This presentation was produced by the Sciencedepartment of Temasek Secondary School.Redistribution or reproduction of this resource isprohibited by copyright regulations. Thisresource should be used for educationalpurposes.
D.C. CircuitsIn this section, we will learn to• draw circuit diagrams,• calculate the current, potential differenceand resistance in a series and parallelcircuit.
18.1 Electric Circuits1. An electric circuit is a closed path in whichelectric charges can flow from one terminalto another.2. In every electric circuit, the following mustbe present:• A e.m.f. source that drives charge andproduces a current, e.g a battery.
18.1 Electric Circuits• Components that “feed” on current and do ajob, e.g. a bell that makes a sound.• Conductors that join the source and variouscomponents together, e.g. copper wires.• Switches that break or complete a circuit.
18.2 Circuit DiagramsA circuit diagram is used to represent anelectric circuit. Symbols are used torepresent the devices or components usedin electric circuits.(Ref. textbook pg 321)
18.5 Current In SeriesCircuit1. In a series circuit, the current throughevery component is the SAME.AI1 I4I3I2I1 = I2 = I3 = I4
Example 112 V3 AQ1: What is the name given to the device X?XA1: LampQ2: The current passing through X is 3A,what is the current passing through theresistor and the cell?A2: 3A
Recall: All the resistors are connected inseries, the current passing through eachresistor is the same.R1 R2 R3 R4V1 V2 V3 V4I I18.5 P.D In Series Circuit
18.5 P.D In Series Circuit2. Let V be the potential difference acrossthe combination of the 4 resistors inseries.V = V1 + V2 + V3 + V4R1 R2 R3 R4V1 V2 V3 V4I IV
18.5 P.D In Series CircuitPoints to note:3. V1 = V2 = V3 = V4 if only the resistorsare identical i.e. having the sameresistance.4. In a series circuit, the component thathas the LARGEST resistance will havethe LARGEST potential differenceacross it.
Example 212 VQ1: What is the total p.d across the resistorand the lamp?A1: P.d across resistor and lamp= e.m.f of cell, E = 12 VQ2: If the p.d across the resistor is 9 V, whatis the p.d across the lamp?A2: P.d across lamp, V = 12 V - 9 V = 3 V
18.5 Resistance InSeries CircuitLet R be the combined/total/effectiveresistance of the 4 resistors.R1 R2 R3 R4V1 V2 V3 V4I IV
18.5 Resistance InSeries CircuitHow is R related to R1, R2, R3 and R4?RI IVThe arrangement can then be simplifiedfrom this …R1 R2 R3 R4V1 V2 V3 V4I IVto…
18.5 Resistance InSeries CircuitUsing equation V = IR, we have…V1 = IR1 V2 = IR2 V3 = IR3 V4 = IR4and… V = IRR1 R2 R3 R4V1 V2 V3 V4I IVRI IV
18.5 Resistance InSeries CircuitR = R1 + R2 + R3 + R46. For resistors in series, the combined/total/effective resistance is the SUMof the individual resistances.7. The effective resistance is GREATERthan any of the individual resistance.
Example 312 V1) Calculate the effective resistance of thecircuit.2) What is the current passing through theresistor?0.5 Ω 3 Ω 2.5 ΩR = 0.5 + 3 + 2.5 = 6 ΩI = V/R = 12/6 = 2 A
Examples on Series CircuitExample 4:Find the combined resistance of the three resistorsin series given that R1 = 1 Ω, R2 unknown and R3= 2 Ω The current I recorded in the ammeter is 1 Aand the voltmeter reading V across R2 = 3 V.IR2AR3R1V
Examples on Series CircuitExample 5:A cell of e.m.f. 1.5 V was connected in serieswith two resistors, as shown below1.5 V6 Ω4 Ω
Calculatei) the effective resistance of the circuit,ii) the current flowing in the circuit,iii) the potential difference across the 4 Ωresistor.1.5 V6 Ω4 Ω
i) Effective resistance, R = R1 + R2= 4 + 6 = 10 Ωii) Current flowing in circuit, I = V / R= 1.5 / 10= 0.15 A1.5 V6 Ω4 Ω
iii) P.d across 4 Ω resistor,V4Ω = IR= 0.15 × 4= 0.6 V1.5 V6 Ω4 Ω0.15 AWAIT! That is not all….What is the p.d across the 6 Ωresistor?
18.6 Current InParallel CircuitIn a parallel circuit, the current from thesource is shared by 2 or more branches.Lamp 1Current from source is shared by lamp 1 and 2Lamp 2A B
Current In ParallelCircuitI1 = I2 + I3 = I4Lamp 1Lamp 2I1I2I3I4
Current In ParallelCircuit1. The SUM of the currents in theseparate branches of a parallel circuitis EQUAL to the current from thesource.I1 = I2 + I3 = I4
Current In ParallelCircuitPoints to note:2. In a parallel circuit, the componentwith the SMALLEST resistance willallow the LARGEST current to pass.3. I1 = I2 if only the lamps are identical i.e.having the same resistance.
Example 8Which ammeter shows a faulty reading?4A2A2A4A2APSRQT
P.D In Parallel CircuitV3V2V14. Each component joined in parallel havethe same potential difference across it.V1 = V2 = V3
Example 9Fill in the blanks using = , > or <.V1 _____V2 I1 ______I2I1I210 Ω5 ΩV1V2
Resistance In ParallelCircuitI = I1 + I2V = V1 = V2Let R be thecombined/total/effective resistanceof the 2 resistors.R1R2V2V1VII1I2
Resistance In ParallelCircuitThe arrangement can then be simplifiedfrom this …to…RVVII R1R2V2V1VII1I2How is R related to R1 and R2?
Resistance In ParallelCircuit4. I = I1 + I2R1R2V2V1VII1I2RVVII2211RVRVRV+=
Resistance In ParallelCircuit21111RRR+=5. For resistors in parallel, the reciprocal ofthe combined /total/effective resistance isthe SUM of the reciprocal of individualresistances.6. The effective resistance is SMALLER thanany of the individual resistance.
Example 10What is the effectiveresistance of the threeresistors?Let effective resistance be R1/R = 1/3 +1/6 + 1/6 R = 3/2 = 1.5Ω3 Ω6Ω6Ω
Examples on ParallelCircuit3.0 Ω6.0 Ω4.0 VThe circuit diagram shows a 6.0 Ω resistorand a 3.0 Ω resistor in parallel andconnected to a 4.0 V battery.Example 11:
3.0 Ω6.0 Ω4.0 V1) Calculate the effective resistance of theparallel resistors.1/R = 1/3.0 + 1/6.0= 1/2.0R = 2.0 Ω
3.0 Ω6.0 Ω4.0 VEffective resistance = 2.0 Ω2) Calculate the current flowing throughthe battery.Ibattery = V/R = 4.0/2.0= 2.0 A
3.0 Ω6.0 Ω4.0 V3) Calculate the current flowing through ineach resistor.P.d across the resistors = 4.0 VI3.0 = V/R = 4.0/3.0 = 1.3 AI6.0 = Ibattery - 1.3 = 2.0 – 1.3 = 0.7 A
3.0 Ω6.0 Ω4.0 V4) A third resistor is connected in parallelwith the original pair. Is the currentthrough the battery larger, smaller or thesame as before. Explain.Current will be larger because the effectiveresistance of the circuit is smaller thanbefore.
Examples on ParallelCircuitExample 12:Textbook pg 328 Example 18.3
Series + Parallel CircuitAR1 R2R3R2 is connected in parallel with R3.R1 is connected in series with the ammeter andthe resistor combination of R2 and R3.
Resistance In Series +Parallel CircuitExample 13:What is the effective resistance of the circuitbelow?5 Ω3 Ω3 Ω
Resistance In Series +Parallel CircuitFirst, add the resistors in parallel.Let R1 be the effective resistance of theresistors in parallel.1/R1 = 1/3 + 1/3R1 = 1.5 Ω5 Ω3 Ω3 Ω
Resistance In Series +Parallel CircuitThis is then reduced to two resistors inseries.5 Ω3 Ω3 Ω5 Ω R1 = 1.5 Ω
Resistance In Series +Parallel CircuitEffective resistance of circuit,R = 5 +1.5= 6.5 Ω5 Ω R1 = 1.5 Ω
Examples on Series +Parallel CircuitExample 14:The diagram shows three resistorsconnected to a 6.0 V battery supplyA6.0 Ω6.0 Ω8.0 Ω6.0 V
1) Calculate the combined resistance of the8.0 Ω and the 6.0 Ω resistors in series.Let combined resistance be RcRc = 8.0 + 6.0 = 14.0 ΩA6.0 Ω6.0 Ω8.0 Ω6.0 V
2) Calculate the effective resistance of thecircuit.Let effective resistance of circuit be R1/R = 1/14.0 + 1/6.0 = 5/21R = 21/5 = 4.2 ΩA6.0 Ω6.0 Ω8.0 Ω6.0 V
3) Calculate the current through the ammeter.Effective resistance of circuit = 4.2 ΩCurrent through ammeter = Current in circuit= V/R = 6.0/4.2A6.0 Ω6.0 Ω8.0 Ω6.0 V
Examples on Series +Parallel CircuitExample 15:Textbook pg 329 Example 18.4
Examples on Series +Parallel CircuitExample 16:Textbook pg 334 Q3Example 17:Textbook pg 335 Q4
Short CircuitShort Circuit occurs in a closed circuitwhen there is an alternative path forcurrent to flow. This alternative path isof a much lower resistance than theoriginal path.
Short CircuitThe lamp will not light up because,compared to the copper wire, it offers moreresistance and thus current by-passed it.ALampAmmeterCellI IIICopper wire
Short CircuitDuring a short circuit, the currentflowing is very large as the resistance isvery small. This can be dangerous as alarge current may cause heating resultingin a fire.
Example 18:Textbook pg 334 Q2Examples on Series +Parallel Circuit