Advertisements
Advertisements
Question
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
Solution
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
RELATED QUESTIONS
What is meant by "electric power"?
How many joules are there in one kilowatt-hour?
The unit for expressing electric power is:
(a) volt
(b) joule
(c) coulomb
(d) watt
State the S.I. unit of electrical power.
An electric heater is rated 220 V, 550 W. Electrical energy is also measured in kWh. What do you understand by a kilo-watt-hour?
A bulb is connected to a battery of e.m.f. 6V and internal resistance 2Ω A steady current of 0.5A flows through the bulb. Calculate the total energy supplied by the battery in 10 minutes.
A battery of e.m.f. 12V and internal resistance 1.6 Ω is connected to two resistors of 4Ω and 6Ω connected in parallel. Calculate:
(i) the current drawn from the battery,
(ii) the power dissipated in each resistor,
(ii)the total power supplied by the battery.
If a heater coil rated lkW, 220 V is connected in series with an electric bulb of 100 W, 220 V and are supplied 200V, power consumed by the bulb in this circuit is ______.
