#### Page No 127:

#### Question 3.1:

The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.4Ω, what is the maximum current that can be drawn from the battery?

#### Answer:

Emf of the battery, *E* = 12 V

Internal resistance of
the battery, *r* = 0.4 Ω

Maximum current drawn
from the battery = *I*

According to Ohm’s law,

The maximum current drawn from the given battery is 30 A.

#### Question 3.2:

A battery of emf 10 V and internal resistance 3 Ω 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?

#### Answer:

Emf of the battery, *E* = 10 V

Internal resistance of
the battery, *r* = 3 Ω

Current in the circuit, *I* = 0.5 A

Resistance of the
resistor = *R*

The relation for current using Ohm’s law is,

Terminal voltage of the
resistor = *V*

According to Ohm’s law,

*V* = *IR*

= 0.5 × 17

= 8.5 V

Therefore, the resistance of the resistor is 17 Ω and the terminal voltage is

8.5 V.

#### Question 3.3:

**(a)** Three resistors 1 Ω,
2 Ω, and 3 Ω 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 negligible internal resistance, obtain the potential drop across
each resistor.

#### Answer:

**(a)** Three resistors of resistances 1 Ω,
2 Ω, and 3 Ω are
combined in series. Total resistance of the combination is given by
the algebraic sum of individual resistances.

Total resistance = 1 + 2 + 3 = 6 Ω

**(b)** 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 = *V*_{1}

From
Ohm’s law, the value of *V*_{1 }can be obtained as

*V*_{1} = 2 × 1= 2 V
… (i)

Potential
drop across 2 Ω resistor = *V*_{2}

Again,
from Ohm’s law, the value of *V*_{2 }can be
obtained as

*V*_{2} = 2 × 2= 4 V
… (ii)

Potential
drop across 3 Ω resistor = *V*_{3}

Again,
from Ohm’s law, the value of *V*_{3 }can be
obtained as

*V*_{3} = 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.

#### Question 3.4:

**(a) ** Three resistors 2 Ω,
4 Ω and
5 Ω 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 negligible internal resistance, determine the current through
each resistor, and the total current drawn from the battery.

#### Answer:

**(a)** There are three resistors of resistances,

*R*_{1} = 2 Ω, *R*_{2} = 4 Ω, and *R*_{3} = 5 Ω

They
are connected in parallel. Hence, total resistance (*R*) of the
combination is given by,

Therefore, total resistance of the combination is.

**(b)** Emf of the battery, *V* = 20 V

Current
(*I*_{1}) flowing through resistor *R*_{1} is given by,

Current
(*I*_{2}) flowing through resistor *R*_{2} is given by,

Current
(*I*_{3}) flowing through resistor *R*_{3} is given by,

Total
current, *I* = *I*_{1} + *I*_{2} + *I*_{3} = 10 + 5 + 4 = 19 A

Therefore, the current through each resister is 10 A, 5 A, and 4 A respectively and the total current is 19 A.

#### Question 3.5:

At room temperature (27.0 °C) the resistance of a heating element is 100 Ω. 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 *T*_{1} is the increased temperature of the filament.

Resistance of the
heating element at *T*_{1}, *R*_{1} = 117 Ω

Temperature co-efficient of the material of the filament,

Therefore, at 1027°C, the resistance of the element is 117Ω.

#### Question 3.6:

A negligibly small
current is passed through a wire of length 15 m and uniform
cross-section 6.0 × 10^{−7 }m^{2}, and
its resistance is measured to be 5.0 Ω.
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
of the wire, *a* = 6.0 × 10^{−7 }m^{2}

Resistance of the
material of the wire, *R* = 5.0 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.

#### Question 3.7:

A silver wire has a resistance of 2.1 Ω at 27.5 °C, and a resistance of 2.7 Ω at 100 °C. Determine the temperature coefficient of resistivity of silver.

#### Answer:

Temperature, *T*_{1} = 27.5°C

Resistance of the
silver wire at *T*_{1}, *R*_{1} = 2.1 Ω

Temperature, *T*_{2} = 100°C

Resistance of the
silver wire at *T*_{2}, *R*_{2} = 2.7 Ω

Temperature coefficient
of silver = *α*

It is related with temperature and resistance as

Therefore, the
temperature coefficient of silver is 0.0039°C^{−1}.

#### Question 3.8:

Aheating
element using nichrome connected to a 230 V supply draws an initial
current of 3.2 A which settles after a few seconds toa steady
value of 2.8 A. What is the steady temperature of the heating element
if the room temperature is 27.0 °C? Temperature coefficient of
resistance of nichrome averaged over the temperature range involved
is 1.70 × 10^{−4 }°C ^{−1}.

#### Answer:

Supply voltage,* V* = 230 V

Initial current drawn, *I*_{1} = 3.2 A

Initial resistance = *R*_{1}, which is given by the relation,

Steady state value of
the current, *I*_{2} = 2.8 A

Resistance at the
steady state = *R*_{2}, which is given as

Temperature
co-efficient of nichrome, *α* = 1.70 × 10^{−4 }°C ^{−1}

Initial temperature of
nichrome, *T*_{1}= 27.0°C

Study state temperature
reached by nichrome = *T*_{2}

*T*_{2 }can
be obtained by the relation for *α*,

Therefore, the steady temperature of the heating element is 867.5°C

#### Page No 128:

#### Question 3.9:

Determine the current in each branch of the network shown in fig 3.30:

#### Answer:

Current flowing through various branches of the circuit is represented in the given figure.

*I*_{1} =
Current flowing through the outer circuit

*I*_{2} =
Current flowing through branch AB

*I*_{3} =
Current flowing through branch AD

*I*_{2} − *I*_{4} = Current flowing through branch BC

*I*_{3} + *I*_{4} = Current flowing through branch CD

*I*_{4} =
Current flowing through branch BD

For the closed circuit ABDA, potential is zero i.e.,

10*I*_{2} + 5*I*_{4} − 5*I*_{3} = 0

2*I*_{2} + *I*_{4} −*I*_{3} = 0

*I*_{3} =
2*I*_{2} + *I*_{4} … (1)

For the closed circuit BCDB, potential is zero i.e.,

5(*I*_{2} − *I*_{4}) − 10(*I*_{3} +* I*_{4}) − 5*I*_{4 }= 0

5*I*_{2} +
5*I*_{4} − 10*I*_{3} − 10*I*_{4 }− 5*I*_{4 }= 0

5*I*_{2} −
10*I*_{3} − 20*I*_{4 }= 0

*I*_{2} =
2*I*_{3} + 4*I*_{4 }… (2)

For the closed circuit ABCFEA, potential is zero i.e.,

−10 + 10 (*I*_{1})
+ 10(*I*_{2}) + 5(*I*_{2} −* I*_{4})
= 0

10 = 15*I*_{2} + 10*I*_{1 }− 5*I*_{4}

3*I*_{2} +
2*I*_{1 }− *I*_{4} = 2 …
(3)

From equations (1) and (2), we obtain

*I*_{3} =
2(2*I*_{3 }+ 4*I*_{4}) +* I*_{4}

*I*_{3} =
4*I*_{3 }+ 8*I*_{4} +* I*_{4}

− 3*I*_{3} = 9*I*_{4 }

− 3*I*_{4} = + *I*_{3 }… (4)

Putting equation (4) in equation (1), we obtain

*I*_{3} =
2*I*_{2 }+ *I*_{4}

− 4*I*_{4} = 2*I*_{2 }

*I*_{2} =
− 2*I*_{4} … (5)

It is evident from the given figure that,

*I*_{1} = *I*_{3 }+ *I*_{2} …
(6)

Putting equation (6) in equation (1), we obtain

3*I*_{2} +2(*I*_{3 }+ *I*_{2}) −* I*_{4} = 2

5*I*_{2} +
2*I*_{3 }− *I*_{4} = 2 …
(7)

Putting equations (4) and (5) in equation (7), we obtain

5(−2* I*_{4})
+ 2(− 3 *I*_{4})_{ −} *I*_{4} = 2

− 10*I*_{4} − 6*I*_{4 }− *I*_{4} = 2

17*I*_{4} = − 2

Equation (4) reduces to

*I*_{3} =
− 3(*I*_{4})

Therefore, current in branch

In branch BC =

In branch CD =

In branch AD

In branch BD =

Total current =

#### Question 3.10:

**(a)** In a metre 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 Ω. 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 *are
interchanged.

**(c)** What happens if the galvanometer and cell are
interchanged at the balance point of the bridge? Would the
galvanometer show any current?

#### Answer:

A metre bridge with
resistors *X* and *Y* is represented in the given figure.

**(a)** Balance
point from end A, *l*_{1} = 39.5 cm

Resistance
of the resistor *Y* = 12.5 Ω

Condition for the balance is given as,

Therefore,
the resistance of resistor *X* is 8.2 Ω.

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.

**(b) **If *X *and *Y* are interchanged, then *l*_{1} and 100−*l*_{1} get interchanged.

The
balance point of the bridge will be 100−*l*_{1} from A.

100−*l*_{1 }= 100 − 39.5 = 60.5 cm

Therefore, the balance point is 60.5 cm from A.

**(c) **When the galvanometer and cell are interchanged at the
balance point of the bridge, the galvanometer will show no
deflection. Hence, no current would flow through the galvanometer.

#### Question 3.11:

A storage battery of emf 8.0 V and internal resistance 0.5 Ω is being charged by a 120 V dc supply using a series resistor of 15.5 Ω. 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* = 8.0 V

Internal resistance of
the battery, *r* = 0.5 Ω

DC supply voltage, *V* = 120 V

Resistance of the
resistor, *R* = 15.5 Ω

Effective voltage in
the circuit =* V*^{1}

*R *is connected
to the storage battery in series. Hence, it can be written as

*V*^{1} = *V* − *E*

*V*^{1 }=
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 × 15.5 = 108.5 V

DC supply voltage =
Terminal voltage of battery + Voltage drop across *R*

Terminal voltage of battery = 120 − 108.5 = 11.5 V

A series resistor in a charging circuit limits the current drawn from the external source. The current will be extremely high in its absence. This is very dangerous.

#### Question 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?

#### Answer:

Emf of the cell, *E*_{1} = 1.25 V

Balance point of the
potentiometer, *l*_{1}= 35 cm

The cell is replaced by
another cell of emf *E*_{2}.

New balance point of
the potentiometer, *l*_{2} = 63 cm

Therefore, emf of the second cell is 2.25V.

#### Question 3.13:

The number density of
free electrons in a copper conductor estimated in Example 3.1 is 8.5
× 10^{28} m^{−3}. 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 × 10^{−6 }m^{2} and it
is carrying a current of 3.0 A.

#### Answer:

Number density of free
electrons in a copper conductor, *n* = 8.5 × 10^{28} m^{−3} Length of the copper wire, *l* = 3.0 m

Area of cross-section
of the wire, *A* = 2.0 × 10^{−6 }m^{2}

Current carried by the
wire, *I* = 3.0 A, which is given by the relation,

*I *= *nA*e*V*_{d}

Where,

e = Electric charge =
1.6 × 10^{−19} C

*V*_{d} =
Drift velocity

Therefore, the time
taken by an electron to drift from one end of the wire to the other
is 2.7 × 10^{4} s.

#### Question 3.14:

The earth’s
surface has a negative surface charge density of 10^{−9} C m^{−2}. 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 × 10^{6} m.)

#### Answer:

Surface charge density
of the earth, *σ* =
10^{−9} C m^{−2}

Current over the entire
globe, *I* = 1800 A

Radius of the earth, *r* = 6.37 × 10^{6} m

Surface area of the earth,

*A* = 4π*r*^{2}

=
4π × (6.37 ×
10^{6})^{2}

=
5.09 × 10^{14} m^{2}

Charge on the earth surface,

*q *= *σ* × *A*

= 10^{−9} × 5.09 × 10^{14}

=
5.09 × 10^{5 }C

Time taken to
neutralize the earth’s surface =* t*

Current,

Therefore, the time taken to neutralize the earth’s surface is 282.77 s.

#### Page No 129:

#### Question 3.15:

**(a)** Six lead-acid type of secondary cells each of emf 2.0 V
and internal resistance 0.015 Ω are
joined in series to provide a supply to a resistance of 8.5 Ω.
What are 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 internal resistance of 380 Ω.
What maximum current can be drawn from the cell? Could the cell drive
the starting motor of a car?

#### Answer:

**(a) **Number of
secondary cells, *n* = 6

Emf
of each secondary cell, *E* = 2.0 V

Internal
resistance of each cell, *r* = 0.015 Ω

series resistor is connected to the combination of cells.

Resistance
of the resistor, *R* = 8.5 Ω

Current
drawn from the supply = *I*, which is given by the relation,

Terminal
voltage,* V* = *IR* = 1.39 × 8.5 = 11.87 A

Therefore, the current drawn from the supply is 1.39 A and terminal voltage is

11.87 A.

**(b) **After a long
use, emf of the secondary cell, *E* = 1.9 V

Internal
resistance of the cell, *r* = 380 Ω

Hence, maximum current

Therefore, the maximum current drawn from the cell is 0.005 A. Since a large current is required to start the motor of a car, the cell cannot be used to start a motor.

#### Question 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. (*ρ*_{Al} = 2.63 × 10^{−8} Ω m, *ρ*_{Cu} = 1.72 × 10^{−8} Ω m, Relative density of
Al = 2.7, of Cu = 8.9.)

#### Answer:

Resistivity of
aluminium, *ρ*_{Al} = 2.63 × 10^{−8} Ω
m

Relative density of
aluminium, *d*_{1} = 2.7

Let *l*_{1} be the length of aluminium wire and *m*_{1} be its mass.

Resistance of the
aluminium wire = *R*_{1}

Area of cross-section
of the aluminium wire =* A*_{1}

Resistivity of copper, *ρ*_{Cu} = 1.72
× 10^{−8} Ω
m

Relative density of
copper, *d*_{2} = 8.9

Let *l*_{2} be the length of copper wire and *m*_{2} be its mass.

Resistance of the
copper wire =* R*_{2}

Area of cross-section
of the copper wire =* A*_{2}

The two relations can be written as

It is given that,

And,

Mass of the aluminium wire,

*m*_{1} =
Volume × Density

= *A*_{1}*l*_{1} × *d*_{1 }= *A*_{1} *l*_{1}*d*_{1} …
(3)

Mass of the copper wire,

*m*_{2} =
Volume × Density

= *A*_{2}*l*_{2} × *d*_{2 }= *A*_{2} *l*_{2}*d*_{2} …
(4)

Dividing equation (3) by equation (4), we obtain

It can be inferred from
this ratio that *m*_{1} is less than *m*_{2}.
Hence, aluminium is lighter than copper.

Since aluminium is lighter, it is preferred for overhead power cables over copper.

#### Question 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

#### Answer:

It can be inferred from the given table that the ratio of voltage with current is a constant, which is equal to 19.7. Hence, manganin is an ohmic conductor i.e., the alloy obeys Ohm’s law. According to Ohm’s law, the ratio of voltage with current is the resistance of the conductor. Hence, the resistance of manganin is 19.7 Ω.

#### Question 3.18:

Answer the following questions:

**(a) ** A steady current flows in a metallic conductor of
non-uniform cross- section. Which of these quantities is constant
along the conductor: current, current density, electric field, drift
speed?

**(b)** Is Ohm’s
law universally applicable for all conducting elements?

If not, give examples of elements which do not obey Ohm’s law.

**(c) ** A low voltage supply from which one needs high currents
must have very low internal resistance. Why?

**(d) ** A high tension (HT) supply of, say, 6 kV must have a very
large internal resistance. Why?

#### Answer:

**(a) **When a steady current flows in a metallic conductor of
non-uniform cross-section, the current flowing through the conductor
is constant. Current density, electric field, and drift speed are
inversely proportional to the area of cross-section. Therefore, they
are not constant.

**(b) **No, Ohm’s law is not universally applicable for all
conducting elements. Vacuum diode semi-conductor is a non-ohmic
conductor. Ohm’s law is not valid for it.

**(c) **According to
Ohm’s law, the relation for the potential is *V *=* IR*

Voltage
(*V*) is directly proportional to current (*I*).

*R* is the internal resistance of the source.

If *V* is low, then *R* must be very low, so that high current
can be drawn from the source.

**(d) **In order to prohibit the current from exceeding the safety
limit, a high tension supply must have a very large internal
resistance. If the internal resistance is not large, then the current
drawn can exceed the safety limits in case of a short circuit.

#### Question 3.19:

Choose the correct alternative:

**(a) ** Alloys of metals usually have (greater/less) resistivity
than that of their constituent metals.

**(b) ** Alloys usually have much (lower/higher) temperature
coefficients of resistance than pure metals.

**(c) ** The resistivity of the alloy manganin is nearly
independent of/increases rapidly with increase of temperature.

**(d) ** The resistivity of a typical insulator (e.g., amber) is
greater than that of a metal by a factor of the order of (10^{22}/10^{3}).

#### Answer:

**(a) **Alloys of metals usually have greater resistivity than
that of their constituent metals.

**(b) **Alloys usually have lower temperature coefficients of
resistance than pure metals.

**(c) **The resistivity of the alloy, manganin, is nearly
independent of increase of temperature.

**(d) **The resistivity of a typical insulator is greater than
that of a metal by a factor of the order of 10^{22}.

#### Question 3.20:

**(a)** 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?

**(b)** 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)
(11/5) Ω, (iii) 6 Ω,
(iv) (6/11) Ω?

**(c)** Determine
the equivalent resistance of networks shown in Fig. 3.31.

#### Answer:

**(a) **Total number
of resistors = *n*

Resistance
of each resistor = *R*

**(i)** When *n* resistors are connected in series, effective
resistance *R*_{1}is the maximum, given by the product *nR*.

Hence, maximum resistance of the combination, *R*_{1 }= *nR*

**(ii)** When *n* resistors are connected in parallel, the
effective resistance (*R*_{2}) is the minimum, given by
the ratio.

Hence,
minimum resistance of the combination, *R*_{2} =

**(iii)** The ratio of the maximum to the minimum resistance is,

**(b) **The
resistance of the given resistors is,

*R*_{1} = 1 Ω, *R*_{2} = 2 Ω, *R*_{3} = 3 Ω2

- Equivalent resistance,

Consider the following combination of the resistors.

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,

**(iii)** Equivalent
resistance, *R*^{’} = 6 Ω

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

Equivalent resistance of the circuit is given by the sum,

*R*^{’} = 1 + 2 + 3 = 6 Ω

**(iv)** Equivalent
resistance,

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

Equivalent resistance of the circuit is given by,

**(c) (a)** It can be observed from the given circuit that in the
first small loop, two resistors of resistance 1 Ω
each are connected in series.

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.

Hence, their equivalent resistance = (2 + 2) = 4 Ω.

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 (*R*^{’})
of each loop is given by,

The circuit reduces to,

All the four resistors are connected in series.

Hence, equivalent resistance of the given circuit is

**(b)** It can be observed from the given circuit that five
resistors of resistance *R* each are connected in series.

Hence,
equivalent resistance of the circuit = *R *+ *R* +* R* + *R* + *R*

=
5 *R*

*2*

#### Page No 130:

#### Question 3.21:

Determine the current drawn from a 12 V supply with internal resistance 0.5 Ω by the infinite network shown in Fig. 3.32. Each resistor has 1 Ω resistance.

#### Answer:

The resistance of each
resistor connected in the given circuit, *R* = 1 Ω

Equivalent resistance
of the given circuit = *R*^{’}

The network is infinite. Hence, equivalent resistance is given by the relation,

Negative value of *R*^{’ }cannot be accepted. Hence, equivalent resistance,

Internal resistance of
the circuit, *r* = 0.5 Ω

Hence, total resistance of the given circuit = 2.73 + 0.5 = 3.23 Ω

Supply voltage, *V* = 12 V

According to Ohm’s Law, current drawn from the source is given by the ratio, = 3.72 A

#### Question 3.22:

Figure 3.33 shows a
potentiometer with a cell of 2.0 V and internal resistance 0.40
Ω 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, a very high
resistance of 600 kΩ 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 *ε* and
the balance point found similarly, turns out to be at 82.3 cm length
of the wire.

**(a) ** What is the
value *ε* ?

**(b) ** What
purpose does the high resistance of 600 kΩ 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
thermo-couple)? If not, how will you modify the circuit?

#### Answer:

**(a) **Constant emf
of the given standard cell, *E*_{1} = 1.02 V

Balance
point on the wire, *l*_{1 }= 67.3 cm

A
cell of unknown emf, *ε*,replaced the standard cell. Therefore, new balance point on the
wire, *l* = 82.3 cm

The relation connecting emf and balance point is,

The value of unknown emfis 1.247 V.

**(b) **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.

**(c) **The balance
point is not affected by the presence of high resistance.

**(d) **The point is
not affected by the internal resistance of the driver cell.

**(e) **The method would not work if the driver cell of the
potentiometer had an emf of 1.0 V instead of 2.0 V. 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.

**(f) **The circuit would not work well for determining an
extremely small emf. As the circuit would be unstable, the balance
point would be close to end A. Hence, there would be a large
percentage of error.

The given circuit can be modified if a series resistance is connected with the wire AB. The potential drop across AB is slightly greater than the emf measured. The percentage error would be small.

#### Page No 131:

#### Question 3.23:

Figure 3.34 shows a
potentiometer circuit for comparison of two resistances. The balance
point with a standard resistor *R* = 10.0 Ω is
found to be 58.3 cm, while that with the unknown resistance *X *is 68.5 cm. Determine the value of *X*.
What might you do if you failed to find a balance point with the
given cell of emf *ε*?

#### Answer:

Resistance of the
standard resistor, *R* = 10.0 Ω

Balance point for this
resistance, *l*_{1} = 58.3 cm

Current in the
potentiometer wire = *i*

Hence, potential drop
across *R*, *E*_{1} = *iR*

Resistance of the
unknown resistor = *X*

Balance point for this
resistor, *l*_{2} = 68.5 cm

Hence, potential drop
across *X*, *E*_{2} = *iX*

The relation connecting emf and balance point is,

Therefore, the value of
the unknown resistance, *X*, is 11.75 Ω.

If we fail to find 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. Only if the potential drop
across *R* or *X* is smaller than the potential drop across
the potentiometer wire AB, a balance point is obtained.

#### Question 3.24:

Figure 3.35 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 Ω 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.

#### Answer:

Internal resistance of
the cell = *r*

Balance point of the
cell in open circuit, *l*_{1} = 76.3 cm

An external resistance
(*R*) is connected to the circuit with *R* = 9.5 Ω

New balance point of
the circuit, *l*_{2 }= 64.8 cm

Current flowing through
the circuit = *I*

The relation connecting resistance and emf is,

Therefore, the internal resistance of the cell is 1.68Ω.

**NCERT Solutions for Class 12 Physics Chapters**

**Physics Part I**

- Chapter 1 – Electric Charges And Fields
- Chapter 2 – Electrostatic Potential And Capacitance
- Chapter 3 – Current Electricity
- Chapter 4 – Moving Charges And Magnetism
- Chapter 5 – Magnetism And Matter
- Chapter 6 – Electromagnetic Induction
- Chapter 7 – Alternating Current
- Chapter 8 – Electromagnetic Waves

**Physics Part II**

- Chapter 1 – Ray Optics And Optical Instruments
- Chapter 2 – Wave Optics
- Chapter 3 – Dual Nature Of Radiation And Matter
- Chapter 4 – Atoms
- Chapter 5 – Nuclei
- Chapter 6 – Semiconductor Electronics: Materials Devices and Simple Circuits
- Chapter 7 – Communication Systems

**NCERT Solutions for Class 12-science:**