Advertisements
Advertisements
प्रश्न
An electric bulb of resistance 500 Ω draws current 0.4 A from the source. Calculate:
- the power of bulb and
- the potential difference at its end.
Advertisements
उत्तर
- Resistance of electric bulb (R) = 500Ω
current drawn from the source (I) = 0.4 A
Power of the bulb (P) = VI
V = I × R
V = 0.4 × 500
V = 200 V - The potential difference at its end is 200 V.
Hence,
Power (P) = VI
P = 200 × 0.4
P = 80 W
The power of the bulb is 80 Watt.
APPEARS IN
संबंधित प्रश्न
A 2 kWh heater, a 200 W TV and three 100 W lamps are all switched on from 6 p.m. to 10 p.m. What is the total cost at Rs 5.50 per kWh?
How many joules of electrical energy are transferred per second by a 6 V; 0.5 A lamp?
(a) 30 J/s
(b) 12 J/s
(c) 0.83 J/s
(d) 3 J/s
An electric iron is rated at 750 W, 230 V. Calculate the electrical energy consumed by the iron in 16 hours .
An electric bulb, when connected across a power supply of 220 V, consumes a power of 60 W. If the supply drops to 180 V, what will be the power consumed? If the supply is suddenly increased to 240 V, what will be the power consumed?
What is the current in the circuit shown (Fig. )

The diagram 31 shows a 3 terminal plug socket.
(i) What is the purpose of the terminal E?
(ii) To which part of the appliance is the terminal E connected?
(iii) To which wire L or N, is the fuse connected and why?

With reference to the diagram shown below calculate:

(i) The equivalent resistance between P and Q.
(ii) The reading of ammeter.
(iii) The electrical power between P and Q.
A bulb is connected to a battery of e.m.f. 6V and internal resistance 2Ω A steady current of 0.5A flows through the bulb. Calculate the resistance of the bulb.
A bulb is connected to a battery of e.m.f. 6V and internal resistance 2Ω A steady current of 0.5A flows through the bulb. Calculate the heat energy dissipated in the bulb in 10 minutes.
