Saturday, May 9, 2015

Physics 9702 Doubts | Help Page 136

  • Physics 9702 Doubts | Help Page 136

Question 675: [Alternating Current > Rectification]
A sinusoidal alternating voltage is to be rectified.
(a) Suggest one advantage of full-wave rectification as compared with half-wave rectification.

(b) The rectification is produced using the circuit of Fig.1.

All the diodes may be considered to be ideal.
Variation with time t of the alternating voltage applied to the circuit is shown in Fig.2 and in Fig.3.

(i) On axes of Fig.2, draw a graph to show variation with time t of the potential difference across diode A.
(ii) On axes of Fig.3, draw a graph to show variation with time t of the potential difference across diode B.

(i) On Fig.1, draw symbol for a capacitor, connected into the circuit so as to provide smoothing.
(ii) Fig.4 shows variation with time t of the smoothed potential difference across the resistor R in Fig.1.

1. State how amount of smoothing may be increased.
2. On Fig.4, draw variation with time t of the potential difference across resistor R for increased smoothing.

Reference: Past Exam Paper – November 2009 Paper 41 Q7

Solution 675:
(a) Example:
More (output) power is available
There is less ripple for same smoothing capacitor

(i) The curve should be drawn showing half-wave rectification

{The function of a diode is to control the direction of current flow. The current flowing through a diode can only go in one direction, called the forward direction.
If the voltage across the ideal diode is negative, the current is blocked and cannot flow – the diode is said to be reversed-biased. The ideal diode looks like an open circuit.
When conducting a current, an ideal diode would act as a short circuit (the voltage across it is zero). A current-conducting diode is said to be forward-biased.

This is easy to understand: if the diode is ideal, it should not be affecting the other components of the circuit when conducting. Thus, the p.d. across it should be zero. All the voltage is used by the other components.}  

(ii) The curve is similar to (i) but with a phase shift of 180°

(i) A correct symbol, connected in parallel with R    

1. By using a larger capacitor / second capacitor in parallel with R

2. The curve has the same peak values, with a correct shape giving less ripple

Question 676: [Current of Electricity]
Two resistors A and B have resistances R1 and R2 respectively. Resistors are connected in series with a battery, as shown in Fig.1.
Battery has electromotive force (e.m.f.) E and zero internal resistance.
(a) State energy transformation that occurs in
(i) battery,
(ii) resistors.

(b) Current in the circuit is Ι.
State the rate of energy transformation in
(i) battery,

(ii) resistor A.

(c) Resistors are made from metal wires. Data for the resistors are given in Fig.2.

resistor                                     A                     B
resistivity of metal                  ρ                      ρ / 2
length of wire                          l                       l
diameter of wire                      d                      2d

Use information from Fig.2 to determine the ratio
power dissipated in A / power dissipated in B

(d) Resistors A and B are connected in parallel across the same battery of e.m.f. E.
Determine the ratio
power dissipated in A / power dissipated in B

Reference: Past Exam Paper – June 2013 Paper 21 Q6

Solution 676:
(i) Chemical to electrical

(ii) Electrical to thermal / heat or heat and light

{Rate of energy transformation = power}
(PB =) EI or I2(R1 + R2)

(ii) (PR =) I2R1

Equation for resistance R = ρl/A        or clear from the following equation
Resistance of A = ρl / [πd2/4] = 4ρl / [πd2]
Resistance of B = (ρ/2)l / [πd2] = ρl / 2[πd2]

Ratio = I2RA / I2RB = RA / RB = {l / [πd2]} / {ρl / 2[πd2]} = 8 or 8:1

Power P dissipated = V2 / R or E2 / R
There is the same p.d. is across resistors A and B when they are in parallel. Hence,
Ratio = {V2 / RA} / {V2 / RB} = RB / RA = 1/8 or 1:8 = 0.125

Question 677: [Electric field]
Which path shows a possible movement of an electron in electric field shown?

Reference: Past Exam Paper – June 2002 Paper 1 Q36

Solution 677:
Answer: A.
The direction of the electric field is from positive to negative.
An electron, which is negatively charged, will be attracted towards the positive plate. Since the electric field is downwards, the positive plate must be the upper plate since electric field is from positive towards negative. [C and D are incorrect]

The strength of the electric field is indicated by how close together the field lines are. In this case, comparing the separation of the plates with the amount of electric field lines along the length of the plates, it is reasonable to assume that the electric field is quite strong (a weak electric field would be indicated by, say, only 3 field lines along the length of the plates).

Electric force = Eq
Since the electric field strength is strong, the force on the electron is greater. So, the electron is more likely to hit the plate.


  1. Question 675 part b(i) and b(ii) are wrong
    The graph for b(i) should be flat first and then a risen curve and the flat again
    and for b(ii) it should be like what is shown in b(i) right now.
    Both the diodes are forward biased so the potential difference will not be negative
    If you check the examiner report for this question, they also say that many drew an inverse graph in b(ii) however this is incorrect.

    1. Yeah, it was wrong initially. Sorry.

      And thanks, comments and suggestions are most welcomed.
      The correction has been made. I have added a few lines above. If you think of any additional information that would be important, just let me know and I'll add it up there.

    2. No worries
      The answer is perfect now with a clear explanation.

      Your blog indeed is very helpful in not only obtaining the answers but also to understand the concepts behind it.
      Keep up the great work! :)

    3. in question 675 b(i) can you please explain why is there a flat line first?

    4. the diode conducts in only one direction. when in the opposite direction, it would not conduct. no current would flow. thus a horizontal line


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