Lecture 10-Lecture 10-11 Capacitor Examples

1 0 2 01 2

0 1 2

/ 2 / 2

2

A AC C C

d dA

d

2C

C

CC/2

C C

1 2 0 0

0

1 1 1 / 4 3 / 4

1 3 1

4 4

d d

C C C A A

d

A

d/4

3d/4

Lecture 10-Lecture 10-22 Electric Current

Current = charges in motion

0limx

q dqI

t dt

Magnitude

rate at which net positive charges move across a cross sectional surface

Units: [I] = C/s = A (ampere)

Current is a scalar, signed quantity, whose

sign corresponds to the direction of motion of net positive charges by convention

J = current density (vector) in A/m²

A

AdJI

Lecture 10-Lecture 10-33Microscopic View of Electric Current in Conductor

All charges move with some velocity ve

random motion with high speeds (O(106)m/s) but with a drift in a certain direction on average if E is present

Drift velocity vd is orders of magnitudes less than the actual velocity of charges,

A

• thermal energy

• scattering off each other, defects, ions, …

Lecture 10-Lecture 10-44Current and Drift Velocity in Conductor

dI nAv q where n =carrier density

d

Iv

nAq

Drift velocity vd is orders of magnitudes less than the actual velocity of charges.

In the following condition:I = 1.0A, copper: n ~1029atoms/m3

1mm radius wire.

Vd~0.01mm/s

Lecture 10-Lecture 10-55Ohm’s Law

Current-Potential (I-V) characteristic of a device may or may not obey Ohm’s Law:

or V IR with R constant

I V

Resistance V V

RI A

(ohms)

gas in fluorescent tubetungsten wire diode

Lecture 10-Lecture 10-66Resistance and Resitivity for Ohmic Material

J E

L

AI L

V EL L IA A

E J

resistivity

R (in )resistance

I VJI EV

Lecture 10-Lecture 10-77Resistance

constant R Ohm’s Law

Resistance (definition)

VR

I

V

R

I

LR

A

Lecture 10-Lecture 10-88Warm up

There are 2x1014 electrons across a resistor in 10 seconds. What is the current through the resistor?

a) 3.2Ab) 1.6 Ac) 3.2 Ad) 1.6Ae) 3.2 mA

Note: e = 1.6x10-19 C

V

R

I

Lecture 10-Lecture 10-99Temperature Dependence of Resistivity

Material 0 (m) (K-1)

Ag 1.6x10-8 3.8x10-3

Cu 1.7x10-8 3.9x10-3

Si 6.4x102 -7.5x10-2

glass 1010 ~ 1014

sulfur 1015

0 01 ( )T T

Copper

• Usually T0 is 293K (room temp.)

• Usually > 0 (ρ increases as T )

Lecture 10-Lecture 10-1010

• Free electrons in a conductor gains kinetic energy due to an externally applied E.

Electric Current and Joule Heating

• Scattering from the atomic ions of the metal and other electrons quickly leads to a steady state with a constant current I.

Transfers energy to the atoms of the solid (to vibrate), i.e., Joule heating.

Lecture 10-Lecture 10-1111Energy in Electric Circuits

dUP IV

dt

So, Power dissipation = rate of decrease of U =

• Steady current means a constant amount of charge Q flows past any given cross section during time t, where I= Q / t.

Energy lost by Q is

( )a bU Q V V I t V

2 2 /I R V R

=> heat

baIa b I

∆Q ∆Q

Lecture 10-Lecture 10-1212 EMF – Electromotive Force

• An EMF device is a charge pump that can maintain a potential difference across two terminals by doing work on the charges when necessary.

Examples: battery, fuel cell, electric generator, solar cell, fuel cell, thermopile, …

• Converts energy (chemical, mechanical, solar, thermal, …) into electrical energy.

Within the EMF device, positive charges are lifted from lower to higher potential.

If work dW is required to lift charge dq,

dWVolt

dq EMF

Lecture 10-Lecture 10-1313Internal Resistance of a Battery

0i r i R , b ai V V irR r

R

R r

internal resistance terminal

voltage

load

Lecture 10-Lecture 10-1414Energy Conservation

A circuit consists of an ideal battery (B) with emf , a resistor R, and two connecting wires of negligible resistance.

Energy conservation

Work done by battery is equal to energy dissipated in resistor

2 2dW i Rdt or i dt i R d

iR

t

• Ideal battery: no internal energy dissipation

• Real battery: internal energy dissipation exists

dW > i2Rdt or > iR=V

Lecture 10-Lecture 10-1515Lecture quiz A

There are1014 electrons across a resistor with potential drop of 3.2V in 10 seconds. What is the resistance of the resistor?

a) 2.0 Ωb) 1.0 Ωc) 2.5 Ωd) 3.0 Ωe) 4.0 Ω

Note: e = 1.6x10-19 C

V

R

I

Lecture 10-Lecture 10-1616Lecture quiz B

There are 1014 electrons across a resistor of resistance 1.0Ω in 10 seconds. What is the potential drop across the resistor?

a)3.2 mVb)8.0 Vc)2.5 Vd)1.6 Ve)1.6mV

Note: e = 1.6x10-19 C

V

R

I

Lecture 10-Lecture 10-1717Lecture quiz C

The potential drop is 6.4V across a resistor of resistance 1.0Ω. How many electrons enter the wire in 10 seconds?

a)3.2×1019

b)8.0×1015

c)2.5×1012

d)4.0x1014

e)1.6×1019

Note: e = 1.6x10-19 C

V

R

I