1.
Current flow versus ElectronflowConventionalcurrent flowsthis way.Electronsflow thisway.
2.
What formula relatesCharge, Current and Time?A current of 1 Ampere is flowing when 1 Coulomb of chargeflows past a point in a circuit in 1 second.Charge = current x time(C) (A) (s)If a current of 5 A is flowing then 5 C of charge pass a pointin 1 second.In general, if a steady current I (amperes) flows for time t(seconds) the charge Q (coulombs) passing any point isgiven byQ = I x t
3.
Worked exampleA current of 150 mA flows around a circuit for1minute. How much electrical charge flows past apoint in the circuit in this time?SolutionSubstituting into Q = Itgives Q = 0.15 A x 60 s= 12 C
4.
1. Convert the following currents into amperes:a) 400 mA b) 1500 mA.Ans. = a) 400 mA = 0.4 A b) 1500 mA = 1.5 A2. What charge is delivered if a current of 6A flows for10 seconds?Ans. = 60 C3. What charge is delivered if a current of 300 mA flowsfor 1 minute(60 seconds)?Ans. = 18 CFor you to do!!
5.
What is Ohm’s Law?The voltage dropped across a resistor is directlyproportional to the current flowing through it,provided the temperature remains constant.Voltage (V) = Current (A) x resistance (Ω)V = I x RWhat is the formula for Ohm’s law?
6.
Worked example on Ohm’s Law2 A 8 ΩV = ?V IxR== 2A x 8= 16 VΩ
7.
Ammeters and VoltmetersAmmeters measure current and are placed in seriesin a circuit.Voltmeters measure voltage and are placedVoltmeters measure voltage and are placedin parallel in a circuit.in parallel in a circuit.AV
8.
Rules forResistors in SERIESRTotal = + +R R R1 2 3
14.
Rules for SERIES CIRCUITS• Same current but ……• split voltage between them.
15.
18 V6 V 6 V 6 V?Equal resistors share thevoltage between them!!
16.
Rules for PARALLELCIRCUITS• Same voltage but ……• split current between them.
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? A? A4 A? AEqualresistorsWhat will be the currents flowingthrough each ammeter?
18.
Electrical PowerE.g. A study lamp is rated at 60 W, 240 V.How much current is the bulb carrying?Solution60 W = 240 V * Current60 WCurrent = ----------- = 0.25 A240 VElectricalElectrical Power = Potential difference * currentPower = Potential difference * currentWatts Volts AmpsWatts Volts Amps
19.
A transformer is a device for increasing or decreasingan a.c. voltage.
23.
All transformers have three parts:1. Primary coil – the incoming voltage Vp(voltage across primary coil) is connectedacross this coil.2. Secondary coil – this provides the outputvoltage Vs(voltage across the secondary coil)to the external circuit.3. Laminated iron core – this links the two coilsmagnetically.Notice that there is no electrical connection between the two coils,which are constructed using insulated wire.
24.
Two Types of TransformerA step-up transformer increases the voltage -there are more turns on the secondary than on theprimary.A step-down transformer decreases the voltage- there are fewer turns on the secondary than onthe primary.To step up the voltage by a factor of 10, theremust be 10 times as many turns on the secondarycoil as on the primary. The turns ratio tells usthe factor by which the voltage will be changed.
25.
Formula for Transformervoltage across the primary coilvoltage across the secondary coilnumber of turns on primarynumber of turns on secondaryVVNNpsps==Where Vp= primary voltageVs = secondary voltageNp= Number of turns in primary coilNs= Number of turns in a secondary coil.
26.
Worked example No. 1The diagram shows a transformer. Calculate thevoltage across the secondary coil of this transformer.Step-up transformer!
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SolutionVVNNSubstituting12V180540Crossmultiplying180.V 12 x 540V12 x 540180V 36 VPSPSSSSS===∴ =∴ =
28.
Worked example No. 2A transformer which has 1380 turns in its primary coil is to be used toconvert the mains voltage of 230 V to operate a 6 V bulb. How manyturns should the secondary coil of this transformer have?VP = 230 VNP = 1380VS = 6 VNS = ?Obviously, a Step-down transformer!!
29.
SolutionVVNNSubstituting23061380NCrossmultiplying2300.N 6 x 13800N6 x 1380230N 36 turnsPSPSSSSS===∴ =∴ =
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