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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