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Question 3
...
If the internal resistance of the battery is
0
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4 Ω
Maximum current drawn from the battery = I
According to Ohm’s law,

The maximum current drawn from the given battery is 30 A
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2:
A battery of emf 10 V and internal resistance 3 Ω is connected
connected to a resistor
...
5 A, what is the resistance of the resistor? What is the terminal voltage
of the battery when the circuit is closed?
Answer

Emf of the battery, E = 10 V
Internal resistance of the battery, r = 3 Ω

Current in the circuit, I = 0
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5 × 17
= 8
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5 V
...
3:
Three resistors 1 Ω, 2 Ω, and 3 Ω are combined in series
...

Answer

Three resistors of resistances 1 Ω, 2 Ω, and 3 Ω are combined in series
...


Total resistance = 1 + 2 + 3 = 6 Ω
Current flowing through the circuit = I
Emf of the battery, E = 12 V
Total resistance of the circuit, R = 6 Ω
The relation for current using Ohm’s law is,

Potential drop across 1 Ω resistor = V1
From Ohm’s law, the value of V1 can be obtained as
V1 = 2 × 1= 2 V … (i)
Potential drop across 2 Ω resistor = V2
Again, from Ohm’s law, the value of V2 can be obtained as
V2 = 2 × 2= 4 V … (ii)
Potential drop across 3 Ω resistor = V3
Again, from Ohm’s law, the value of V3 can be obtained as
V3 = 2 × 3= 6 V … (iii)
Therefore, the potential drop across 1 Ω, 2 Ω, and 3 Ω resistors are 2 V, 4 V, and 6 V
respectively
...
4:
Three resistors 2 Ω, 4 Ω and 5 Ω are combined in parallel
...

Answer

There are three resistors of resistances,
R1 = 2 Ω, R2 = 4 Ω, and R3 = 5 Ω
They are connected in parallel
...


Therefore, the current through each resister is 10 A, 5 A, and 4 A respectively and the
total current is 19 A
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5:
At room temperature (27
...
What is the
temperature of the element if the resistance is found to be 117 Ω, given that the
temperature coefficient of the material of the resistor is
Answer

Room temperature, T = 27°C
Resistance of the heating element at T, R = 100 Ω
Let T1 is the increased temperature of the filament
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Question 3
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0 × 10−7 m2, and its resistance is measured to be 5
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What is the resistivity of
the material at the temperature of the experiment?
Answer

Length of the wire, l =15 m
Area of cross-section
section of the wire, a = 6
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0 Ω
Resistivity of the material of the wire = ρ
Resistance is related with the resistivity as

Therefore, the resistivity of the material is 2 × 10−7 Ω m
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7:
A silver wire has a resistance of 2
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5 °C, and a resistance of 2
...

Determine the temperature coefficient of resistivity of silver
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5°C
Resistance of the silver wire at T1, R1 = 2
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7 Ω
Temperature coefficient of silver = α
It is related with temperature and resistance as

Therefore, the temperature coefficient of silver is 0
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Question 3
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2 A which settles after a few seconds toa steady value of 2
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What is the steady
temperature of the heating element if the room temperature is 27
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70 × 10−4 °C −1
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2 A
Initial resistance = R1, which is given by the relation,

Steady state value of the current, I2 = 2
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70 × 10−4 °C −1
Initial temperature of nichrome, T1= 27
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5°C

Question 3
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30:

Answer

Current flowing through various branches of the circuit is represented in the given figure
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e
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e
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e
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10:
In a metre bridge [Fig
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27], the balance point is found to be at 39
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5 Ω
...
Why are the connections
between resistors in a Wheatstone or meter bridge made of thick copper strip
strips?
Determine the balance point of the bridge above if X and Y are interchanged
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Balance point from end A, l1 = 39
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5 Ω
Condition for the balance is given as,

Therefore, the resistance of resistor X is 8
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The connection between resistors in a Wheatstone or metre bridge is made of thick copper
strips to minimize the resistance, which is not taken into consideration in the bridge
formula
...

The balance point of the bridge will be 100−l1 from A
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5 = 60
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5 cm from A
...
Hence, no current would flow through the
galvanometer
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11:
A storage battery of emf 8
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5 Ω is being charged by a 120 V
dc supply using a series resistor of 15
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What is the terminal voltage of the battery
during charging? What is the purpose of having a series resistor in the charging circuit?
Answer

Emf of the storage battery, E = 8
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5 Ω
DC supply voltage, V = 120 V
Resistance of the resistor, R = 15
...
Hence, it can be written as
V1 = V − E
V1 = 120 − 8 = 112 V
Current flowing in the circuit = I, which is given by the relation,

Voltage across resistor R given by the product, IR = 7 × 15
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5 V
DC supply voltage = Terminal voltage of battery + Voltage drop across R
Terminal voltage of battery = 120 − 108
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5 V

A series resistor in a charging circuit limits the current drawn from the external source
...
This is very dangerous
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12:
In a potentiometer arrangement, a cell of emf 1
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0 cm
length of the wire
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0 cm, what is the emf of the second cell?
Answer

Emf of the cell, E1 = 1
...

New balance point of the potentiometer, l2 = 63 cm

Therefore, emf of the second cell is 2
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Question 3
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1 is
8
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How long does an electron take to drift from one end of a wire 3
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0 × 10−6 m2 and it is carrying a
current of 3
...

Answer

Number density of free electrons in a copper conductor, n = 8
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0 m
Area of cross-section
section of the wire, A = 2
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0 A, which is given by the relation,
I = nAeVd
Where,
e = Electric charge = 1
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7 × 104 s
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14:
The earth’s surface has a negative surface charge density of 10−9 C m−2
...
If there were no mechanism of sustaining
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 ddifferent
ifferent parts of the globe)
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37 × 106
m
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37 × 106 m
Surface area of the earth,
A = 4πr2
= 4π × (6
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09 × 1014 m2
Charge on the earth surface,
q=σ×A
= 10−9 × 5
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09 × 105 C
Time taken to neutralize the earth’s surface = t

Current,

Therefore, the time taken to neutralize the earth’s surface is 282
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Question 3
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0 V and internal resistance 0
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5 Ω
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9 V and a large internal resistance of 380
Ω
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0 V
Internal resistance of each cell, r = 0
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Resistance of the resistor, R = 8
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39 × 8
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87 A
Therefore, the current drawn from the supply is 1
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87 A
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9 V
Internal resistance of the cell, r = 380 Ω

Hence, maximum current
Therefore, the maximum current drawn from the cell is 0
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Since a large current is
required to start the motor of a car, the cell cannot be used to start a motor
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16:
Two wires of equal length, one of aluminium and the other of copper have the same
resistance
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(ρAl = 2
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72 × 10−8 Ω m,
Relative density of Al = 2
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9
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63 × 10−8 Ω m
Relative density of aluminium, d1 = 2
...

Resistance of the aluminium wire = R1
Area of cross-section of the aluminium wire = A1
Resistivity of copper, ρCu = 1
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9
Let l2 be the length of copper wire and m2 be its mass
...
Hence, aluminium is lighter than
copper
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Question 3
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2

3
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0

59
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4

7
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0

78
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6

11
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0

98
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8

15
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0

118
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0

19
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0

138
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0

39
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0

158
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7
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e
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According to Ohm’s law, the ratio of voltage with current is the re
resistance of
the conductor
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7 Ω
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18:
Answer the following questions:
A steady current flows in a metallic conductor of non-uniform
non
cross- section
...

A low voltage supply from which one needs high currents must have
have very low internal
resistance
...

Why?
Answer

When a steady current flows in a metallic conductor of non-uniform
non
cross-section,
section, the
current flowing through the conductor is constant
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section
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No, Ohm’s law is not universally applicable for all conducting elements
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Ohm’s law is not valid for it
...

(
R is the internal resistance of the source
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In order to prohibit the current from exceeding the safety limit, a high tension supply
must have a very large internal resistance
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Question 3
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Alloys usually have much (lower/higher) temperature coefficients of resistance than pure
metals
...

The resistivity of a typical insulator (e
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, amber) is greater than that of a metal by a factor
of the order of (1022/103)
...

Alloys usually have lower temperature coefficients of resistance than pure metals
...

The resistivity of a typical insulator is greater than that of a metal by a factor of the order
of 1022
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20:
Given n resistors each of resistance R,, how will you combine them to get the (i)
maximum (ii) minimum effective resistance? What is the ratio of the maximum to
minimum resistance?
Given the resistances of 1 Ω, 2 Ω, 3 Ω, how will be combine them to get an equivalent
resistance of (i) (11/3) Ω (ii)
ii) (11/5) Ω, (iii) 6 Ω, (iv) (6/11) Ω?
Determine the equivalent resistance of networks shown in Fig
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31
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Hence, maximum resistance of the combination, R1 = nR
When n resistors are connected in parallel, the effective resistance (R2) is the minimum,
given by the ratio


...


Equivalent resistance of the circuit is given by,

Equivalent resistance,
Consider the following combination of the resistors
...


Equivalent resistance of the circuit is given by the sum,
R’ = 1 + 2 + 3 = 6 Ω

Equivalent resistance,
Consider the series combination of the resistors, as shown in the given circuit
...

Hence, their equivalent resistance = (1+1) = 2 Ω
It can also be observed that two resistors of resistance 2 Ω each are connected in series
...

Therefore, the circuit can be redrawn as

It can be observed that 2 Ω and 4 Ω resistors are connected in parallel in all the four
loops
...


Hence, equivalent resistance of the given circuit is
It can be observed from the given circuit that five resistors of resistance R each are
connected in series
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21:
Determine the current drawn from a 12 V supply with internal resistance 0
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3
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Each resistor has 1 Ω resistance
...
Hence, equivalent resistance is given by the relation,

Negative value of R’ cannot be accepted
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5 Ω
Hence, total resistance of the given circuit = 2
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5 = 3
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72 A

=

Question 3
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33 shows a potentiometer with a cell of 2
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40 Ω
maintaining a potential drop across the resistor wire AB
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02 V (for very moderate currents up to a few mA) gives a balance
point at 67
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To ensure very low currents drawn from the standard
cell, a very high resistance of 600 kΩ is put in series with it, which is shorted close to the
balance point
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3 cm length of the wire
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0 V instead of 2
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02 V
Balance point on the wire, l1 = 67
...
Therefore, new balance point on the
wire, l = 82
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247 V
...

The balance point is not affected by the presence of high resistance
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The method
thod would not work if the driver cell of the potentiometer had an emf of 1
...
0 V
...

The
he circuit would not work well for determining an extremely small emf
...
Hence, there would be a
large percentage of error
...

The potential drop across AB is slightly greater than the emf measured
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Question 3
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34 shows a potentiometer circuit for comparison of two resistances
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0 Ω is found to be 58
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5 cm
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What might you do if yyou
failed to find a balance point with the given cell of emf ε?

Answer

Resistance of the standard resistor, R = 10
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3 cm
Current in the potentiometer wire = i
Hence, potential drop across R, E1 = iR
Resistance of the unknown resistor = X

Balance point for this resistor, l2 = 68
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75 Ω
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Only if the
potential drop across R or X is smaller than the potential drop across the potentiometer
wire AB,
B, a balance point is obtained
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24:
Figure 3
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0 V potentiometer used for the determination of internal resistance
of a 1
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The balance point of the cell in open circuit is 76
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When a resistor of
9
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8 cm length
of the potentiometer wire
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Answer

Internal resistance of the cell = r
Balance point of the cell in open circuit, l1 = 76
...
5 Ω
New balance point of the circuit, l2 = 64
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68Ω

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