Chapter 16 - Static Electricity

1.    How does an ebonite rod become negatively charged when it is rubbed with a duster? What is the charge of the duster and how does it acquire that charge?

• The ebonite rod gains electron from the duster and becomes negatively charged as it now has an excess of electrons.

• The duster lose electrons to the ebonite and becomes positively charged as it now has more protons than electrons.

2.    Two plastic rods repel each other after they have been rubbed with a cloth. Suggest why the plastic rods repel each other.

• Rubbing the rods with a cloth cause them to acquire the same charges.

• Since like charges repel each other, the two rods will repel each other.

3.    A positively charged rod is brought near a piece of paper. Explain why the paper is attracted to the rod and sticks to it.

• When the positively charged rod is brought near the neutral paper, it induces a negative charge on the side of the paper nearer to the rod.

• Since unlike charges attract each other, the paper will be attracted to the rod.

• When the paper touches the rod, it does not get discharge as it is an insulator.

• The paper remains attracted to the rod.

4.    A negatively charge metal ball suspended on an insulating thread is brought near a neutral metal sphere held on an insulated stand. State and explain what will happen to the metal ball.

• The negatively charged metal ball will repel the electrons in the sphere to the far right side, hence inducing a positive charge in the near left side.

• Since unlike charges attract, the negatively charge ball will now be attracted to the positively charged side of the sphere.

State and explain what will happen to the ball if the metal ball is allowed to touch the sphere.

• Since both are conductors, the excess negative charges (electrons ) in the ball will flow to the sphere, and be redistributed among the ball and sphere.

• Both the ball and sphere will now be negatively charge.

• Since unlike charges repel, the ball will repel from the sphere.

5.    A positively charged rod is brought near a metal sphere held on an insulated stand. State and explain the movement of electrons in the sphere that occurs as the rod is brought near.

• The electrons, being negatively charged, are attracted to the positively charged rod.

• This leaves the left side of the sphere negatively charged and right side positively charged.

The metal sphere is earthed and the metal ball removed thereafter. State and explain what happens.

• Electrons will flow from the ground, through the earth wire to the right side of the sphere, discharging the right side of the sphere.

• When the positively charged rod is removed, the negative charges will be redistributed. A negatively charged sphere is obtained.

      *Name this process.

• Charging by induction.

6.    Describe how the spray painting of objects makes use of electrostatic principle. State the advantages of using this method of spray painting.

• As the paint leaves the nozzle, the particles of paint become charged by friction.

OR

The nozzle is connected to a positive (negative) terminal so that paint that passes through it will become positively (negatively) charged.

• The paint particles being likely charged, will repelled one another, spreading out when being sprayed on an object.

• The charged paint particles will be attracted to the earthed object by induction.

• This method of spray painting ensures uniform coating of paint and reduces wastage of paint.

Chapter 17 - Current Electricity

1.    Describe an experiment to determine the resistance of a resistor.

• Set up the apparatus as show in the diagram above.

• Adjust the rheostat to the maximum so that the smallest possible current flows in the circuit.

• Record the ammeter reading (I) and the voltmeter reading (V).

• Adjust the rheostat to allow a larger current to flow in the circuit. Record the values of I and V.

• Repeat the procedure to obtain five sets of I and V readings.

• Plot a graph of V against I. Draw a best fit line.

• The gradient of the graph gives the resistance of the resistor.

2.    Define Ohm’s law.

• Ohm’s law states that for metallic conductor, the current passing through it is directly proportional to the potential difference provided physical conditions such as the temperature remains constant

3.    Explain why the component with a voltage-current graph as shown below obeys Ohm’s law (is an ohmic conductor).

• For an Ohmic conductor, the current passing through it is directly proportional to the potential difference provided that the temperature remains constant.

• The graph shown is a straight line through the origin which indicates that V is directly proportional to I.

4.    Draw the voltage versus current graph of a filament lamp and explain why Ohm’s law does not apply to the filament lamp. Give a reason for the relationship.

• The graph is a curve which indicates that V is not directly proportional to I. Therefore, it does not obey Ohm’s Law.

• When the current flows through the filament, its temperature increases. This causes its resistance to increase.

Chapter 18 - D.C. Circuits

1.

Explain why electric lights in a building are connected in parallel rather than in series.

• When the bulbs are connected in parallel, the current through each bulb will be higher and hence each bulb will glow more brightly.

• Parallel arrangement also allows the bulbs to work independently. That is, when one bulb is fused or blown, the other bulbs can still work.

2. 

Resistor A and B are in parallel. Would the addition of an extra resistor parallel to A and B increase or decrease the current flowing through the battery? Explain your answer.

• Adding a resistor in parallel decreases the effective resistance of a circuit.

• Using I=V/R, the current through the battery will increases.

Chapter 19 - Practical Electricity

1.   

Explain why the metal case of electrical equipment which is operated from the mains supply should be earthed.

• If the live wire touches the metal casing accidentally, current will flow through the casing and the user touching the casing will get an electrical shock.

• The earth wire has a resistance much lower than the human body.This will allow the current to flow through it to the ground instead of through the user’s body.

• This protects the user from getting electrocuted.

2.   

Explain why thin wires should not be used in electrical systems, even through this would be cheap.

• Since resistance is inversely proportional to cross-sectional area of wire, thin wires have high resistance.

• Since the heating effect of the wire is given by P=I2R, a thin wire with large R will result in excessive heating. This results in loss of energy and in more serious cases, overheating which may lead to a fire hazard.

3.   

Explain why switches should be placed in the live wire, not the neutral wire, in household electrical systems.

• Switches must be fitted onto the live wire so that opening it will disconnect (isolate) the appliance from the high voltage live wire. (If the switch is placed in the neutral wire, the electrical appliance is still connected to the high voltage live wire even when the switch is opened.)

• This may cause the user to get an electric shock if he touches the live wire accidentally.

4.   

Explain why a fuse should have a rating a little higher than the current normally expected in the circuit.

• This is to ensure that the appliance will operate when the normal operating current flows in the circuit.

• The fuse will only melt when the current is higher than the rated value of the fuse.

5.   

What is the function of a fuse?

·         A fuse will prevent excessive current from flowing in a circuit. This protects the electrical appliance from being damaged.

6.   

How does a fuse work?

• The wire in a fuse will melt and break the circuit if the current exceeds the rating of the fuse.

• This ensures that the circuit is broken.

7.   

Why should a fuse be placed in the live wire of a mains circuit?

• The fuse is connected to the live wire so that the appliance will not become charged (have a potential difference of 230 V) after the fuse has melted due to excessive current.

• Fuses must be fitted onto the live wire so that when it blows, it will disconnect (isolate) the appliance from the high voltage live wire. (If the fuse is placed in the neutral wire, the electrical appliance is still connected to the high voltage live wire even when the switch is opened.)

• This may cause the user to get an electric shock if he touches the live wire accidentally.

Chapter 20 - Magnetism

1.   

State the properties of a magnet.

• A magnet can attract another magnetic material such as iron, steel, cobalt and nickel.

• A magnet has two poles, the North pole and South pole.

• Like poles repel and unlike poles attract.

• A freely suspended magnet always points in a North-South direction.

 2.   

With the aid of a diagram, describe how a magnetic material such as a steel bar can be magnetised using the electrical method.

• A steel bar to be magnetised is placed in a solenoid (wire coil) as shown.

• The solenoid is connected to a direct current (d.c.) supply and switched on. The current flowing through the solenoid produces a magnetic field that will magnetise the steel bar.

• The resistance of the rheostat can be reduced to increase the current flowing in the circuit hence increasing the strength of the magnet.

3.   

With the aid of a diagram, describe how a magnet can be demagnetised using the electrical method.

• A magnet to be demagnetised is placed in a solenoid.

• An alternating current is allowed to flow through the solenoid.

• The magnet is withdrawn in an east-west direction, far away from the solenoid while the current is still flowing.

 4.   

With the aid of a diagram, describe how the magnetic field lines can be plotted with a compass.

• Place a bar magnet on a sheet of paper. Mark the outline of the magnet.

• Put a small compass near to the North-pole of the magnet.

• Using a pencil, mark dots 1 and 2, the position of the South and North pole of the compass respectively.

• Move the compass so that the S-pole of the compass is on dot 2 and mark dot 3 against the N-pole of the compass.

• Repeat the above method to plot other lines of force on either side of the magnet to obtain a magnetic field pattern around the magnet.

5.   

State the differences between the magnetic properties of steel and iron.

• Iron can be magnetised and demagnetised easily while steel is harder to magnetised and demagnetised.

6.   

You are given three steel bars. Only two of the bars are magnets. Explain how you would identify the magnets without using any other equipment.

• Using two of the three bars, check for repulsion between the ends of the two bars. The bars whose ends repel are magnets.

7.   

You are given three apparently identical metal bars, one steel, one soft iron, and one copper. How would you identify them without damaging them in any way?

• Place all 3 bars in a solenoid with direct current flowing through it to magnetise it.

• Bring an iron paper clip close to one end of each bar.

• The bar that does not attract the paper clip at all is the copper.

• Switch off the current.

• The bar that still attracts the paper clip is the steel (as it retained its magnetism).

Chapter 21 - Electromagnetism

Chapter 21: Electromagnetism

 1.    In an electromagnet, adjusting the variable resistor alters the size of the current. Explain the effect on the strength of the electromagnet.

• Increasing the resistance of the variable resistance will decrease the current flowing through the wire which will decrease the size of the magnetic strength of the electromagnet.

       What other methods are there to increase the strength of the electromagnet.

• Increasing the number of turns in the coil.

• Inserting a soft iron core.

 2.    *State the function of the split-ring commutator in a d.c. motor.

• The purpose of the split-ring commutator is to reverse the direction of the current in the coil every half a revolution to ensure that the coil will always rotate in one direction.

Chapter 22 - Electromagnetic Induction

1.   

Describe an experiment to demonstrate electromagnetic induction. Explain the principles behind the phenomena. What variations in your experiment can you include to demonstrate the factors which affect the magnitude and the direction of the induced emf.

• When a magnet is pushed into the solenoid, there is a change in magnetic field lines linking the solenoid, which produces an induced e.m.f. This induced e.m.f. drives a current in the circuit, causing the pointer of the galvanometer deflects momentarily.

• When the magnet is stationary, there is no change in magnetic field lines linking the circuit. There is no induced e.m.f. and hence no current detected by the galvanometer.

• The experiment show that induced current (or induced emf) is produced in the coil due to the changing magnetic field in the solenoid. This process is called electromagnetic induction.

• The factors affecting the magnitude of the induced emf can be demonstrated as follows:

o   When the magnet moves at a faster speed in or out of the coil, the magnitude of the induced current is increased.

o   When a stronger magnet is used, the magnitude of the induced current is increased.

o   When the number of turns in the coil is increased, the magnitude of the induced current is increased.

2.   

With the aid of a diagram, describe the structure of an a.c. generator and explain how it works.

Structure

• An a.c. generator consists of a rectangular coil of wire connected to a pair of slip rings. Each slip ring is in contact with a carbon brush and the carbon brushes are connected to an external circuit.

• The rectangular coil is placed between the opposite poles of a magnet.

How it works

• When a coil is rotated, the magnetic field flux through the coil changes and e.m.f is induced in the coil.

• The induced e.m.f generated drives a current through the external circuit.

3.   

What are the functions of the slip rings and the carbon brush of an a.c. generator?

• The carbon brushes provide constant, sliding contact with the rotating slip rings.

• This ensures constant electrical contact between the coil and the external circuit. In this way, it allows the transfer of alternating e.m.f. induced in the rotating coil to the external circuit.

4.        

With the aid of a diagram, describe the structure of a step-down transformer and explain how the application of an alternating voltage produces a lower output voltage.

Structure

• The transformer consists of a primary coil and a secondary coil of wire wound round a laminated soft iron core.

• The primary coil is connected to an alternating current source and the secondary coil is connected to an external circuit.

How it works

• The a.c. supply produces a changing magnetic field in the primary coil.

• The changing magnetic field is linked to the secondary coil via the soft iron core.

• The changing magnetic field will induce an alternating e.m.f in the secondary coil and hence an alternating current will flow.

In a step-down transformer

• The number of turns in the secondary coil, Ns, is less than the number of turns in the primary coil, Np.

• The voltage across the secondary coil, Vs, is less than the voltage across the primary coil, Vp.

• Formula used in calculation:

5.   

What is an ideal transformer?

• In an ideal transformer , all power supplied to the primary coil is transferred to the secondary coil i.e. there is no power lost (100% efficient).

• Pp = Ps à Vp Ip = Vs Is

6.           

In the transmission of electricity from power stations to homes, why is it necessary to step up the voltage?

• The voltage is stepped up in the secondary coil so that the secondary current is decreased.

• In this way, the loss of power due to Joule Heating (P=I2R) in the cables can be reduced.

7.       

In a step-down transformer, suggest why the wire used for the secondary coil is thicker than that used for the primary coil.

• As the voltage is stepped down in the secondary coil, the secondary current is increased and that caused a significant heating effect in the coil.

• Since the loss of power due to Joule Heating is P=I2R , a thicker wire has a lower resistance and the power lost can be reduced.