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A torch bulb is labelled 2.5 V, 0.3 A. Calculate the power of the bulb.

Physics

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284. A torch bulb is labelled 2.5 V, 0.3 A. Calculate the power of the bulb.

  1. Use the formula for electrical power, P=V×IP = V \times I.

    P=2.5 V×0.3 AP = 2.5 \text{ V} \times 0.3 \text{ A} P=0.75 WP = 0.75 \text{ W}

    The power of the bulb is 0.75 W\boxed{\text{0.75 W}}.

    285. An electric bulb is labelled '40 W, 240 V'. Calculate:

    a) the resistance of the filament used in the bulb.

  2. Use the formula for resistance in terms of power and voltage, R=V2PR = \frac{V^2}{P}.

    R=(240 V)240 WR = \frac{(240 \text{ V})^2}{40 \text{ W}} R=57600 V240 WR = \frac{57600 \text{ V}^2}{40 \text{ W}} R=1440 ΩR = 1440 \, \Omega

    The resistance of the filament is 1440 Ω\boxed{\text{1440 } \Omega}.

    b) the current through the filament when the bulb works normally.

  3. Use the formula for current in terms of power and voltage, I=PVI = \frac{P}{V}.

    I=40 W240 VI = \frac{40 \text{ W}}{240 \text{ V}} I=0.1666... AI = 0.1666... \text{ A} I≈0.17 AI \approx 0.17 \text{ A}

    The current through the filament is 0.17 A\boxed{\text{0.17 A}}.

    286. A 3 kW immersion heater is used to heat water. Calculate the electrical energy converted into heat energy in 40 minutes.

  4. Convert power to Watts and time to seconds.

    P=3 kW=3×1000 W=3000 WP = 3 \text{ kW} = 3 \times 1000 \text{ W} = 3000 \text{ W} t=40 minutes=40×60 s=2400 st = 40 \text{ minutes} = 40 \times 60 \text{ s} = 2400 \text{ s}
  5. Use the formula for energy, E=P×tE = P \times t.

    E=3000 W×2400 sE = 3000 \text{ W} \times 2400 \text{ s} E=7,200,000 JE = 7,200,000 \text{ J}

    The electrical energy converted is 7,200,000 J\boxed{\text{7,200,000 J}}.

    287. What is the power dissipated in a 6 Ω\Omega resistor when the current through it is?

    The current value is missing from the question. Please provide the current to calculate the power dissipated.

    288. In 5 seconds, an electric iron takes 10 000 joules of energy from the main supply. What is its power:

    a) In Watts?

  6. Use the formula for power, P=EtP = \frac{E}{t}.

    P=10000 J5 sP = \frac{10000 \text{ J}}{5 \text{ s}} P=2000 WP = 2000 \text{ W}

    The power in Watts is 2000 W\boxed{\text{2000 W}}.

    b) In Kilowatts?

  7. Convert Watts to Kilowatts (1 kW=1000 W1 \text{ kW} = 1000 \text{ W}).

    P=2000 W1000P = \frac{2000 \text{ W}}{1000} P=2 kWP = 2 \text{ kW}

    The power in Kilowatts is 2 kW\boxed{\text{2 kW}}.

    289. How much electric energy in joules does a 150 watts lamp convert to heat and light energy in:

    a) 15 seconds

  8. Use the formula for energy, E=P×tE = P \times t.

    E=150 W×15 sE = 150 \text{ W} \times 15 \text{ s} E=2250 JE = 2250 \text{ J}

    The energy converted is 2250 J\boxed{\text{2250 J}}.

    b) 5 seconds

  9. Use the formula for energy, E=P×tE = P \times t.

    E=150 W×5 sE = 150 \text{ W} \times 5 \text{ s} E=750 JE = 750 \text{ J}

    The energy converted is 750 J\boxed{\text{750 J}}.

    c) 1 min

  10. Convert time to seconds.

    t=1 min=60 st = 1 \text{ min} = 60 \text{ s}
  11. Use the formula for energy, E=P×tE = P \times t.

    E=150 W×60 sE = 150 \text{ W} \times 60 \text{ s} E=9000 JE = 9000 \text{ J}

    The energy converted is 9000 J\boxed{\text{9000 J}}.

    290. Define electrical power as used in electrical circuit.

    Electrical power is the rate at which electrical energy is converted into other forms of energy (like heat, light, or mechanical energy) or transferred within an electrical circuit. It is measured in Watts (W).

    291. Calculate the (a) current through (b) resistance of the filament of

    a) a bulb rated at 240 V, 60 W

  12. Calculate the current using I=PVI = \frac{P}{V}.

    I=60 W240 VI = \frac{60 \text{ W}}{240 \text{ V}} I=0.25 AI = 0.25 \text{ A}

    The current is 0.25 A\boxed{\text{0.25 A}}.

  13. Calculate the resistance using R=V2PR = \frac{V^2}{P}.

    R=(240 V)260 WR = \frac{(240 \text{ V})^2}{60 \text{ W}} R=57600 V260 WR = \frac{57600 \text{ V}^2}{60 \text{ W}} R=960 ΩR = 960 \, \Omega

    The resistance is 960 Ω\boxed{\text{960 } \Omega}.

    b) an electric kettle rated at 2 kW, 240 V.

  14. Convert power to Watts.

    P=2 kW=2000 WP = 2 \text{ kW} = 2000 \text{ W}
  15. Calculate the current using I=PVI = \frac{P}{V}.

    I=2000 W240 VI = \frac{2000 \text{ W}}{240 \text{ V}} I≈8.33 AI \approx 8.33 \text{ A}

    The current is 8.33 A\boxed{\text{8.33 A}}.

  16. Calculate the resistance using R=V2PR = \frac{V^2}{P}.

    R=(240 V)22000 WR = \frac{(240 \text{ V})^2}{2000 \text{ W}} R=57600 V22000 WR = \frac{57600 \text{ V}^2}{2000 \text{ W}} R=28.8 ΩR = 28.8 \, \Omega

    The resistance is 28.8 Ω\boxed{\text{28.8 } \Omega}.

    292. a) A washing machine is marked 240 V, 3 kW. What does this mean?

    This means that the washing machine is designed to operate safely and efficiently when connected to a 240 Volt electrical supply. When operating normally, it will consume electrical energy at a rate of 3 kilowatts (3000 Watts).

    b) Calculate the electrical energy consumed by this machine in 1 hour.

  17. Use the formula for energy, E=P×tE = P \times t.

    E=3 kW×1 hE = 3 \text{ kW} \times 1 \text{ h} E=3 kWhE = 3 \text{ kWh}

    The electrical energy consumed is 3 kWh\boxed{\text{3 kWh}}.

    293. a) Which bulb in Fig. below is the brightest? Explain your answer.

    The figure shows bulb P (2000 Ω\Omega) and bulb Q (500 Ω\Omega) connected in series.

  18. In a series circuit, the current (II) flowing through both bulbs is the same.

  19. The brightness of a bulb is determined by the power it dissipates, given by P=I2RP = I^2 R.

  20. Since II is the same for both, the bulb with the higher resistance (RR) will dissipate more power and thus be brighter.

  21. Comparing the resistances, Bulb P has 2000 Ω2000 \, \Omega and Bulb Q has 500 Ω500 \, \Omega.

    Therefore, Bulb P is the brightest because it has a higher resistance and will dissipate more power (P=I2RP = I^2 R) for the same current flowing through it.

    b) Both the bulbs in Fig. below are dimmer compared to the normal brightness, when each bulb is connected in turn to the same power supply. Why?

    The bulbs are dimmer because the voltage supplied by the "same power supply" is lower than their individual rated voltage required for normal brightness. When connected in series, the voltage across each bulb is even lower than the supply voltage, making them even dimmer than if connected individually to that same supply.

    294. A 2 kW immersion water heater is switched on for 3 hours. Calculate the amount of heat energy given off by the heater.

  22. Use the formula for energy, E=P×tE = P \times t.

    E=2 kW×3 hE = 2 \text{ kW} \times 3 \text{ h} E=6 kWhE = 6 \text{ kWh}

    The amount of heat energy given off is 6 kWh\boxed{\text{6 kWh}}.

    295. If an electric heater consumes a current of 4 A when connected to a 240 V supply, what is its power.

  23. Use the formula for power, P=V×IP = V \times I.

    P=240 V×4 AP = 240 \text{ V} \times 4 \text{ A} P=960 WP = 960 \text{ W}

    The power of the heater is 960 W\boxed{\text{960 W}}.

    296. A bulb rated 200 W is used for 12 hours. Calculate the energy it consumed in kWh.

  24. Convert power to kilowatts.

    P=200 W=0.2 kWP = 200 \text{ W} = 0.2 \text{ kW}
  25. Use the formula for energy, E=P×tE = P \times t.

    E=0.2 kW×12 hE = 0.2 \text{ kW} \times 12 \text{ h} E=2.4 kWhE = 2.4 \text{ kWh}

    The energy consumed is 2.4 kWh\boxed{\text{2.4 kWh}}.

    297. Define a fuse and state its function in an electrical circuit.

    A fuse is a safety device that contains a thin wire designed to melt and break an electrical circuit if the current exceeds a safe level. Its function is to protect electrical appliances and wiring from damage due to overcurrents or short circuits, thereby preventing fires and electrical hazards.

    298. Name any four devices that have electric heating elements.

    • Electric kettle
    • Toaster
    • Electric iron
    • Immersion heater

    299. How much current flows through a bulb rated 150 W when a potential difference of 240 V is supplied to the bulb when operating normally?

  26. Use the formula for current, I=PVI = \frac{P}{V}.

    I=150 W240 VI = \frac{150 \text{ W}}{240 \text{ V}} I=0.625 AI = 0.625 \text{ A}

    The current flowing through the bulb is 0.625 A\boxed{\text{0.625 A}}.

    300. Explain why a magnetic compass deflects when brought near a conductor carrying current.

    A magnetic compass deflects because a conductor carrying electric current produces a magnetic field around it. This magnetic field interacts with the magnetic field of the compass needle, causing the needle to align itself with the resultant magnetic field, thus deflecting from its original orientation (usually pointing north). This phenomenon is known as Oersted's discovery.

    301. Define the following terms:

    The terms to be defined are missing from the question. Please provide the terms you would like defined.

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