Q. Mention one reason why tungsten is used for making filament of electric lamp. [CBSE 2014]
Ans: tungsten does not oxidize, i. e. burn easily at high temperatures as it has higher melting and boiling point than metals
Q. List two characteristics of the material to be used in fuse wire. Name the material it is made of. A fuse is always connected in series in an electric circuit ? Justify this statement giving reason [CBSE 2014]
Answer:
two characteristics of the material to be used in fuse wire
(a) Low melting and boiling point (b) high electric resistance
Fuse is generally made up of an alloy of tin and lead
In series connection the current for the entire house pass through the fuse. So, when fuse melts , it breaks down the entire home circuit and no current flows to the household circuitry. Thus, a fuse is always connected in series
Q. A circuit has a line of 5 A. How many lamps of rating 40 W; 220 V can simultaneously run on this line safely ?[CBSE 2014]
Solution : P = VI so, I = P/V = 40/220
No of lamp can simultaneously run on this line safely = 5A/(40/220 )A = 27
Q. Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A?
Solution :
Resistance R 1 of the bulb = V^2/p1 = (220)2/10 = 4840 Ohm
Let lamps can be connected in parallel with each other
According to O hm’s law, V = I R
Where, R is the total resistance of the circuit for x number of electric bulbs
R = V/I = 220/5 = 44 Ohm
Number of lamp = 4840/44 = 110
Q. Express work done in an electric field in terms of charge and potential difference. Calculate the amount of work done in carrying a charge of 5 mC against a potential difference of 100 V [CBSE 2014]
Solution : Work dome = charge x potential difference
the amount of work done in carrying a charge of 5 mC against a potential difference of 100 V = 5 × 10^3 × 100 = 0. 5 J
Q. Distinguish between kilowatt and kilowatt hour. For a heater rated at 4.4 kW; 220 V . Calculate the -
(i) current drawn by the heater (ii) resistance of the heater element
(iii) energy consumed by the heater in 5 hours (iv) cost of running the heater if 1 kWh costs Rs. 6.50 [CBSE 2014]
Answer: Kilowatt is the unit of Power where as KiloWatt Hour is the commercial unit of energy.
Given, P = 4.4 kW = 4400 W and V = 220 volts
(i) As P = V /I So, I = P/V = 4400/220 = 20 A
(ii) As P = I^2R So, R = P/I^2 = 4400/(20) 2 = 11 ohms
(iii) since, electric energy = electric power X time = 4.4 kW X 4 hours = 17.6 kWh
(iv) Since, 1KwH costs Rs. 6.50, hence, costs of 17.6 kWh = 17.6 X 6.50 = Rs. 114.4
Q.Distinguish between resistance and resistivity of a conductor.
The resistors are generally made of thin wires of Eureka or Manganin while the wires used in connections are made comparatively thicker and are of copper or aluminium. Why ? Give reason.
What would happen to the resistance of a wire if it is stretched to double its length ? Justify your answer.[CBSE 2014]
Solution :
Resistivity
1. It is the measure of a material’s ability to oppose the flow of current.
2. It depends on the property of the material. Its value is constant for a particular range of temperature.
3. Unit of resistivity is Ohm meter
Resistance
1. It is a measure of the opposition that a circuit (or an electrical element) offers to the flow of electric current.
2. Resistance of a conductor is directly proportional to its length and inversely proportional to its area of cross section.
3. Unit of resistance is Ohm
The resistors are generally made of thin wires of Eureka or Manganin while the wires used in connections are made comparatively thicker and are of copper or aluminium. This is because thicker wire has less resistance than thinner wire
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Ans: tungsten does not oxidize, i. e. burn easily at high temperatures as it has higher melting and boiling point than metals
Q. List two characteristics of the material to be used in fuse wire. Name the material it is made of. A fuse is always connected in series in an electric circuit ? Justify this statement giving reason [CBSE 2014]
Answer:
two characteristics of the material to be used in fuse wire
(a) Low melting and boiling point (b) high electric resistance
Fuse is generally made up of an alloy of tin and lead
In series connection the current for the entire house pass through the fuse. So, when fuse melts , it breaks down the entire home circuit and no current flows to the household circuitry. Thus, a fuse is always connected in series
Q. A circuit has a line of 5 A. How many lamps of rating 40 W; 220 V can simultaneously run on this line safely ?[CBSE 2014]
Solution : P = VI so, I = P/V = 40/220
No of lamp can simultaneously run on this line safely = 5A/(40/220 )A = 27
Q. Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A?
Solution :
Resistance R 1 of the bulb = V^2/p1 = (220)2/10 = 4840 Ohm
Let lamps can be connected in parallel with each other
According to O hm’s law, V = I R
Where, R is the total resistance of the circuit for x number of electric bulbs
R = V/I = 220/5 = 44 Ohm
Number of lamp = 4840/44 = 110
Q. Express work done in an electric field in terms of charge and potential difference. Calculate the amount of work done in carrying a charge of 5 mC against a potential difference of 100 V [CBSE 2014]
Solution : Work dome = charge x potential difference
the amount of work done in carrying a charge of 5 mC against a potential difference of 100 V = 5 × 10^3 × 100 = 0. 5 J
Q. Distinguish between kilowatt and kilowatt hour. For a heater rated at 4.4 kW; 220 V . Calculate the -
(i) current drawn by the heater (ii) resistance of the heater element
(iii) energy consumed by the heater in 5 hours (iv) cost of running the heater if 1 kWh costs Rs. 6.50 [CBSE 2014]
Answer: Kilowatt is the unit of Power where as KiloWatt Hour is the commercial unit of energy.
Given, P = 4.4 kW = 4400 W and V = 220 volts
(i) As P = V /I So, I = P/V = 4400/220 = 20 A
(ii) As P = I^2R So, R = P/I^2 = 4400/(20) 2 = 11 ohms
(iii) since, electric energy = electric power X time = 4.4 kW X 4 hours = 17.6 kWh
(iv) Since, 1KwH costs Rs. 6.50, hence, costs of 17.6 kWh = 17.6 X 6.50 = Rs. 114.4
Q.Distinguish between resistance and resistivity of a conductor.
The resistors are generally made of thin wires of Eureka or Manganin while the wires used in connections are made comparatively thicker and are of copper or aluminium. Why ? Give reason.
What would happen to the resistance of a wire if it is stretched to double its length ? Justify your answer.[CBSE 2014]
Solution :
Resistivity
1. It is the measure of a material’s ability to oppose the flow of current.
2. It depends on the property of the material. Its value is constant for a particular range of temperature.
3. Unit of resistivity is Ohm meter
Resistance
1. It is a measure of the opposition that a circuit (or an electrical element) offers to the flow of electric current.
2. Resistance of a conductor is directly proportional to its length and inversely proportional to its area of cross section.
3. Unit of resistance is Ohm
The resistors are generally made of thin wires of Eureka or Manganin while the wires used in connections are made comparatively thicker and are of copper or aluminium. This is because thicker wire has less resistance than thinner wire
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