National 5 Physics Solutions to Electricity & Energy exam questions

National 5 Physics Solutions to Electricity & Energy exam questions

1. (a) [number and unit must be correct] (b) (i) [number and unit must be correct] (ii) [number and unit must be correct] [or calculate individual power of each heating element and add together] (iii) S (only) 1 (iii) Greatest value of resistance lowest current lowest power 1

2. (a) for correct current [no mark for reuse of Ohm s Law] [number and unit must be correct] (b) Transistor (switch) 1 (c) R of LDR increases V across LDR increases (above 0 7V) Transistor switches ON Relay coil is energised (which closes the relay switch and activates the motor) 4

. (a) c = 4180 (J Kg -1 C -1 ) full marks only possible when correct value from datasheet is used. E h = c m ΔT = 4180 x 1.6 x 80 = 55040 J [number and unit must be correct] 4 (b) (i) Eh = ml Eh = 0.9 22.6 10 5 for correct L value from datasheet both substitutions correct Eh = 2.04 10 6 J. [number and unit must be correct] 4 (ii) P =!! 2000 =!.!"#!"!! t = 1017 s [number and unit must be correct]

4. E p = m g h = 25 x 9.8 x 1.2 = 290 J [number and unit must be correct]

5 (a) (i) (-21) = 12 C [number and unit must be correct] 1 (ii) (120,000-12,000) = 108,000 J [number and unit must be correct] 1 (iii) E h = cmδt 108,000 = c x 2.0 x 12 c = 4,500 J kg -1 C -1 [number and unit must be correct] (b) (i) Any two of the following; Measured value of Eh too large ΔT too small Heat lost to surroundings (or similar) water not evenly heated (or similar) 2 (ii) Insulate beaker Put lid on beaker Stir water Fully immerse heater 1 (c) E = P t 108,000 = P x (5 x 60) P = 60 W [number and unit must be correct]

6 (a) [number and unit must be correct] (b) RT = R1 + R2 = 1 + 6 = 7 Ω [number and unit must be correct] (c) (Voltage across 2 Ω resistor = Voltage across 4 Ω resistor) V = IR = 0 1 4 (or 0 2 2) = 0 4 V [number and unit must be correct]

7 (a) dc electrons flow (or current flows) around a circuit in one direction only ac electrons (or current) direction changes/reverses after a set time 2

8. (a) To reduce current in LED To reduce voltage across LED 1 (b) V = 6 2 = 4 V V = IR 4= 0.1 x R R = 40 Ω [number and unit must be correct] 4 (c) P = I 2 R P = V 2 /R = (0 1) 2 40 =!!!" = 0 4 W = 0 4 W [number and unit must be correct] P = IV = 0 1 4 = 0 4 W [number and unit must be correct]

9. D 10. E 11. B 12. A 1. E 14. A

15. (a) (i) P = I V 6 = I x 12 I = A [number and unit must be correct] (ii) 48 = 12 + 12 + V R V R = 24 V [number and unit must be correct] 1 (iii) V = I R 24 = x R R = 8 Ω [number and unit must be correct] (b) (i) [number and unit must be correct] (ii) A. The reading decreases/gets smaller/reduces 1 B The resistance increases (so the current decreases) 1

16. (a) Use Ohm s Law twice. Once to calculate the current, then once to find V R. V = I R 0.6 = I x 2000 I = 0.00018 (A) V = I R [no mark for using equation again] = 0.00018 x 4800 for both substitutions = 8.64 V [number and unit must be correct] (b) [number and unit must be correct]

17. (a) (b) (c) Protect the LED prevent damage to the LED limits the current reduces voltage across LED [number and unit must be correct] 1 1 4

18. [number and unit must be correct]

19. (a) [number and unit must be correct] (b) [number and unit must be correct] (c) Heat is Lost Radiated escapes 1 from the sole plate

20. C 21. D 22. D 2. E

24. (a) (i) Ep = m g h Ep = 0.50 x 9.8 x 19. Ep = 95 J (ii) Ec = c m ΔT [number and unit must be correct] 95 = 86 x 0.50 x ΔT ΔT = 0.5 C [number and unit must be correct] (iii) Less than. Some heat is lost to surroundings/ or equivalent. 2 (b) E! = ml E! = 0 5 (2 05 10! ) E! = 1 025 10! J [number and unit must be correct]

25. (a) Ammeter in series Voltmeter in parallel with resistor Battery (not cell) symbol (b) [number and unit must be correct] (c) Power rating of resistor = W 1 Power developed in resistor is P = IV P = 0.6 5.7 P =.42 W The power rating of the resistor is too low for these current & voltage values. 4 (d) No, the student is not correct. In parallel the voltage across each resistor is still the same 6V across each resistor so power is the same 2

26 (a) MOSFET 1 (b) Voltage decreases 1 (c) (i) V! = V! + V!" 12 = 2 4 + V!" V!" = 9 6 V [number and unit must be correct] (ii)!!!!" =!!!!!! =!"##!!!! R V = 22400 V [number and unit must be correct] (d) The lamp stays on. When temperature decreases, R T increases Increase in R T will increase voltage across the thermistor (V T ). (so MOSFET does not switch off)

27. (a) Parallel 1 (b) P = I V 00 = I x 20 I = 1. A [number and unit must be correct] P = I V 900 = I x 20 I =.9 A Current in one mat=.9 I = 1.A [number and unit must be correct] (c) P total = x 00W = 900W P = V 2 / R 900 = 20 2 / R R = 59 Ω [number and unit must be correct] I total = x 1. =.9 A P = I 2 R 900 =.9 2 x R R = 59 Ω [number and unit must be correct]

28. (a) 1 (b) Vr = Vs - Vmotor = 24-18 = 6 (V) Vr = I R 6 = I x 2.1 I = 2.9 A [number and unit must be correct] 4 (c) Q = I x t =.2 x (10 x 60 x 60) = 115 200 C [number and unit must be correct]

29. t = 1/250 = 0 004(s) E = P t 60 x 10 - = P x 0.004 P = 15 W [number and unit must be correct] 4 E Total = 250 60 10 (J) E = P t 15 = P x 1 P = 15 W [number and unit must be correct]

0. (a) Transistor 1 (b) (As temp increases,) input voltage to transistor increases (above 0 7V) switching transistor on Current in the (relay) coil produces magnetic field to close switch. (c) 1 = 1 + 1 Rt R1 R2 1 = 1 + 1 Rt 16 16 Rt = 8 Ω [number and unit must be correct]

1. (a) Eh = cmδt = 420 x 82 x 125 = 44 280 000 J [number and unit must be correct] (b) Eh = 60% of the heat energy is used Eh = 44 280 000 0.6 = 26 568 000 J Eh = ml 26 568 000 = m (.42 10 5 ) m = 77.7 kg [number and unit must be correct]

2. (a) Lamp A It has the lowest resistance/highest current/greatest power 2 (b) P = V 2 /R = 24 2 /2 = 20 W [number and unit must be correct] (c) 1 (d) (i) 12 V 1 (ii) 1/Rp = 1/R1 + 1/R2 = 1/8 + 1/24 Rp = 6 Ω [number and unit must be correct] (e) The motor speed will reduce The (combined) resistance (of the circuit) is now higher current is lower Voltage across motor is less Motor has less power 2

. (a) (i) transistor 1 (ii) To act as a switch 1 (b) Resistance of LDR reduces, so voltage across LDR reduces Voltage across variable resistor/r increases When voltage across variable resistor/r reaches 0 7 V transistor switches buzzer on. (c) 80 units: resistance of LDR = 2500 (Ω) Total resistance = 2500 + 570 = 070 (Ω) -------------------------------------------------------- I = V/R = 5/070 = 1 6 10 - A or 1 6 ma [number and unit must be correct] 4 (d) To set the light level at which the transistor will switch on To set the level at which the buzzer will sound 1

4. A 5. A 6. C 7. B 8. C 9. C

40. P =!! 1 01 10! =!"!! A = 2 59 10!! m 2 [number and unit must be correct]

41. B 42. C 4. A 44. B

45. (a) (i) P =!! 4 6 10! = F = 1 800 N!!!"!! [number and unit must be correct] (ii) P! V! = P! V! (4 6 10! ) (1 6 10!! ) = (1 0 10! ) V! V! = 7 6 10!! m [number and unit must be correct]

46. When the volume of a gas decreases, the distance to the walls of the container decreases gas particles collide with the walls more often/frequently the increased collision rate increases the force on the walls, so pressure is increased.

47. When the temperature of a gas is increased, the gas particles gain kinetic energy and collide with the walls more often/frequently AND each collision exerts a greater force on the container walls the increased force results in increased gas pressure.

48. D

48. (a) P! V! = P! V! (750 10! ) (8 0 10!! ) = (125 10! ) V! V! = 0 48 m [number and unit must be correct] (b) Volume of cylinder = 8 0 10!! m total volume of gas available to fill balloons = 0.48 0.08 = 0.4 m number of balloons filled = 0.4 0.02 = 20 balloons