how to answer this past year physics questions

Question

1.1. A resitance of 1.5ohm is connected in parallel with a resistance of 3 ohm. This combination is connected in series with a resistance of unknown resistance of value R. The circuit is then connected acress a 10 V supply. Calculate: 1.1.1. The value of the resistor R when a 2A current is drawn from the supply 1.1.2. The powe sdissipated in the circuit. 1.2. The field coil of a motor has a resistance of 100 ohms at 10 degrees C. By how much will the resistance increase of the motor attains a temperature of 35 degrees C when running? Take the temperature coefficient as 0.004 per degrees C at 10 degrees C. 3. Define the ampere 4. An aluminum conductor, 1 km long, is connected in parallel with a copper conductor having the same length. When a current of 220A passed through the combination it is found that the current through the copper conductor is 120A. The diameter of the aluminum is 10mm. Calculate: 4.1. The diameter of the copper conductor if the specific resistivity of copper is 0.015 coefficient ohm/m and the aluminuim is 0.028 coefficient ohm/m. 4.2. The voltage drop across conductors. 5. A DC generator of EMF 60V and internal resistance of 0.5 ohm are connected in parallel with a battery of EMF 30V and internal resistance 0.6ohm. The combination is used ro supplt a load having a resistance of 10ohm. Using kirchhoff's laws to determing: 5.1. The value and direction of current through the generator. 5.2. The value and direction of current through the batter 5.3. The terminal voltage across the load. 6. A resistance of 6ohm is connected in parallel with resistance of 9ohm. The combination is connected in series with a third resistance of 2ohm. If the whole circuit is connected across a battery having an EMF of 18V and an interbal resistance of 0.4ohm, calculate: 6.1. The terminal voltage of the battery. 6.2. The current through each resistor

Answer 1

3).

Definition of ampere :

• An ampere is a unit of measure of the rate of electron flow or current in an electrical conductor.
• One ampere of current represents one coulomb of electrical charge (6.24 x 10¹⁸ charge carriers) moving past a specific point in one second.
• Physicists consider current to flow from relatively positive points to relatively negative points,  this is called conventional current or Franklin current.

Answer 2

Step 1:

(1.1.1):

A resistance of is connected in parallel with resistor.

If two resistors    connected in parallel, then equivalent resistor is given by,

Substitute and in the above formula.

 

The resistance is connected in series with unknown resistor and circuit is connected to 10 V supply.

Now the two resistors are in series, then equivalent resistor is given by,

.

Substitute and in the above formula.

.

 

2 A current is drawn from the supply of 10 V.

From the ohms law: .

Substitute and in above formula.

This resistance is equal to the equivalent resistance.

Therefore value of the unknown resistor is when 2 A current is drawn from the source.

Step 2:

(1.1.2):

Power dissipated in the circuit:

Power dissipated in the circuit with current drop and supply voltage is given by,

.

Substitute and in the above formula.

Power dissipated in the circuit is 20 watt.

Solution:

Unknown resistance is .

Power dissipated in the circuit is 20 watt.

Answer 3

(6)

Step 1:

A resistance of is connected in parallel with resistance of .

Find the equivalent resistance.

Consider is the equivalent resistance.

is connected in parallel to .

The equivalent resistance is .

Step 2:

The equivalent resistance is connected in series with a resistance of .

Now find the total resistance.

Consider is the total resistance.

So is connected in series with .

The total resistance is .

Step 3:

The above circuit is connected across a battery having an EMF of 18 V and an internal resistance of .

Now find the total equivalent resistance.

Consider is the total equivalent resistance.

So is connected in series with .

The total equivalent resistance is .

Find the current in the circuit.

Consider is the current in the circuit.

From ohms law :

Then current is

Substitute in the current.

The current in the circuit is .

Comments

Contd....

Step 4:

(6.1)

Find the terminal voltage of the battery.

Suppose that the battery with EMF  and internal resistance  supplies a current  through an external load resistor , then the terminal voltage across the battery is .

Substitute in the terminal voltage.

The terminal voltage of the battery is .

Step 5:

(6.2)

Find the current through each resistance.

The current through resistance is .

The voltage across resistance is

Voltage across resistance is .

Now find the voltage across the parallel resistance loop .

The terminal voltage is .

The voltage across the resistance is .

The voltage is same in the parallel loop.

So voltage across the and is .

Find the current .

Find the current .

Solution :

(6.1) The terminal voltage of the battery is .

(6.2) The current through resistance is .

The current through resistance is .

The current through resistance is .

Answer 4

Step 1 :

1.2)

Linear approximation : .

Where, is the final resistance,

is the initial resistance,

is the temperature difference and

is the temperature coefficient.

Step 2 :

From the given data :

,

,

.

Substitute corresponding values in .

Increase resistance at .

Thus, increase resistance of the motor at is .

Solution :

Increase resistance of the motor at is .

Answer 5

Step 1:

4.1

The diameter of the aluminum wire is d₁=10 mm.

The length of the aluminum wire is l₁ = 1 km.

The length of the copper wire is l₂ = 1 km.

Specific resistivity of the aluminum .

Specific resistivity of the copper .

Total current in the circuit is i_t = 220 A.

Current through copper wire is i₂ = 120 A.

The aluminum wire and copper wire are connected in parallel.

Total Current = i₁ + i₂ .

220 = i₁ + 120

i₁ = 220-120

i₁ = 100 A.

Current through aluminum wire is i₁ = 100 A.

Step 2:

Law of Resistivity:

Resistance offered by a conductor is given by .

Resistance offered by aluminum wire is .

Area of the aluminum wire is .

Resistance offered by aluminum wire is .

Resistance offered by copper wire is .

Area of the copper wire is .

Resistance offered by copper wire is .

Ratio of the resistance is

Step 3:

In a parallel combination, the voltage across both the elements are same.

Substitute in equation (1).

Substitute the corresponding values in the above formula.

The diameter of the copper wire is 8.02 mm.

Comments

Contd..

Step 2:

(b)

Specific resistivity of the aluminum .

Specific resistivity of the copper .

Voltage drop across the aluminum wire is .

Resistance offered by aluminum wire is .

Resistance offered by aluminum wire is .

Voltage drop across the aluminum wire:

Voltage drop across the aluminum wire is 3.57 v.

In a parallel combination, the voltage across both the elements are same.

Voltage drop across the copper wire is 3.57 v.

Solution:

(a) The diameter of the copper wire is 8.02 mm.

(b) Voltage drop across the conductors is 3.57 v.

Answer 6

(5)

Step 1:

A DC generator has emf of 60 V.

Internal resistance of the DC generator is 0.5 .

A battery has emf of 30 V.

Internal resistance of the battery is 0.6 .

DC generator and battery are connected in parallel to the load of 10 .

Draw a circuit with above specifications.

Step 2:

Find the current through generator :

Redraw the circuit with current directions and nodes.

Apply Junction rule at node b.

.

Apply KVL to a loop abcda.

 

Apply KVL to a loop befcb.

 

Substitute equation (1) in equation (2) and (3).

Solve the equations (4) and (5).

Add the equations (4) and (5).

Substitute in equation (4).

Substitute in .

So the value of current through the generator is 29.7 A.

The direction of the current is toward the node b.

Step 3:

(b)

The value of current through the battery is .

So the value of current through the battery is 25.19  A.

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

Step 4:

Find the terminal voltage across the load.

The current across the load is .

Use ohms law : .

Where i is current through load,

           R is resistance offered by load.

The voltage across the load is 45.1 V.

Solution:

(a)

The value of current through the generator is 29.7 A.

The direction of the current is toward the node b.

(b)

The value of current through the battery is 25.19  A.

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

(c) The voltage across the load is 45.1 V.