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.

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.

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 = 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.

ANSWER:

(a) There are three resistors of resistances,

R₁ = 2 Ω, R₂ = 4 Ω, and R₃ = 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 (I₁) flowing through resistor R₁ is given by,

Current (I₂) flowing through resistor R₂ is given by,

Current (I₃) flowing through resistor R₃ is given by,

Total current, I = I₁ + I₂ + I₃ = 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.

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.

Resistance of the heating element at T₁, R₁ = 117 Ω

Temperature co-efficient of the material of the filament,

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

ANSWER:

Length of the wire, l =15 m

Area of cross-section of the wire, a = 6.0 × 10⁻⁷ m²

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.

ANSWER:

Temperature, T₁ = 27.5°C

Resistance of the silver wire at T₁, R₁ = 2.1 Ω

Temperature, T₂ = 100°C

Resistance of the silver wire at T₂, R₂ = 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.

ANSWER:

Supply voltage, V = 230 V

Initial current drawn, I₁ = 3.2 A

Initial resistance = R₁, which is given by the relation,

Steady state value of the current, I₂ = 2.8 A

Resistance at the steady state = R₂, which is given as

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

Initial temperature of nichrome, T₁= 27.0°C

Study state temperature reached by nichrome = T₂

T₂ can be obtained by the relation for α,

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

ANSWER:

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

I₁ = Current flowing through the outer circuit

I₂ = Current flowing through branch AB

I₃ = Current flowing through branch AD

I₂ − I₄ = Current flowing through branch BC

I₃ + I₄ = Current flowing through branch CD

I₄ = Current flowing through branch BD

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

10I₂ + 5I₄ − 5I₃ = 0

2I₂ + I₄ −I₃ = 0

I₃ = 2I₂ + I₄ … (1)

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

5(I₂ − I₄) − 10(I₃ + I₄) − 5I₄ = 0

5I₂ + 5I₄ − 10I₃ − 10I₄ − 5I₄ = 0

5I₂ − 10I₃ − 20I₄ = 0

I₂ = 2I₃ + 4I₄ … (2)

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

−10 + 10 (I₁) + 10(I₂) + 5(I₂ − I₄) = 0

10 = 15I₂ + 10I₁ − 5I₄

3I₂ + 2I₁ − I₄ = 2 … (3)

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

I₃ = 2(2I₃ + 4I₄) + I₄

I₃ = 4I₃ + 8I₄ + I₄

− 3I₃ = 9I₄

− 3I₄ = + I₃ … (4)

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

I₃ = 2I₂ + I₄

− 4I₄ = 2I₂

I₂ = − 2I₄ … (5)

It is evident from the given figure that,

I₁ = I₃ + I₂ … (6)

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

3I₂ +2(I₃ + I₂) − I₄ = 2

5I₂ + 2I₃ − I₄ = 2 … (7)

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

5(−2 I₄) + 2(− 3 I₄) ₋ I₄ = 2

− 10I₄ − 6I₄ − I₄ = 2

17I₄ = − 2

Equation (4) reduces to

I₃ = − 3(I₄)

Therefore, current in branch 

In branch BC = 

In branch CD = 

In branch AD 

In branch BD = 

Total current = 

ANSWER:

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

(a) Balance point from end A, l₁ = 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 l₁ and 100−l₁ get interchanged.

The balance point of the bridge will be 100−l₁ from A.

100−l₁ = 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.

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 = V¹

R is connected to the storage battery in series. 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 × 15.5 = 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.

ANSWER:

Emf of the cell, E₁ = 1.25 V

Balance point of the potentiometer, l₁= 35 cm

The cell is replaced by another cell of emf E₂.

New balance point of the potentiometer, l₂ = 63 cm

Therefore, emf of the second cell is 2.25V.

ANSWER:

Number density of free electrons in a copper conductor, n = 8.5 × 10²⁸ m⁻³ Length of the copper wire, l = 3.0 m

Area of cross-section of the wire, A = 2.0 × 10⁻⁶ m²

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

I = nAeV_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 × 10⁴ s.

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 × 10⁶ m

Surface area of the earth,

A = 4πr²

= 4π × (6.37 × 10⁶)²

= 5.09 × 10¹⁴ m²

Charge on the earth surface,

q = σ × A

= 10⁻⁹ × 5.09 × 10¹⁴

= 5.09 × 10⁵ C

Time taken to neutralize the earth’s surface = t

Current, 

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

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.

ANSWER:

Resistivity of aluminium, ρ_Al = 2.63 × 10⁻⁸ Ω m

Relative density of aluminium, d₁ = 2.7

Let l₁ be the length of aluminium wire and m₁ be its mass.

Resistance of the aluminium wire = R₁

Area of cross-section of the aluminium wire = A₁

Resistivity of copper, ρ_Cu = 1.72 × 10⁻⁸ Ω m

Relative density of copper, d₂ = 8.9

Let l₂ be the length of copper wire and m₂ be its mass.

Resistance of the copper wire = R₂

Area of cross-section of the copper wire = A₂

The two relations can be written as

It is given that,

And,

Mass of the aluminium wire,

m₁ = Volume × Density

= A₁l₁ × d₁ = A₁ l₁d₁ … (3)

Mass of the copper wire,

m₂ = Volume × Density

= A₂l₂ × d₂ = A₂ l₂d₂ … (4)

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

It can be inferred from this ratio that m₁ is less than m₂. Hence, aluminium is lighter than copper.

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

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.

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.

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 10²².

ANSWER:

(a) Total number of resistors = n

Resistance of each resistor = R

(i) When n resistors are connected in series, effective resistance R₁is the maximum, given by the product nR.

Hence, maximum resistance of the combination, R₁ = nR

(ii) When n resistors are connected in parallel, the effective resistance (R₂) is the minimum, given by the ratio.

Hence, minimum resistance of the combination, R₂ = 

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

(b) The resistance of the given resistors is,

R₁ = 1 Ω, R₂ = 2 Ω, R₃ = 3 Ω2

1. Equivalent resistance, 

Consider the following combination of the resistors.

Equivalent resistance of the circuit is given by,

2. 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

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

ANSWER:

(a) Constant emf of the given standard cell, E₁ = 1.02 V

Balance point on the wire, l₁ = 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.

ANSWER:

Resistance of the standard resistor, R = 10.0 Ω

Balance point for this resistance, l₁ = 58.3 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 = 68.5 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 11.75 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.

ANSWER:

Internal resistance of the cell = r

Balance point of the cell in open circuit, l₁ = 76.3 cm

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

New balance point of the circuit, l₂ = 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.