Lecture 2F d l f El i lFundamentals of Electrical

EngineeringEngineering

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Introduction1. Define current, voltage, and power, including their units.including their units.

2. Calculate power and energy, as well as p gy,determine whether energy is supplied or absorbed by a circuit element.y

3. State and apply basic circuit laws.

4. Solve for currents, voltages, and powers in simple circuits.

Electrical CurrentElectrical Current

Electrical current is 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

d (C/ )second (C/s).

Electrical Current

d )(=

dttdqti )()(

∫ +=t

tqdttitq

dt

)()()( ∫ +=t

tqdttitq0

)()()( 0

Direct Current Alternating Current

When a current is constant with time, we th t h di t tsay that we have direct current,

abbreviated as dc. On the other hand, a t th t i ith ti icurrent that varies with time, reversing

direction periodically, is called alternating c rrent bb i t dcurrent, abbreviated as ac.

.

VoltagesVoltages

The voltage associated with a circuit element is the energy transferred per unit of charge that flows through the element. The units of voltage are volts (V), which are

i l t t j l l b (J/C)equivalent to joules per coulomb (J/C).

“downhill: resistor”“uphill: battery”

Power and Energy

)()()(

gy

= )()()( titvtp Watts

∫=2

)(t

dttpw Joules∫=1

)(t

dttpw Joules

Current is flowing in the passivethe passive

configuration

If the current flows opposite to the passive configuration, the power is given by p = -vi

Pa = iava = (2A)*(12V) = 24W, energy is being absorbed

P i (1A)*(12V) 12W i b i li dPb = -ibvb = -(1A)*(12V) = -12W, energy is being supplied

Pc = icvc = (-3A)*(12V) = -36W, energy is being supplied

Power and Energygy

v(t) = 12 Vv(t) 12 Vi(t) = 2e-t Ap(t) = v(t)i(t) = 24e-tp(t) v(t)i(t) 24e

∞ ∞

Jeedtedttpw t 24)24(2424)( 0 =−−−=== ∫ ∫∞ ∞

∞−−

0 0

?

• An ideal voltage source has a voltage v =12 V• An ideal voltage source has a voltage vx =12 V independent of the load

• An ideal conductor requires that v = 0An ideal conductor requires that vx 0

Resistors and Ohm’s LawResistors and Ohm’s Lawa

iRv =a

Riv Riv abab =b

The units of resistance are Volts/Amp which are called “ohms” The symbol for ohms is omega: Ωohms . The symbol for ohms is omega: Ω

George Simon Ohm, 1789-1854g

VIR

I =

In 1805 Ohm entered the University of Erlangen but he became rather carried away with student life. Rather than concentrate on his studies he spent much time dancing, ice skating and playing billiards.p g, g p y g

http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Ohm.html

ConductanceConductance

G =1R

G =

Gvi =The units of conductance are 1/Ω or Ω-1. The units

are called “siemens”are called siemens

Resistance Related to Physical Parameters

LR ρA

R ρ=

ρ is the resistivity of the material used to fabricateρ is the resistivity of the material used to fabricate the resistor. The units of resitivity are ohm-meters

(Ω-m)(Ω m)

Power dissipation in a resistor

v2

RvRivip 2 ===R

p

KIRCHHOFF’S CURRENTKIRCHHOFF S CURRENT LAW

• The net current entering a node is zero.g

• Alternatively, the sum of the currents entering a node equals the sum of the currents leaving a node.

Gustav Kirchhoff (1824-1887)

Kirchhoff's Current LawThe principle of conservation of electric charge implies that:

At any point in an electrical circuit where charge density is not changing in time, the sum of currents flowing towards that point is equal to the sum of currents flowing away from that point.

A charge density changing in time would mean the accumulation of a net positive or negative charge, which typically cannot happen to any significant degree because of the strength ofsignificant degree because of the strength of electrostatic forces: the charge buildup would cause repulsive forces to disperse the charges.

http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Kirchhoff.html

(a) Currents into the node = 1A+3A = 4A current out of the node ia

(b) C i h d 3A 1A i f h d(b) Currents into the node = 3A+1A+ib = current out of the node = 2A so ib=2A-4A=-2A

(c) Currents into the node = 1A+i +3A+4A = current out of the(c) Currents into the node = 1A+ic+3A+4A = current out of the node= 0 amps so ic =-8A

Series Connection

Which elements are in series?

KIRCHHOFF’S VOLTAGEKIRCHHOFF S VOLTAGE LAW

Th l b i f h l lThe algebraic sum of the voltages equals zero for any closed path (loop) in an l i l i ielectrical circuit.

Loop 1: -va+vb+vc = 0Loop 1: va vb vc 0

Loop 2: -vc-vd+ve = 0

Loop 3: v v +v v 0Loop 3: va-ve+vd-vb = 0

Parallel Connection

KVL through A and B: -va+vb = 0 → va = vb

KVL through A and C: -va - vc = 0 → va = -vc

-3V - 5V + vc = 0 → vc = 8Vc c

-8V - (-10V) + ve = 0 → ve = -2V

Which elements are in series? Which elements are in parallel?Which elements are in series? Which elements are in parallel?

Find the current, voltage and power for each element:

Power

P = i v = (2A)(10V) = 20W (power dissipated)PR iRvR (2A)(10V) 20W (power dissipated)= iR

2R2/= VR2/R

PS = -vSiS = -(10V)(2A) = -20W (power supplied)

PR+ PS = 0PR+ PS 0

Example

• What is the current iR flowing through the resistor?R g g

• What is the power for each element in the circuit?

• Which elements are absorbing power?• Which elements are absorbing power?

• Since all of the elements are in series, the same current iR= 2A runs through each of themruns through each of them

• The voltage drop across the resistor: vR= iR = (2A)(5Ω) = 10VR ( )( )

• Apply KVL:vc = vR + 10 = 20V

Th di i d i h l• The power dissipated in each element:pR = iR

2R = (2A)2(5Ω) = 20W (absorbing)pvs= iv = (2A)(10V) = 20W (absorbing)pvs iv (2A)(10V) 20W (absorbing)pcs= -iv = -(2A)(20V) = 40W (supplying)

Example

Use Ohm’s law, KVL and KCL to find Vx

R

IT

IR2 IR3

R3R2

R1

R4

R2 R3

Oh ’ LOhm’s Law:

VR4 = (1A)(5Ω) = 5V = VR2 = VR3

IR2 = 5V/10Ω = 0.5A

IR3 = 5V/5Ω = 1A

KCL: IT = IR2 + IR3 + IR4 = 0.5A + 1A + 1A = 2.5A

Ohm’s Law: VR1= ITR1 = (2.5A)(5Ω) = 12.5VR1 T 1 ( )( )

KVL: Vx = VR1 + VR4 = 12.5V+5V = 17.5V