Advertisements
Advertisements
Question
A. bulb is rated at 100W, 250V and another one at 60W, 250V. What is the current flowing in the circuit if the two bulbs are put in series across a 220 V mains supply?
Advertisements
Solution
The resistance of first bulb = `"V"^2/"P"_1 = (250 xx 250)/100` Ω = 625 Ω
Resistance of second bulb = `"V"^2/"P" = (250 xx250)/60` Ω = `3125/3` Ω
∴ Resistance of the circuit containing the two bulbs in series
`= (625 + 3125/3) Omega = 5000/3 Omega`
∴ Current flowing `= 200 div (5000/3)`A
`=(220 xx 3)/5000` A `= (220 xx 3)/5000 xx 1000 "mA" = 132 "mA"`
RELATED QUESTIONS
What is the SI unit of (i) electric energy, and (ii) electric power?
A bulb is rated as 250 V; 0.4 A. Find its : (i) power, and (ii) resistance.
The commercial unit of energy is :
(a) watt
(b) watt-hour
(c) kilowatt-hour
(d) kilo-joule
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?
Name the S.I unit of electrical energy. How is it related to Wh?
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.
What is the function of the split rings in a d.c. motor?
An immersion rod having resistance of 50 Ω is connected to 220V main supply. Assuming that all the energy generated goes to heat the water, calculate the time taken to heat 5 kg water from 30°C to 100°C.
A house is provided with 15 bulbs of 40W, 5 bulbs of 100W, 5 fans of 80 W, and one heater of 1.0 kW. Each day bulbs are used for 4h, fans for lOh, and heater for 2h. The voltage of mains is 220 V. Calculate:
(i) Maximum power of the circuit in the house,
(ii) maximum current capacity of the main fuse in the house,
(iii) the electrical energy consumed in a week,
(iv) cost of electricity consumed at 1.25 Rs. per kWh.
