Announcements First lab this afternoon SHL 016 –Introduction to EWB Tuesday: Matthew Bihler Gaurav Pandey Philip Zandona Trisha Wednesday: Joseph Brosch.

Announcements

• First lab this afternoon SHL 016– Introduction to EWB

Tuesday:Matthew BihlerGaurav PandeyPhilip ZandonaTrisha

Wednesday:Joseph BroschErin GraceAli JafriLei ChenHalise CelikFatihJingliang ZhangEric

http://www.physics.udel.edu/~jholder/Phys645/index.htm

Lecture 2 Overview

• Kirchoff’s Laws• Power and Resistance• Practical sources• Voltage/ Current Dividers • DC circuit analysis

– Element combination– Using Kirchoff’s Voltage Law– Using Kirchoff’s Current Law– Mesh Analysis

Gustav Robert Kirchhoff (12 March 1824 – 17 October 1887)

Kirchhoff's three laws of spectroscopy

1. A hot solid object produces light with a continuous spectrum. 2. A hot tenuous gas produces light with spectral lines at discrete wavelengths (i.e. specific colors) which depend on the energy levels of the atoms in the gas. (emission spectrum) 3. A hot solid object surrounded by a cool tenuous gas (i.e. cooler than the hot object) produces light with an almost continuous spectrum which has gaps at discrete wavelengths depending on the energy levels of the atoms in the gas. (absorption spectrum)

Kirchoff’s Current Law

The sum of the current at any node must equal zero: i.e., the current flowing into a node must equal the current flowing out of the node

At Node 1: -i+i1+i2+i3=0 i=i1+i2+i3

Conservation of Charge(Current = rate of change of charge)

Note convention: current flows from positive terminalIn order for current to flow, there must exist a closed circuit

Kirchoff’s Voltage Law

The sum of the voltages around a closed loop is zero

Around the loop: -v1+v2=0 v1=v2

Potential: at a=va

b=vb

Potential difference: v2=va-vb

Note: potential measured relative to ground: true ground (earth)or chassis ground (enclosure)

Circuit elements and their i-v characteristics: Resistor

Ohm’s law: Current is proportional to applied Voltage, and inversely proportional to the Resistance: V=IR

1

.2

m

mm

A

LR

R = resistance: depends on materials and geometry ρ = resistivity: depends only on materials σ = conductivity

v

Resistor Color codehttp://www.dannyg.com/examples/res2/resistor.htm

560KΩ ± 5%560KΩ ± 5%

Electric Power

Electric power = amount of work done/unit timeVoltage V = work/unit chargeSo, to move charge Q, work done=VQPower = VQ/t =VI

P=IV =I2R =V2/R

Units = Joules/sec = Watts (W)What do we pay for in the electricity bill?kWh=energy/time × time = energy

More on Power and ResistanceWhat do the power ratings of appliances mean?e.g. what does a 1000W hair dryer tell us?• Assume 120V (USA) if voltage not specified.

Never exceed the rated powerCan you use this appliance on a 240V (UK) line?

4.141000

120

A33.8120

1000

222

P

VR

R

VP

V

PiiVP Max current

W40004.14

24022

R

VP Destroys the appliance!

Resistance Limits: Open and Short Circuits

• Short Circuit: A wire! R=0, V=0 for any i.– Particularly bad for any voltage source

• Open circuit: A break! R→, i=0 for any V.– Particularly bad for current source

Series Resistors and the Voltage Divider Rule

For N resistors in series:

V)5.1(22

EQR

Rv Voltage Divider:

N

nnN

N

nnEQ vviiiiRR

121

1

,...,

REQ>(R1,R2,…..,RN)

V)5.1(11

EQR

Rv

V)5.1(33

EQR

Rv

R3

Practical Voltage Sources

Modelled with an ideal source and a series resistor

Ideal voltage source: rS=0Current is the same at all points

So for a practical voltage source, the output voltage depends upon RL

If rS<< RL , vL = vS , independent of RL

LS

S

EQ

SS Rr

v

R

vi

SLS

LLSL v

Rr

RRiv

voltage divider

Parallel Resistors and the Current Divider Rule

For N resistors in parallel:

Current Divider:

N

nnSN

N

n nEQ

iivvvRR 1

11

,.....,11

REQ<(R1,R2,…..,RN)

SEQ i

R

R

R

vi

111 S

EQ iR

R

R

vi

222 S

EQ iR

R

R

vi

333

Large current through smaller R.Advantage of a parallel circuit; a broken branch will not affect other branches

Practical Current SourcesModelled with an ideal source and a parallel resistor

Ideal current source: rS=

The output current now depends upon RL

If rS>> RL , iL = iS , independent of RL

SL

EQL i

R

Ri

LS

LS

LS

EQ Rr

Rr

Rr

R

11

1

SLS

SL i

Rr

ri

current divider

Circuit analysis method 1:Apply element combination rules

Series resistors

Parallel resistors

Series voltage sources

Parallel current sources

Circuit analysis method 1: element combination

R1=10ΩR2=20ΩR3=30ΩV=10VFind the equivalent resistance and the current at I

R=22Ω

I= 10V/22Ω = 0.45A

Circuit analysis method 2a: KVL and KCL

Kirchoff’s Voltage LawThe sum of the voltages around a closed loop must be zero

• Draw the current direction (arbitrary) and label the voltage directions (determined by the defined current direction). Voltage on a voltage source is always from positive to negative end.

• Define either clockwise or counter-clockwise as positive direction for summing voltages. Once the direction is defined, use the same convention in every loop.Voltage across a resistor is +’ve if voltage direction the same as current direction, -’ve otherwise

• Apply KVL

Kirchoff’s Voltage Law: Multiloop

The sum of the voltages around a closed loop must be zero

• Draw the current direction (arbitrary) and label the voltage directions (determined by the defined current direction).

• Define either clockwise or counter-clockwise as positive voltage direction. Once the direction is defined, use the same convention in every loop.

• Apply KVL

R3

0

0

32

21

VV

VVVr03322

221

RIRI

RIIRIr

Kirchoff’s Current LawThe sum of the current at a node must be zero: Iin=Iout

R3

I=I2+I3 (1)

ε=Ir+IR1+I2R2 (2)

I3R3-I2R2 =0 (3)

I- I2- I3 = 0 (4)

4I+5I2+ 0I3= 3 (5)

Set r=1Ω, R1=3Ω, R2=5Ω, R3=10Ω, ε=3V

0I- 5I2+10I3= 0 (6)

Last note on KCL KVL analysis

• If solutions to currents or voltages are negative, this means the real direction is opposite to what you originally defined

• To deal with current sources: current is known, but assign a voltage across it which has to be solved

Sample Problem