# A Level Physics Electricity Complete Circuits Answers OCR

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Visit http://www.mathsmadeeasy.co.uk/ for more fantastic resources.OCRA LevelA Level PhysicsELECTRICAL CIRCUITS:Complete Circuits 1 (Answers)Name:Total Marks:Maths Made Easy Complete Tuition Ltd 2017/30

1.Total for Question 1: 8(a) Define electrical work, W, in terms of potential difference, V, and charge, Q. Using this relationship,show that P I 2 RSolution: W V QBut, Q It and P W/t P V It/t V IFrom Ohm’s law, V IR P I 2 R(b) The P.D. across a 5.0 Ω resistor is measured as 6.0 V. What power is it dissipating?Solution: 7.2 W(c) An LED is connected in series with an ammeter and a power supply. A voltmeter is connectedacross the LED. They read 2.2 A and 4.6 V. If it is left on for 1 hour and 15 minutes, how muchwork is done by the LED?Solution: 45 kjPage 2

(d) Sketch how the electrical work done by the resistor at a given point in time would vary with theresistance of the resistor. Assume the P.D. across the resistor is constant.Solution: y 1/x graph i.e. nonlinear decreasePage 3

2. This question exploits Kirchoff’s laws to determine the resistances of several components in Figure 1.Total for Question 2: 10Figure 1: A circuit containing two resistors, a voltmeter, an ammeter, a cell and a bulb.Tom notes that the the bulb has an effective resistance of 5.0 Ω, that the voltmeter reads 2.0 V and thatthe ammeter reads 3.5 A.(a) State Kirchoff’s First Circuit Law. What implications does it have for the charge entering andleaving a circuit junction?Solution: Sum of currents entering a junction equals the sum of currents leaving a junctioni.e. ΣIintojunction ΣIoutof junctionSince Q It, the same conservation applies at a junction for charge.(b) State Kirchoff’s Second Circuit Law.Solution: In a given closed loop, the sum of the potential differences is equal to the sum ofthe EMFs: ΣP Di ΣEM Fi(c) Calculate R1 .Solution: 0.57 ΩPage 4

(d) Calculate R2 .Solution: 6.7 Ω(e) Calculate the power dissipated by the bulb.Solution: 20 W(f) The bulb dissipates 75% of its power as heat and converts the rest to light. What is the efficiencyof this circuit as a means of lighting?Solution: 12%Page 5

3. Based on the conservation of charge and of energy, it is possible to derive several laws that dictate howthe total effective resistance in a circuit varies when a combination of resistors are used in series and/orparallel.Total for Question 3: 8(a) Use Kirchoff’s and Ohm’s laws to derive an expression for the total effective resistance of tworesistors, R1 2 , in series.Solution: From KSL and Ohm’s : IR I1 R1 I2 R2From KFL I is the same for all R R1 R2(b) Using a similar technique, show that for two resistors in parallel,Solution: KFL: I I1 I2V1 .Incorporating Ohms: VR R1KSL: V of each loop is the same V V1 V2 1R 1R1 1R 1R1 1R2 .1R2(c) Two resistors (1.0 Ω and 2.0 Ω) connected in parallel are linked in series to a 3.0 Ω resistor. Allof this is in parallel with a fourth resistor. If the total effective resistance is 1.0 Ω, what is theresistance of the fourth resistor?Solution: 1.4 ΩPage 6

4. Draw the symbols for the following circuit components:Total for Question 4: 4(a) An LED.Solution:(b) A variable resistor.Solution:(c) A thermistor.Solution:(d) An LDR.Solution:Page 7

ELECTRICAL CIRCUITS: Complete Circuits 1 (Answers) Name: Total Marks: /30 . 1. Total for Question 1: 8 (a) De ne electrical work, W, in terms of potential di erence, V, and charge, Q. Using this relationship,  show that P I2R Solution: W VQ But, Q It and P W t !P VIt t VI From Ohm’s law, V IR !P I2R (b) The P.D. across a 5:0 resistor is measured as 6.0 V. What power is it .

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