Higher - Electricity Powerpoint Answers 1. Electrical current is defined as the number of coulombs of charge that pass a point per second. 2. Potential difference is defined as the energy given to each coulomb of charge. 3. Electrical power is defined as the electrical energy transformed per second. 4. R XY = 5Ω. 5. When S closes we have two resistors in parallel The total resistance decreases and therefore the current will increase The voltage across the parallel resistors decreases and therefore the voltage across the series resistor increases 6. R XY = 4Ω. 7. R T = 31.8Ω. 8. S Open: I = 2.45x10-2 A and V = 0.68V. S Closed: I = 2.51x10-2 A and V = 0.55V. 9. Wheatstone Bridge Circuits allow you to find the resistance of an unknown resistor accurately regardless of the magnitude of the resistance.
10. A balanced Wheatstone Bridge Circuit will Have a zero reading on the meter used (ammeter, voltmeter etc) The ratio of the resistances on each arm of the bridge will be the same 11. V against ΔR will be an SLTO. 12. Reading on the voltmeter is 2V. 13. The EMF is the energy given to each coulomb of charge passing through the source. 14. A terminal potential difference of 5V means that 5J of electrical energy are given to each coulomb of charge. 15. E = V + Ir, E = IR + Ir and E = I (R + r) 16. Y axis => EMF = 1.4V. X axis => Short Circuit Current, I SC = 2.5A. 17. r = 1.6Ω. 18. S Open: I = 1A. S Closed: I = 1.5A. 19. The short circuitcurrent is the current flowing when the external or load resistance equals zero. 20. Maximum power transfer occurs when the internal resistance of the battery equals the load or external resistance connected to it. (r = R)
21. a) Switches Closed Current (A) Network Resistance(Ω) Terminal PD (V) S1 only 0.354 24 8.5 S1 and S2 only 0.657 12 7.9 S1,S2 and S3 0.920 8 7.4 b) i) E = 9.3V and ii) r = 2Ω.
c) As the current decreases the lost volts decreases As the lost volts decreases the teminal PD increases as E = V + Ir 22. a) EMF = 4.5V => 4.5J of electrical energy is given to each coulomb of charge as it passes through the source. b) i) I = 0.40A. ii) V = 4V. When S 2 closes the 15Ω resistor is now part of the circuit Total resistance decreases as the load resistors are in parallel From E = V + Ir, the reading on the ammeter increases As the ammeter reading increases the lost volts increases As the lost volts increases the V tpd which is the reading on the voltmeter decreases 23. UK Mains Voltage = 230V ac. UK Mains Frequency = 50Hz. 24. Oscilloscope a) Peak Voltage => Y- Gain control. Frequency => Time Base setting. b) AC pattern => Current changes direction every half cycle and the current continuously changes in magnitude. 25. Vp = 5V and Vrms = 3.54V. 26. f = 50Hz.
27. Time base setting x 4 8 waves now on the screen instead of 2 The amplitude of the waves stays constant 28. P = 60W. 29. Vp = 8.49V and Ip = 2.83A. 30. a) Y Gain setting = 4V div -1. b) Vrms = 8.49V. c) f = 400Hz. 31. a) Vp = 10V. b) Vrms = 7.07V. c) f = 25Hz. 32. Capacitors store electrical charge and electrical energy Capacitors are used in time delay circuits (t = RC) 33. Capacitance is measured in Farads (F). 1F = 1CV -1. 34. 3000µF = 3000µCV -1. 3000µC of charge stored per volt. 35. a) Q max = 7.2x10-4 C. b) E max = 2.16x10-3 J. 36. V-t graph is the same shape, but it will take longer than 4s to reach 10V.
37. Ecap = 1.6x10-2 J. 38. Options 2,3 and 4 are correct. 39. Resistive ac circuits => No change on ammeter. Capacitive ac circuits => Ammeter reading increases. 40. a) I = 6x10-4 A. b) I = 2x10-4 A. c) E extra = 4.95x10-2 J. 41. a) Imax = 5x10-5 A.
b) 42. P-type and N-type semiconductors are produced from a process called doping. 43. Two types of bias: Forward Bias => Negative side of the supply is connected to the n-type material and vice-versa. Reverse Bias => Negative side of the supply is connected to the p-type material and vice-versa. 44. P type and N type semiconductor materials are both electrically neutral. However the p-type material prefers holes (+ve charge carriers) and the N-type material prefers electrons (-ve charge carriers).
45. X Photovoltaic Mode. The photodiode acts as a voltage supply to provide the energy required to turn the electric motor When the light energy falls on to the photodiode it will split up the electron-hole pairs at the junction, to create free charge carriers. This results in a voltage being set up across the p-n junction. The voltage produced will operate and turn the electric motor in the circuit. 46. Semiconductors have a much smaller band gap between the conduction band and the valence band than an insulator. 47. The band gap decreases in a semiconductor when the temperature increases. 48. a) Material 1 => Insulator. Material 2 => Semiconductor. Material 3 => Conductor. b) The conduction band is the highest energy band. c) An Insulator has no electrons in the conduction band. A semiconductor has a very small quantity of electrons in the conduction band. A conductor is the most populated with electrons in the conductioin band. 49. Frequency minimum = > f = 3.16x10 14 Hz. Energy minimum = > f = 2.09x10-19 Hz. 50. An applied voltage causes the electrons to move away from the conduction band of the n-type material towards the junction The electrons then drop from the conduction band to the valence band When the electrons drop to the valence band they produce electron-hole pairs in the valence band with photons being emitted.