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प्रश्न
A bulb is connected to a battery of p.d. 4 V and internal resistance 2.5 Ω . A steady current of 0.5 A flows through the circuit. Calculate:
The energy dissipated in the bulb in 10 minutes.
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उत्तर
Given ,
Voltage , V = 4 V
Resistance of the battery , RB = 2.5 Ω
Current , I = 0.5 A
Energy dissipated in the bulb in 10 min , E = I2Rt
E = (0.5)2 × 5.5 × 600 = 825 J
संबंधित प्रश्न
Which of the following units could be used to measure electric charge?
(a) ampere
(b) joule
(c) volt
(d) coulomb
Which type of circuit, series or parallel, is preferred while connecting a large number of bulbs:
(a) for decorating a hotel building from outside?
For a heater rated at 4 kW and 220 V, calculate:
(a) the current,
(b) the resistance of the heater,
(c) the energy consumed in 2 hours, and
(d) the cost if 1 kWh is priced at Rs 4.60.
An electric kettle has a heating element which is rated at 2 kW when connected to a 250 V electric supply. Calculate the
(a) Current which flows in the element when it is connected to a 250 V supply,
(b) Resistance of the filament,
(c) Heat produced in 1 minute by the element,
(d) Cost of running the kettle for 10 minutes a day for 30 days of the month, at the rate of Rs. 3.00 per unit.
Electrical power P is given by the expression P = (Q × V) ÷ time
(a) What do the symbols Q and V represent?
(b) Express the power P in terms of current and resistance explaining the meanings of symbols used there in.
Three heaters each rated 250 W, 100 V are connected in parallel to a 100 V supply. Calculate :
The total current taken from the supply
State expression for Cells connected in series.
Problems for practice:
If 5A current flows through a circuit, then convert the current in terms of microampere?
For current to flow, one needs art open circuit.
