The Lead Acid Electric Battery
I
Sulfuric Acid Electrolyte:
-
Terminals
+
Sulfuric
Acid
Solution
H2SO4
Spongy
Lead (Pb)
Lead Oxide
(PbO2)
Cell: 2 V
Battery: Multiple cells
H 2SO 4  2H 2O
2H 3O   SO 42
Oxidation at the Negative
Plate (Electrode:Anode):
Pb  SO42  PbSO4  2e
Reduction at the Positive
Plate (Electrode:Cathode):
PbO2  4H3O   SO 42  2e   PbSO 4  6H 2O
Batteries
Vab    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
Apply the Loop Rule
Vac  Vab  Vbc  0
+
Vac
Vac  Vab  Vbc  IR1  IR 2  I  R1  R 2   IR eq
R eq  R1  R 2  .....
Parallel Circuits
I
I1
I3
+
Apply the
Junction Rule
I2
V
 1
V V V
1
1  V
I  I1  I2  I3 


 V 


R1 R 2 R 3
 R1 R 2 R 3  R eq
1
1
1
1



 ....
R eq R1 R 2 R 3
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
Given:
R1  1690
R 3  1000
R 4  3000
V = 3 Volts
Find: current in each resistor
Example Problem 2
Given:
10V
5
10
20
20V
Find: current in the 20  resistor
Alternating Current
V  t   Vo
I  t   Io
V  t   Vo sin t
V  t  Vo
I(t) 

sin t  Io sin t
R
R
AC Power
P  I2 R  I2o R sin 2 t
1 2
1 Vo2
P  Io R 
2
2 R
T/2
1
1
2
sin  t dt  ?

T T / 2
2
Root Mean Square (rms)
V  t   Vo sin t
I(t)  Io sin t
2
V
V2  o
2
Vrms
2
I
I2  o
2
2
V
Vo
2
o
 V 

2
2
I rms 
1 2
2
P  I o R  I rms R
2
2
1 Vo2 Vrms
P

2 R
R
2
I
Io
2
o
I 

2
2
The Wheatstone bridge
a simple Ohmmeter
I3R 3  I1R1
I3R x  I1R 2
R2
Rx 
R3
R1
Charging a capacitor in an RC circuit
Same
Symbol

At t = 0, Qo = 0 and I o 
R
Solving the charging differential
equation
Kirchoff’s loop rule
Q
  IR   0
C
Convert to a simple equation
in Current by taking the first
derivative w.r.t. time
dI 1 dQ
R 
0
dt C dt
dI
1
R  I
dt
C
Separate variables
dI
1

dt
I
RC
Integrate the results
I t 

Io
t
dI
1
 
dt
I
RC
0
t
ln  I  t    ln  I o   
RC
It
t
ln 

RC
 Io 
t

It
 e RC
Io
I  t   Io e

t
RC
Charge buildup
dQ
It 
 Io e
dt
dQ  Io e
Q t 

0

t
RC
t
dQ  Io  e

t
RC
dt

t
RC
dt
0

Q  t   Io  RC  e

t

RC
t
t




RC

I
RC
1

e



o
0


Discharging the capacitor in an RC
circuit
At t = 0, Q = Qo
Solving the discharging differential
equation
Q
 IR  0
C
dQ 1
R
 Q
dt C
Kirchoff’s loop rule
dQ
I
since charge is decreasing
dt
dQ
1

dt
Q
RC
Separate variables
Q t 
Integrate

Qo
t
dQ
1
 
dt
Q
RC
0
Charge and current decay
t
ln Q  t    ln  Q o   
RC
Qt
t
ln 

RC
 Qo 
t

Qt
 e RC
Qo
Q  t   Qo e

t
RC
Charge and current decay
dQ
d
I
  Qo e
dt
dt
Qo  RCt
I
e
RC
I  Io e

t
RC

t
RC
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