CBSE NCERT Solutions for Class 12 Physics Chapter 3...Class- XII-CBSE-Physics Current Electricity...
Transcript of CBSE NCERT Solutions for Class 12 Physics Chapter 3...Class- XII-CBSE-Physics Current Electricity...
Class- XII-CBSE-Physics Current Electricity
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CBSE NCERT Solutions for Class 12 Physics Chapter 3 Back of Chapter Questions
3.1. The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.4 Ξ©, what is the maximum current that can be drawn from the battery?
Solution:
Given:
emf of the battery (πΈπΈ) = 12 V
Internal resistance of the battery (ππ) = 0.4 Ξ©
Let the maximum current drawn from the battery is I
By Ohmβs law,
πΈπΈ = πΌπΌππ
πΌπΌ =πΈπΈππ
πΌπΌ =120.4
= 30 A
β΄ 30 A is the maximum current drawn from the battery.
3.2. A battery of emf 10 V and internal resistance 3 Ξ© is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?
Solution:
Given:
emf of the battery, πΈπΈ, is 10 V
Internal resistance of the battery, ππ, = 3 Ξ©,
Current in the circuit, πΌπΌ, = 0.5 A,
Resistance of the resistor ππππ π π .
According to Ohmβs law,
πΌπΌ =πΈπΈ
π π + ππ
β π π + ππ =πΈπΈπΌπΌ
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=100.5
= 20 Ξ©
β΄ π π = 20 β 3 = 17 Ξ©
The terminal voltage of the resistor = ππ
According to Ohmβs law,
ππ = πΌπΌπ π
= 0.5 Γ 17
= 8.5 V.
β΄ The terminal voltage is 8.5,ππand the resistance of the resistor is 17 Ξ©.
3.3. (a) Three resistors 1 Ξ©, 2 Ξ©, and 3 Ξ©, are combined in series. What is the total resistance of the combination?
(b) If the combination is connected to a battery of emf 12 V and negligibleinternal resistance, obtain the potential drop across each resistor.
Solution:
(a) The resistors 1 Ξ©, 2 Ξ©, and 3 Ξ© are combined in series.
Hence, total resistance = 1 + 2 + 3 = 6 Ξ©
(b) Current flowing through the circuit = πΌπΌ
Emf of the battery, πΈπΈ ππππ 12 V
The total resistance of the circuit in series combination, π π ππππ 6 Ξ©
According to Ohmβs law
πΌπΌ =πΈπΈπ π
=126
= 2 A
Let the potential drop across 1Ξ© resistor ππππ ππ1The value of ππ1 obtained by Ohmβs law is:
ππ1 = 2 Γ 1 = 2 V β¦ (ππ)
Let the potential drop across 2 Ξ© resistor ππππ ππ2Again, from Ohmβs law, the value of ππ2 can be obtained as
ππ2 = 2 Γ 2 = 4 V β¦ (ππππ)
Let the potential drop across 3 Ξ© resistor ππππ ππ3
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Again, from Ohmβs law, the value of ππ3 can be obtained as
ππ3 = 2 Γ 3 = 6 V β¦ (ππππππ)
β΄ The potential drop across 1 Ξ© is 2 V, 2 Ξ© is 4 V, and 3 Ξ© is 6 V.
3.4. (a) Three resistors 2 Ξ©, 4 Ξ© and 5 Ξ©, are combined in parallel. What is the total resistance of the combination?
(b) If the combination is connected to a battery of emf 20 V and negligibleinternal resistor, and the total current drawn from the battery.
Solution:
Given:
(a) Given:
π π 1 = 2 Ξ©,π π 2 = 4 Ξ©, and π π 3 = 5 Ξ©
The total resistance (R) of the parallel combination is given by.
1π π
=1π π 1
+1π π 2
+1π π 3
=12
+14
+15
=10 + 5 + 4
20=
1920
β΄ π π =2019
Ξ©
The total resistance of the combination is 2019
Ξ©
(b) Emf of the battery, ππ = 20 V
Current flowing through the resistor π π 1 is given by,
πΌπΌ1 =πππ π 1
=202
= 10 A
Current flowing through the resistor π π 2 is given by,
πΌπΌ2 =πππ π 2
=204
= 5 A
Current flowing through the resistor π π 3 is given by,
πΌπΌ3 =πππ π 3
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=205
= 4 A
Total Current, πΌπΌ = πΌπΌ1 + πΌπΌ2 + πΌπΌ3 = 10 + 5 + 4 = 19 A
β΄ The current through each resistor π π 1,π π 2, and π π 3 are 10 A, 5 A, and 4 A respectively and the total current is 19 A.
3.5. At room temperature (27. 0ππ C) the resistance of a heating element is 100 Ξ©. What is the temperature of the element if the resistance is found to be 117 Ξ©, given that the temperature coefficient of the material of the resistor is 1.70 Γ 10β4 Cβ1
ππ
Solution:
Given:
Room temperature, ππ ππππ 27 ππC
The resistance of the heating element at ππ,π π ππππ 100 Ξ©
Suppose, ππ1 is the increased temperature of the filament.
The resistance of the heating element at ππ1,π π 1 ππππ 117 Ξ©
Temperature Coefficient of filament material is, πΌπΌ ππππ 1.70 Γ 10β4 Cβ1 ππ
πΌπΌ is given by the relation,
πΌπΌ =π π 1 β π π
π π (ππ1 β ππ)
ππ1 β ππ =π π 1 β π π π π πΌπΌ
ππ1 β 27 =117 β 100
100(1.7 Γ 10β4)
ππ1 β 27 = 1000
ππ1 = 1027ππC
β΄ At 1027πππΆπΆ, the resistance of the element is 117 Ξ©.
3.6. A negligibly small current is passed through a wire of length 15 m and uniform cross-section 6.0 Γ 10β7 m2, and its resistance is measured to be 5.0 Ξ© . What is the resistivity of the material at the temperature of the experiment?
Solution:
Given:
Length of the wire, ππ ππππ 15 m
Area of a cross-section of the wire, ππ ππππ 6.0 Γ 10β7 m2
The resistance of the wire's material, π π ππππ 5.0 Ξ©
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ππ is the resistivity of the material/
Resistance is related to the resistivity as
π π = πππππ΄π΄
ππ =π π π΄π΄ππ
=5 Γ 6 Γ 10β7
15= 2 Γ 10β7 Ξ© ππ
β΄ The resistivity of the material is 2 Γ 10β7 Ξ© m.
3.7. A silver wire has a resistance of 2.1 Ξ© at 27.5 πΆπΆ ππ , and a resistance of 2.7 Ξ© at 100 πΆπΆ ππ . Determine the temperature coefficient of resistivity of silver.
Solution:
Given:
Temperature, ππ1 ππππ 27.5 C o
The Resistance of the silver wire at ππ1,π π 1 ππππ 2.1 Ξ©
Temperature, ππ2 = 100 C o
The Resistance of the silver wire at ππ2,π π 2 ππππ 2.7 Ξ©.
Temperature coefficient of silver ππππ πΌπΌ
It is related to the temperature and resistance as
πΌπΌ =π π 2 β π π 1
π π 1(ππ2 β ππ1)
=2.7 β 2.1
2.1(100 β 27.5) = 0.0039 Cβ1 o
Temperature coefficient of silver is 0.0039 Cβ1 o .
3.8. A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds to a steady state value of 2.8 A. What is the steady temperature of the heating element if the room temperature is 27.0 πΆπΆ
ππ ? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is 1.70 Γ 10β4 πΆπΆ
ππ β1.
Solution:
Given:
Supply voltage, ππ ππππ 230 V
The Initial current drawn, πΌπΌ1 ππππ 3.2 A
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Initial resistance = π π 1, which is given by the relation,
π π 1 =πππΌπΌ
=2303.2
= 71.87 Ξ©
Steady state value of the current, πΌπΌ2 ππππ 2.8 A
Resistance at the steady state = π π 2, which is given as
π π 2 =2302.8
= 82.14 Ξ©
Temperature coefficient of nichrome, ππ ππππ 1.70 Γ 10β4 Cβ1 o
The Initial temperature of nichrome, ππ1 ππππ 27.0 C o
Study state temperature reached by nichrome ππππ ππ2
ππ2 can be obtained by the relation for πΌπΌ,
πΌπΌ =π π 2 β π π 1
π π 1(ππ2 β ππ1)
ππ2 β 27 πΆπΆ ππ =
82.14 β 71.8771.87 Γ 1.7 Γ 10β4
= 840.5
ππ2 = 840.5 + 27 = 867.5 πΆπΆ ππ
The steady state temperature of the heating element is 867.5 C o
3.9. Determine the current in each branch of the network shown in fig 3.30:
Solution:
The current flowing through different branches of the circuit is shown in the figure
Provided
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πΌπΌ1 = Current flowing through the outer circuit
πΌπΌ2 = Current flowing through branch AB
πΌπΌ3 = Current flowing through branch AD
πΌπΌ2 β πΌπΌ4 = Current flowing through branch BC
πΌπΌ3 + πΌπΌ4 = Current flowing through branch DC
πΌπΌ4 = Current flowing through branch BD
For the closed circuit ABDA, the potential drop is zero, i.e.,
10πΌπΌ2 + 5πΌπΌ4 β 5πΌπΌ3 = 0
β 2πΌπΌ2 + πΌπΌ4 β πΌπΌ3 = 0
β πΌπΌ3 = 2πΌπΌ2 + πΌπΌ4 β¦ (1)
For the closed circuit BCDB, the potential drop is zero, i.e.,
5(πΌπΌ2 β πΌπΌ4) β 10(πΌπΌ3 + πΌπΌ4) β 5πΌπΌ4 = 0
β 5πΌπΌ2 + 5πΌπΌ4 β 10πΌπΌ3 β 10πΌπΌ4 β 5πΌπΌ4 = 0
β 5πΌπΌ2 β 10πΌπΌ3 β 20πΌπΌ4 = 0
β πΌπΌ2 = 2πΌπΌ3 + 4πΌπΌ4 β¦ (2)
For the closed circuit ABCFEA, the potential drop is zero, i.e.,
β10 + 10(πΌπΌ1) + 10(πΌπΌ2) + 5(πΌπΌ2 β πΌπΌ4) = 0
β 10 = 15πΌπΌ2 + 10πΌπΌ1 β 5πΌπΌ4β 3πΌπΌ2 + 2πΌπΌ1 β πΌπΌ4 = 2 β¦ (3)
From equations (1) and (2), we obtain
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πΌπΌ3 = 2(2πΌπΌ3 + 4πΌπΌ4) + πΌπΌ4
β πΌπΌ3 = 4πΌπΌ3 + 8πΌπΌ4 + πΌπΌ4β3πΌπΌ3 = 9πΌπΌ4β3πΌπΌ4 = + πΌπΌ3 β¦ (4)
Putting equation (4) in equation (1), we can obtain
πΌπΌ3 = 2πΌπΌ2 + πΌπΌ4β β4πΌπΌ4 = 2πΌπΌ2β πΌπΌ2 = β2πΌπΌ4 β¦ (5)
From the given figure,
πΌπΌ1 = πΌπΌ3 + πΌπΌ2 β¦ (6)
From equation (6) and equation (1), we obtain
3πΌπΌ2 + 2(πΌπΌ3 + πΌπΌ2) β πΌπΌ4 = 2
β 5πΌπΌ2 + 2πΌπΌ3 β πΌπΌ4 = 2 β¦ (7)
β 5(β2 πΌπΌ4) + 2(β3 πΌπΌ4) β πΌπΌ4 = 2
β β10πΌπΌ4 β 6πΌπΌ4 β πΌπΌ4 = 2
17πΌπΌ4 = β2
πΌπΌ4 =β217
π΄π΄
Equation (4) reduces to
πΌπΌ3 = β3(πΌπΌ4)
= β3 οΏ½β217οΏ½ =
617
A
πΌπΌ2 = β2(πΌπΌ4)
= β2 οΏ½β217οΏ½ =
417
π΄π΄
πΌπΌ2 β πΌπΌ4 =4
17β οΏ½
β217οΏ½ =
617
A
πΌπΌ3 + πΌπΌ4 =6
17+ οΏ½
β217οΏ½ =
417
A
πΌπΌ1 = πΌπΌ3 + πΌπΌ2
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=6
17+
417
=1017
A
Therefore, current in branch π΄π΄π΄π΄ = 417
A
In branch π΄π΄πΆπΆ = 617
A
In branch πΆπΆπΆπΆ = β417
A
In branch π΄π΄πΆπΆ = 617
A
In branch π΄π΄πΆπΆ = β217
A
Total current = 417
+ 617
+ β417
+ 617
+ β217
= 1017
A
3.10. (a) In a meter bridge [Fig. 3.27], the balance point is found to be at 39.5 cm from the end A when the resistor Y is of 12.5 Ξ© . Determine the resistance of X. Why are the connections between resistors in a Wheatstone or meter bridge made of thick copper strips?
(b) Determine the balance point of the bridge above if X and Y areinterchanged.
(c) What happens if the galvanometer and cell are interchanged at the balancepoint of the bridge? Would the galvanometer show any current?
Solution:
The figure shows a meter bridge with resistors X and Y.
(a) Balance point from end π΄π΄, πΌπΌ1 ππππ 39.5 cm
The Resistance of the resistor ππ ππππ 12.5 Ξ©
Condition for the balance is given as,
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ππππ
=100 β ππ1
ππ1
ππ =100 β 39.5
39.5Γ 12.5 = 8.2 Ξ©
The resistance of resistor ππ is 8.2 Ξ© .
The thick copper strips which are used in a Wheat stone bridge is used to minimize the resistance of connecting wires
(b) If ππ and ππ are interchanged, then ππ1 and 100 β ππ1 get interchanged.
The balance point of the bridge is 100 β ππ1 from π΄π΄.
100 β ππ1 = 100 β 39.5 = 60.5 cm
The balance point is 60.5 cm from A.
(c) The galvanometer will show no deflection when the galvanometer and cellare interchanged at the balance point. Therefore, no current would flowthrough the galvanometer.
3.11. A storage battery of emf 8.0 ππ and internal resistance 0.5 Ξ© is being charged by a 120 ππ DC supply using a series resistor of 15.5 Ξ© . What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit?
Solution:
Given:
Emf of the storage battery, πΈπΈ, ππππ 8.0 V
The Internal resistance of the battery, ππ, ππππ 0.5 Ξ©
DC supply voltage, ππ ππππ 120 V
The Resistance of the resistor, π π , ππππ 15.5 Ξ©
Effective voltage in the circuit = ππβ²
R is connected in series to the storage battery. It can therefore be written as
ππβ² = ππ β πΈπΈ
β ππβ² = 120 β 8 = 112 V
Current flowing in the circuit = πΌπΌ, which is given by the relation, Reference the sentence
πΌπΌ =ππβ²
π π + ππ
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=112
15.5 + 5=
11216
= 7 A
The Voltage across resistor R given by the product, πΌπΌπ π = 7 Γ 15.5 = 108.5 ππ
DC voltage supply = Voltage drop across R+Terminal voltage of the battery
The terminal voltage of battery = 120 β 108.5 = 11.5 V
The Current will be very large in the absence of a series resistor in a charging circuit. The Work of a series resistor in a charging circuit is to reduce the amount of current drawn from the external circuit.
3.12. In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63.0 cm, what is the emf of the second cell?
Solution:
Given:
Emf of the cell, πΈπΈ1ππππ 1.25 V
The Balance point of the potentiometer, ππ1 ππππ 35 cm
The cell is substituted by another cell of emf πΈπΈ2
The new balance point of the potentiometer, ππ2 ππππ 63 cm
The balance condition is given by the relationship,
πΈπΈ1πΈπΈ2
=ππ1ππ2
πΈπΈ2 = πΈπΈ1 Γππ2ππ1
= 1.25 Γ6335
= 2.25 V
Emf of the second cell is 2.25 V.
3.13. The number density of free electrons in a copper conductor estimated in Example 3.1 is 8.5 Γ 1028 mβ3. How long does an electron take to drift from one end of a wire 3.0 m long to its other end? The area of cross-section of the wire is 2.0 Γ 10β6 m2 and it is carrying a current of 3.0 A.
Solution:
Given:
The number density in a copper conductor of free electrons, ππ ππππ 8.5 Γ 1028 mβ3
Length of the copper wire, πΌπΌ ππππ 3.0 m
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Area of a cross-section of the wire, π΄π΄ ππππ 2.0 Γ 10β6 m2
The Current carried by the wire, πΌπΌ = 3.0 A,
πΌπΌ = πππ΄π΄ππππππ
Where,
e = Electric charge = 1.6 Γ 10β19 C
ππππ = Drift velocity = Length of the wire (ππ)
Time taken to cover (t)
πΌπΌ = ππ π΄π΄πππππ‘π‘
π‘π‘ =πππ΄π΄πππππΌπΌ
= 3Γ8.5Γ1028Γ2Γ10β6Γ1.6Γ10β19
3.0
= 2.7 Γ 104 s
The Time taken by an electron to drift from one end of the wire to the other is given by 2.7 Γ 104 s.
3.14. The earth's surface has a negative surface charge density of 10β9 C mβ2. The potential difference of 400 kV between the top of the atmosphere and the surface results (due to the low conductivity of the lower atmosphere) in a current of only 1800 A over the entire globe. If there were no mechanism of sustaining atmospheric electric field, how much time (roughly) would be required to neutralise the earth's surface? (This never happens in practice because there is a mechanism to replenish electric charges, namely the continual thunderstorms and lightning in different parts of the globe). (Radius of earth = 6.37 Γ 106 m. )
Solution:
Given:
The Surface charge density of the earth (ππ) ππππ 10β9 C mβ2
The Current over the entire globe (πΌπΌ) ππππ 1800 A
The Radius of the earth (ππ) ππππ 6.37 Γ 106 m
The Surface area of the earth,
π΄π΄ = 4ππππ2
= 4 β Γ (6.37 Γ 106)2
= 5.09 Γ 1014 m2
Charge on the earth surface,
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ππ = ππ Γ π΄π΄
= 10β9 Γ 5.09 Γ 1014
= 5.09 Γ 105 C
Time taken to neutralize the earth's surface = π‘π‘
Current, πΌπΌ = πππ‘π‘
π‘π‘ =πππΌπΌ
=5.09 Γ 105
1800= 282.77 s
The Time taken to neutralize the earth's surface is 282.77 s.
3.15. (a) Six lead-acid type of secondary cells each of emf 2.0 V and internal resistance 0.015 Ξ© are joined in series to provide a supply to the resistance of 8.5 Ξ©. What is the current drawn from the supply and its terminal voltage?
(b) A secondary cell after long use has an emf of 1.9 V and a large internalresistance of 380 Ξ©.What maximum current can be drawn from the cell?Could the cell drive the starting motor of a car?
Solution:
Given:
(a) Number of secondary cells, ππ ππππ 6
Emf of each secondary cell, πΈπΈ ππππ 2.0 V
The internal resistance of each cell, ππ ππππ 0.015 Ξ©
The Series resistor is connected to the combination of cells.
The Resistance of the resistor, π π ππππ 8.5 Ξ©
Current drawn from the supply = πΌπΌ, which is given by the relation,
πΌπΌ = πππππ π +ππππ
= 6Γ28.5+6Γ0.015
= 128.59
= 1.39 A
Terminal voltage, ππ = πΌπΌπ π = 1.39 Γ 8.5 = 11.87 A
A current of 1.39 A, is drawn from the supply and the terminal voltage is 11.87 A.
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(b) After a long use, emf of the secondary cell, πΈπΈ ππππ 1.9 V
The Internal resistance of the cell, ππ ππππ 380 Ξ©
A maximum current of 0.005 A can be drawn from the cell.
Since a large current is required to start the motor of a car, it is not possibleto use the cell to start a motor.
3.16. Two wires of equal length, one of aluminium and the other of copper, have the same resistance. Which of the two wires is lighter? Hence explain why aluminium wires are preferred for overhead power cables (ππAl = 2.63 Γ 10β8 Ξ© m, ππCu =1.72 Γ 10β8 Ξ© m, Relative density of Al = 2.7, of Cu = 8.9)
Solution:
Given:
Resistivity of aluminium πππ΄π΄π΄π΄ ππππ 2.63 Γ 10β8 Ξ© m
The Relative density of aluminium, ππ1, ππππ 2.7
Let ππ1 be the length of aluminium wire and ππ1 be its mass.
The Resistance of the aluminium wire ππππ π π 1
Area of a cross-section of the aluminium wire ππππ π΄π΄1
The Resistivity of copper, πππΆπΆπΆπΆ = 1.72 Γ 10β8 Ξ© m
The Relative density of copper, ππ2, ππππ 8.9
Let ππ2 be the length of copper wire and ππ2 be its mass.
The Resistance of the copper wire ππππ π π 2
Area of the cross-section of the copper wire ππππ π΄π΄2
The two relations can be written as
π π 1 = ππ1π΄π΄1π΄π΄1
β¦(1)
π π 2 = ππ2π΄π΄2π΄π΄2
β¦(2)
It is given that π π 1 = π π 2 β ππ1π΄π΄1π΄π΄1
= ππ2π΄π΄2π΄π΄2
β π΄π΄1π΄π΄2
= ππ1ππ2
= 2.63Γ10β8
1.72Γ10β8= 2.63
1.72
Mass of the aluminium wire, ππ1 = volume Γ Density
π΄π΄1ππ1 Γ ππ1 = π΄π΄1ππ1ππ1 (3)
Mass of the copper wire, ππ2 = volume Γ Density
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π΄π΄2ππ2 Γ ππ2 = π΄π΄2ππ2ππ2 (4)
By Dividing equation (3) by equation (4), we obtain ππ1ππ2
= π΄π΄1π΄π΄1ππ1π΄π΄2π΄π΄2ππ2
For ππ1 = ππ2 ππ1ππ2
= π΄π΄1ππ1π΄π΄2ππ2
For π΄π΄1π΄π΄2
= 2.631.72
,
ππ1ππ2
= 2.631.72
Γ 2.78.9
= 0.46
It can be inferred from this ratio that ππ1 is less than ππ2. Hence, aluminium is lighter than copper. Because aluminium is lighter, it is preferred for overhead power cables over copper.
3.17. What conclusion can you draw from the following observations on a resistor made of alloy manganin?
Current A Voltage V Current A Voltage V
0.2 3.94 3.0 59.2
0.4 7.87 4.0 78.8
0.6 11.8 5.0 98.6
0.8 15.7 6.0 118.5
1.0 19.7 7.0 138.2
2.0 39.4 8.0 158.0
Solution:
From the given table, it can be inferred that the voltage-to-current ratio is a constant equal to 19.7. Therefore, manganin is an ohmic conductor, i.e., the alloy obeys Ohm's law. According to Ohm's law, the conductor's resistance is the voltage proportion with the current. Therefore, manganin's resistance is 19.7 Ξ©.
3.18. Answer the following questions:
(a) A steady current flow in a metallic conductor of non - uniform cross- section. Which of these qualities is constant along the conductor: Current,Current density, electric field, drift speed?
(b) Is Ohmβs law is universally applicable for all conducting elements?
(c) A low voltage supply from which one needs high currents must have verylow internal resistance. Why?
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(d) A high tension (HT) Supply of, say, 6 KV must have a very large internalresistance. Why?
Solution:
(a) Current density, drift speed and electric field are inversely proportional tothe area of cross section, and hence, they are not constant. The currentflowing through the conductor is constant when a steady current flow in ametallic conductor of the non-uniform cross-section.
(b) Ohmβs law cannot be applied universally to all conducting elements. Ohmβslaw is not applicable for Vacuum diode semiconductor.
(c) By Ohmβs law, the relation for the potential is ππ = πΌπΌπ π
Voltage (ππ) is directly proportional to the Current (I).
π π is the internal resistance of the source.
πΌπΌ = ππ π π
If ππ is low, then R must be very low, so that high current can be drawn fromthe source.
(d) To prevent the electrical current from exceeding the safety threshold, a hightension supply must have a very high internal resistance. If the internalresistance is not high, the current drawn may exceed the safety limits in thecase of a short circuit.
3.19. Choose the correct alternative:
(a) Alloys of metals usually have (greater/less) resistivity than that of theirconstituent metals.
(b) Alloys usually have much (lower/higher) temperature coefficients ofresistance than pure metals.
(c) The resistivity of the alloy manganin is nearly independent of/increasesrapidly with the increase of temperature.
(d) The resistivity of a typical insulator (e.g., amber) is greater than that of ametal by a factor of the order of (1022 1033β ).
Solution:
(a) Alloys of metals usually have greater resistivity than that of their constituentmetals.
(b) Alloys usually have lower temperature coefficients of resistance than puremetals.
(c) The resistivity of the alloy, manganin is nearly independent of increase oftemperature.
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(d) The resistivity of a typical insulator is greater than that of a metal by a factorof the order of 1022.
3.20. (a) Given ππ resistors each of resistance π π , how will you combine them to getthe (i) maximum (ii) minimum effective resistance? What is the ratio of themaximum to minimum resistance?
(b) Given the resistances of 1 Ξ©, 2 Ξ©, 3 Ξ©, how will you combine them to get anequivalent resistance of (i) (11 3β ) Ξ© (ii) (11 5β ) Ξ©, (iii) 6 Ξ©, (iv)(6 11β ) Ξ©?
(c) Determine the equivalent resistance of networks shown in Fig.
Solution:
Given:
(a) Total number of resistors = ππ
The resistance of each resistor = π π
(i) The effective resistance π π 1 maximum when n resistors are connectedin series, given by the product πππ π .
Hence, maximum resistance of the combination, π π 1 = πππ π
(ii) When ππ resistors are connected in parallel, the effective resistance(π π 2) is the minimum, given by the ratio π π
ππ.
Hence, minimum resistance of the combination, π π 2 = π π ππ
(iii) The ratio of the maximum to the minimum resistance is,π π 1π π 2
= πππ π π π ππ
= ππ2
(b) The resistance of the given resistors is,
π π 1 = 1 Ξ© ,π π 2 = 2 Ξ© ,π π 3 = 3 Ξ©
(i) Equivalent resistance, π π β² = 113
Ξ©
Considering the following combination of the resistors.
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The Equivalent resistance of the circuit is given by,
π π β² = 2Γ12+1
+ 3 = 23
+ 3 = 113
Ξ©
(ii) Equivalent resistance, π π β² = 115
Ξ©
Considering the following combination of the resistors.
The Equivalent resistance of the circuit is given by,
π π β² = 2Γ32+3
+ 1 = 65
+ 1 = 115
Ξ©
(iii) Equivalent resistance, π π β² = 6 Ξ©
The series combination of the resistors is as shown in the givencircuit.
The Equivalent resistance of the circuit is given by the sum,
π π β² = 1 + 2 + 3 = 6 Ξ©
(iv) Equivalent resistance, π π β² = 611
Ξ©
Considering the series combination of the resistors, as shown in thegiven circuit.
The Equivalent resistance of the circuit is given by,
π π β² = 1Γ2Γ31Γ+2Γ3+3Γ1
= 611
Ξ©
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(c) It can be observed from the given circuit that in the first small loop, tworesistors of resistance 1 Ξ© each are connected in series.
Hence, their equivalent resistance = (1 + 1) = 2 Ξ©
It can also be observed that two resistors of resistance 2 Ξ© each areconnected in series.
Hence, their equivalent resistance = (2 + 2) = 4 Ξ©.
Therefore, the circuit can be redrawn as
It can be observed that 2 Ξ© and 4 Ξ© resistors are connected in parallel in all the four loops.
Hence, the equivalent resistance (π π β²) of each loop is given by,
π π β² = 2Γ42+4
= 86
= 43
Ξ©
The circuit reduces to,
All the four resistors are connected in series.
Hence, the equivalent resistance of the given circuit is 43
Γ 4 = 163
Ξ©
(b) The five resistors of resistance π π each are connected in series hence, theequivalent resistance of the circuit = π π + π π + π π + π π + π π = 5π π
3.21. Determine the current drawn from a 12 V supply with internal resistance 0.5 Ξ© by the infinite network shown in Fig. Each resistor has 1 Ξ© resistance.
Solution:
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Given:
The resistance of each resistor connected in the circuit, π π ππππ 1 Ξ©
Let the equivalent resistance of the given circuit is π π β²and the network is infinite.
Hence, the equivalent resistance is given by the relation,
π π β² = 2 + π π β²
(π π β²+1)
β (π π β²)2 β 2π π β² β 2 = 0
π π β² = 2Β±β4+82
πΊπΊ
= 2Β±β122
= οΏ½1 Β± β3οΏ½ Ξ©
Negative value of π π β² cannot be accepted.
Hence, equivalent resistance,
π π β² = οΏ½1 + β3οΏ½ πΊπΊ = (1 + 1.73) Ξ© = 2.73 Ξ©
The Internal resistance of the circuit,
ππ = 0.5 Ξ©
The total resistance of the given circuit is given by
= 2.73 + 0.5 = 3.23 Ξ©
Supply voltage, V = 12 V
Current drawn from the source is given according, to Ohm's law by 123.23
= 3.72 A
3.22. The figure shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Ξ© maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for very moderate currents up to a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, very high resistance of 600 kΞ© is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf Ξ΅, and the balance point found similarly, turns out to be at 82.3 cm length of the wire.
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(a) What is the value Ξ΅?
(b) What purpose does the high resistance of 600 kΞ© have?
(c) Is the balance point affected by this high resistance?
(d) Is the balance point affected by the internal resistance of the driver cell?
(e) Would the method work in the above situation if the driver cell of the potentiometer had an emf of 1.0 V instead of 2.0 V?
(f) Would the circuit work well for determining an extremely small emf, say of the order of a few mV (such as the typical emf of a thermocouple)? If not, how will you modify the circuit?
Solution:
Given:
(a) A Constant emf of the given standard cell, πΈπΈ1 ππππ 1.02 V
The Balance point on the wire, ππ1 ππππ 67.3 cm
The standard cell is replaced by a cell of unknown emf, Ξ΅
β΄ The new balance point on the wire, ππ ππππ 82.3 cm
The relation connecting balance point and emf is, ππ1π΄π΄1
= πππ΄π΄
ππ = π΄π΄π΄π΄1
Γ πΈπΈ1
= 82.367.3
Γ 1.02 = 1.247 V
The value of the unknown emf is 1.247 V.
(b) High resistance of 600 kΞ© is used to lower the current flow through the galvanometer when the movable contact is far from the balance point.
(c) The presence of high resistance does not affect the balance point.
(d) The internal resistance of the driver cell does not affect the balance point.
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(e) If the potentiometer's driver cell emf is less than the other cell's emf, then there would be no balance point on the wire, so the method won't operate, if the potentiometer's driver cell has an emf of 1 V instead of 2 V.
(f) The circuit cannot work well for determining an extremely small emf as there will be a very high percentage of error. This is because the circuit would be unstable; the balance point would be close to end A.
If a series resistance is connected to the wire AB, the given circuit can be modified. AB's potential drop is slightly higher than the measured emf. It would be a small percentage error.
3.23. The figure shows a 2.0 V potentiometer used for the determination of internal resistance of a 1.5 V cell. The balance point of the cell in open circuit is 76.3 cm. When a resistor of 9.5 Ξ© is used in the external circuit of the cell, the balance point shifts to 64.8 cm length of the potentiometer wire. Determine the internal resistance of the cell.
Solution:
Given:
The internal resistance of the cell is r
The balance point of the cell in the open circuit, ππ1 is 76.3 cm
An external resistance of resistance R = 9.5 Ξ© is connected to the circuit.
The new balance point of the circuit, ππ2 ππππ 64.8 cm
The current (πΌπΌ) flowing through the circuit is
The relation between emf and resistance is,
ππ = οΏ½π΄π΄1βπ΄π΄2π΄π΄2οΏ½π π
= 76.3 cmβ64.8 cm64.8 cm
Γ 9.5 Ξ© = 1.68 Ξ©
The internal resistance of the cell is 1.68 Ξ© .
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