Current, Resistance and Power

Battery

+-Negative Electrode

Positive Electrode

electrolyte

Battery

+-Negative Electrode

Positive Electrode

electrolyte

Current

Simple flow of charge

-

I

dt

dQI

(Note Convention)

MKS unit - Ampere

Current

See Active Figure 27.09

Current

+

++

++

+

n - concentration of charges per unit volume

vd – drift velocity

A – cross sectional area of conducting wire

q – charges carried by each particle

A

Charge carriers

+

++

++

+

tvl d

A

After some time t, the particles will pass beyond a particular point on the wire

The volume contains the passing charges can be found with vd, A and t

tAvAlVol d

+

++

++

+

tvl d

A

Volume of passing charges

qVolnQ

NqQ

)(

+

++

++

+

qnAvt

QI

tqnAvQ

d

d

Current Density and Ohm’s Law

A

IJ

Current Density

Current per unit Area

Current Density and Ohm’s Law

EJ

Ohm’s Law

- conductivity

AdJI

A more familiar form

A

lIV

l

VE

A

I

EJ

1

Resistivity

IRV

A

lR

where

Resistance

Volt/amp= MKS Unit

Ohm’s Law: Macroscopic form

Note Dependencies

• If you double the area (ie. Adding an addition wire) the effective resistance halves

• If you add the wire to the length the effective resistance doubles

• The resistivity is an intrinsic property of the material the resistor is made of. If you change material keeping physical geometry the same, the resistance changes

Ohmic (or linear) device

V

V

Non-Ohmic (or nonlinear) device

Slope = 1/R

Microscopic View of Conductor

smmCmmatoms

Amps

nAq

Iv

qnAvt

QI

d

d

/37.0106.110/1048.8

51926328

e

d

m

en

Em

qEnqnqvJ

EJ

2

Independent of Electric Field: OhmicBut can depend on conditions which effect , such as temperature

copper Cross sectional area Electron Charge

See Active Figure 27.09

acceleration

Time between collisions

Resistivity vs. Temperature

(T) – characteristic of material

T

T

metallic semi-conductors

insulators

T

))(1( 00 TT

T0

0 – resistivity at T0

– thermal coefficient of resistivity

Superconductors

• Many materials will below a specific characteristic temperature, Tc, have a pronounced decrease in resistivity.

Power

Battery – “works” to push current through circuit

Powersource = VI

V – Potential Source

I – Current sent from source through circuit

V

I

Thermal energy dissipated through resistors

RI

Rate of Thermal Energy dissipation through Resistor

R

VRIIVP

dt

Vdq

dt

dU

dt

dWP

22

Voltage drop across resistor

Example Problem: Suppose we wanted to design a small heater for your to work before your car warmed up. We want 500Watts using the 12V of your car battery. How much Nichrome wire with a crossectional area of 0.1 cm2 do we need?

29.0

12500

2

22

RR

Vwatts

R

VRIIVP

cm

cm

cmA

R

290

1.0

10029.0

2

r

Battery (Source)

- actual potential difference between electrodes of battery (EMF)

r – internal resistance of battery

R

By attaching the battery to a circuit including a load resistor R, the current drawn through the battery will

effect the actual potential difference in the battery

r

Battery (Source)

I

Kirchoff’s Voltage Loop Theorem

• The algebraic sum of the changes in electric potential encountered in a complete traversal of the circuit must be zero.

• A circuit is closed path through which current (electrons) may be forced to move through circuit elements (resistors).

V = - Ir

r

I

R

I

Jumping from the negative to the positive end of the battery, the potential increases by , but after going

through the resistor, the potential drops by IR

Battery voltage terminal to terminal

r

I

R

I

)(

0

RrI

IRIr

To find the current…

Kirchoff’s Voltage Loop

Resistors in Series and Parallel:Equivalent Resistance

Resistors in Series

R2

R1

R3

I

I

0)(

0

321

321

RRRI

IRIRIR

Resistors in Series

Rseries

I

I

)(

0

321 RRRR

IR

series

series

Kirchoff’s Junction Theorem

• At any junction (point where current can split) the algebraic sum of the currents into and out of the wires of the junction must add to zero.

• By convention the current into a junction is positive and the current out of a junction is negative.

Resistors in Parallel

I

I

R3R2R1

Resistors in Parallel

I

I

R3R2R1

I2I1 I3

321 IIII

Resistors in Parallel

I

I

R3R2R1

I2I1 I3

1111 0R

IRI

Resistors in Parallel

I

I

R3R2R1

I2I1 I3

2222 0R

IRI

Resistors in Parallel

I

I

R3R2R1

I2I1 I3

3333 0R

IRI

Resistors in Parallel

I

I

Rparallel

I

parallelparallel R

IIR 0

Resistors in Parallel

321

321

321

1111

RRRR

RRRR

IIII

parallel

parallel

42 V

2

12

4

2 1

Solve for the currents going through each of the resistors by circuit reduction (equivalent resistance)

Break circuit down into series and parallel resistors

42 V

2

12

4

2 1

I1

I2I

Currents in the various branches

42 V

2

12

4

2 1

I1

I2I

Find equivalent resistance for the Series Resistors

42 V

12

6 3

I1

I2I

Find Equivalent Parallel Resistance

42 V

12

2

I

Find Equivalent Series Resistance

42 V14

I

Find Equivalent Series Resistance

42 V14

I

AI

I

3

01442

42 V

2

12

4

2 1

I1

I2I

AI

VII

III

2

612423

0211242

1

1

11

42 V

2

12

4

2 1

I1

I2I

AIII

III

112

21

0

0)2()6(

0)4()6(

321

311

2211

III

II

II

Kirchoff’s Analysis

Solve simultaneously for the unknown currents