1.3)

The magnitude of the hysteresis loss depends on the composition of the spicemen, on the heat treatment and mechanical handling to which the specimen has been subjected.

1.4)

They are used to make standard resistors because,

1) They have high value of resistivity.

2) Temperature cofficient of resistance is less.

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

(1.1)

Step 1:

A DC generator has emf of 60 V.

Internal resistance of the DC generator is 0.2 .

Battery A has emf of 50 V.

Internal resistance of the battery A is 0.4 .

Battery B has emf of 30 V.

Internal resistance of the battery B is 1.2 .

Draw a circuit with above specifications.

Step 2:

(1.1.1)

Find the current through generator :

Redraw the circuit with current directions and nodes.

Apply KVL to a loop abeda.

Apply KVL to a loop bcfeb.

Solve the equations (1) and (2).

Multiply equation (1) with 4.

Subtract the equations (2) and (3).

A.

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

The direction of the current is toward the node b.

Step 3:

(1.1.2)

Find the value of current through battery A.

Current through battery A is .

Find .

Substitute in equation (1).

A.

The value of current through the battery is .

So the value of current through the battery is 10 A.

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

Step 4:

(1.1.3)

Find the value of current through battery B.

Current through battery B is .

Find .

From (1.1.2), the value of current through the battery is 20 A.

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

Solution:

(1.1.1) The value of current through the generator is 30 A.

(1.1.2) The value of current through the battery is 10 A.

(1.1.3) The value of current through the battery is 20 A.

(1.2)

Step 1:

A 50 resistor is connected across a rheostat.

Let R_h be the rheostat resistance.

A heater is connected in series to it.

Power dissipated by heater is 500 W.

Supply voltage is 150 V.

Current in the circuit is 10 A.

Draw a circuit with above specifications :

The power across the heater is .

Current in a series loop is same.

Resistance provided by the heater is 5.

Step 2:

Here R₁ and R_h are in parallel, then the equivalent resistance is

Apply Kirchoff voltage law.

The value of the rheostat must be 12.5 so that heater draws 10 A of current.

Solution:

Resistance provided by rheostat is 12.5 .