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Question

Answer 1

2.4)

Alloys are used to make standard resistors because of the following reasons:

1) They have high value of resistivity.

2) Temperature coefficient of resistance is less.

Special alloys such as Invar steel used in measuring instruments because of its negligible coefficient of thermal expansion.

Answer 2

2.3)

An ampere is a unit measure of the rate of electron flow or current  in an electrical conductor.

The ampere is the base SI unit of electrical current.

The ampere is defined as the amount of electrical current required to maintain a force of newtons per meter between two infinitely long parallel wires of negligible cross section held one meter apart.

The symbol for ampere is a capital letter A.

Answer 3

(2.1)

Step 1:

A DC generator has emf of 75 V.

Internal resistance of the DC generator is 0.5 .

A battery has emf of 41 V.

Internal resistance of the battery is 0.3 .

DC generator and battery are connected in parallel to the load of 2.5 .

Draw a circuit with above specifications.

Step 2:

(2.1.1)

Find the current through generator.

Redraw the circuit with current directions and nodes.

Apply Junction rule at node b.

Apply KVL to a loop abeda.

Apply KVL to a loop bcfeb.

Solve the equations (2) and (3).

Multiply equation (2) with 0.3.

Multiply equation (3) with 0.8.

Subtract the equations (4) and (5).

A.

Substitute in equation (2).

So the value of current through the generator is 50 A.

The direction of the current is toward the node b.

Solution:

(2.1.1)

The value of current through the generator is 50 A.

The direction of the current is toward the node b.

Answer 4

(2.1.2)

Step 1:

Find the value of current flowing through the battery.

The value of current through the battery is .

So the value of current through the battery is 30  A.

The direction of the current is toward the node c.  (In opposite direction with respect to generator)

Solution:

(2.1.2)

The value of current through the battery is 30  A.

The direction of the current is toward the node c.

Answer 5

(2.1.3)

Step 1:

Find the potential difference across the load.

The current across the load is A.

Use ohms law : .

Where i is current through load,

           R is resistance offered by load.

The voltage across the load is 50 V.

Solution:

(2.1.3) The voltage across the load is 50 V.

Answer 6

(2.2)

Step 1:

A 3 resistor is connected in parallel with resistor of 3.

This is connected in series to resistor R.

The applied voltage to the network is 45 V.

The current through the circuit is 10 A.

Draw a circuit with above specifications.

R₁ parallel to R₂ then the parallel resistance is 

is in series with the R, hence equivalent resistance is

Find the equivalent resistance.

Equivalent resistance : .

.

Substitute in equation (1).

The series resistance is 3.

Step 2:

(2.2.1)

Find the voltage across the resistors.

Current in a series network is same.

Equilvalent parallel resistance is .

Voltage drop across the parallel as .

Voltage drop across the parallel network is .

Voltage drop across a resistor in parallel network is same.

Hence Voltage drop each 3 resistor is 15 V.

Step 3:

Voltage drop across series resistance R.

Voltage drop across the series resistance is .

Voltage drop across the series resistance is .

Solution:

(2.2.1)

Voltage drop each 3 resistor in parallel network is 15 V.

Voltage drop across the series resistance is 30 V.

Answer 7

(2.2.2)

Step 1:

Find the current through the parallel resistors.

Voltage drop across a resistor in parallel network is same.

Voltage drop across each 3 resistor in parallel network is 15 V.       (From 2.2.1)

Current through the resistance is .

Current through the resistance is 5 A.

Similarly current through the resistance is 5 A.

Solution :

(2.2.2) Current through the resistance and is 5 A.