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प्रश्न
An electric bulb is rated 240V-60W and is working at 100% efficiency.
(i) Calculate the resistance of bulb.
(ii) (a) Draw the circuit diagram.
(b) What is the rate of conversion of energy in each bulb?
(c) Total power used by the bulbs.
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उत्तर
(i) From P = `"V"^2/"R"`
Resistance of bulb (R) = `"V"^2/"P" = (240 xx 240)/60 = 960 Omega`
(ii) (a)
(b) Resistance of bulbs in series = (960 + 960) Ω = 1920 Ω
∴ Current in series circuit I = `"V"/"R" = 240/1920 = 0.125` A
∴ Rate of conversion of energy in any one bulb in 1 second = I2Rt
= (0.125)2 × 960 × 1 = 15.5
(c) Power of 1 bulb in series circuit = 15 W
So poewr of 2 bulb in series circuit = 2 × 15 W = 30 W.
संबंधित प्रश्न
The table below shows the current in three different electrical appliances when connected to the 240 V mains supply:
| Appliance | Current |
| Kettle | 8.5 A |
| Lamp | 0.4 A |
| toaster | 4.8 A |
- Which appliance has the greatest electrical resistance? How does the data show this?
- The lamp is connected to the main supply by using a thin, twin-cored cable consisting of live and natural wires. State two reasons why this cable should not be used for connecting the kettle to the mains supply.
- Calculate the power rating of the kettle when it is operated from the 240 V mains supply.
- A main takes the kettle abroad where the mains supply is 120 V. What is the current is the kettle when it is operated from the 120 V supply?
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