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RAE practice questions

Chapter 7: Power in DC Circuits

Free practice questions on Power in DC Circuits, chapter 7 of the SARL's Class A manual, for South Africa's Radio Amateur Examination (RAE). Every question has the right answer and an explanation. Sections covered:

  • 7.1 Power dissipation in resistances
  • 7.2 Using Ohm's Law with the formula for power
  • 7.3 Electrical sources
  • 7.4 Matching source and load
  • 7.5 Efficiency

80 questions

  1. 1. Electrically, what is the heating element of a kettle or bar heater?
    Show answer

    A resistance built to carry the power and shed heat

    Current through any resistance turns electrical energy into heat, and the element is a resistance chosen to carry the power and pass its heat on. Heating by magnetism (induction) is a different method.

  2. 2. In electrical terms, what is power?
    Show answer

    How much energy is used each second

    Power is energy per second: one watt is one joule every second. The total energy also depends on how long the power is used, so a small power used for a long time can add up to more energy.

  3. 3. Multiplying a voltage by a current gives a power. Why?
    Show answer

    Volts are energy per unit charge; amperes are charge per second

    A volt is a joule per coulomb and an ampere a coulomb per second, so V × I gives joules per second, which is watts. Charge per second describes a current, not a voltage, so the swapped pair is wrong.

  4. 4. A soldering iron is switched on for 2 hours and uses 120 watt-hours of energy. What is its power?
    Show answer

    60 W

    Power is energy per unit time: 120 watt-hours ÷ 2 hours = 60 W. 120 W mistakes the energy for the power; 0,017 W divides watt-hours by seconds, mixing units.

  5. 5. A kettle normally runs on AC mains. What does the manual say about using P = V × I for it?
    Show answer

    The same formula still applies, as the section on AC shows

    P = V × I holds for AC too when the RMS values of voltage and current are used, and mains values are quoted that way. A different formula for the kettle is therefore not needed.

  6. 6. A VHF mobile transceiver draws 5,5 A from a 13,8 V supply while transmitting. What power does it take from the supply?
    Show answer

    75,9 W

    P = V × I = 13,8 × 5,5 = 75,9 W. Dividing instead, 13,8 ÷ 5,5 = 2,51, gives the resistance of the transceiver in ohms, not its power.

    Try it: Ohm's Law & Power
  7. 7. An electric kettle draws 9,5 A from the 230 V mains. Which is nearest to its power?
    Show answer

    2,2 kW

    P = V × I = 230 × 9,5 = 2185 W, nearest 2,2 kW; the formula holds on AC mains too. Writing 2185 W as 2,2 W drops the kilo: 2185 W is 2,185 kW.

    Try it: Ohm's Law & Power
  8. 8. Electrical power is the current multiplied by which 'force'?
    Show answer

    Electromotive force

    Electromotive force (EMF) is the voltage a source produces, so power is EMF × current; despite the name it is not a force. Magnetomotive force belongs to coils and magnetic circuits and gives no power.

  9. 9. A valve linear amplifier's output stage works from 800 V DC and draws 250 mA. What DC power does it consume?
    Show answer

    200 W

    P = V × I = 800 V × 0,25 A = 200 W. Using 250 as if it were amperes gives 200 000 W, or 200 kW: convert the milliamperes to amperes first.

    Try it: Ohm's Law & Power
  10. 10. An electric motorcycle's 72 V battery supplies 50 A to the motor. What power is the motor taking?
    Show answer

    3,6 kW

    P = V × I = 72 × 50 = 3600 W, which is 3,6 kW. 360 W loses a zero, and 0,69 W divides the current by the voltage instead of multiplying.

    Try it: Ohm's Law & Power
  11. 11. A lightning stroke briefly carries 30 kA while 200 MV appears across its path. What is the peak power?
    Show answer

    6 × 10¹² W

    P = V × I = (200 × 10⁶ V) × (30 × 10³ A) = 6000 × 10⁹ = 6 × 10¹² W. 6 × 10⁹ W takes the 30 kA as 30 A, leaving out the kilo.

    Try it: Ohm's Law & Power
  12. 12. A 6 V bicycle lamp is rated at 1,8 W. What current does it draw from a 6 V battery?
    Show answer

    300 mA

    From P = V × I, I = P ÷ V = 1,8 ÷ 6 = 0,3 A, which is 300 mA. Dividing the other way, 6 ÷ 1,8 = 3,3, gives the 3,3 A slip.

    Try it: Ohm's Law & Power
  13. 13. A 28 V DC supply delivers 140 W to a solid-state amplifier. What current does the amplifier draw?
    Show answer

    5 A

    I = P ÷ V = 140 ÷ 28 = 5 A. Dividing the other way, 28 ÷ 140 = 0,2, gives the 200 mA slip, and multiplying gives 3920 A.

    Try it: Ohm's Law & Power
  14. 14. A solid-state amplifier takes 1,5 kW from a 48 V DC supply. Which is nearest to the current it draws?
    Show answer

    31 A

    I = P ÷ V = 1500 ÷ 48 = 31,25 A, nearest 31 A; the 1,5 kW is the DC input, not the RF output. Dividing volts by watts, 48 ÷ 1500 = 0,032, gives the 32 mA slip.

    Try it: Ohm's Law & Power
  15. 15. A 2,5 kV supply delivers 550 W to a valve amplifier. What current does it supply?
    Show answer

    220 mA

    I = P ÷ V = 550 W ÷ 2500 V = 0,22 A, which is 220 mA. Using 2,5 instead of 2500 leaves out the kilo and gives 220 A.

    Try it: Ohm's Law & Power
  16. 16. A load uses 100 W while drawing 4 A. What voltage is across it?
    Show answer

    25 V

    V = P ÷ I = 100 ÷ 4 = 25 V. Dividing the other way, 4 ÷ 100 = 0,04, gives the 40 mV slip, and multiplying gives 400 V.

    Try it: Ohm's Law & Power
  17. 17. A 5 V circuit board draws 70 mA. How much power does it take?
    Show answer

    350 mW

    P = V × I = 5 V × 0,07 A = 0,35 W, which is 350 mW. Treating the 70 mA as 70 A gives the 350 W slip.

    Try it: Ohm's Law & Power
  18. 18. A 3,3 V sensor board draws 27 mA. Which is nearest to the power it takes?
    Show answer

    90 mW

    P = V × I = 3,3 V × 0,027 A = 0,0891 W = 89,1 mW, nearest 90 mW. Taking the 27 mA as 27 A gives 89,1 W, the 90 W slip.

    Try it: Ohm's Law & Power
  19. 19. A 2 W resistor has 40 V across it. What is the largest current it may carry without exceeding its rating?
    Show answer

    50 mA

    I = P ÷ V = 2 ÷ 40 = 0,05 A, which is 50 mA; more current would exceed the 2 W rating. Dividing the other way, 40 ÷ 2 = 20, gives the 20 A slip.

    Try it: Ohm's Law & Power
  20. 20. A 450 mW LED module draws 150 mA. What voltage does it need?
    Show answer

    3 V

    V = P ÷ I = 0,45 W ÷ 0,15 A = 3 V. Taking the 450 mW as 450 W gives 450 ÷ 0,15 = 3000 V, the 3 kV slip.

    Try it: Ohm's Law & Power
  21. 21. A 5 W resistor is fitted in a circuit. Which combination of voltage across it and current through it would overload it?
    Show answer

    45 V and 160 mA

    P = V × I for each: 45 × 0,16 = 7,2 W, over the 5 W rating. The others give 4,8 W, 3 W and 0,24 W; 4,8 W comes close but stays within it.

    Try it: Ohm's Law & Power
  22. 22. A resistor has 18 V across it and carries 30 mA. Which is the smallest of these power ratings that will do?
    Show answer

    1 W

    P = V × I = 18 × 0,03 = 0,54 W. A 0,5 W resistor would be overloaded, so 1 W is the smallest rating that copes; 2 W copes too but is bigger than needed.

    Try it: Ohm's Law & Power
  23. 23. What power does a 180 Ω load take from a 42 V DC supply?
    Show answer

    9,8 W

    P = V² ÷ R = 42² ÷ 180 = 1764 ÷ 180 = 9,8 W. Leaving out the square, 42 ÷ 180 = 0,23, gives the current in amperes, not the power.

    Try it: Ohm's Law & Power
  24. 24. If the current through a fixed resistor is halved, what happens to the power it dissipates?
    Show answer

    It drops to a quarter

    P = I² × R, and I² means I × I, so half the current gives 0,5 × 0,5 = 0,25: a quarter of the power, as the voltage halves too. Merely halving leaves out the square.

    Try it: Ohm's Law & Power
  25. 25. A voltage of 70 V is measured across a 50 Ω dummy load. What power is it dissipating?
    Show answer

    98 W

    P = V² ÷ R = 70² ÷ 50 = 4900 ÷ 50 = 98 W. Leaving out the square, 70 ÷ 50 = 1,4, gives the current in amperes, not the power.

    Try it: Ohm's Law & Power
  26. 26. A 1,5 kΩ resistor has 30 V across it. How much power does it dissipate?
    Show answer

    0,6 W

    P = V² ÷ R = 30² ÷ 1500 = 900 ÷ 1500 = 0,6 W, with 1,5 kΩ taken as 1500 Ω. Leaving out the square, 30 ÷ 1500 = 0,02, gives the 20 mW slip.

    Try it: Ohm's Law & Power
  27. 27. A 24 V, 21 W truck lamp is lit. Which is nearest to the resistance of its hot filament?
    Show answer

    27 Ω

    R = V² ÷ P = 24² ÷ 21 = 576 ÷ 21 = 27,4 Ω, nearest 27 Ω. Leaving out the square, 24 ÷ 21 = 1,1, gives the slip. A cold filament measures much lower.

    Try it: Ohm's Law & Power
  28. 28. Which is nearest to the highest voltage a 4,7 kΩ, 0,25 W resistor can take without exceeding its rating?
    Show answer

    34 V

    From P = V² ÷ R, V = √(P × R) = √(0,25 × 4700) = √1175 = 34,3 V, nearest 34 V. Forgetting the square root leaves 1175, the 1,2 kV slip.

    Try it: Ohm's Law & Power
  29. 29. Which formula gives power directly from the current through a resistor and its resistance?
    Show answer

    P = I² × R

    Putting V = I × R from Ohm's Law into P = V × I gives P = I × R × I = I² × R. Leaving out the square, I × R, gives the voltage, not the power.

    Try it: Ohm's Law & Power
  30. 30. An 82 Ω resistor carries a current of 2 A. How much power is it dissipating?
    Show answer

    328 W

    P = I² × R = 2² × 82 = 4 × 82 = 328 W. Leaving out the square, 2 × 82 = 164, gives the voltage across the resistor, not its power.

    Try it: Ohm's Law & Power
  31. 31. A 2,2 kΩ resistor in a transistor stage carries 5 mA. What power does it dissipate?
    Show answer

    55 mW

    P = I² × R = 0,005² × 2200 = 0,000 025 × 2200 = 0,055 W = 55 mW. Leaving out the square gives 5 mA × 2,2 kΩ = 11 V, a voltage, not 11 mW.

    Try it: Ohm's Law & Power
  32. 32. Which is nearest to the largest current a 3,3 kΩ, 0,5 W resistor can carry without exceeding its rating?
    Show answer

    12 mA

    From P = I² × R, I = √(P ÷ R) = √(0,5 ÷ 3300) = √0,000 152 = 0,0123 A, nearest 12 mA. Forgetting the square root leaves 0,000 15 A, the 0,15 mA slip.

    Try it: Ohm's Law & Power
  33. 33. A 470 Ω resistor carries 35 mA. Which is nearest to the power it dissipates?
    Show answer

    0,58 W

    P = I² × R = 0,035² × 470 = 0,001 225 × 470 = 0,576 W, nearest 0,58 W. Leaving out the square, 0,035 × 470 = 16,45, gives the 16 W slip.

    Try it: Ohm's Law & Power
  34. 34. A 20 Ω resistor carries 3 A. Using Ohm's Law first, what are the voltage across it and the power it dissipates?
    Show answer

    60 V and 180 W

    V = I × R = 3 × 20 = 60 V; P = V × I = 60 × 3 = 180 W, as I² × R = 9 × 20 confirms. 1,2 kW wrongly uses V × R.

    Try it: Ohm's Law & Power
  35. 35. A 100 Ω and a 220 Ω resistor are connected in series across a supply. Which one dissipates more power?
    Show answer

    The 220 Ω resistor

    In series both carry the same current, and P = I² × R, so the larger resistance takes more power: 2,2 times as much. The 100 Ω would pass more current only if it had the supply to itself.

    Try it: Ohm's Law & Power
  36. 36. A 150 Ω and a 680 Ω resistor are connected in parallel across a supply. Which one dissipates more power?
    Show answer

    The 150 Ω resistor

    In parallel both have the same voltage, and P = V² ÷ R, so the smaller resistance takes more power. Power rises with resistance only when the current is the same, as in series.

    Try it: Ohm's Law & Power
  37. 37. Two 14 V lamps are rated 5 W and 20 W. Which has the higher resistance when lit?
    Show answer

    The 5 W lamp, four times as much

    At the same voltage R = V² ÷ P, so the lower-power lamp has the higher resistance, 20 ÷ 5 = 4 times. The bigger lamp needs more current, so it must have the lower resistance.

    Try it: Ohm's Law & Power
  38. 38. A cranking starter motor has 9,9 V across it while it draws 220 A. What is its effective resistance?
    Show answer

    45 mΩ

    R = V ÷ I = 9,9 ÷ 220 = 0,045 Ω = 45 mΩ; 450 mΩ slips the decimal comma. A turning motor is not a plain resistor, so this V ÷ I value is only its effective resistance.

    Try it: Ohm's Law & Power
  39. 39. A 36 V cooling-fan motor draws 0,45 A while running. What is its effective resistance?
    Show answer

    80 Ω

    R = V ÷ I = 36 ÷ 0,45 = 80 Ω. Dividing the other way, 0,45 ÷ 36 = 0,0125, gives the 12,5 mΩ slip. A running motor's V ÷ I is an effective value, not its winding resistance.

    Try it: Ohm's Law & Power
  40. 40. A source gives VS with no load and has an internal resistance RSource. Which expression gives the voltage at its terminals while it delivers a current I to a load RLoad?
    Show answer

    VOut = VS − I × RSource

    The current drops I × RSource across the internal resistance, so VOut = VS − I × RSource, and VOut = VS with no current. VS − I × RLoad is what is left for RSource: the internal drop.

  41. 41. A battery has an open-circuit voltage of 12,6 V and an internal resistance of 0,05 Ω. What is its terminal voltage when it supplies 20 A?
    Show answer

    11,6 V

    The drop is I × RSource = 20 × 0,05 = 1 V, so VOut = 12,6 − 1 = 11,6 V. Adding the drop gives 13,6 V, but the drop is lost inside the battery.

    Try it: Ohm's Law & Power
  42. 42. A battery reads 12,8 V with no load and 12,2 V while it supplies 15 A. What is its internal resistance?
    Show answer

    40 mΩ

    The drop is 12,8 − 12,2 = 0,6 V, so RSource = 0,6 ÷ 15 = 0,04 Ω = 40 mΩ. Dividing the loaded 12,2 V by 15 A gives 0,81 Ω, the load's resistance instead.

    Try it: Ohm's Law & Power
  43. 43. Why does the terminal voltage of a real battery fall as more current is drawn from it?
    Show answer

    The drop across its internal resistance increases

    The model's VS stays constant; the current drops I × RSource across the internal resistance, leaving less at the terminals. The immediate fall when a load is switched on comes from this drop, not from VS changing.

  44. 44. Which power supply will keep its output voltage most constant as the load current changes?
    Show answer

    One with a very low internal resistance

    The drop I × RSource is smaller when RSource is small, so a low internal resistance keeps the output steady as the load changes. A high internal resistance does the opposite, and thin leads add to it.

  45. 45. A battery's voltage collapses badly whenever the transmitter is keyed. What is the most likely cause?
    Show answer

    A high internal resistance

    A high internal resistance, as in a worn battery, drops a large I × RSource at the transmit current, so the terminal voltage collapses. A low internal resistance would keep it steady.

  46. 46. Which two kinds of supply does the manual name as having a low internal resistance?
    Show answer

    Well-regulated supplies and good-quality batteries

    Good-quality batteries and well-regulated supplies keep a low internal resistance, so their output stays steady under load. Nearly flat batteries and unregulated supplies sag far more.

  47. 47. Two batteries each read 12,7 V with no load. With the same 25 A load, battery A reads 12,3 V and battery B 11,5 V. Which has the lower internal resistance?
    Show answer

    Battery A, whose voltage drops less

    The drop I × RSource is smaller where the internal resistance is lower: at 25 A battery A loses 0,4 V and B 1,2 V, so A has a third of the resistance. Equal no-load readings say nothing about it.

  48. 48. The manual describes two models of a real supply. Which pair of descriptions is right?
    Show answer

    Thevenin: voltage source with series resistance; Norton: current source with shunt resistance

    Thevenin: a constant-voltage source with the internal resistance in series. Norton: a constant-current source with its resistance connected across it. Both describe a supply; the manual keeps to Thevenin, as easier to picture.

  49. 49. In the Norton model, a load is connected across the output. What happens to the constant source current IS?
    Show answer

    It divides between RShunt and the load

    IS is constant and is shared between RShunt and the load: the more the load takes, the less flows in RShunt, so VOut falls. RShunt keeps its share while it is connected.

  50. 50. For a Thevenin model and a Norton model to describe the same supply, what must be true?
    Show answer

    RSource equals RShunt, and VS = IS × RSource

    The models behave the same when the resistances are equal and VS = IS × R, the voltage IS develops across that resistance. A larger RShunt would keep the no-load voltage but change the fall under load.

  51. 51. A supply is modelled as a 15 V source in series with 0,5 Ω. What current IS would the equivalent Norton model need?
    Show answer

    30 A

    IS = VS ÷ R = 15 ÷ 0,5 = 30 A, with RShunt = 0,5 Ω, so that IS × RShunt gives the same 15 V. Multiplying, 15 × 0,5 = 7,5, is the slip.

    Try it: Ohm's Law & Power
  52. 52. A Norton model has IS = 2,5 A and RShunt = 6 Ω. What is its output voltage with no load connected?
    Show answer

    15 V

    With no load, all of IS flows through RShunt, so VOut = IS × RShunt = 2,5 × 6 = 15 V. Dividing, 2,5 ÷ 6 = 0,42, is the slip.

    Try it: Ohm's Law & Power
  53. 53. A 9 V source with 0,6 Ω internal resistance feeds a 2,4 Ω load. What voltage appears across the load?
    Show answer

    7,2 V

    The current is 9 ÷ (0,6 + 2,4) = 3 A, so the load gets 3 × 2,4 = 7,2 V. The other 1,8 V is dropped inside the source, across its internal resistance, not across the load.

    Try it: Voltage Divider
  54. 54. A 4,8 V battery pack with 0,8 Ω internal resistance is connected to a 2,2 Ω load. What current flows?
    Show answer

    1,6 A

    The internal resistance is in series with the load, so I = VS ÷ (RSource + RLoad) = 4,8 ÷ 3 = 1,6 A. Leaving out the 0,8 Ω gives 4,8 ÷ 2,2 = 2,2 A.

    Try it: Ohm's Law & Power
  55. 55. A 3,7 V lithium cell has 0,09 Ω internal resistance. Which is nearest to the current if a coin shorts its terminals?
    Show answer

    41 A

    With the terminals shorted only the internal resistance limits the current: I = VS ÷ RSource = 3,7 ÷ 0,09 = 41,1 A, nearest 41 A. Multiplying, 3,7 × 0,09 = 0,33, is the slip.

    Try it: Ohm's Law & Power
  56. 56. A 9,6 V battery pack has 0,7 Ω internal resistance. What load resistance will draw 1,2 A from it?
    Show answer

    7,3 Ω

    The whole circuit needs 9,6 ÷ 1,2 = 8 Ω, and the internal resistance supplies 0,7 Ω of that, so RLoad = 8 − 0,7 = 7,3 Ω. 8 Ω forgets the internal resistance.

    Try it: Ohm's Law & Power
  57. 57. When talking about power transfer, what is the 'load'?
    Show answer

    Whatever the circuit is there to feed

    A load is anything the circuit feeds with power: an antenna, a motor or a lamp, for example. The internal resistance dissipates power too, but it is part of the source, not the load.

  58. 58. A car battery lights the headlamps through a switch and thick cables. In the manual's terms, which is the load?
    Show answer

    The headlamps

    The load is what the source delivers power to, here the headlamps. The battery's internal resistance also heats up, but it belongs to the source.

  59. 59. For the greatest power transfer from a source to a load, what should the load resistance be?
    Show answer

    Equal to the source's internal resistance

    The load takes the most power when RLoad = RSource. A lower load draws more current, but less of the voltage appears across it, so P = V × I in the load falls.

  60. 60. A battery pack has an internal resistance of 0,3 Ω. Which load resistance takes the most power from it?
    Show answer

    0,3 Ω

    Maximum power transfer needs RLoad = RSource = 0,3 Ω. A short circuit draws the most current, but with no voltage across it the load takes no power at all.

  61. 61. A source feeds a load resistance equal to its internal resistance. How does the power heating the source compare with the power in the load?
    Show answer

    The source heats exactly as much

    The same current flows through two equal resistances, so each dissipates the same power: the source heats as much as the load receives, so cooling matters. Half as much would need a smaller internal resistance.

  62. 62. If the load resistance is much lower than the source resistance, what happens?
    Show answer

    A large current flows, but little voltage appears across the load

    A very low RLoad draws a large current, but most of the voltage is dropped inside the source, so little is left across the load. A large voltage across a tiny resistance would need an enormous current.

  63. 63. If the load resistance is much higher than the source resistance, what happens?
    Show answer

    The load voltage is high, but little current flows

    A very high RLoad takes nearly all of VS, but the current is small, so P = V × I in the load is small too. Both cannot be high, as the high resistance limits the current.

  64. 64. A 16 V source with 2 Ω internal resistance feeds a 2 Ω load. How much power does the load receive?
    Show answer

    32 W

    With RLoad = RSource only half of V reaches the load, so P = V² ÷ (4 × RLoad) = 256 ÷ 8 = 32 W. Leaving out the 4 gives 128 W, as if all 16 V reached it.

    Try it: Ohm's Law & Power
  65. 65. A 20 V source with 4 Ω internal resistance feeds a 4 Ω load. What power does the load dissipate?
    Show answer

    25 W

    P = V² ÷ (4 × RLoad) = 400 ÷ 16 = 25 W: 2,5 A flows, with 10 V across the load. Leaving out the 4 gives 100 W, as if all 20 V reached the load.

    Try it: Ohm's Law & Power
  66. 66. A 13,5 V source with 0,65 Ω internal resistance feeds a 0,65 Ω load. Which is nearest to the power in the load?
    Show answer

    70 W

    P = V² ÷ (4 × RLoad) = 182,25 ÷ 2,6 = 70,1 W, nearest 70 W. Leaving out the 4 gives 280,4 W, the 280 W slip.

    Try it: Ohm's Law & Power
  67. 67. A 39 V source with 5 Ω internal resistance feeds an 8 Ω load. What power does the load take?
    Show answer

    72 W

    The current is 39 ÷ (5 + 8) = 3 A, so the load takes I² × R = 3² × 8 = 72 W. Ignoring the internal resistance gives 39² ÷ 8 = 190 W.

    Try it: Ohm's Law & Power
  68. 68. What is the purpose of the fins on a heat sink?
    Show answer

    To give more area in contact with the air

    Fins add surface area, so more heat passes from the metal into the air around it. A heat sink holds a little heat, but its job is to pass the heat on all the time.

  69. 69. Which list contains only methods used to cool power devices?
    Show answer

    Fans, heat sinks and liquid cooling

    Fans and finned heat sinks carry heat away, and very high-power equipment may use liquid cooling. Extra series resistance would only add another source of heat.

  70. 70. The load on a source with internal resistance is adjusted until it takes the greatest possible power. What is the efficiency then?
    Show answer

    50%

    The greatest power needs RLoad = RSource, and then half the power heats the internal resistance, so the efficiency is 50%. Maximum power is not maximum efficiency: a larger load is more efficient but takes less.

  71. 71. A source with 1 Ω internal resistance feeds a 3 Ω load. What fraction of the total power reaches the load?
    Show answer

    75%

    The same current flows in both, so the power divides like the resistances: 3 ÷ (1 + 3) = 0,75 = 75%. 25% is the share that heats the source.

  72. 72. How does the manual define the efficiency of a system?
    Show answer

    Useful output power divided by input power

    Efficiency is the useful power out as a fraction of the power put in, η = POut ÷ PIn, so it lies between 0 and 1. Input divided by output is upside down and would come out above 1.

  73. 73. An antenna system is fed with 80 W but radiates only 52 W. What is its efficiency?
    Show answer

    65%

    η = POut ÷ PIn = 52 ÷ 80 = 0,65, and multiplying by 100% gives 65%. 35% is the share lost as heat, not the share radiated.

  74. 74. A solid-state amplifier draws 22 A from a 50 V supply and delivers 600 W of RF. Which is nearest to its efficiency?
    Show answer

    55%

    PIn = V × I = 50 × 22 = 1100 W, so η = 600 ÷ 1100 = 0,545, nearest 55%. 45% is the share turned into heat, and 180% divides the wrong way.

    Try it: Ohm's Law & Power
  75. 75. An amplifier is 60% efficient. How much DC input power does it need to deliver 420 W of RF?
    Show answer

    700 W

    PIn = POut ÷ η = 420 ÷ 0,6 = 700 W. Multiplying instead, 420 × 0,6 = 252 W, would make the input smaller than the output.

  76. 76. An antenna system is 80% efficient. If the transmitter feeds it 50 W, how much is radiated?
    Show answer

    40 W

    POut = η × PIn = 0,8 × 50 = 40 W. Dividing instead, 50 ÷ 0,8 = 62,5 W, would radiate more power than is fed in.

  77. 77. Which symbol is used for efficiency?
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    η (eta)

    Efficiency is written η. λ stands for wavelength, Ω for ohms and π for pi.

  78. 78. A transmitter takes 250 W of DC input and delivers 90 W of RF. How much power is turned into heat?
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    160 W

    The power not delivered as RF becomes heat: 250 − 90 = 160 W. Adding the two, 340 W, would mean more heat than the whole input.

  79. 79. A transmitter delivers 115 W of RF while 135 W of its input is lost as heat. What DC input power does it take?
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    250 W

    Every watt taken in is either delivered or lost, so PIn = POut + losses = 115 + 135 = 250 W. Subtracting, 135 − 115 = 20 W, would leave less input than output.

  80. 80. What becomes of the power fed to an antenna system that is lost rather than radiated?
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    It heats the antenna conductors and the ground nearby

    Losses in the antenna conductors and in the earth around the antenna turn the unradiated power into heat there; the manual jokes that earthworms enjoy it. It does not go back to heat the transmitter.

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