Advertisements
Advertisements
प्रश्न
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 total energy supplied by the battery in 10 minutes.
उत्तर
Given: E = 6V, r = 2Ω, i = 0.5A, t = 10 minute = 10 × 60s = 600s.
The total energy supplied by the battery = Eit
= 6 × 0.5 × 600 = 1800 J
APPEARS IN
संबंधित प्रश्न
What is the maximum power in kilowatts of the appliance that can be connected safely to a 13 A ; 230 V mains socket?
Why is an electric light bulb not filled with air? Explain why argon or nitrogen is filled in an electric bulb.
Name the S.I. unit of electrical energy. How is it related to Wh?
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.
If the rating of an electric bulb is 100W – 230V, Explain its meaning.
Three 250 W heaters are connected in parallel to a 100 V supply, Calculate the energy supply in kWh to the three heaters in 5 hour.
An Electric bulb is marked 100 W, 230 V. What how long would this lamp take to consume one unit of electricity?
Two resistances R1 = 4Ω and R2 = 6Ω are connected in series. The combination is connected with a battery of e.m.f. 6V and negligible resistance. Calculate:
(i) the heat produced per minute in each resistor,
(ii) the power supplied by the battery.
1 mV is equal to:
