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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
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
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.
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 -- 80W
IF THEY ARE RUNNING FOR 8 HOURS,
THEN TOTAL WATT HOURS = 80 x 8 = 640Wh PER DAY
TOTAL WATT HOURS = 480 Wh + 640 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. 134.40
POWER RATING, FUSE RATING OR CURRENT RATING
 ELECTRIC IRON ----- 750W
 T.V. SET
----- 120W
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.
ESSENTIAL COMPONENTS OF A HOUSE ELECTRICAL SYSTEM
 FUSE : SAFETY DEVICE TO LIMIT THE CURRENT IN A CIRCUIT. IT IS A WIRE
THINK AND ANSWER
1. What is the fuse wire made of?
2. Why is a fuse wire not made of copper or aluminum?
3. What is the characteristic of a fuse wire?
4. 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
POWER DISTRIBUTION FROM
GENERATING STATION TO CONSUMER
POWER
STATION
STEP UP
TRANSFORMER
TRANSMISSION
WIRES
11 kV to 132 kV
consumers
Sub
Substation
station
11 kV to 220V
STEP DOWN
TRANSFORMER
132 kV to 11 kV
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




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
 EACH APPLIANCE HAS ITS OWN FUSE
 LENGTH OF WIRING IS SMALL
 EASY TO INSTALL
http://users.wfu.edu/matthews/misc/switches/4WayA
nimation.html
Bulb glowing
neutral
P
1
line
X
Q
1
222
2
3
3
Y
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
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