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
Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V?
The diagram below shows a circuit containing a lamp L, a voltmeter and an ammeter,. The voltmeter reading 3 V and the ammeter reading is 0.5 A.
What is the resistance of the lamp?
What is the power of the lamp?
When an electric lamp is connected to 12 V battery, it draws a current of 0.5 A. The power of the lamp is:
(a) 0.5 W
(b) 6 W
(c) 12 W
(d) 24 W
The SI unit of energy is :
(a) joule
(b) coulomb
(c) watt
(d) ohm-metre
A geyser is rated 1500 W, 250 V. This geyer is connected to 250 V mains. Calculate:
(i) the current drawn
(ii) the energy consumed in 50 hours, and
(iii) the cost of energy consumed at Rs. 4.20 per kWh
The coil of an electric bulb takes 40 watts to start glowing. If more than 40 W are supplied, 60% of the extra power is converted into light and the remaining into heat. The bulb consumes 100 W at 220 V. Find the percentage drop in the light intensity at a point if the supply voltage changes from 220 V to 200 V.
Which material is the calorimeter commonly made of? Give one reason for using this material.
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:
1 mV is equal to:
