energy appears as heat in the component.

Resistance is measured in ohms, the symbol for ohm is an omega Ω.

1 Ω is quite small for electronics so resistances are often given in kΩ and M Ω. 1 k Ω = 1000 1 M Ω = 1000000 .

Resistors connected in Series

When resistors are connected in series their combined resistance is equal to the individual resistances added together.For example:

Combined resistance in series: R = R1 + R2Note that the combined resistance in series will always be greater than any of the individual resistances.

Resistors connected in Parallel When resistors are connected in parallel their combined

resistance is less than any of the individual resistances. There is a special equation for the combined resistance

of two resistors R1 and R2:Combined resistance of two resistors in parallel: R = R1 × R2

R1 + R2

For more than two resistors connected in parallel: 1 = 1 + 1 + 1 + ...

R R1 R2 R3Note that the combined resistance in parallel will always be less than any of the individual resistances.

Conductors, Semiconductors and Insulators resistance of an object depends on its shape and

the material from which it is made

For a given material, objects with a smaller cross-section or longer length will have a greater resistance.

Materials can be divided into three groups: Conductors which have low resistance.

Examples: metals (aluminium, copper, silver etc.) and carbonMetals are used to make connecting wires, switch contacts and lamp filaments. Resistors are made from carbon or long coils of thin wire.

Semiconductors which have moderate resistance. Examples: germanium, silicon. Semiconductors are used to make diodes, LEDs, transistors and integrated circuits (chips).

Insulators which have high resistance. Examples: most plastics such as polythene and PVC (polyvinyl chloride), paper, glass. PVC is used as an outer covering for wires to prevent them making contact.

DC Electric Power The electric power in watts associated with a complete electric

circuit or a circuit component represents the rate at which energy is converted from the electrical energy of the moving charges to some other form, e.g., heat, mechanical energy, or energy stored in electric fields or magnetic fields. For a resistor in a D C Circuit the power is given by the product of applied voltage and the electric current:

P = VI Power = Voltage x Current

The details of the units are as follows:

Power Dissipated in ResistorConvenient expressions for the power dissipated in a resistor can be obtained by the use of Ohm's Law.

These relationships are valid for AC applications also if the voltages and currents are rms or effective values. The resistor is a special case, and the AC power expression for the general case includes another term called the power factor which accounts for phase differences between the voltage and current.

DC Power in Series and Parallel CircuitsThe power relationship is one of the main tools for the analysis of electric circuits, along with Ohm's Law, the voltage law and the current law. Applying the current law to the above circuits along with Ohm's law and the rules for combining resistors gives the numbers shown below. The determining of the voltages and currents associated with a particular circuit along with the power allows you to completely describe the electrical state of a direct current circuit.

Ohm’s Law

Ohm’s law deals with the relationship between voltage and current

in an ideal conductor.

This relationship states that the potential difference (voltage)

across an ideal conductor is proportional to the current through it.

The constant of proportionality is called the "resistance", R.

Resistance Current Voltage

DefinitionThe opposition to the flow of

charges

The flow of electrons through a

circuit

Potential difference(the push

behind electricity)

Symbol R I V

Equation R = V / I I = V / R V = IR

Resistance Current Voltage

Units ω - ohms A - amperes V - volts

Depends on

The wire sizeThick wire – Less

resistanceThin wire – More

resistanceLong wire – more

resistanceShort wire- less

resistance

The resistance in the circuit

Greater resistance- less current

Less resistance - the greater the

current

The voltagesourceGreaterPotential

difference =greatervoltage

Practice Problems:1. In a circuit, 0.5 A is flowing through the bulb. The voltage across

the bulb is 4.0 V. What is the bulbs resistance?

Solution:R = V / IR = 4.0 / 0.5R = 8R = 8 Ω

2. You light a light bulb with a 1.5 volt battery. If the bulb has a

resistance of 10 ohms, how much current is flowing?

Solution:I = V / R

I =1.5 / 10

I = 0.15

I = 0.2 A

Types of Circuits Series

Parallel

Combination (Series and Parallel)

Series Circuits

each component has the same current

Example:

- The battery voltage is divided between the two lamps. Each lamp

will have half the battery voltage if the lamps are identical.

- If several lamps are connected in series they will all be switched

on and off together by a switch connected anywhere in the circuit. The

supply voltage is divided equally between the lamps (assuming they are all

identical). If one lamp blows all the lamps will go out because the circuit is

broken.

Lamps in series

- If several on-off switches are connected in series they must

all be closed (on) to complete the circuit. The diagram shows a

simple circuit with two switches connected in series to control a

lamp. Switch S1 AND Switch S2 must be closed to light the lamp.

Switches in series

each component has the same voltage

Example:

- Both lamps have the full battery voltage across them. The

battery current is divided between the two lamps.

Parallel circuits

- If several lamps are connected in parallel each one has the full supply

voltage across it. The lamps may be switched on and off independently by

connecting a switch in series with each lamp as shown in the circuit diagram.

This arrangement is used to control the lamps in buildings.

Lamps in parallel

- If several on-off switches are connected in parallel only one

needs to be closed (on) to complete the circuit. The diagram shows a

simple circuit with two switches connected in parallel to control a lamp.

Switch S1 OR Switch S2 (or both of them) must be closed to light the

lamp.

Switches in parallel

Node Junction of two or more circuit elements Trivial node – 2 elements are connected Essential node – 3 or more elements are connected

Branch Portions of the circuit where in two terminals are connected to it

having only a single element.

Loop Any closed path containing nodes and branches and circuit

elements like batteries and resistors.

Mesh A loop that does not contain any other loop.

Example 1Determine the total no. of nodes branches, loops and mesh

Solution

Answers to example 1 Nodes = 7 Branches = 9 Loops :

1 = 32 = 23 = 1total = 6

Meshes = 3

Example 2Determine the no. of nodes, branches, loops and mesh

Solution

Answers to example 2 Nodes = 4 Branches = 6 Loops :

1 = 22 = 1Total = 3

Mesh = 2

Example 3 (quiz)Determine the no. of nodes, branches, loops and mesh

At any node (junction) in a circuit,the algebraic sum of currents entering and leaving a node at any instant of time must beequal to zero.

Here currents entering and currents leaving the node must be assigned opposite algebraic signs:

Kirchhoff’s Current Law (KCL)

Kirchhoff’s Current Law (KCL)

The sum of the currents in all branches which converge to a common node is equal to zero.The sum of the currents in all branches which converge to a common node is equal to zero.

I1 − I2 + I3 − I4 + I5 − I6 = 0

(a)The sum of the currents into a node equals zero.

(b)The sum of the currents leaving the node is zero.

(c) The sum of the currents entering a node equals the sum of the currents leaving the node.

Various ways to state Kirchhoff’s current law

Various ways to state Kirchhoff’s current law

In a clockwise manner, passing a resistor will give it a negative voltage while passing a voltage source (crossing from negative to positive) will give it a positive voltage.

Going in a clockwise direction starting from the voltage source V1 and returning to the same point, the equation goes this way:

The sum of all voltages between successive nodes in a closed path in the circuit is equal to zero.The sum of all voltages between successive nodes in a closed path in the circuit is equal to zero.

Kirchhoff’s Voltage Law (KVL)

Kirchhoff’s Voltage Law (KVL)

V1 − IR1 − IR2 − V2 − IR3 − IR4 + V3 − IR5 − V4 = 0

Consider the circuit shown:

Three loops in this circuit gives us three equation:

The original loop:v1 + v2 + v3 + v4 = 0

Loop abcd considering R5:v1 + v2 + v5 + v4 = 0

Loop cd: v3 + v5 = 0

Summary of steps

1. Draw the circuit on a flat surface with no conductor crossovers.2. Label the mesh currents ( i I ) carefully in a clockwise direction.3. Write the mesh equations by inspecting the circuit (No. of independent mesh

(loop) equations=no. of branches (b) - no. of principle nodes (n) + 1).

Note: A resistive network containing voltage and current sources using ‘mesh’equations method the following steps are essential to note for analysis.• If possible, convert current source to voltage source.• Otherwise, define the voltage across the current source and write the meshequations as if these source voltages were known. Augment the set of equationswith one equation for each current source expressing a known mesh current ordifference between two mesh currents.• Mesh analysis is valid only for circuits that can be drawn in a two-dimensionalplane in such a way that no element crosses over another.

Summary of steps

Summary of steps

Consider a simple DC network as shown find the currents through different branches using loop current method.

Example 1

Solution:

Applying KVL around loop 1:

Applying KVL around loop 2:

Applying KVL around loop 3:

Example 2

Find Iab and Vcg.

Solution:

Assume voltage across the current source is v1.Applying KVL around Loop 1:

Assume that the voltage across the current source is v1 and its top end is assigned with a positive sign.Applying KVL around Loop 2:

Solving simultaneously, we’ll get

Applying KVL around Loop 3:

Note: negative sign of current means that the current flows in reverse direction (in our case, the current flows through 4Ω resistor from ‘c’ to ‘b’ ).

From the KCL equation in Loop 2, one can get v1= 6.27 volts.

Find Vx using the mesh current method.

Example 3

Solution:

One can easily convert the extreme right current source (6 A) into a voltage source. Note that the current source magnitude is 6 A and its internal resistance is 6 Ω .The given circuit is redrawn and shown below.

Applying KVL around Loop 1: (note that I1 =12 A)

Applying KVL around Loop 2:

Solving for Vx gives Vx = 48 volts.

Example 4Solve for the currents through R2 and R3 in the circuit shown.

Solution:

Redrawing the circuit into a more recognizable form, we see that the 2-mA current source is in parallel with a 6-kV resistor.

Redrawing the circuit is further simplified by labelling some of thenodes, in this case a and b. After performing a source conversion, we have the two-loop circuit shown.

The loop equations are

In loop 1, both voltages are negative since they appear as voltage drops whenfollowing the direction of the loop current.

Rewriting gives

Loop 1: (21 kΩ)I1 – (5kΩ)I2 = -22VLoop 2: -(5kΩ)I1 + (21kΩ)I2 = 18V

Solving simultaneously gives

I1 = -0.894 mAI2 = -0.644 mA

Loop 1: (6 kΩ + 10kΩ + 5 kΩ)I2 = - 12V – 10VLoop 2: -(5 kΩ)I1 + (5 kΩ + 12 kΩ + 4 kV)I2 = 10V + 8V

The current through resistor R2 is easily determined to be

I2 - I1 = 0.644 mA - (-0.894 mA) = 1.54 mA

The current through R3 is not found as easily. A common mistake is to say thatthe current in R3 is the same as the current through the 6-kV resistor of the circuitin Figure 8–28. This is not the case. Since this resistor was part of thesource conversion it is no longer in the same location as in the original circuit.Although there are several ways of finding the required current, themethod used here is the application of Ohm’s law. Observe that the voltage across R3 is equal to Vab. We determine Vab by using the calculated value of I1.

Vab = -(6 kΩ)I1 - 12 V = -(6 kΩ)(-0.894 mA) - 12 V = -6.64 V

The above calculation indicates that the current through R3 is upward (sincepoint a is negative with respect to point b). The current has a value of

IR3 = (6.64V)/(6kΩ) = 1.11 mA

Example 4 (quiz)Use mesh analysis to find the loop currents in this circuit.