engr 101: engineering fundamentals

ENGR 101: Engineering Fundamentals

Agenda

Circuits Basic Electricity Ohm’s Law Combining Resistors Kirchhoff’s Laws DC current Power

Electrical Circuits – Electric Charge

The smallest amount of electric charge that can exist is that of a single electron which has a charge of 1.602 x 10-19 C

A constant charge is usually denoted by Q, while a charge that changes with time is written as q or q(t).

Charge can be either positive or negative.

Dissimilar charges attract

Similar charges repel

By the Electrostatic Force

Electric Circuits – Electric Charge

When a particle has equal parts positive and negative charge it is electrically neutral.

When there is an imbalance, the particle is electrically charged.

In conductors, a significant number of charged particles are free to move.

When charge moves through a material an electric current exists in the material.

However, unless an external force is applied, the electric charges in a conductor move about at random in an electric field.

Electric Circuits - Voltage

Once an electromotive force is applied, charges move in a unified manner.

The electric potential difference, or voltage V, between two points is work done in moving a charge from one point to the other.

The instantaneous voltage v is described as a derivative

In SI the unit for electric potential is the volt, V

CJ111 =V

dqdwv =

Electrical Circuits - Current

Electric current, I, is the rate at which charge flows through an area.

The instantaneous current is written as a derivative

tqIavg Δ

Δ=

dtdqi =

The SI unit for electric current is the ampere (A) and it is defined as

sCA111 =

Electric Circuits - Current

Conventional Current – based on the flow of positive charges, flows from positive to negative.

Electron Current – the movement of free electrons from negative to positive

Conventional current is generally used in circuit analysis.

+-

+

-

Electric Circuits - Current

Direct Current (DC)- direction of charge flow is always the same

Alternating Current (AC)- direction of charge flow alternates in direction, often sinusoidal

Other forms include Pulsating DC Exponential Sawtooth Transient effects

Electrical Circuits – Ohm’s Law

Ohm’s law states that the potential difference across a conductor is directly proportional to the current

Where V is the potential difference and I is the current. Introducing a constant of proportionality R, denoting resistance.

IV ∝

RIV =

Electrical Circuits - Resistance

Electrical resistance, R, is an impedance to current flow through a circuit.

By rearranging Ohm’s law, the resistance between any two points of a conductor is determined by applying a potential difference (V) between those two points and measuring the current, I, that results.

The SI unit for resistance is ohm (Ω)

A111 V

=Ω

IVR =


It is often necessary to find the equivalent or total resistance of two or more resistors connected together.

Series

Parallel

nt RRRRR ++++= ...321

nt RRRRR1...1111

321

++++=


Series Example R1 = 10 Ω, R2 = 20 Ω, R3 = 30 Ω

nt RRRRR ++++= ...321

Ω+Ω+Ω= 302010tR

Ω= 60tR

Rt


Parallel Example R1 = 10 Ω, R2 = 20 Ω, R3 = 30 Ω

nt RRRRR1...1111

321

++++=

Rt

Ω+

Ω+

Ω=

301

201

1011

tR

Ω=

Ω+

Ω+

Ω=

6011

602

603

6061

tR

Ω≈Ω= 45.51160

tR

Electric Circuits - Power

Remember the electric potential difference, or voltage V, between two points is work done in moving a charge from one point to the other.

By definition the rate at which work is done is power, P.

So using the definition of instantaneous voltage, and rearranging for work

For rate, per unit time, thus But remember the definition of current, thus

Power can be written as For this class, we will only be looking at

steady state DC. Thus,

dtdwP = vdqdw =

dtdqv

dtdw

=

viP =

VIP =

dqdwv =

dtdqi =

Electric Circuits - Power

We can check the units,

VIP =

!"

#$%

&!"

#$%

&=sC

CJP

[ ]WsJP =!"

#$%

&=

Electric Circuits – more Power

As current flows through a resistor the absorbed electrical energy is dissipated as thermal energy.

The rate at which this occurs is referred to as power dissipation.

As before

Using Ohm’s law, two alternate forms can be created using

Ohm’s law as V = RI Ohm’s law as I = V/R

VIP =

RIP 2=RVP2

=

Electric Circuits – Schematics

A schematic diagram is a symbolic representation of the devices and interconnections in the circuit.

It can be as simple as

Electric Circuits – Circuit Element

A circuit element is a generic term that refers to an electrical component such as a resistor, capacitor, or inductor.

+ -Circuit Element

V

I IP=V I

Electric Circuits - Schematics

Common circuit elements and their symbols

There are also idealized independent voltage and current sources

Electric Circuits – Kirchhoff’s Laws

Kirchhoff’s Current Law states that the algebraic sum of the currents entering a node is zero.

A node is a point of connection of two or more circuit elements

∑ = 0inI


So to illustrate Kirchhoff’s current law.

Current entering a node has a positive sign, and those leaving have a negative sign

054321=−+−+=∑ IIIIIIin

∑ = 0inI


Kirchhoff’s Voltage Law states that the algebraic sum of the voltages around a loop is zero.

A loop is a closed path in a circuit

So the sum of the voltages about the loop

∑ = 0V

∑ = 0V

∑ =−−+= 010 21 VVV

Kirchhoff’s example

Apply Kirchhoff’s laws to a simple example

At node 2 + I1 - I2 - I3 = 0

At node 3 + I2 + I3 - I4 = 0

At node 1 + I4 - I1 = 0

I3

I1

I4

I2



Large Loop V1 + V3 + V = 0

Small loop V1 + V2 + V = 0

V1

V2

V3



Smallest Loop - V2 + V3 = 0

V2

V3



Large Loop V1 + V3 + V = 0

Small loop V1 + V2 + V = 0

Smallest Loop -V2 + V3 = 0

V1

V2

V3

At node 2 + I1 - I2 - I3 = 0

At node 3 + I2 + I3 - I4 = 0

At node 1 + I4 - I1 = 0


Example Problem #1

Given: V = 12 V R1 = 2 Ω R2 = 3 Ω R3 = 4 Ω

Find:

I1 I2 I3


Example Problem #1

Current Law Σ I = 0

At node 2 + I1 - I2 - I3 = 0

At node 3 + I2 + I3 - I4 = 0

At node 1 + I4 - I1 = 0

I3

I1

I4

I2 Given V = 12 V R1 = 2 Ω R2 = 3 Ω R3 = 4 Ω


Example Problem #1

V3

V1

V2Given: V = 12 V R1 = 2 Ω R2 = 3 Ω R3 = 4 Ω

Voltage Law ΣV = 0

Large Loop - V1 -V3 + V = 0

Small loop -V1 - V2 + V = 0


V


Example Problem #1

I3

I1

I4

I2Given: V = 12 V R1 = 2 Ω R2 = 3 Ω R3 = 4 Ω

Using V=IR

Large Loop - I1R1 - I3R3 + V = 0

Small loop - I1R1 - I2R2 + V = 0

Smallest Loop -I2R2 + I3R3 = 0

Large Loop - V1 - V3 + V = 0

Small loop - V1 - V2 + V = 0



Example Problem #1

I3

I1

I4

I2Given: V = 12 V R1 = 2 Ω R2 = 3 Ω R3 = 4 Ω

Plug in Values

- I1 (2 Ω) - I3 (4 Ω) + (12 V) = 0

- I1 (2 Ω) - I2 (3 Ω) + (12 V) = 0

-I2 (3 Ω) + I3 (4 Ω) = 0

From Current Law at node 2

+ I1 - I2 - I3 = 0


Solve for 3 unknown, I1, I2 and I3

- I1 (2 Ω) - I3 (4 Ω) + (12 V) = 0

- I1 (2 Ω) - I2 (3 Ω) + (12 V) = 0

-I2 (3 Ω) + I3 (4 Ω) = 0

+ I1 - I2 - I3 = 0

Solve for 3 unknown, I1, I2 and I3

I1 = 3.2308 A

I2 = 1.8461 A

I3 = 1.3846 A

I1 = 3.23 A

I2 = 1.85 A

I3 = 1.38 A

engr 101: engineering fundamentals

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