ENE 103 Electrotechnology

Semester 1/52

Dr. Ekapon Siwapornsathain

Outline

• Introduction• Current • Voltage• Kirchhoff’s Laws• Thevenin Equivalent of a circuit• Norton Equivalent of a circuit

Electrical Circuits

An electrical circuit consists of various types of ckt elements Connected in closed paths by conductors.

Electrical Current: the time rate of flow of electrical charge through a conductor or circuit element. The units are amperes (A),which are equivalent to coulombs per second (C/s). (The charge onan electron is -1.602x10-19 C

dt

tdqti

)()(

To find charge given current, we must integrate. Thus we have

t

ttqdttitq

0

)()()( 0

in which t0 is some initial time at which the charge is known

Reference Directions

In analyzing electrical circuits, we may not initially know the actual direction of current flow in a particular element. Therefore,we start by assigning current variables and arbitrarily selecting areference direction for each current of interest.

Direct Current and alternating current

When a current is constant with time, we say that we have directcurrent, abbreviated as dc. On the other hand, a current that varies with time, reversing direction periodically, is called alternating current, abbreviated as ac.

Reference directions can be indicated by labeling the ends ofCircuit elements and using double subscripts on current variables.The reference direction for iab points from a to be. On the otherHand, the reference direction for iba points from b to a

Voltages

Voltage is a measure of the energy transferred per unit of chargewhen charge moves from one point in an electrical ckt to a secondpoint. The units of voltage are volts (V), which are equivalent tojoules per coulomb (J/C).

Voltages are assigned polarities that indicate the direction of energy flow. If positive charge moves from the positive polaritythrough the element toward the negative polarity, the elementabsorbs energy that appears as heat, or some other form.

In ckt analysis, we frequently assign reference polarities forvoltages arbitrarily. If we find at the end of the analysis that thevalue of a voltage is negative, then we know that the true polarityis opposite of the polarity selected initially.

Double-subscript notation for voltages

Another way to indicate the reference polarity of a voltage is touse double subscripts on the voltage variable.

vab = - vba

Power and Energy

p = vi

The product of current and voltage is power:

The physical units of the quantities on the right-hand side ofThis equation are

volts x amperes = (joules/coulomb) x (coulombs/second) = joules/second

= watts

Energy Calculations

2

1

)(t

tdttpw

Find an expression for the power for the voltage source shown. Compute theenergy for the interval from t1 = 0 tot2= ∞

The current reference enters the positive reference polarity.Thus, we compute power as p(t) = v(t)i(t) = 12x2e-t = 24e-t W Subsequently, the energy transferred is given by

0

)( dttpw

J24)24(24]24[

24

00

0

eee

dte

t

t

Kirchhoff’s Laws

Physical Basis for Kirchhoff’s Current Law (KCL)

Node a: i1 + i2 + i3 = 0

Node b: i3 – i4 = 0

Node c: i5 + i6 + i7 = 0

The net current entering a node is zero

All points in a ckt that are connected directly by conductors can be considered to be a single node

ia + ic = ib + id

Series Ckts: When elements are connected end to end, we saythat they are connected in series. In order for elements A and B to be in series, no ther path for current can be connected to the node joining A and B. Thus, all elements in a series ckt have identical currents.

Example, we have ia = ib = ic

Kirchhoff’s Voltage Law (KVL)A loop in an electrical ckt is a closed path starting at a node andproceeding through ckt elements, eventually returning to thestarting node.

Kirchhoff’s voltage law (KVL) states: The algebraic sum of the Voltages equals to zero for any closed path (loop) in an Electrical ckts.

In applying KVL to a loop, voltagesare added or subtracted dependingon their reference polarities relativeto the direction of travel around theloop.

We obtain the following equationsLoop 1: -va + vb + vc = 0Loop 2: -vc – vd + ve = 0Loop 3: va – vb + vd – ve = 0

Parallel circuits

Two circuit elements are connected in parallel if both ends of one element are connected directly (i.e., by conductors) to corresponding ends of the other.

In the above ckt, elements A and B are in parallel. Elements D,E and F form another parallel combination.

Introduction to circuit elements

In this section, we define several types of ideal ckt elements:- Conductors- Voltage sources- Current sources- Resistors

Conductors: ideal conductors are represented in ckt diagrams byunbroken lines between the ends of other ckt elements.

The voltage between the ends of an ideal conductor is zeroregardless of the current flowing through the conductor.

When two points in a ckt are connected together by an idealconductor, we say that the points are shorted together. Anotherterm for an ideal conductor is short ckt.

All points in a ckt that are connected by ideal conductors can beconsidered as a single node.

If no conductors or other ckt elements are connected betweentwo parts of a ckt, we say that an open ckt exists between the two parts of the ckt. No current can flow through an ideal open ckt.

Independent voltage sources: An ideal independent voltagesource maintains specified voltage across its terminals. Thevoltage across the source is independent of other elements thatare connected to it and of the current flowing through it.

Dependent voltage sources

A dependent or controlled voltage source is similar to an independentsource except that the voltage across the source terminals is a function of other voltages or currents in the ckt.

A voltage-controlled voltage source is a voltage source having a voltage equal to a constant times the voltage across a pair of terminals elsewhere in the network. The factor multiplying the voltage is called the gain parameter, which as a unit of V/V

A current-controlled voltage source is a voltage source having a voltage equal to a constant times the current through some other element in the ckt. The gain parameter has a unit of V/A

Independent current sources

An ideal independent current source forces a specified current flow through itself. The current through an independent current source is independent of the elements connected to it and of the voltage across it.

Dependent Current Sources

The current flowing through a dependent current source is Determined by a current or voltage elsewhere in the circuit.

A voltage-controlled current source is a current source having a Current equal to a constant times the voltage through some other element in the ckt. The gain parameter of the source has units of A/V

A current-controlled current source is a current source having a current equal to a constant times the current through some other element in theckt. The gain parameter of the source has units of A/A

Resistors and Ohm’s Law

The voltage v across an ideal resistor is proportional to the current Ithrough the resistor. The constant of proportionality is the resistance R

The voltage and current are related by Ohm’s law: v = iR

If the references for v and I are opposite to the passive configuration, we have v = - iR

Conductance

Solving Ohm’s law for current, we have vR

i1

We call the quantity 1/R a conductance. It is customary to denoteconductances with the letter G:

RG

1

Conductances have the unitsOf inverse ohms ()

Resistance related to physical parameters

The dimensions and geometry of the resistor as well as the particularmaterial used to construct a resistor influence its resistance. Theresistance is approximately given by

A

LR

Power Calculations for resistances

R

vRivip

22

Example: Solve for the source voltage in the ckt of figure below.

First, we use Ohm’s Law to determine the value of iy :

AV

iy 35

15

Next, we apply KCL at the top end of the controlled source:

yxx iii 5.0

Substituting the value found for iy and solving, we determine thatIx = 2A. Then Ohm’s law yields vx = 10ix = 20V.

Applying KCL around the periphery of the ckt give Vs = vx + 15Finally substituting the value found for vx yields Vs = 35V