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I. The Lead Acid Electric Battery. Sulfuric Acid Electrolyte:. -. +. Terminals. Oxidation at the Negative Plate (Electrode:Anode):. Sulfuric Acid Solution H 2 SO 4. Spongy Lead (Pb). Lead Oxide (PbO 2 ). Reduction at the Positive Plate (Electrode:Cathode):. Cell: 2 V - PowerPoint PPT Presentation
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The Lead Acid Electric Battery
+-
Spongy Lead (Pb)
Lead Oxide (PbO2)
Sulfuric Acid SolutionH2SO4
22 4 2 3 4H SO 2H O 2H O SO
Sulfuric Acid Electrolyte:
Oxidation at the Negative Plate (Electrode:Anode):
24 4Pb SO PbSO 2e
Reduction at the Positive Plate (Electrode:Cathode):
22 3 4 4 2PbO 4H O SO 2e PbSO 6H O
I
Terminals
Cell: 2 VBattery: Multiple cells
Batteries
abV Ir
Electromotive Force
r Batterey internal resistance
Kirchoff’s Rules
• Conservation of charge • Junction (Node) Rule: At any junction point, the
sum of all currents entering the junction must equal the sum of the currents leaving the junction.
• Conservation of energy • Loop Rule: The some of the changes in potential
around any closed path of a circuit must be zero.
Energy in a circuit
Series Circuit
+acV
ac ab bc 1 2 1 2 eqV V V IR IR I R R IR
Apply the Loop Rule
eq 1 2R R R .....
ac ab bcV V V 0
Parallel Circuits
+
V
1I 2I 3I
Apply the Junction Rule
I
1 2 31 2 3 1 2 3 eq
V V V 1 1 1 VI I I I VR R R R R R R
eq 1 2 3
1 1 1 1 ....R R R R
Rule Set – Problem Solving Strategy• A resistor transversed in the direction of assumed
current is a negative voltage (potential drop)• A resistors transversed in the opposite direction of
assumed current is a positive voltage (potential rise)• A battery transversed from – to + is a positive voltage.• A battery transversed from + to - is a negative voltage.• Ohm’s Law applies for resistors.• Both the loop rule and junction rule are normally
required to solve problems.
More about the Loop Rule
• Traveling around the loop from a to b
• In (a), the resistor is traversed in the direction of the current, the potential across the resistor is – IR
• In (b), the resistor is traversed in the direction opposite of the current, the potential across the resistor is is + IR
Loop Rule, final• In (c), the source of emf is
traversed in the direction of the emf (from – to +), and the change in the electric potential is +ε
• In (d), the source of emf is traversed in the direction opposite of the emf (from + to -), and the change in the electric potential is -ε
Example Problem 1
1R 1690
3R 1000
4R 3000
Given:
Find: current in each resistor
V = 3 Volts
Example Problem 2
10V 20V
5 10
20
Given:
Find: current in the 20 resistor
Alternating Current oV t V
oV t V sin t
oo
V t VI(t) sin t I sin tR R
oI t I
AC Power
o
2 2 2P I R I R sin t
22 oo
V1 1P I R2 2 R
T / 2
2
T / 2
1 1sin t dtT 2
?
Root Mean Square (rms)
oV t V sin t oI(t) I sin t
22 oVV
2
22 o o
rmsV VV V2 2
22 oII
2
22 o o
rmsI II I2 2
2 2o rms
1P I R I R2
2 2o rmsV V1P
2 R R
The Wheatstone bridgea simple Ohmmeter
3 3 1 1I R I R
3 x 1 2I R I R
2x 3
1
RR RR
Charging a capacitor in an RC circuit
At t = 0, Qo = 0 and oIR
SameSymbol
Solving the charging differential equation
QIR 0C
Kirchoff’s loop rule
Convert to a simple equation in Current by taking the first
derivative w.r.t. time
dI 1 dQR 0dt C dt
dI 1R Idt C
dI 1 dtI RC
Separate variables
Integrate the results
o
I t t
I 0
dI 1 dtI RC
otln I t ln I
RC
tRC
o
I te
I
t
RCoI t I e
o
I t tlnI RC
Charge buildup
t
RCo
dQI t I edt
tRC
odQ I e dt
Q t t tRC
o0 0
dQ I e dt
tt t
RC RCo o
0
Q t I RC e I RC 1 e
Discharging the capacitor in an RC circuit
At t = 0, Q = Qo
Solving the discharging differential equation
Q IR 0C
Kirchoff’s loop rule
dQ 1R Qdt C
dQI since charge is decreasingdt
dQ 1 dtQ RC
o
Q t t
Q 0
dQ 1 dtQ RC
Separate variables
Integrate
Charge and current decay
otln Q t ln Q
RC
o
Q t tlnQ RC
tRC
o
Q te
Q
t
RCoQ t Q e
Charge and current decay
tRC
odQ dI Q edt dt
to RCQI e
RC
tRC
oI I e
Electrical Safety
• Current kills, not voltage (70 mA)• Normal body resistance = 105
But could be less than 1000 • Take advantage of insulators, remove
conductors• Work with one hand at a time• Shipboard is more dangerous• Electrical safety is an officer responsibility