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Question 3
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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
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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
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3:
Three resistors 1 Ω, 2 Ω, and 3 Ω are combined in series
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Answer

Three resistors of resistances 1 Ω, 2 Ω, and 3 Ω are combined in series
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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
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4:
Three resistors 2 Ω, 4 Ω and 5 Ω are combined in parallel
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Answer

There are three resistors of resistances,
R1 = 2 Ω, R2 = 4 Ω, and R3 = 5 Ω
They are connected in parallel
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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
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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 r
a constant emf of 1
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3 cm length of the wire
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The standard cell is then replaced
re
by a cell of unknown emf ε and the
balance point found similarly, turns out to be at 82
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What is the value ε ?
What purpose does the high resistance of 600 kΩ have?
Is the balance point affected by this high resistance?
Is the balance point affected by the internal resistance of the driver cell?
Would the method work in the above situation if the driver cell of the
potentiometer had an emf of 1
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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 thermo-couple)? If not, how will you modify
the circuit?
Answer

Constant emf of the given standard cell, E1 = 1
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3 cm
A cell of unknown emf, ε,replaced the standard cell
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3 cm
The relation connecting emf and balance point is,

The value of unknown emfis 1
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The purpose of using the high resistance of 600 kΩ is to reduce the current through the
galvanometer when the movable contact is far from the balance point
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The point is not affected by the internal resistance of the driver cell
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0 V
instead of 2
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This is because if the emf of the driver cell of the potentiometer is less
than the emf of the other cell, then there would be no balance point on the wire
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As the circuit
would be unstable, the balance point would be close to end A
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The given circuit can be modified if a series resistance is connected with the wire AB
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The percentage
error would be small
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23:
Figure 3
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The balance
point with a standard resistor R = 10
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3 cm, while that with the
unknown resistance X is 68
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Determine the value of X
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0 Ω
Balance point for this resistance, l1 = 58
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5 cm
Hence, potential drop across X, E2 = iX
The relation connecting emf and balance point is,

Therefore, the value of the unknown resistance, X, is 11
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If we fail to find
ind a balance point with the given cell of emf, ε,, then the potential drop
across R and X must be reduced by putting a resistance in series with it
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Question 3
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35 shows a 2
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5 V cell
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3 cm
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5 Ω is used in the external circuit of the cell, the balance point shifts to 64
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Determine the internal resistance of the cell
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3 cm
An external resistance (R)) is connected to the circuit with R = 9
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8 cm
Current flowing through the circuit = I
The relation connecting resistance and emf is,

Therefore, the internal resistance of the cell is 1
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