Advertisements
Advertisements
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.
Advertisements
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
What is the SI unit of (i) electric energy, and (ii) electric power?
The diagram below shows a magnetic kept just below the conductor AB which is kept in North South direction.

(i) In which direction will the needle deflect when the key is closed ?
(ii) Why is the deflection produced ?
(iii) What will be the change in the deflection if the magnetic needle is taken just above the conductor AB ?
(iv) Name one device which works on this principle.
Define 1 kilowatt hour.
Household wiring for lamp connections can either be done in parallel or in series.Which one would you prefer? Give a reason for your answer.
The resistance of filament of an electric heater is 500 Ω It is operated at 200V for 1 hour daily. Calculate the current drawn by the heater and the energy consumed in kWh, by the heater in a month of 30 days.
An electrical appliance is rated 1500 W, 250 V. This appliance is connected to 250 V mains. Calculate:
(i) the current drawn,
(ii) the electrical energy consumed in 60 hours,
(iii) the cost of electrical energy consumed at Rs. 2.50 per kWh.
A cooler of 1500 W, 200 volts and a fan of 500 W, 200 volts are to be used from a household supply. The rating of fuse to be used is:
If a heater coil rated lkW, 220 V is connected in series with an electric bulb of 100 W, 220 V and are supplied 200V, power consumed by the bulb in this circuit is ______.
