Updated By SaraNextGen

On March 11, 2024, 11:35 AM

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.

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 = 

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, 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.

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 = 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.

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, 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.

Question 3.13:

The number density of free electrons in a copper conductor estimated in Example 3.1 is 8.5 × 10²⁸ 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 m² and it is carrying a current of 3.0 A.

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.

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 × 10⁶ 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 × 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.