Advertisements
Advertisements
प्रश्न
An electric lamp A of 40 W and another electric lamp B of 100 W are connected to 220 V supply. Calculate the ratio of their filament resistances?
Advertisements
उत्तर
P = `"V.I" = "V.V"/"R"`
`40 = (220)^2/"R"_"A"`
RA = `(220 xx 220)/40`
RA = 110 × 11
P = `"V"^2/"R"`
`100 = (220)^2/"R"_"B"`
`"R"_"B" = (220 xx 220)/100`
RB= 22 × 22
`"R"_"A"/"R"_"B" = (110 xx 11)/(22 xx 22) = 5/2 = 5 : 2`
`"R"_"A"/"R"_"B" = 5: 2`
संबंधित प्रश्न
Write the SI units for:
(a) Electric current
(b) Potential difference
(c) Charge.
What does the unit kilowatt hour measure? Define it.
Why is heating element wound on a long porcelain rod in a room heater.
Why is the heating element of an electric oven wound on a helix? State the reason.
What changes in energy occur in an electric heater?
An electric filament lamp is connected to a supply of voltage higher than the recommended value. Give reasons, why compared with a normal performance, the lamp emits a brighter light and its life is shortened.
The resistance of filament of an electric heater is 500 Ω It is operated at 200V for 1 hour daily. Calculate the current drawn by the heater and the energy consumed in kWh, by the heater in a month of 30 days.
In the given figure, the emf of the cell is 2.2 V, and if internal resistance is 0.6 Ω. Calculate the power dissipated in the whole circuit.

