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q2 past year paper

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Q1 1.1. three capacitors of 2 microfarads, 4 microfarads and 6 microfarads respectively are connected in series a 100 volt supply. Calculate: a. the total capacitance b. the pd across each capacitor. 1.2. define farad 1.3. two capacitors of 3 microfarads and 6 microfarads respectively are connected in series. This combination is then connected in parallel with a 4 microfarads capacitor across a 200 volts DC supply. Calculate the total change, capacitance, the charge across the series capacitors and potential difference (PD) across each of the series capacitors. Q2 2.1. name three principle sources of EMF. 2.2. a battery having EMF of 50 volts and internal resistance of 0,3 ohms is connected in parallel with a direct-current generator of EMF 60 Volts and internal resistance of 0,4 ohms. The combination is used to supply a load having a resistance of 10,6 ohms. Use Kirchoff’s theorem to determine: 2.2.1. the value and direction of the current through the battery 2.2.2. the value and direction of the current through the generator 2.2.3. the potential difference across the load Q3 An aluminum conductor of unknown length with a voltage of 75V is connected in parallel with a copper conductor with same length. When a current of 10A is passed combination, it is found that Determine: 3.1. the length of the copper conductor if the resistively of copper is 0,017 micro-ohm meters and that of aluminum is 2,5mm. micro-ohm meters. 3.2. the current through the copper conductor is 2,5A. the diameter of the aluminum is 2,5mm. the resistance and the diameter of the copper conductor. Q4 4.1. the field coils of motor have a resistance of 50 ohms at 10 degrees Celsius. After a run at full load, the resistance increases to 100 ohms and the temperature of the coils to 201 degrees Celsius. Calculate the temperature coefficient of resistance at 0 degrees Celsius. 4.2. the field-coil of motor has a resistance of 100 ohms at 15 degrees Celsius. By how much will the resistance increase if the motor reaches a temperature of 65 degrees Celsius when running? Take the temperature coefficient of resistance to be 0,004 per degrees Celsius at 15 degrees Celsius. 4.3. difference between positive and negative coefficient of resistance.
asked Mar 29, 2015 in PHYSICS by anonymous

11 Answers

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1.1)

The capacitors are in series:

1/Ceq = 1/C1 + 1/C2 +1/C3

Substitute C1 = 2 microfarads, C2 = 4 microfarads, and C3 = 6 microfarads.

1/Ceq = 1/2 + 1/4 +1/6

           = (6 + 3 + 2) /12

           = 11/12

Ceq = 12/11 

        = 1.091 microfarads

answered Mar 29, 2015 by bradely Mentor
0 votes

Step 1:

image

The three capacitors are connected in series with a supply.

(a)

Find the total capacitance.

Consider be the total capacitance.

are in series.

So

Therefore the total capacitance is .

Step 2:

(b)

Find the potential difference across each capacitance.

Now find the total charge .

.

Substitute and in the total charge.

The total charge is image.

Find the voltage across the capacitance .

Find the voltage across the capacitance .

Find the voltage across the capacitance .

The voltage across the capacitors are .

Solution :

(a) The total capacitance is .

(b) The voltage across the capacitors are .

answered Mar 30, 2015 by joseph Apprentice
edited Mar 30, 2015 by joseph
0 votes

(1.2)

Define Farad :

The farad is the standard unit of capacitance in the SI unit and it is denoted by F.

Farad is defined by the charge of 1 coulomb increase the potential difference between its plates by 1 volt.

image.

answered Mar 30, 2015 by joseph Apprentice
0 votes

(1.3)

Step 1:

The two capacitors  are in series.

Consider be the total series capacitance.

Now Find the total capacitance.

Consider be the total capacitance.

Here are in parallel.

So .

The total capacitance is .

Step 2:

Find the total charge image on the circuit.

Voltage across the circuit is 200 V.

image

The total charge is image.

In parallel circuit, voltage is same and charge will change.

Find the charge across the series capacitance.

So image.

image

The charge across the series capacitance is image.

Step 3:

Find the potential difference across the each of the series capacitance.

In series circuit, charge is same and voltage will change.

Find the voltage across the capacitor image.

image

Find the voltage across the capacitor image.

image

The voltage across the capacitors are image.

Solution :

The total capacitance is .

The total charge is image.

The charge across the series capacitance is image.

The voltage across the capacitors are image.

answered Mar 30, 2015 by joseph Apprentice
0 votes

Step 1:

(2.2)

A DC generator has emf of 60 V.

Internal resistance of the DC generator is 0.4 .

A battery has emf of 50 V.

Internal resistance of the battery is 0.3 .

DC generator and battery are connected in parallel to the load of 10.6 .

Draw a circuit with above specifications.

Step 2:

(2.2.1)

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.

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

Step 3:

Solve the equations (4) and (5).

Add the equations (4) and (5).

Substitute image in equation (4).

image

Substitute imagein image.

image

So the value of current through the generator is 17.2043 A.

The direction of the current is toward the node b.

answered Mar 30, 2015 by Lucy Mentor

Contd....

Step 4:

(2.2.2)

The value of current through the battery is .

So the value of current through the battery is 12.968  A.

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

Step 5:

(2.2.3)

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 54.1395 V.

Solution:

(2.2.1)

The value of current through the generator is 17.2043 A.

The direction of the current is toward the node b.

(2.2.2)

The value of current through the battery is 12.968  A.

The direction of the current is toward the node c.

(2.2.3)

The voltage across the load is 54.1395 V.

0 votes

Q4

Step 1:

(4.1)

The resistance of the field coil of motor at 10° C is 50 .

The resistance of the field coil of motor at 201° C is 100 .

Find temperature coefficient at 0° C.

Temperature coefficient at 0° C : image.

image

Temperature coefficient at 0° C is image/C.

Solution:

Temperature coefficient at 0° C is image/C.

answered Mar 31, 2015 by Lucy Mentor
0 votes

Step 1:

(4.2)

The resistance of the motor at 15° C is 100 .

Temperature coefficient is 0.004/C.

Final temperature is 65° C

Find the increase in resistance.

Temperature coefficient at 0° C : image.

image

The increase in resistance is 20 .

Solution:

The increase in resistance is 20 .

answered Mar 31, 2015 by Lucy Mentor
0 votes

Step 1:

(4.3)

Negative coefficient of resistance :

The resistance of the material decreases with increase in temperature,this phenomenon is said to be negative temperature coefficient of semiconductor.

Semiconductors and insulators have negative temperature coefficient. These are used as thermistors.

Positive coefficient of resistance :

The resistance of the material increases with increase in temperature,this phenomenon is said to be positive temperature coefficient of semiconductor.

Usually all metals and their alloys have positive temperature coefficient.

Solution:

The difference between between Negative coefficient of resistance and positive coefficient of resistance is the change in the resistance of the material with temperature.

answered Mar 31, 2015 by Lucy Mentor
0 votes

(3.1)

Step 1:

The diameter of the aluminum wire is d1= 2.5 mm.

Let the length of the aluminum wire is l.

Since the length are same, the length of the copper wire is l.

Specific resistivity of the aluminum is .

Specific resistivity of the copper .

Total current in the circuit is it = 10 A.

Current through copper wire is i2 = 2.5 A.

The aluminum wire and copper wire are connected in parallel.

Total Current = i1 + i2 .

10 = i1 + 2.5

i1  =10 - 2.5

i1 = 7.5 A.

Current through aluminum wire is i1 = 7.5 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

answered Apr 1, 2015 by yamin_math Mentor
reshown Apr 1, 2015 by casacop
0 votes

Contd....(3.1)

Step 3:

In a parallel combination, the voltage across the element are same.

Substitute in equation (1).

Substitute the corresponding values in the above formula.

image

The diameter of the copper wire is 1.1902 mm.

Step 4:

Find the length of the aluminum conductor .

Resistance of the aluminum conductor .

Voltage drop across the aluminum is .

An aluminum conductor has a voltage of 75V.

Now find the resistance of aluminum conductor.

Substitute resistance of aluminum the .

image

Length of aluminum conductor is 1.96 km.

So the length of copper conductor is 1.96 km.

Solution :

The length of copper conductor is 1.96 km.

answered Apr 1, 2015 by yamin_math Mentor

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