Advertisements
Advertisements
प्रश्न
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.
उत्तर
Given: R1 = 4Ω, R2 = 6Ω, E = 6V, t = 1 minute = 60s.
Total resistance of the circuit R = R1 + R2 = 4 + 6 = 10Ω
Current through the battery I = `"E"/"R" = 6/10 = 0.6` A
The same current flows through each resistor.
(i) The heat produced in resistor R1 is H1 = I2R1t
= (0.6)2 × 4 × 60 = 86.4 J
The heat produced in resistor R2 is H2 = I2R2t
= (0.6)2 × 6 × 60 = 129.6 J
(ii) The power supplied by the battery = E × I
= 6 × 0.6 = 3.6 W.
APPEARS IN
संबंधित प्रश्न
Name the commercial unit of electric energy.
What is the meaning of the symbol kWh? Which quantity does it represent?
The unit for expressing electric power is:
(a) volt
(b) joule
(c) coulomb
(d) watt
An electric bulb is rated ‘100 W, 250 V’. How much current will the bulb draw if connected to a 250 V supply?
What is the resistance, under normal working conditions of a 240 V electric lamp rated at 60 W? If two such lamps are connected in series across a 240 V mains supply, explain why each one appears less bright.
Why does not the heating coil produce any visible light? in an electric stove.
If the rating of an electric bulb is 100W – 230V, Explain its meaning.
An electric heater is rated 220 V, 550 W. Calculate the electrical energy consumed in 3 hours?
A fuse is rated 5 A. Can it be used with a geyser rated 1540 W, 220 V Write Yes or No. Give supporting calculations to justify your answer.