1. ELECTRICAL ENERGY AND POWER
WHEN CURRENT FLOWS THROUGH A CONDUCTOR, HEAT IS PRODUCED. IT DEPENDS ON
RESISTANCE OF THE CONDUCTOR
CURRENT FLOWING THROUGH THE CONDUCTOR
TIME OF FLOW OF CURRENT
SUPPOSE THERE ARE ‘Q’ CHARGES FLOWING IN A CIRCUIT WHOSE POTENTIAL DIFFERENCE IS ‘V’, THEN WORK DONE = QV
WE KNOW I = Q/t; current is the rate of flow of charge
THEREFORE WORK DONE = V x I x T ( REMEMBER THE UNIT OF WORK IS JOULES)
WE ALSO KNOW THAT FROM OHM’S LAW I = V/R
SUBSTITUTING
ELECTRIC WORK = V x V x t R
WE CAN ALSO WRITE ELECTRIC WORK = I x I x R x t

2. POWER = I x I x R POWER = V x I POWER = V x V R
ELECTRIC POWER: IT IS THE RATE OF DOING ELECTRIC WORK
(i.e) POWER (P) = W/t; DIVIDE BY t.
POWER = I x I x R
POWER = V x I
POWER = V x V R
UNIT OF POWER IS IN WATTS
ALSO REMEMBER 1WATT = 1J/s

3. HOW TO ATTEMPT NUMERICAL PROBLEMS?
TYPE 1 PROBLEM ---- STRAIGHT APPLICATION OF THE FORMULAE
REMEMBER THE FORMULAE AND APPLY
WRITE WHAT IS GIVEN AS PER THE SYMBOLS AND THE CORRECT ANSWER WITH UNITS
TYPE 2 PROBLEM EQUATING ELECTRICAL AND HEAT ENERGY
(i.e) REMEMBER HEAT ENERGY = mCΔt ( unit is in joules )
you know electrical energy is given by one of the three formulae
W = VIt
W = (V x V x t)/ R
W = (I x I x t)/ R
EQUATING BOTH ENERGIES BY THE PRINCIPLE OF CALORIMETRY, WE CAN FIND THE UNKNOWN QUANTITY.

4. TYPE THREE PROBLEM: TO FIND THE MONTHLY ELECTRICITY BILL
(e.g) IN A HOUSE IF THERE ARE 2 BULBS OF 60W , TWO FANS OF 40W.
BULBS ARE ON FOR 4 HOURS PER DAY
FANS ARE RUNNING FOR 8 HOURS PER DAY.
WHAT IS THE ELECTRICITY BILL FOR ONE MONTH? (cost of 1 unit = Rs.4)
ANSWER:
1 BULB – 60W; THEREFORE FOR 2 BULBS – 120W
IF IT IS FOR 4 HOURS, THEN TOTAL WATT HOURS = 120 x 4 = 480Wh
1 FAN – 40W; THEREFORE FOR 2 FANS W
IF THEY ARE RUNNING FOR 8 HOURS,
THEN TOTAL WATT HOURS = 80 x 8 = 640Wh PER DAY
TOTAL WATT HOURS = 480 Wh Wh = 1120 Wh PER DAY;
1 kWh = 1000 Wh = 1 unit
1120 Wh = 1120/1000 = 1.12 kWh = 1.12 x 4 = Rs.4.48/day
For 30 days 4.48 x 30 = Rs

5. POWER RATING, FUSE RATING OR CURRENT RATING
ELECTRIC IRON W
T.V. SET W
WHAT DOES THIS MEAN?
It means that the rated power is 750W for an electric iron. This is called power rating.
FUSE OR CURRENT RATING
The maximum current that can flow in an appliance which the circuit can tolerate. (e.g) for an electric iron the power is 750W and the voltage is 220V
we know P = V x I (or) I = P/V = 750/220 = 3.4 A
So the fuse rating is 3.4 amps. which is the maximum current allowed to flow in the appliance without getting damaged.

6. ESSENTIAL COMPONENTS OF A HOUSE ELECTRICAL SYSTEM
FUSE : SAFETY DEVICE TO LIMIT THE CURRENT IN A CIRCUIT. IT IS A WIRE
THINK AND ANSWER
What is the fuse wire made of?
Why is a fuse wire not made of copper or aluminum?
What is the characteristic of a fuse wire?
Where is the fuse wire connected?
SWITCH : IT IS AN ON-OFF DEVICE FOR CURRENT IN A CIRCUIT.
MINIATURE CIRCUIT BREAKERS: USED INSTEAD OF A FUSE: SWITCHES WITHIN 25 MILLI SECOND. IT AVOIDS THE INCONVENIENCE OF REPLACING A FUSE WIRE.
EARTHING: CONNECTED TO THE APPLIANCE. USED TO SAVE THE APPLIANCE FROM EXCESSIVE CURRENT. EARTHWIRE IS CONNECTED TO THE BODY OF THE APPLIANCE.
THREE PIN PLUG: TOP IS EARTHING; RIGHT ONE IS FOR LIVE AND LEFT ONE IS FOR NEUTRAL.
COLOUR CODING OF WIRES: BROWN – LIVE; LIGHT BLUE – NEUTRAL; GREEN OR YELLOW -- EARTH

7. POWER DISTRIBUTION FROM GENERATING STATION TO CONSUMER
POWER STATION
STEP UP TRANSFORMER
TRANSMISSION WIRES
11 kV to 132 kV
STEP DOWN
TRANSFORMER
consumers
Sub station
Sub station
132 kV to 11 kV
11 kV to 220V

8. POWER DISTRIBUTION IN A HOUSE
FROM SUB STATION TO HOUSE, MOSTLY OVERHEAD WIRES ARE USED
BASICALLY THERE ARE THREE WIRES:
LIVE (OR PHASE) WIRE (L)
NEUTRAL WIRE (N)
EARTH WIRE (E)
kWh meter
Main switch
HOUSE

9. TREE SYSTEM RING SYSTEM
ALL CIRCUITS ORIGINATE FROM THE MAIN BOARD
LONGER LENGTH OF WIRE REQUIRED
IF ONE FUSE MELTS DUE TO HIGH CURRENT, ALL THE APPLIANCES SWITCH OFF
EXPENSIVE
RING SYSTEM
EACH APPLIANCE HAS ITS OWN FUSE
LENGTH OF WIRING IS SMALL
EASY TO INSTALL

10. ACTUAL DUAL SWITCH CIRCUIT

11. Dual switch circuits P Q X Y Bulb glowing neutral 1 1 22 line 2 2 3 3
WHEN AN ELECTRICAL APPLIANCE IS PLACED IN SERIES, BY A PAIR OF SWITCHES, IT CAN BE SWITCHED ON OR OFF FROM ANY OF THE DUAL SWITCHES

12. SUMMARY OF THE LESSON
UNDERSTAND THE FORMULAE AND USE THEM AS PER THE PROBLEMS
POWER RATING AND CURRENT RATING
POWER DISTRIBUTION (BASICS)
ESSENTIAL ELECTRICAL COMPONENTS IN A HOUSE
DUAL SWITCH CIRCUITS
COMMON HOUSEHOLD ELECTRIC APPLIANCES
(PRODUCING HEAT,LIGHT), CHARACTERISTICS

13. ALL THE BEST
TO MY STUDENTS