Electricity NCERT Exercise Questions
Part 3
Question 14: Compare the power used in the 2 Ω resistor in each of the following circuits:
(a) A 6 V battery in series with 1 Ω and 2 Ω resistors, and
Answer: Potential difference = 6 V
Resistance of resistors = 1 Ω and 2 Ω
Power used through resistors of 2 Ω =?
Since, resistors are connected in series, thus, total effective resitance, `R = 1 Ω + 2 Ω = 3 Ω`
We know; `I = V/R = 6 V ÷ 3 Ω = 2 A`
Since current remains same when resistors are connected in series, hence current through resistor of 2 Ω = 2 A
Thus, power `(P) = I^2 xx R`
`= (2 A)^2 xx 2 Ω = 8 W`
(b) A 4 V battery in parallel with 12 Ω and 2 Ω resistors.
Answer: Potential difference, V = 4V
Resistance of resistors connected in parallel = 12 Ω and 2 Ω
Power used by resistors having resistance of 2 Ω = ?
Since, voltage across the circuit remains same if resistors are connected in parallel.
Thus, power (P) used by resistance of 2 Ohm
`= (V^2)/(R) = ((4 V)^2)/(2 Ω)`
`= 16/2 = 8 W`
Thus, power used by resistance of 2 Ω in both the case = 8 W
Question 15: Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V?
Answer: Since, both the lamps are connected in parallel, thus, potential difference will be equal
Thus, potential difference = 220 V
Power of one lamp, P1 = 100W
Power of second lamp, P2 = 60W
We know, `P = VI`
Or, `I = P/V`
Thus, total current through the circuit
`I=(P_1)/(V)+(P_2)/(V)`
Or, `I=(100W)/(220V)+(60W)/(220V)`
`=(100+60)/(220)A=(160)/(220)A=0.727A`
Question 16: Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?
Answer: Given, power of TV (P) = 250W,
time (t) = 1 hr = 60 × 60 s = 3600 s
Thus, energy used by it =?
We know that energy used by appliance = Power x time
Thus, energy used by TV `= 250 W xx 3600 s = 900000 J = 9 xx 10^5 J`
Power of toaster = 1200W
Time (t) = 10 minute `= 60 xx 10 = 600 s`
Thus, energy used by toaster `= P xx t = 1200W xx 600 s = 720000 J = 7.2 xx 10^5 J`
Thus, given TV set will use more energy than toaster.
Question 17: An electric heater of resistance 8 Ω draws 15 A from the service mains 2 hours. Calculate the rate at which heat is developed in the heater.
Answer: Given, Resistance, R = 8 Ω
Electric current ( I ) = 15A
Time (t) `= 2 h = 2 xx 60 xx 60 s = 7200 s`
Rate at which heat is developed in heater =?
We know that rate of heat produced `= I^2R = (15A)^2 xx 8 Ω = 225 xx 8 J//s = 1800 J//s` Answer
Question 18: Explain the following.
Why is tungsten used almost exclusively for filament of electric lamps?
Answer: The melting point of tungsten is very high, i.e. 3380°C, which enables it not to melt at high temperature and to retain most of the heat. The heating of tungsten makes it glow. This is the cause that tungsten is used almost exclusively for filament of electric lamps.
Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal?
Answer: To produce more heat, the high melting point of conductors is necessary. The alloys of metal have higher melting points than pure metals. Thus, to retain more heat alloy is used rather than pure metal in electrical heating devices, such as bread-toaster, electric iron, etc.
Why is the series arrangement not used for domestic circuits?
Answer: There is loss of voltage in the series arrangement in the circuits because of add on effect of resistances. So, series arrangement is not used for domestic circuits.
How does the resistance of a wire vary with its area of cross-section?
Answer: Resistance of a wire is indirectly proportional to the area of cross section. Resistance increases with decrease in area of cross section and vice versa.
Why are copper and aluminium wires, usually employed for electricity transmission?
Answer: The resistivity of copper and aluminium wires are lower than that of iron but more than that of silver. These wires are cheaper than silver, that’s why copper and aluminium wires usually employed for electricity transmission.