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 V

R 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 t

R R

oI t I

AC Power

o

2 2 2P I R I R sin t

22 oo

V1 1P I R

2 2 R

T / 2

2

T / 2

1 1sin t dt

T 2

?

Root Mean Square (rms)

oV t V sin t oI(t) I sin t

22 oV

V2

22 o o

rms

V VV V

2 2

22 oI

I2

22 o o

rms

I II I

2 2

2 2o rms

1P I R I R

2

2 2o rmsV V1

P2 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 R

R

Charging a capacitor in an RC circuit

At t = 0, Qo = 0 and oIR

SameSymbol

Solving the charging differential equation

QIR 0

C Kirchoff’s loop rule

Convert to a simple equation in Current by taking the first

derivative w.r.t. time

dI 1 dQR 0

dt C dt

dI 1R I

dt C

dI 1dt

I RCSeparate variables

Integrate the results

o

I t t

I 0

dI 1dt

I RC

o

tln I t ln I

RC

t

RC

o

I te

I

t

RCoI t I e

o

I t tln

I RC

Charge buildup

t

RCo

dQI t I e

dt

t

RCodQ I e dt

Q t t t

RCo

0 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

QIR 0

C Kirchoff’s loop rule

dQ 1R Q

dt C dQ

I since charge is decreasingdt

dQ 1dt

Q RC

o

Q t t

Q 0

dQ 1dt

Q RC

Separate variables

Integrate

Charge and current decay

o

tln Q t ln Q

RC

o

Q t tln

Q RC

t

RC

o

Q te

Q

t

RCoQ t Q e

Charge and current decay

t

RCo

dQ dI Q e

dt dt

to RCQ

I eRC

t

RCoI 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