A abampere (aA) The unit of electric current in the CGSeniu system, defined as that current that, if flowing through two parallel conductors of negligible cross section and infinite length, placed 1 cm apart in vacuo, would produce on each conductor a force of 1 dyne per centimeter of length. 1 abampere = 1 abcoulomb/s = r statampere (where c = speed of light in cm/s) = 10 ampere. aberration Imperfect image formation due to geometric imperfections in the optical elements of a system ablation 1 . The wasting of glacier ice by any process (calving, melting, evaporation, etc.). 2. The shedding of molten material from the outer sur- face of a meteorite or tektite during its flight through the atmosphere. absolute age The age of a natural substance, of a fossil or living organism, or of an artifact, obtained by means of an absolute dating method. See absolute dating method. absolute density Density in kg/m' or, more commonly, in g/cm\ both at STP. Cf. density, relative density abso
Electricity class 10 NCERT physics MCQ & SAQ
Electricity class 10 NCERT physics NCERT SOLUTION:
1. What does an electric circuit mean?
A continuous and closed path of an electric current is called an electric circuit. An electric circuit consists of electric devices, a source of electricity and wires that are connected with the help of a switch.
2. Define the unit of current.
Answer
The unit of electric current is ampere (A). 1 A is defined as the flow of 1 C of charge through a wire in 1 s.
3. Calculate the number of electrons constituting one coulomb of charge.
Answer
One electron possesses a charge of 1.6 ×10-19C, i.e., 1.6 ×10-19C of charge is contained in 1 electron.
∴ 1 C of charge is contained in 1/1.6 x 10-19 = 6.25 x 1018 = 6 x 1018
Therefore, 6 x 1018 electrons constitute one coulomb of charge.
Page No: 202
1. Name a device that helps to maintain a potential difference across a conductor.
Answer
Any source of electricity like battery, cell, power supply, etc. helps to maintain a potential difference across a conductor.
2. What is meant by saying that the potential difference between two points is 1 V?
Answer
If 1 J of work is required to move a charge of amount 1 C from one point to another, then it is said that the potential difference between the two points is 1 V.
3. How much energy is given to each coulomb of charge passing through a 6 V battery?
Answer
The energy given to each coulomb of charge is equal to the amount of work which is done in moving it.
Now we know that,
Potential difference = Work Done/Charge
∴ Work done = Potential difference × charge
Where, Charge = 1 C and Potential difference = 6 V
∴ Work done = 6×1
= 6 Joule.
Page No: 209
1. On what factors does the resistance of a conductor depend?
Answer
The resistance of a conductor depends upon the following factors:
→ Length of the conductor
→ Cross-sectional area of the conductor
→ Material of the conductor
→ Temperature of the conductor
2. Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?
Answer
The current will flow more easily through thick wire. It is because the resistance of a
conductor is inversely proportional to its area of cross - section. If the wire is thicker, less resistance and hence more easily the current flows.
3. Let the resistance of an electrical component remain constant while the potential difference across the two ends of the component decreases to half of its former value. What changes will occur in the current through it?
Answer
According to Ohm’s law
V = IR
⇒ I=V/R ... (1)
Now Potential difference is decreased to half
∴ New potential difference Vʹ=V/2
Resistance remains constant
So the new current Iʹ = Vʹ/R
= (V/2)/R
= (1/2) (V/R)
= (1/2) I = I/2
Therefore, the amount of current flowing through the electrical component is reduced by half.
4. Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?
Answer
The resistivity of an alloy is higher than the pure metal. Moreover, at high temperatures, the alloys do not melt readily. Hence, the coils of heating appliances such as electric toasters and electric irons are made of an alloy rather than a pure metal.
5. Use the data in Table 12.2 to answer the following -
Table 12.2 Electrical resistivity of some substances at 20°C
Answer
(a) Resistivity of iron = 10.0 x 10-8 Ω
Resistivity of mercury = 94.0 x 10-8 Ω
Resistivity of mercury is more than that of iron. This implies that iron is a better conductor than mercury.
(b) It can be observed from Table 12.2 that the resistivity of silver is the lowest among the listed materials. Hence, it is the best conductor.
Page No: 213
1. Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 Ω resistor, an 8 Ω resistor, and a 12 Ω resistor, and a plug key, all connected in series.
Answer
Three cells of potential 2 V, each connected in series, therefore the potential difference of the battery will be 2 V + 2 V + 2 V = 6V. The following circuit diagram shows three resistors of resistances 5 Ω, 8 Ω and 12 Ω respectively connected in series and a battery of potential 6 V and a plug key which is closed means the current is flowing in the circuit.
2. Redraw the circuit of question 1, putting in an ammeter to measure the current through the resistors and a voltmeter to measure potential difference across the 12 Ω resistor. What would be the readings in the ammeter and the voltmeter?
Answer
An ammeter should be connected in the circuit in series with the resistors. To measure the potential difference across the resistor it should be connected in parallel, as shown in the following figure.
The resistances are connected in series.
Ohm’s law can be used to obtain the readings of ammeter and voltmeter. According to Ohm’s law,
V = IR,
Where,
Potential difference, V = 6 V
Current flowing through the circuit/resistors = I
Resistance of the circuit, R = 5 + 8 + 12 = 25Ω
I = V/R = 6/25 = 0.24 A
Potential difference across 12 Ω resistor = V1
Current flowing through the 12 Ω resistor, I = 0.24 A
Therefore, using Ohm’s law, we obtain
V1 = IR = 0.24 x 12 = 2.88 V
Therefore, the reading of the ammeter will be 0.24 A.
The reading of the voltmeter will be 2.88 V.
Page No: 216
1. Judge the equivalent resistance when the following are connected in parallel − (a) 1 Ω and 106Ω, (b) 1 Ω and 103Ω and 106Ω.
Answer
(a) When 1 Ω and 106 Ω are connected in parallel:
Let R be the equivalent resistance.
Therefore, equivalent resistance ≈ 1 Ω
(b) When 1Ω, 103 Ω and 106 Ω are connected in parallel:
Let R be the equivalent resistance.
Therefore, equivalent resistance = 0.999 Ω
2. An electric lamp of 100 Ω, a toaster of resistance 50 Ω, and a water filter of resistance 500 Ω are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?
Answer
Resistance of electric lamp, R1 = 100 Ω
Resistance of toaster, R2 = 50 Ω
Resistance of water filter, R3 = 500 Ω
Potential difference of the source, V = 220 V
These are connected in parallel, as shown in the following figure.
Let R be the equivalent resistance of the circuit.
3. What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?
Answer
There is no division of voltage among the appliances when connected in parallel. The potential difference across each appliance is equal to the supplied voltage.
The total effective resistance of the circuit can be reduced by connecting electrical appliances in parallel.
4. How can three resistors of resistances 2 Ω, 3 Ω and 6 Ω be connected to give a total resistance of (a) 4 Ω, (b) 1 Ω?
Answer
There are three resistors of resistances 2 Ω, 3 Ω, and 6 Ω respectively.
(a) The following circuit diagram shows the connection of the three resistors.
Here, 6 Ω and 3 Ω resistors are connected in parallel.
Therefore, their equivalent resistance will be given by
This equivalent resistor of resistance 2 Ω is connected to a 2 Ω resistor in series.
Therefore, the equivalent resistance of the circuit = 2 Ω + 2 Ω = 4 Ω
Hence the total resistance of the circuit is 4 Ω.
(b) The following circuit diagram shows the connection of the three resistors.
All the resistors are connected in series. Therefore, their equivalent resistance will be given as
Therefore, the total resistance of the circuit is 1 Ω.
5. What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 Ω, 8 Ω, 12 Ω, 24 Ω?
Answer
There are four coils of resistances 4 Ω, 8 Ω, 12 Ω and 24 Ω respectively.
(a) If these coils are connected in series, then the equivalent resistance will be the highest, given by the sum 4 + 8 + 12 + 24 = 48 Ω
(b) If these coils are connected in parallel, then the equivalent resistance will be the lowest, given by
Therefore, 2 Ω is the lowest total resistance.
Page No: 218
1. Why does the cord of an electric heater not glow while the heating element does?
Answer
The heating element of the heater is made up of alloy which has very high resistance so when current flows through the heating element, it becomes too hot and glows red. But the resistance of cord which is usually of copper or aluminum is very low so it does not glow.
2. Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.
Answer
Given Charge, Q = 96000C
Time, t= 1hr = 60 x 60= 3600s
Potential difference, V= 50volts
Now we know that H= VIt
So we have to calculate I first
As I= Q/t
∴ I = 96000/3600 = 80/3 A
Therefore, the heat generated is 4.8 x 106 J.
3. An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat developed in 30 s.
Answer
The amount of heat (H) produced is given by the joule's law of heating asH= Vlt
Where,
Current, I = 5 A
Time, t = 30 s
Voltage, V = Current x Resistance = 5 x 20 = 100 V
H= 100 x 5 x 30 = 1.5 x 104 J.
Therefore, the amount of heat developed in the electric iron is 1.5 x 104 J.
Page No: 220
1. What determines the rate at which energy is delivered by a current?
Answer
The rate of consumption of electric energy in an electric appliance is called electric power. Hence, the rate at which energy is delivered by a current is the power of the appliance.
2. An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.
Answer
Power (P) is given by the expression,P = VI
Where,
Voltage,V = 220 V
Current, I = 5 A
P= 220 x 5 = 1100 W
Energy consumed by the motor = Pt
Where,
Time, t = 2 h = 2 x 60 x 60 = 7200 s
∴ P = 1100 x 7200 = 7.92 x 106 J
Therefore, power of the motor = 1100 W
Energy consumed by the motor = 7.92 x 106 J
Page No: 221
Exercise
1. A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R', then the ratio R/R' is -(a) 1/25
(b) 1/5
(c) 5
(d) 25
► (d) 25
2. Which of the following terms does not represent electrical power in a circuit?
(a) I2R
(b) IR2
(c) VI
(d) V2/R
► (b) IR2
3. An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be -
(a) 100 W
(b) 75 W
(c) 50 W
(d) 25 W
► (d) 25 W
4. Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be -
(a) 1:2
(b) 2:1
(c) 1:4
(d) 4:1
► (c) 1:4
5. How is a voltmeter connected in the circuit to measure the potential difference between two points?
Answer
To measure the potential difference between two points, a voltmeter should be connected in parallel to the points.
6. A copper wire has a diameter of 0.5 mm and a resistivity of 1.6 × 10−8 Ω m. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?
Answer
Area of cross-section of the wire, A =π (d/2) 2
Diameter= 0.5 mm = 0.0005 m
Resistance, R = 10 Ω
We know that
Therefore, the length of the wire is 122.7 m and the new resistance is 2.5 Ω.
7. The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below −
Plot a graph between V and I and calculate the resistance of that resistor.
Answer
The plot between voltage and current is called the IV characteristic. The voltage is plotted on the x-axis and the current is plotted on the y-axis. The values of the current for different values of the voltage are shown in the given table.
The IV characteristic of the given resistor is plotted in the following figure.
The slope of the line gives the value of resistance (R) as,
Slope = 1/R = BC/AC = 2/6.8
R= 6.8/2 = 3.4 Ω
Therefore, the resistance of the resistor is 3.4 Ω.
8. When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.
Answer
Resistance (R) of a resistor is given by Ohm's law as,V= IR
R= V/I
Where,
Potential difference, V= 12 V
Current in the circuit, I= 2.5 mA = 2.5 x 10-3 A
Therefore, the resistance of the resistor is 4.8 kΩ
9. A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω, respectively. How much current would flow through the 12 Ω resistor?
Answer
There is no current division occurring in a series circuit. Current flow through the component is the same, given by Ohm’s law as
V= IR
I= V/R
Where,
R is the equivalent resistance of resistances 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω. These are connected in series. Hence, the sum of the resistances will give the value of R.
R= 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 Ω
Potential difference, V= 9 V
I= 9/13.4 = 0.671 A
Therefore, the current that would flow through the 12 Ω resistor is 0.671 A.
10. How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?
Answer
For x number of resistors of resistance 176 Ω, the equivalent resistance of the resistors connected in parallel is given by Ohm's law asV= IR
R= V/I
Where,
Supply voltage, V= 220 V
Current, I = 5 A
Equivalent resistance of the combination = R,given as
Therefore, four resistors of 176 Ω are required to draw the given amount of current.
11. Show how you would connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of (i) 9 Ω, (ii) 4 Ω.
If we connect the resistors in series, then the equivalent resistance will be the sum of the resistors, i.e., 6 Ω + 6 Ω + 6 Ω = 18 Ω, which is not desired. If we connect the resistors in parallel, then the equivalent resistance of 6/2 = 3 Ω is also not desired. Hence, we should either connect the two resistors in series or parallel.
(a) Two resistors in parallel
Two 6 Ω resistors are connected in parallel. Their equivalent resistance will be
The third 6 Ω resistor is in series with 3 Ω. Hence, the equivalent resistance of the circuit is 6 Ω+ 3 Ω = 9 Ω.
(b) Two resistors in series
Two 6 Ω resistors are in series. Their equivalent resistance will be the sum 6 + 6 = 12 Ω.
The third 6 Ω resistor is in parallel with 12 Ω. Hence, equivalent resistance will be
Therefore, the total resistance is 4 Ω.
12. 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?
Resistance R1 of the bulb is given by the expression,
Supply voltage, V = 220 V
Maximum allowable current, I = 5 A
Rating of an electric bulb P=10 watts
Because R=V2/P
∴ Number of electric bulbs connected in parallel are 110.
13. A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 Ω resistances, which may be used separately, in series, or in parallel. What are the currents in the three cases?
Supply voltage, V= 220 V
Resistance of one coil, R= 24 Ω
(i) Coils are used separately
According to Ohm's law,
V= I1R1
Where,
I1 is the current flowing through the coil
I1 = V/R1 = 220/24 = 9.166 A
Therefore, 9.16 A current will flow through the coil when used separately.
(ii) Coils are connected in series
Total resistance, R2 = 24 Ω + 24 Ω = 48 Ω
According to Ohm's law,V = I2R2
Where,
I2 is the current flowing through the series circuit
I2 = V/R2 = 220/48 = 4.58 A
Therefore, 4.58 A current will flow through the circuit when the coils are connected in series.
(iii) Coils are connected in parallel
Total resistance, R3 is given as =
According to Ohm's law,
V= I3R3
Where,
I3 is the current flowing through the circuit I3 = V/R3 = 220/12 = 18.33 A
Therefore, 18.33 A current will flow through the circuit when coils are connected in parallel.
14. Compare the power used in the 2 Ω resistor in each of the following circuits: (i) a 6 V battery in series with 1 Ω and 2 Ω resistors, and (ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.
(i) Potential difference, V = 6 V
1 Ω and 2 Ω resistors are connected in series. Therefore, equivalent resistance of the circuit, R = 1 + 2 = 3 Ω
According to Ohm’s law,
V = IR
Where,
I is the current through the circuit
I= 6/3 = 2 A
This current will flow through each component of the circuit because there is no division of current in series circuits. Hence, current flowing through the 2 Ω resistor is 2 A. Power is given by the expression,
P= (I)2R = (2)2 x 2 = 8 W
(ii) Potential difference, V = 4 V
12 Ω and 2 Ω resistors are connected in parallel. The voltage across each component of a parallel circuit remains the same. Hence, the voltage across the 2 Ω resistor will be 4 V.
Power consumed by 2 Ω resistor is given by
P= V2/R = 42/2 = 8 W
Therefore, the power used by the 2 Ω resistor is 8 W.
15. Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to the electric mains supply. What current is drawn from the line if the supply voltage is 220 V?
Answer
Both the bulbs are connected in parallel. Therefore, the potential difference across each of them will be 220 V, because no division of voltage occurs in a parallel circuit.
Current drawn by the bulb of rating 100 W is given by,Power = Voltage x Current
Current = Power/Voltage = 60/220 A
Hence, current drawn from the line = 100/220 + 60/220 = 0.727 A
16. Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?
Energy consumed by an electrical appliance is given by the expression,H= Pt
Where,
Power of the appliance = P
Time = t
Energy consumed by a TV set of power 250 W in 1 h = 250 ×3600 = 9 ×105 J
Energy consumed by a toaster of power 1200 W in 10 minutes = 1200 ×600
Energy consumed by a toaster of power 1200 W in 10 minutes = 1200 ×600
= 7.2×105 J
Therefore, the energy consumed by a 250 W TV set in 1 h is more than the energy consumed by a toaster of power 1200 W in 10 minutes.
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.
Rate of heat produced by a device is given by the expression for power as, P= I2R
Where,
Resistance of the electric heater, R= 8 Ω
Current drawn, I = 15 A
P= (15)2 x 8 = 1800 J/s
Therefore, heat is produced by the heater at the rate of 1800 J/s.
18. Explain the following.
(a) Why is tungsten used almost exclusively for filament of electric lamps?
(b) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal?
(c) Why is the series arrangement not used for domestic circuits?
(d) How does the resistance of a wire vary with its area of cross-section?
(e) Why are copper and aluminum wires usually employed for electricity transmission?
Answer
(a) The melting point of Tungsten is an alloy which has very high melting point and very high resistivity so does not burn easily at a high temperature.
(b) The conductors of electric heating devices such as bread toasters and electric irons are made of alloy because the resistivity of an alloy is more than that of metals which produce a large amount of heat.
(c) In series circuits voltage is divided. Each component of a series circuit receives a small voltage so the amount of current decreases and the device becomes hot and does not work properly. Hence, series arrangement is not used in domestic circuits.
(d) Resistance (R) of a wire is inversely proportional to its area of cross-section (A), i.e. when area of cross section increases the resistance decreases or vice versa.
(e) Copper and aluminum are good conductors of electricity also they have low resistivity. So they are usually used for electricity transmission.
Electricity class 10 NCERT physics MCQ
1. Resistivity of a metallic wise depends on
(a) its length
(b) its shape
(c) its thickness
(d) nature of material
► (d) nature of material
2. A battery of 10 volt carries 20,000 C of charge through a resistance of 20 Ω. The work done in 10 seconds is
(a) 2 × 103 joule
(b) 2 × 105joule
(c) 2 × 104 joule
(d) 2 × 102 joule
► (b) 2 × 105joule
3. To get 2 Ω resistance using only 6 Ω resistors, the number of them required is
(a) 2
(b) 3
(c) 4
(d) 6
► (b) 3
4. The least resistance obtained by using 2 Ω, 4 Ω, 1 Ω and 100 Ω is
(a) < 100 Ω
(b) < 4 Ω
(c) < 1 Ω
(d) > 2 Ω
► (c) < 1 Ω
5. A metallic conductor has loosely bound electrons called free electrons. The metallic
conductor is
(a) negatively charged
(b) positively charged
(c) neutral
(d) Either positively charged or negatively charged
► (c) neutral
6. Calculate the current flows through the 10 Ω resistor in the following circuit.
(a) 1.2 A
(b) 0.6 A
(c) 0.2 A
(d) 2.0 A
► (b) 0.6 A
7. If R1 and R2 be the resistance of the filament of 40 W and 60 W respectively operating 220 V, then
(a) R1 < R2
(b) R2 < R1
(c) R1 = R2
(d) R1 ≥ R2
► (b) R2 < R1
8. Resistivity of a metallic wise depends on
(a) its length
(b) its shape
(c) its thickness
(d) nature of material
► (d) nature of material
9. A coil in the heater consumes power P on passing current. If it is cut into halves and joined in parallel, it will consume power
(a) P
(b) P/2
(c) 2P
(d) 4P
► (d) 4P
10. Two resistors are connected in series gives an equivalent resistance of 10 Ω. When connected in parallel, gives 2.4 Ω. Then the individual resistance are
(a) each of 5 Ω
(b) 6 Ω and 4 Ω
(c) 7 Ω and 4 Ω
(d) 8 Ω and 2 Ω
► (b) 6 Ω and 4 Ω
11. If the current I through a resistor is increased by 100% the increase in power dissipation will be (assume temperature remain unchanged)
(a)100%
(b) 200%
(c) 300%
(d) 400%
► (c) 300%
12. The nature of the graph between potential difference and the electric current flowing through a conductor is
(a)parabolic
(b) circle
(c) straight line
(d) hyperbolic
► (c) straight line
13. A cooler of 1500 W, 200 volt and a fan of 500 W, 200 volt are to be used from a household supply. The rating of fuse to be used is
(a) 2.5 A
(b) 5.0 A
(c) 7.5 A
(d) 10 A
► (d) 10 A
14. The resistance of the hot filament of the bulb is about 10 times the cold resistance. What will be the resistance of a 100 -220 V lamp, when not in use?
(a) 48 Ω
(b) 400 Ω
(c) 484 Ω
(d) 48.4 Ω
► (c) 484 Ω
15. A student says that the resistance of two wires of the same length and same area of cross section is the same. This statement is correct if
(a) Both wires are of different materials
(b) Both wires are made of the same material and are at different temperatures.
(c) Both wires are made of the same material and are at the same temperature.
(d) Both wires are made of different materials and are at the same temperature.
► (c) Both wires are made of the same material and are at the same temperature.
16. A current of 1 A is drawn by a filament of an electric bulb. Number of electrons passing through a cross-section of the filament in 16 seconds would be roughly
(a) 1020
(b) 1016
(c) 1018
(d) 1023
► (a) 1020
17. The effective resistance between A and B is
(a) 4Ω
(b) 6Ω
(c) May be 10 Ω
(d) Must be 10 Ω
► (a) 4Ω
18. What is the maximum resistance which can be made using five resistors each of 1/5 W?
(a) 1/5 Ω
(b) 10 Ω
(c) 5 Ω
(d) 1 Ω
► (d) 1 Ω
19. Work done to move 1 coulomb charge from one point to another point on a charged
conductor having potential 10 volt is
(a) 1 Joule
(b) 10 Joule
(c) zero
(d) 100 Joule
► (c) zero
20. Electric potential is a:
(a) scalar quantity
(b) vector quantity
(c) neither scalar nor vector
(d) sometimes scalar and sometimes vector
► (a) scalar quantity
21. Coulomb is the SI unit of:
(a) charge
(b) current
(c) potential difference
(d) resistance
► (a) charge
22. A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R’, then the ratio R/R’ is:
(a) 1/25
(b) 1/5
(c) 5
(d) 25
► (d) 25
23. The heating element of an electric iron is made up of:
(a) copper
(b) nichrome
(c) aluminum
(d) iron
► (b) nichrome
24. 1 mV is equal to:
(a) 10 volt
(b) 1000 volt
(c) 10-3 volt
(d) 10-6 volt
► (c) 10-3 volt
25. When electric current is passed, electrons move from:
(a) high potential to low potential.
(b) low potential to high potential.
(c) in the direction of the current.
(d) against the direction of the current.
► (b) low potential to high potential.
26. An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power
consumed will be:
(a) 100 W
(b) 75 W
(c) 50 W
(d) 25 W
► (d) 25 W
27. The electrical resistance of insulators is
(a) high
(b) low
(c) zero
(d) infinitely high
► (d) infinitely high
28. Electric power is inversely proportional to
(a) resistance
(b) voltage
(c) current
(d) temperature
► (a) resistance
29. An electric bulb is connected to a 220V generator. The current is 0.50 A. What is the power of the bulb?
(a) 440 W
(b) 110 W
(c) 55 W
(d) 0.0023 W
► (b) 110 W
30. Which of the following gasses are filled in electric bulbs?
(a) Helium and Neon
(b) Neon and Argon
(c) Argon and Hydrogen
(d) Argon and Nitrogen
► (d) Argon and Nitrogen
31. The resistivity of insulators is of the order of
(a) 10-8 Ω-m
(b) 101 Ω-m
(c) 10-6 Ω-m
(d) 106 Ω-m
►(a) 10-8 Ω-m
32. Three resistors of 1 Ω, 2 ft and 3 Ω are connected in parallel. The combined resistance of the three resistors should be
(a) greater than 3 Ω
(b) less than 1 Ω
(c) equal to 2 Ω
(d) between 1 Ω and 3 Ω
► (b) less than 1 Ω
33. 100 J of heat is produced each second in a 4Ω resistor. The potential difference across the resistor will be:
(a) 30 V
(b) 10 V
(c) 20 V
(d) 25 V
► (b) 10 V
34. A fuse wire repeatedly gets burnt when used with a good heater. It is advised to use a fuse wire of
(a) more length
(b) less radius
(c) less length
(d) more radius
► (d) more radius
35. Two devices are connected between two points say A and B in parallel. The physical quantity that will remain the same between the two points is
(a) current
(b) voltage
(c) resistance
(d) None of these
► (b) voltage
36. A boy records that 4000 joule of work is required to transfer 10 coulomb of charge between two points of a resistor of 50 Ω. The current passing through it is
(a) 2 A
(b) 4 A
(c) 8 A
(d) 16 A
► (c) 8 A
37. Electrical resistivity of any given metallic wire depends upon
(a) its thickness
(b) its shape
(c) nature of the material
(d) its length
► (c) nature of the material
38. What is the commercial unit of electrical energy?
(a) Joules
(b) Kilojoules
(c) Kilowatt-hour
(d) Watt-hour
► (c) Kilowatt-hour
39. Assertion: In an open circuit, the current passes from one terminal of the electric cell to another.
Reason: Generally, the metal disc of a cell acts as a positive terminal.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
(e) Both A and R are false.
► (d) A is false but R is true.
40. Assertion: Bending of wire decreases the resistance of electric wire.
Reason: The resistance of a conductor depends on length, thickness, nature of material and temperature of the conductor.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
(e) Both A and R are false.
► (a) Both A and R are true and R is the correct explanation of A.
41. Assertion: When a battery is short circuited, the terminal voltage is zero.
Reason: In short circuit, the current is zero.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
(e) Both A and R are false.
► (c) A is true but R is false.
42. Assertion: Conductors allow the current to flow through themselves.
Reason: They have free charge carriers.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
(e) Both A and R are false.
► (a) Both A and R are true and R is the correct explanation of A.
Electricity class 10 NCERT physics SAQ
Q.1. On what factors does the resistance of a conductor depend ?
Or
List the factors on which the resistance of a conductor in the shape of a wire depends. [CBSE 2019]
Ans: Resistance of a conductor:
(i) is directly proportional to its length
(ii) inversely proportional to its cross-section area and depends on the material of the conductor.Resistance also depends on the temperature.
Q.2. (a) A bulb is rated 40 W, 220 V. Find the current drawn by it when it is connected to a 220 V supply. Also find its resistance.
(b) If the given bulb is replaced by a bulb of rating 25 W, 220 V, will there be any change in the value of current and resistance? Justify your answer and determine the change. [CBSE 2019]
Ans: (a) Here power of bulb P = 40 W and voltage V = 220 V
∴ Current drawn by the bulb
and resistance of the bulb R =
(b) On taking another bulb of power P' = 25 W and voltage V = 220 V, there is a change in the value of current and resistance because their values depend on the power of the bulb.
New current
and new resistance
Q.3. Derive the expression for power P consumed by a device having resistance jR and potential difference V.
Or
A device of resistance R is connected across a source of V voltage and draws a current I . Derive an expression for power in terms of voltage (or current) and resistance. [CBSE 2019]
Ans: Amount of work done in carrying a charge Q through a potential difference V is
W = QV.
But Q = It ∴ W = VIt
As power is defined as the rate of doing work, hence
If R be the value of resistance of the conductor, then V - RI and hence
P = VI = (RI)I = I2R
Again
Thus, in general we can say that electric power is given by
Q.4. Compare the power used in the 2 Ω resistor in each of the following circuits shown in Fig. 12.36 (a) and (b) respectively.
Ans: When a 2 Ω resistor is joined to a 6 V battery in series with 1 Ω and 2 Ω resistors, total resistance of the combination Rs = 2 + 1 + 2 = 5Ω
∴ Current in the circuit I1 =
∴ Power used in the 2Ω resistor P1 =
(ii) When a 2 Ω resistor is joined to a 4 V battery in parallel with 12 Ω and 2 Ω resistors, current flowing in the 2 Ω resistor is independent of the other resistors.
∴ Current flowing through 2Ω resistor
∴ Power used in the 2Ω resistor
∴
Q.5. Derive the relation R = R1 + R2 + R3 when resistors are joined in series. [CBSE 2019]
Ans:
In series combination, the same current flows in all the resistances but the potential difference across each of the resistance is different.
According to Ohm ’s law, we have
If the total potential difference between A and B is V, then
Let the equivalent resistance be R, then
V = IR
and hence
⇒
Q.6. When do we say that the potential difference between two points of a circuit in 1 volt ? [CBSE 2019]
Ans: Potential difference between two points of an electric circuit is said to be 1 volt, when a work of 1 J is to be done for moving a charge of 1 C between these two points.
Q.7. Define resistance. Give its SI unit. [CBSE 2019]
Ans. Resistance of a conductor is the measure of opposition offered by it for the flow of electric charge through it. The SI unit of resistance is ‘ohm’ (Ω).
Q.8. State ohm’s law. [CBSE 2019]
Ans: According to Ohm’s law, temperature remaining constant, the current passing through a conductor is directly proportional to the potential difference across its ends, i.e.,
V ∝ I or V = IR
Here, constant R is known as the resistance of a given conductor.
Q.9. (a) Write the relation between resistance and electrical resistivity of the material of a conductor in the shape of a cylinder of length 'l' and area of crosssection 'A' Hence derive the SI unit of electrical resistivity.
(b) Resistance of a metal wire of length 5 m is 100 Ω. If the area of crosssection of the wire is 3 x 10-7 m2, calculate the resistivity of the material. [CBSE 2019]
Ans: (a) The resistance R of a conductor, the shape of a cylinder, of length l and area of cross-section A is given as:
where ρ is a constant, which is known as the electrical resistivity of the material of the conductor.
Thus, resistivity ρ = RA/l
∴ SI unit of resistivity ρ shall be = Ω-m
(b) Here l = 5 m, R = 100 Ω and A = 3 x 10-7 m2
∴ Resistivity of the material of wire ρ = RA/l =
Q.10. (a) With the help of a suitable circuit diagram prove that the reciprocal of the equivalent resistance of a group of resistors joined in parallel is equal to the sum of the reciprocals of the individual resistances.
(b) In an electric circuit two resistors of 12Ω each are joined in parallel to a 6 V battery. Find the current drawn from the battery. [CBSE 2019]
Ans: (a)
In a parallel combination of three resistances R1 R2 and R3, the current in each of the resistances is different. If I is the current drawn from the cell then it is divided into branches I1 I2 and I3. Thus,
I = I1 + I2 + I3.
The potential difference across each of these resistances is the same.
Thus, from Ohm’s law
If R is the equivalent resistance then,
I = V/R
∴
and
(b) Here R1 = R2 = 12 Ω and V = 6 V
Net resistance R o f the parallel grouping is :
∴ Current drawn by the circuit from the battery I =
Q.11. Study the circuit of Fig. 12.24 and find out : (i) Current in 12 Ω, resistor (ii) difference in the readings of A1 and A2 if any. [CBSE 2019]
Ans: (i) In the circuit the resistor of R1 = 12 Ω is connected in series with a parallel combination of two resistors R2 and R3 of 24 Ω each.
The effective resistance of parallel combination of R2 and R3 is given as :
∴ Net resistance of the circuit R = R1 + R23 = 12 + 12 = 24 Ω
∴ Current through 12 Ω resistor = Circuit current
(ii) Both ammeters A1 and A2 give the same reading of 0.25 A and there is no difference in their readings.
Q.12. (i) Draw a schematic diagram of a circuit consisting of a battery of five 2 V cells, a 5 Ω resistor, a 10 Ω resistor and a 15 Ω resistor and a plug key all connected in series.
(ii) Calculate the electric current passing through the above circuit when the key is closed. (iii) Potential difference across 15 Ω resistors. [CBSE 2019]
Ans:
(i) The schematic diagram is given in Fig. 12.25.
(ii) Here total voltage V = 5 x 2 = 10 V
and total resistance R = R1 + R2 + R3
= 5 + 10 + 15 = 30 Ω
∴ Current passing through the circuit when the key is closed
(iii) Potential difference across resistor R3 of 15 Ω
Q.13. Show how you would join three resistors, each of resistance 9 Ω so that the equivalent resistance of the combination is (i) 13.5 Ω, (ii) 6 Ω ? [CBSE 2018]
Ans: Here resistances R1 - R2 = R3 = 9Ω
(i) To obtain an equivalent resistance Req = 13.5Ω, we connect one resistor R1 in series to the parallel combination o f R2 and R3 as shown in figure (i). Then
(ii) To obtain equivalent resistance Req = 6 Ω, we connect resistor R1 in parallel to the series combination of R2 and R3 as shown in figure (ii). Then
Q.14. State Joule’s law of heating. [CBSE 2018]
Ans: As per Joule’s law the heat produced in a resistor is (i) directly proportional to square of current flowing through it, (ii) directly proportional to resistance, and (iii) directly proportional to time. Mathematically,
Heat H = I2Rt
Q.15. Why are metals good conductors of electricity whereas glass is a bad conductor of electricity ? Give reason. [CBSE 2018]
Ans: Metals are good conductors of electricity because they have a large number of free (conduction) electrons which can easily conduct electricity. On the other hand, glass has no free electrons. Thus conduction of charge is not easily possible in glass and so glass is a bad conductor of electricity.
Q.16. Nichrome is used to make the element of an electric heater. Why ? [CBSE 2017]
Ans: Nichrome is an alloy of high resistivity and high melting point and does not oxidize easily.
Q.17. Derive the relation when resistors are joined in parallel. [CBSE 2017]
Ans:
In a parallel combination of three resistances R1 R2 and R3, the current in each of the resistances is different. If I is the current drawn from the cell then it is divided into branches I1 I2 and I3. Thus,
I = I1 + I2 + I3.
The potential difference across each of these resistances is the same.
Thus, from Ohm’s law
If R is the equivalent resistance then,
I = V/R
∴
and
Q.18. Why do electricians wear rubber hand gloves while working ? [CBSE 2016]
Ans: Rubber is an electrical insulator. Hence an electrician can work safely while working on an electric circuit without a risk of getting any electric shock.
Q.19. How are two resistors with resistances R1Ω and R2Ω are to be connected to a battery of emf 3 volts to obtain maximum current flowing through it ? [CBSE 2016]
Ans: For maximum current flow, the net resistance of the circuit must be least possible. Hence resistors R1 and R2 should be connected in parallel.
Q.20. What potential difference is needed to send a current of 5 A through the electrical appliance having a resistance of 18 Ω ? [CBSE 2016]
Ans: Potential difference V = RI - 18 x 5 = 90 V.
Q.21. Power of a lamp is 60 W. Find the energy in the SI unit consumed by it in 1 s. [CBSE 2016]
Ans: Energy consumed E = Pt = 1 W x 1 s - l J .
Q.22. (a) What do you mean by resistance of a conductor ? Define its unit.
(b) In an electric circuit with a resistance wire and a cell, the current flowing is I . What would happen to this current if the wire is replaced by another thicker wire of same material and same length ? Give reason. [CBSE 2016]
Ans. (a) The resistance of a conductor is a property of the conductor, which affects the flow of current through it on maintaining a potential difference across its ends.
Unit of resistance is ohm. Resistance of a conductor is said to be 1 ohm, if a potential difference of 1 V is to be applied across its ends for maintaining flow of 1 A current.
(b) If given resistance wire is replaced by another thicker wire o f same material and same length then cross-section area of wire is increased and consequently its resistance decreases So the current flowing in the circuit increases.
Q.23. (a) Why are electric bulbs filled with chemically inactive nitrogen or argon gas?
(b) The resistance of a wire of 0.01 cm radius is 10 Ω. If the resistivity of the material of the wire is 50 x 10-8 Ω-m, find the length of the wire. [CBSE 2016]
Ans: (a) Electric bulbs are generally filled with some inert gas like nitrogen or argon. This enables prolonging the life of the filament of an electric bulb.
(b) Here the radius of wire r = 0.01 cm = 0.01 x 10-2 m, resistance R = 10 Ω and resistivity p = 50 x 10-8 Ω-m.
⇒
Q.24. (a) Define electric power. Express it in terms of V, I and R where V stands for potential difference, R for resistance and I for current.
(b) V -I graphs for two wires A and B are shown in the Fig. 12.34. Both of them are connected in series to a battery. Which of the two will produce more heat per unit time ? Give justification for your answer. [CBSE 2016]
Ans: (a) Electric power is defined as the rate of supplying electrical energy for maintaining current flow through a circuit.
Electric power P = VI = I2R = V2/R.
(b) We know that the slope of the V-I graph for a given wire gives its resistance and in the given figure the slope of the graph is more for wire A. It means that RA > RB.
In series arrangement the same current I flows through both the resistance.
As heat produced per unit time is given by I2R, hence it is obvious that more heat will be produced per unit time in wire A.
Q.25. Identify the following symbols of commonly used components in a circuit diagram: [CBSE 2015]
Ans. (a) A rheostat, (b) A closed plug key.
Q.26. Define electric current. [CBSE 2015]
Ans: Time rate of flow of charge through any cross-section of a conductor is called electric current.
Q.27. State the SI unit of electric current and define it. [CBSE 2015]
Ans: An ampere (A), which is defined as the rate of flow of 1 coulomb of charge per second.
Q.28. In an electric circuit, state the relationship between the direction of conventional current and the direction of flow of electrons. [CBSE 2015]
Ans: In an electric circuit, the direction of conventional current is taken as opposite to the direction of flow of electrons.
Q.29. What is meant by the statement "potential difference between points A and B in an electric field is 1 volt” ? [CBSE 2015]
Ans: Amount of work done to bring 1 C charge from point B to point A in the electric field is 1 joule.
Q.30. Define kW h. [CBSE 2015]
Ans. A kilowatt hour (kW h) is the commercial unit of electrical energy. It is the energy consumed when 1 kW (1000 W) power is used for 1 hour.
Q.31. How many joules are equal to 1 kW h ? [CBSE 2015]
Ans: 3.6 x 106 J = 1 kW h.
Q.32. Define an electric circuit. Draw a labeled, schematic diagram of an electric circuit consisting of a cell, a resistor, an ammeter, a voltmeter and a closed switch. Distinguish between an open and a closed circuit. [CBSE 2015]
Ans: A continuous and closed path of an electric current is called an electric circuit.
A labeled, schematic diagram of an electric circuit showing a cell E, a resistor R, an ammeter A, a voltmeter V and a closed switch S is shown here.
An electric circuit is said to be an open circuit when the switch is in ‘off’ mode (or key is unplugged) and no current flows in the circuit.
The circuit is said to be a closed circuit when the switch is in 'on' mode (or key is plugged) and a current flows in the circuit.
Q.33. (a) n electrons, each carrying a charge -e, are flowing across a unit cross section of a metallic wire in unit time from east to west. Write an expression for electric current and also give its direction of flow. Give reason for your answer.
(b) The charge possessed by an electron is 1.6 x 10-19 coulomb. Find the number of electrons that will flow per second to constitute a current of 1 ampere. [CBSE 2015]
Ans: (a) Electric current in a circuit is defined as the time rate of flow of electric charge through any cross-section and its direction is opposite to that of flow of electrons. Hence in present case
electric current I =
As electrons are flowing from east to west, the direction of electric current is from west to east.
(b) Here current I = 1 A, time t = 1 s and charge on each electron e = 1.6 x 10-19 C.
Hence, number of electrons flowing n =
Q.34. (a) List the factors on which the resistance of a cylindrical conductor depends and hence write an expression for its resistance.
(b) How will the resistivity of a conductor change when its length is tripled by stretching it ? [CBSE 2015]
Ans. (a) The resistance of a cylindrical conductor i.e., a wire (R) is (i) directly proportional to its length L, (ii) inversely proportional to its cross-section area A and (iii) depends on the nature of material of wire. Mathematically,
Here, ρ is known as the resistivity of a given material. It is defined as the resistance offered by a unit cube of a given material when current flows perpendicular to the opposite faces. (b) The resistivity of the conductor remains unchanged.
Q.35. (a) V-I graphs for two wires A and B are shown in the Fig. 12.17. If both the wires are made of same material and are of same length, which of the two is thicker? Give justification for your answer.
(b) A wire of length L and resistance R is stretched so that the length is doubled and the area of cross-section halved.
How will (i) resistance change, and (ii) resistivity change ? [CBSE 2015]
Ans. (a) Resistance R of wire A is more than that of B (RA > RB) because the slope of the V-I graph for A is more. The resistance of a wire is inversely proportional to its cross-section area. Hence the area of cross-section of wire B, of smaller resistance, must be more. Thus, wire B is thicker.
(b) (i) The new resistance of wire Thus, resistance increases to four times its original value.
(ii) The resistivity remains unchanged because it does not depend on the dimensions of a conductor of a given material.
Q.36. Study the electric circuit of Fig. 12.21 and find (i) the current flowing in the circuit, and (ii) the potential difference across the 10 Ω resistor. [CBSE 2015]
Ans. (i) Here V = 3 V, R1 = 10 Ω and R 2 = 20 Ω. Since R1 and R2 are connected in series, the effective resistance of the circuit
R = R1 + R2 = 10 + 20 - 30 Ω
∴ Current flowing in the circuit
(ii) The potential difference across the R 1 = 10 Ω resistor is V1 = IR1 = 0.1 x 10 = 1.0 V.
Q.37. Three resistors of 3 Ω each are connected to a battery of 3 V as shown in Fig. 12.22. Calculate the current drawn from the battery. [CBSE 2015]
Ans: In the electric circuit shown the series combination of R1 and R2 is joined in parallel to the resistance R3. Hence equivalent resistance R of the circuit is
⇒ R = 2 Ω
∴ Current drawn from the battery I=
Q.38. Find the equivalent resistance across the two ends A and B of the circuit [Fig. 12.31]. [CBSE 2015]
Ans: Equivalent resistance of R1 and joined in parallel is R2 , where
⇒ R12 = 1Ω
Similarly R34 = R56 = R78 = 1Ω
Now R12 and R34 are in series making a resistance
Rq = R12 + -R34 = 1 + 1 = 2Ω
Similarly R56 and R78 are in series and make a resistance R10 = 2Ω
Finally R9 and R10 are connected in parallel between the points A and B and hence equivalent resistance R is given as
Q.39. A 5 Ω resistor is connected across a battery of 6 volts. Calculate :
(i) the current flowing through the resistor.
(ii) the energy that dissipates as heat in 10 s. [CBSE 2015]
Ans: Here V = 6 V and R = 5 Ω
(i) The current flowing through the resistor I =
(ii) Energy dissipated as heat in time t = 10 s is
∴ H = I2Rt = (1.2)2 x 5 x 10 = 72 J
Q.40. Calculate the amount of heat generated while transferring 90000 coulombs of charge between the two terminals of a battery of 40 V in one hour. Also determine the power expended in the process. [CBSE 2015]
Ans: Here charge transferred Q = 90000 C, potential difference between the terminals o f battery V = 40 V and time t = 1 h = 3600 s.
Current =
Amount of heat generated H = Vlt = 40 x 25 x 3600 = 3600000 J = 3.6 x 106 J
and power expended
Q.41. How many 40 W; 220 V lamps can be safely connected to a 220 V, 5 A line ?
Justify your answer.[CBSE 2015]
Ans. The current drawn by a 40 W, 220 V electric lamp
As the electric line is o f rating 220 V, 5 A, hence we can connect n lamps in parallel where
Thus, we can safely connect 27 lamps of 40 W, 220 V rating to a 220 V, 5 A line.
Electricity class 10 NCERT physics long question
Q.1. An electric lamp of resistance 20 Ω and a conductor of resistance 4 Ω are connected to a 6 V battery as shown in the circuit. Calculate
(а) the total resistance of the circuit,
(b) the current through the circuit,
(c) the potential difference across the (i) electric, lamp and (ii) conductor, and
(d) power of the lamp. [CBSE 2019]
Ans: Here voltage of battery V = 6 V, resistance of electric lamp = R1 = 20 Ω and resistance of conductor R2 = 4 Ω.
(a) Since R1 and R2 are connected in series, the total resistance of the circuit
R = R1 + R2 = 20 + 4 = 24 Ω
(b) The current through the circuit I = V/R = 6/24 = 0.25 A
(c) (i) Potential difference across the electric lamp V1 = IR1 = 0.25 x 20 = 5 V
(ii) Potential difference across the conductor V2 = IR2 = 0.25 x 4 = 1 V
(d) Power of the lamp P = I2R1 = (0.25)2 x 20 = 1.25 W
Q.2. (a ) Three resistors R1, R2 and R3 are connected in parallel and the combination is connected to a battery, ammeter, voltmeter and key. Draw a suitable circuit diagram and obtain an expression for the equivalent resistance of the combination of the resistors.
(b) Calculate the equivalent resistance of the network shown in Fig. 12.38 [CBSE 2019]
Ans:
(a) The arrangement is shown in the circuit diagram of Fig. 12.39.
In a parallel combination of three resistances R1 R2 and R3, the current in each of the resistances is different. If I is the current drawn from the cell then it is divided into branches I1 I2 and I3. Thus,
I = I1 + I2 + I3.
The potential difference across each of these resistances is the same.
Thus, from Ohm’s law
If R is the equivalent resistance then,
I = V/R
∴
and
(b) In the network resistors of 20 Ω and 20 Ω are joined in parallel and make a resistance R1, where
This combined resistance R1 = 10 Ω is joined in series with a given 10 Ω resistance. Hence, the equivalent resistance of the network will be R = 10 + 10 = 20 Ω.
Q.3. (a) Three resistors o f resistances R1, R2 and R3 are connected in (i) series, and (ii) parallel. Write an expression for the equivalent resistance of the combination in each case.
(b) Two identical resistances of 12 Ω each are connected to a battery of 3 V.
Calculate the ratio of the power consumed by the resulting combinations with minimum resistance and maximum resistance. [CBSE 2019]
Ans: (a) (i) In series arrangement, equivalent resistance Rs = R1 + R2 + R3
(ii) In parallel arrangement, equivalent resistance Rp is given as:
(b) Here = R1 = R2 = 12 Ω and V = 3V.
For minimum resistance, two resistors must be connected in parallel so that
Hence power
For maximum resistance, two resistors must be connected in series so that
Rs = R1 + R2 = 12 + 12 = 24 Ω
So, the power
⇒
Q.4. Experimentally prove that in series combination of three resistances :
(а) current flowing through each resistance is same, and
(b) total potential difference is equal to the stun of potential differences across individual resistors. [CBSE 2019]
Ans: Series combination o f resistors : We take three resistors R1 R2 and R3 and join them in series between the points X and Y in an electric circuit as shown in Fig. 12.42.
(a) Plug the key and note the ammeter reading.
Then change the position of the ammeter to anywhere in between the resistors and again note the ammeter reading. We find that ammeter reading remains unchanged. It shows that in series arrangement the same current flows through each resistor.
(b) Insert a voltmeter across the ends X and Y of the series combination of resistors. Plug the key so as to complete the circuit and note the voltmeter reading V across the series combination of resistors.
Take out the plug from key K and disconnect the voltmeter. Now insert the voltmeter across the ends of the first resistor R1 as shown in Fig. 12.43. Plug the key and note the voltmeter reading V1. Similarly, measure the potential difference across the other two resistors R2 and R3 separately. Let these potential differences be V2 and V3, respectively. Experimentally we find that
V = V1 + V2 + V3
It shows that in series arrangement of resistors total potential difference is equal to the sum of potential differences across individual resistors.
Q.5. (a) Establish a relationship to determine the equivalent resistance IS of a combination of three resistors having resistances R1, R2 and R3 connected in parallel.
(b) Three resistors are connected in an electrical circuit as shown. Calculate the resistance between A and B. [CBSE 2016]
Ans:
(a) The arrangement is shown in the circuit diagram of Fig. 12.39.
In a parallel combination of three resistances R1 R2 and R3, the current in each of the resistances is different. If I is the current drawn from the cell then it is divided into branches I1 I2 and I3. Thus,
I = I1 + I2 + I3.
The potential difference across each of these resistances is the same.
Thus, from Ohm’s law
If R is the equivalent resistance then,
I = V/R
∴
and
(b) Here the series combination of R1 and R2 is joined to R3 in a parallel arrangement. Hence the net resistance R between points A and B is given as
⇒ r = 4 Ω
Q.6. (a) Establish a relationship to determine the equivalent resistance R of a combination of three resistors having resistances R1, R2 and R3 connected in series.
(b) Calculate the equivalent resistance R of a combination of three resistors of 2 Ω, 3 Ω and 6 Ω joined in parallel. [CBSE 2016]
Ans: (a)
In series combination, the same current flows in all the resistances but the potential difference across each of the resistance is different.
According to Ohm ’s law, we have
If the total potential difference between A and B is V, then
Let the equivalent resistance be R, then
V = IR
and hence
⇒
(b) Here = R1 = 2 Ω R2 = 3 Ω and R3 = 6 Ω
The equivalent resistance R for parallel combination of resistors will be
Q.7. What is meant by electric current ? Name and define its SI unit. In a conductor electrons are flowing from B to A. What is the direction of the conventional current ? Give justification for your answer.
A steady current of 1 ampere flows through a conductor. Calculate the number of electrons that flow through any section of the conductor in 1 second.
(Charge on electron — 1.6 x 10-19 coulomb) [CBSE 2015]
Ans. Electric current is defined as the rate of flow of electric charge through a cross- section of a conductor. If Q charge passes through a section of a conductor in time t, then current I = Q/t.
The SI unit of electric current is an ampere (A). Current is said to be one ampere if the rate of flow of charge through a cross-section of conductor is 1 coulomb per second.
Direction of conventional current is taken as the direction of flow of positive charge or opposite to the direction of flow of negative charge. If negatively charged in a conductor flow from B to A then the direction of conventional current will be from A to B.
Here current I = 1 A, time t = 1 s and charge on electron e = 1.6 x 10-19 C. Let n electrons flow through a section of conductor so that charge passing through the section is Q = ne.
∴
Q.8. What is the heating effect of electric current ? Find an expression for the amount of heat produced. Name some appliances based on the heating effect of current. [CBSE 2015]
Ans: When a current flows through a conducting wire (resistance wire), heat is developed and the temperature of the wire rises. It is known as the heating effect of electric current.
If V be the potential difference maintained across the ends of a wire then, by definition, the amount of work done for flow of 1 C charge through the wire is V.
∴ Work done for flow of Q charge
W = VQ = VIt [∵ Q = It]
where I is the current flowing in time t.
As V = IR, hence
W= VIt = (IR)It = I2Rt
This work done (i.e., electrical energy dissipated) is converted into heat. Hence, the amount of heat produced,
Q = I2Rt J
This is known as Joule’s law of heating.
Incandescent lamps, electric iron, electric stove, toaster, geyser, electric room heater etc., are the appliances based on the heating effect of electric current.
Q.9. The values of current J flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below:
(i) Plot a graph between V and I.
(ii) Calculate the resistance of that resistor.
(iii) What does the graph represent ? [CBSE 2015]
Ans:(i) The plotted V-I graph is shown in Fig. 12.37.
(ii) We take two points A and B on the graph for which currents are 1.5 A and 2.5 A respectively and the corresponding values of V are 4.6 V and 7.8 V.
∴ Resistance of resistor
(iii) The V-I graph is a linear one. It shows that the given resistor strictly obeys Ohm’s law.
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