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Physics [2 Mark Question Answer]

Electric Power and House hold Circuits · ICSE STD 10 (ICSE - English Medium). 31 questions.

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[2 Mark Question Answer]

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31 questions · timed · auto-graded

Question 12 Marks
Establish the relation that heat produced in a metallic conductor due to the flow of current is $V^2t/R$ joules, where V is the potential difference across the ends of the conductor, R its resistance and t is the time in second for which current flows.
Answer
From the relation $Q =$ Vit joule
Applying Ohm's law, $V = iR$
or i $= V/R$
Substituting the value of i, we get
$Q = V \cdot \frac{ V }{ R } t =\frac{ V ^2 t }{ R }$ joule.
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Question 22 Marks
What is the resistance of an appliance rated at V volt, P watt?
Answer
Let I be the current drawn by the appliance.
Then Power, $P = V \times \mid$ or $I =\frac{ P }{ V }$
Now, Resistance $R =\frac{ V }{ I }$
Hence, $R = V +\frac{ P }{ V }=\frac{ V ^2}{ P }$.
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Question 32 Marks
How can electric energy consumed by an electric appliance be calculated in kilowatt hour (kWh)?
Answer
The electric energy in kWh van be calculated as follows:
Energy in kWh =
$\frac{\text { Power(in watt) } \times \text { Time }(\text { in hour })}{1000}$
or $kWh =\frac{ V (\text { volt }) \times i (\text { ampere }) \times t (\text { hour })}{1000}$
$\therefore$ Energy (in kWh) $=\frac{ Vit }{1000}$
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Question 42 Marks
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.
Answer
The potential difference across the resistor $V = IR$ volt
Charge passed in resistor $Q =$ It coulomb
(i) Work was done (or the electrical energy spend) in passing charge $Q$ coulomb through the resistor at potential difference $V$ volt $= QV$ joule
$= (It) \times (IR) = I^2Rt$ joule.
(ii) Electrical power spent $=\frac{\text { Energy }}{\text { Time }}=\frac{I^2 R t}{t}$
$= I^2R$ watt.
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Question 52 Marks
An electric kettle is rated 3kW, 250V. What should be the current capacity of the fuse with it?
Answer
The maximum current capacity of kettle, I =
$\frac{ P }{ V }=\frac{3000}{250}=12 A$
Thus, the fuse should be of current capacity slightly less than 12A.
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Question 62 Marks
State $(i)$ the $S. I.$ unit, $(ii)$ the household unit, of electrical energy. How are these units related?
Answer
(i) The S.I. unit of electrical energy is joule (J)(ii) The household unit of electrical energy is kilo-watt hour (kWh).
Relationship: $1 kWh = 1 kW \times 1h$
$= 1000W \times 3600 s$
$= 3.6 \times 10^6 J.$
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Question 72 Marks
The equation$ I^2R$ seems to suggest that the rate of heating in a resistor is reduced if resistance decreases, whereas equation $P = V^2/R$ suggests just the opposite. How do you reconcile this problem?
Answer
There is no contradiction in the relation. It $V$ is constant, on decreasing resistance, the current increases and so as $P = I^2R$ the same result we gets from equation $P = V^2/R.$
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Question 82 Marks
An Electric bulb is marked 100 W, 230 V. What current does it take?
Answer
$=\frac{100}{230}=\frac{10}{23} A =0.435 A$
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Question 92 Marks
An Electric bulb is marked 100 W, 230 V. What find the energy consumed by the bulb in one hour.
Answer
Energy consumed in Is is 100 J.
∴ Energy consumed in hour = 100 × 60 × 60J = 360000 J.
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Question 102 Marks
An electric bulb is rated as 100W – 250V. What information does it convey? Calculate the current through the filament of the bulb.
Answer
Given: P = 100 W, V = 250 volt
The current through the filament, I =
$\frac{ P }{ V }=\frac{100}{250}=0.4 A$
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Question 112 Marks
An electric bulb is rated as 100W – 250V. What information does it convey? Calculate the resistance of its filament while glowing.
Answer
Given: P = 100 W, V = 250 volt
The resistance of filament, R =
$\frac{ V ^2}{ P }=\frac{(250)^2}{100}=625 \Omega$
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Question 122 Marks
Find the energy released by a current of 0.25 amperes flowing through a heater for 5 minutes. The p.d. is 230 V.
Answer
V = 230 volts; t = 5 minutes = 60 × 5 = 300 s, I = 0.25 amperes
$W = VIt =230 \times \frac{25}{100} \times 300$
W = 17250 Joules
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Question 132 Marks
In what unit does the electric meter in a house measure the electrical energy consumed? What is its value in $S.I$ unit?
Answer
The domestic electric meter measure the electrical consumption in kilowatt-hour (kwh).$1 kWh = 1 kilowatt \times 1 hour$
$= 1000 Js^{-1} \times (60 \times 60) s$
$= 3.6 \times 10^6 J$ in S.I. unit.
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Question 142 Marks
State other bigger units of electric power and express them in watt.
Answer
The other bigger units of power commonly used are:$1$ kilowatt $= 1000$ watt
$1$ Horse power $(H.P.) = 746$ watt
Megawatt $= 10^6$​​​​​​​ watt
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Question 152 Marks
Why should switches always be connected to the live wire? Give one precaution that should be taken while handling switches.
Answer
The switch should always be connected to the live wire, so that current is cut off to that appliance to which it is connected. A switch should not be touched with wet hands.
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Question 162 Marks
Two electric lamps each rating 100 W, 110 V are connected in series to a 220 V power supply and two other electric lamps each of marking 100 W, 220 V are connected in parallel to the same power supply. Will anyone of the two combinations give more light than the other? Give a reason for your answer.
Answer
Both combinations will give the same light.
Reason: Each combination has the same power namely 200 W.
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Question 172 Marks
Name a metal that is used as an electron emitter. Give one reason for using this metal.
Answer
Thoriated tungsten (tungsten coated with carbon and thorium) is used as an electron emitter. Thoriated tungsten is used since it has a work function of only 2-6 eV and it needs to be heated to only 2000K to emit electrons.
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Question 182 Marks
State two advantages of filling an inert gas in an evacuated electric filament lamp.
Answer
(i) The inert gas is filled at very low pressure to prevent oxidation as well as evaporation and thus increases its life.
(ii) While bulb is working, a convection current is set-up within it which carries the spurting particles of tungsten upward towards the cap and hence the blackening of the lower part (transparent portion) of the bulb is avoided.
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Question 192 Marks
Where is a fuse placed in an electrical circuit? Name a material suitable for making a fuse wire.
Answer
The fuse is placed in the live wire just after the ‘electricity meter’ of the consumer. Fuse wires are generally made of an alloy of tin and lead and have a relatively low melting point.
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Question 202 Marks
Express watt hour and kilowatt hour in joule.
Answer
$1 kWh = 1W \times 1 hour = 1000 W \times (60 \times 60) s$
$= 1W \times 1s \times 10^5 \times 36 = 36 \times 10^5 J = 3.6 \times 10^6 J$
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Question 212 Marks
The electric bulbs used these days have a coil-coiled filament. State its reason.
Answer
On account of having a coil-coiled filament, the area of spread of tungsten is much and so the heat now produced in a very small area, raises the temperatures to a high degree i.e., 2300°C, producting more bright light.
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Question 222 Marks
In a three-pin plug top, one of the pins is thicker and longer than the other two, Give reason. Name this pin.
Answer
The thicker arid longer pin is the earth pin. It is longer so that the earth connection is made first of all and it is thicker so that even by mistake it cannot be inserted into the live socket.
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Question 232 Marks
Why is a concave reflector placed behind the heating element in a room heater.
Answer
The concave reflector, reflects the heat radiations in the forward direction. Thus, large amount of heat radiations are available in one particular direction.
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Question 242 Marks
Why is heating element wound on a long porcelain rod in a room heater.
Answer
This reduces the area over which element of the heater is spread. As heat is produced over smaller area, therefore temperature of its element rises to about 1000°C.
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Question 252 Marks
State the colour coding of wires in the flexible lead used for connecting the appliances to the mains in a house circuiting.
Answer
The flexible lead has three wires with insulation of different colours. The red (or brown) is for live, the black (or light blue) is for neutral and the green (or yellow) is for earth connections.
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Question 262 Marks
A bulb is connected to a battery of e.m.f. $6 V$ and internal resistance $2 \Omega A$ steady current of $0.5 A$ flows through the bulb. Calculate the heat energy dissipated in the bulb in $10$ minutes.
Answer
Given:$ E = 6V, r = 2Ω, i = 0.5A, t = 10$ minute $= 10 \times 60s = 600s.$
The heat energy dissipated in the bulb $= i^2Rt$
$= (0.5)^2 \times 10 \times 600 = 1500 J$
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Question 272 Marks
An electric appliance is rated at 1000 W, 250V. Calculate:
(i) The electrical energy consumed by the appliance in 12 hour.
(ii) The cost of energy consumed at Rs. 1.20 per kWh.
How long would this appliance take to use 1 kWh energy when operated at 250 V?
Answer
Given: $P=1000 W , V =250 V , t =12 hr$
(i) Electrical energy consumed $=\frac{\text { Watt }}{1000} \times$ time $($ in hr. $)$
$
=\frac{1000}{1000} \times 12=12 kWh
$
(ii) Cost $=12 \times 1.20=$ Rs. 14.40
If it takes $t$ hour to consume $1 kWh$ electric energy, then:
$
I=\frac{1000 \times t}{1000}
$
or $t=1 hr$.
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Question 282 Marks
An electric bulb is rated as 100W – 250V. What information does it convey? Calculate the current through the filament of the bulb.
Answer
Given: P = 100 W, V = 250 volt
The current through the filament, I =
$\frac{ P }{ V }=\frac{100}{250}=0.4 A$
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Question 292 Marks
An electric bulb is rated as 100W – 250V. What information does it convey? Calculate the resistance of its filament while glowing.
Answer
Given: P = 100 W, V = 250 volt
The resistance of filament, R =
$\frac{ V ^2}{ P }=\frac{(250)^2}{100}=625 \Omega$
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Question 302 Marks
An electric heater is rated 220 V, 550 W. Calculate the electrical energy consumed in 3 hours?
Answer
Energy consumed in $3 hrs =\frac{550 \times 3}{1000}=1.65 kWh$
= 1.65 × 3600 = 5940 kJ.
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Question 312 Marks
Three 250 W heaters are connected in parallel to a 100 V supply, Calculate the resistance of each heater.
Answer

The resistance of each heater
Since $P=\frac{V^2}{R}$
or $I =\frac{ V ^2}{ R }=\frac{100 \times 100}{250}=40 \Omega$
Therefore, 40 Ω is the resistance of each heater.
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[2 Mark Question Answer] - Physics STD 10 Questions - Vidyadip