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.
Advertisements
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 S.I. unit of electrical energy. How is it related to Wh?
State and define the household unit of electricity.
Point out two main disadvantages of ‘Tree System’.
Which material is the calorimeter commonly made of? Give one reason for using this material.
An electric heater is rated 220 V, 550 W. What does this statement mean?
I ampere of current is passed through a resistor of resistance R ohm for time t second. Obtain expression for
(i) the electrical energy, and
(ii) the electrical power spent.
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?
A 100-watt electric bulb is used for 5 hours daily and four 60 watt bulbs are used for 5 hours daily. Calculate the energy consumed (in kWh) in the month of January.
