Advertisements
Advertisements
प्रश्न
Two bulbs are rated: bulb A 100W, 120 V bulb B 10 W, 120 V. If both are connected across a 120V supply, which bulb will consume more energy, When in parallel? Also calculate the current through each bulb in the above cases.
Advertisements
उत्तर
In the case of bulb A, Power: P = `"V"^2/"R"_"A"`
RA = `"V"^2/"P" = (120 xx 120)/100` = 144 Ω
i.e., Resistance (RA ) of bulb, A = 144 Ω
Similarly resistance RB of bulb, B = `"V"^2/"P"`
`= (120 xx 120)/10 = 1440` Ω
When the bulb are connected in Parallel:
Voltage is the same (120 V) across each bulb.
∴ The current in bulb A or IA = `"V"/"R"_"A" = 120/144 = 0.83` A
∴ The current in bulb B or IB `= 120/1440` = 0.083 A
Hence, power consumed in bulb A = PA = `"V"^2/"R"_"A" = (120 xx 120)/144`
= 100 W
and power consumed in bulb B = PB = `"V"^2/"R"_"B"`
`= (120 xx 120)/1440 = 10 "W"`
Hence, bulb A (100W, 120 V) consumes more energy then bulb B, when both are connected in parallel.
संबंधित प्रश्न
In a house two 60 W electric bulbs are lighted for 4 hours, and three 100 W bulbs for 5 hours everyday. Calculate the electric consumed in 30 days.
Calculate the cost of operating a heater of 500 W for 20 hours at the rate of Rs 3.90 per unit.
State and define the household unit of electricity.
A current of 2 A is passed through a coil of resistance 75Ω for 2 minutes. (a) How much heat energy is produced? (b) How much charge is passed through the resistance?
Two bulbs are rated 60 W, 220 V and 60 W, 110 V respectively. Calculate the ratio of their
resistances
What changes in energy occur in an electric heater?
How is the amount of heat produced calculated due to the passage of current in a metallic conductor? Derive an expression for it.
Find the energy released by a current of 0.25 amperes flowing through a heater for 5 minutes. The p.d. is 230 V.
An electric iron is rated at 230 V, 750 W. What is its resistance? What maximum current can be passed through it?
An electrical appliance is rated 1500 W, 250 V. This appliance is connected to 250 V mains. Calculate:
(i) the current drawn,
(ii) the electrical energy consumed in 60 hours,
(iii) the cost of electrical energy consumed at Rs. 2.50 per kWh.
