Transcript Heat Flow

Conduction and Current
• Polarization vs. Conduction
• Batteries, Current, Resistance
• Ohm’s Law and Examples
• Resistivity and Examples
• Power and Examples
Electrical Properties of Materials
•
Materials can do 2 things:
• Store charge
– Initial alignment of charge
with applied voltage
– Charge proportional to
voltage
– Temporary short-range
alignment
• Conduct charge
– Continuous flow of charge
with applied voltage
– Current proportional to
voltage
– Continuous long-range
movement
Charge Storage vs Conduction
• Storage
• Conduction
• Q = CV
• V= IR (I=GV G=1/R)
• Charge in Coulombs
• Charge flow in
Coulombs/second (amps)
• Energy stored in Joules
• Power created or
expended in Watts
Batteries
• Battery
– Electrochemical
– Source of voltage
– Positive and negative
– 1.5 volt, 3 volt, 9 volt, 12 volt
– Circuit symbol
Current
• Current
–
–
–
–
–
–
Coulombs/second = amps
I = ΔQ / Δt
Example 18-1
Requires complete circuit
Circuit diagram
Positive vs. negative flow
Resistance
• Resist flow of current (regulate)
• Atomic scale collisions dissipate energy
• Energy appears in other forms (heat, light)
• Applications
– Characterize appliance behavior
• Heater (collisions cause heat)
– Regulate current/voltage on circuit board
• Resistors and color code
Resistance and Ohm’s Law
• Storage vs. Conduction
–
–
–
–
Q = CV (storage)
I = GV (conduction)
Current proportional to voltage
Proportionality is conductance
• Use inverse relation
– V = IR
– Resistance
– Units volts/amps = ohms
• Ohm’s law
– If V proportional to I, ohmic
– Otherwise non-ohmic
• Example 18-3
Ohm’s Law Examples
𝐼=
𝑉
𝑅
=
240 𝑉
9.6 Ω
= 25 𝐴
25 𝐴 = 25 𝐶 𝑆
𝐼=
9𝑉
1.6 Ω
= 5.625 𝐴
⋕ 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠 =
(25 𝐶 𝑠)(60 𝑠 min)(50 min) = 75,000 𝐶
(5.625 𝐶 𝑠) 60𝑠 = 337.5 𝐶
337.5 𝐶
1.6∙10−19 𝐶 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛
= 2.11 ∙ 1021
Ohm’s Law Examples
𝑅 = 2.5 ∙ 10−5 Ω 𝑚 .04 𝑚 = 10−6 Ω
𝑉 = 𝐼𝑅 = 2800 𝐴 ∙ 10−6 𝑉 𝐴 = 2.8 𝑚𝑉
OK, but do not touch the other wire!
(heat of vaporization of squirrel)
Resistivity vs. Resistance
• Property of material vs. property of device
• Similar to dielectric constant vs. capacitance
• Becomes resistance vs. resistivity
Storage
𝐾𝜀𝑜 𝐴
𝐶=
𝑑
Conduction
𝜌𝑑
𝑅=
𝐴
𝜎𝐴
𝐺=
𝑑
• We use reciprocals, resistance and resistivity
• ρ published for materials, like K.
• High ρ poor conductor, σ good conductor (similar to K for storage)
Resistivity of materials
Resistivity Example
• Area of wire from resistivity and length
𝑅=
𝜌𝑑
𝐴
𝐴=
𝑝𝑑
𝑅
=
(1.68 ∙10−8 Ω 𝑚)(20 𝑚)
0.1Ω
• Radius of wire
𝐴 = 𝜋𝑟 2
𝑟=
𝐴
𝜋
= 1.04 𝑚𝑚
• Voltage Drop along wire
𝑉 = 𝐼𝑅 = 4𝐴 ∙ 0.1 Ω = 0.4 𝑉
= 3.4 ∙ 10−6 𝑚2
Resistivity examples
𝑅𝑎𝑙 =
𝜌𝑎𝑙 𝑑
𝐴
𝑅𝑐𝑢 =
𝜌𝑐𝑢 𝑑
𝐴
𝑅=
𝜌𝑑
𝐴
=
=
=
2.65∙10−8 Ω 𝑚 (10 𝑚)
3.14∙10−6 𝑚2
1.68∙10−8 Ω 𝑚 (20 𝑚)
4.91∙10−6 𝑚2
1.68∙10−8 Ω 𝑚 (26 𝑚)
2.08∙10−6 𝑚2
= 0.084 Ω
= 0.068 Ω
= 0. 21Ω
𝑉 = 𝐼𝑅 = 12 𝐴 ∙ 0.21 Ω = 2.52 𝑉
Resistivity example
• Along x
𝑅=
𝜌𝑑
𝐴
=
3.0∙10−8 Ω 𝑚 (.01 𝑚)
.02 𝑚 (.04 𝑚)
= 3.8 ∙ 10−4 Ω
• Along y
𝑅=
𝜌𝑑
𝐴
=
3.0∙10−8 Ω 𝑚 (.02 𝑚)
.01 𝑚 (.04 𝑚)
= 1.5 ∙ 10−3 Ω
• Along z
𝑅=
𝜌𝑑
𝐴
=
3.0∙10−8 Ω 𝑚 (.04 𝑚)
.01 𝑚 (.02 𝑚)
= 6.0 ∙ 10−4 Ω
Power
• Work done/ loss of PE for (+) going with field
𝑊 = −∆𝑃𝐸 = 𝑄𝑉
• No ½ because voltage is constant
𝑃𝑜𝑤𝑒𝑟 =
∆𝑃𝐸
∆𝑡
=
𝑞𝑉
∆𝑡
=
• Alternate forms
P=IV = I2R = V2/R
𝑞
𝑡𝑖𝑚𝑒
+
V
𝑉 = 𝐼𝑉
0
Power example
• Calculate current
𝑃 = 𝐼𝑉
𝐼=
𝑃
𝑉
=
40 𝑉 𝐴
12 𝑉
= 3.33 𝐴
• Calculate Resistance
R=
𝑉
𝐼
=
12 𝑉
3.33 𝐴
= 3.6 Ω
• In one step
• 𝑃=
𝑉2
𝑅
𝑅=
𝑉2
𝑃
=
12 𝑉 2
40 𝑉 𝐴
= 3.6Ω
Power example
• Power
𝑃 = 𝐼𝑉 = 15 𝐴 ∙ 120 𝑉 = 1800 𝑊 = 1.8 𝑘𝑊
• Electric Company charges for energy not power
𝐸𝑛𝑒𝑟𝑔𝑦 = 𝑝𝑜𝑤𝑒𝑟 ∙ 𝑡𝑖𝑚𝑒
• But instead of using Joules, they use kW-hours
$ = 90 ℎ 1.8 𝑘𝑊 (.092 $ 𝑘𝑊 ∙ ℎ) = $15
Power examples
• Problems 31, 32, 33, 38,
Power transmission
• All customer cares about to run his home or factory: 𝑃 = 𝑉𝐼 = 620 𝑘𝑊
– Can do low voltage at high current
– High voltage at low current
– Transformers can switch back and forth
Power transmission – 12,000 V
• At 12,000 V current must be
𝐼=
620 𝑘𝑊
12 𝑘𝑉
= 51.67 𝐴
• Voltage drop along wire will be
𝑉 = 𝐼𝑅 = 51.67 𝐴 ∙ 3Ω = 155 𝑉
• Power wasted in wire will be
P = 𝐼𝑉 = 51.67 𝐴 ∙ 155 𝑉 = 8 𝑘𝑊
Power transmission – 50,000 V
• At 50,000 V current must be
𝐼=
620 𝑘𝑊
50 𝑘𝑉
= 12.4 𝐴
• Voltage drop along wire will be
𝑉 = 𝐼𝑅 = 12.4 𝐴 ∙ 3Ω = 37.2 𝑉
• Power wasted in wire will be
P = 𝐼𝑉 = 12.4 𝐴 ∙ 37.2 𝑉 = 0.46 𝑘𝑊
Power transmission – 2,000 V
• At 2,000 V current must be
𝐼=
620 𝑘𝑊
2 𝑘𝑉
= 310 𝐴
• Voltage drop along wire will be
𝑉 = 𝐼𝑅 = 310 𝐴 ∙ 3Ω = 930 𝑉
• Power wasted in wire will be
P = 𝐼𝑉 = 310 𝐴 ∙ 930 𝑉 = 288 𝑘𝑊
<< Half the voltage and
power are wasted!