# 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.

**(c)**

(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

**(b)**

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

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.}

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

**(c)**

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

(ii)

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 R

_{1}and R_{2}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:__**(a)**

(i) Chemical to electrical

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

**(b)**

(i)

{Rate of energy
transformation = power}

(P

_{B}=) EI or I^{2}(R_{1}+ R_{2})
(ii) (P

_{R}=) I^{2}R_{1}**(c)**

Equation for resistance R = ρ

*l*/A or clear from the following equation
Resistance of A = ρ

*l*/ [πd^{2}/4] = 4ρ*l*/ [πd^{2}]
Resistance of B = (ρ/2)

*l*/ [πd^{2}] = ρ*l*/ 2[πd^{2}]
Ratio = I

^{2}R_{A}/ I^{2}R_{B}= R_{A}/ R_{B}= {4ρ*l*/ [πd^{2}]} / {ρ*l*/ 2[πd^{2}]} = 8 or 8:1**(d)**

Power P dissipated = V

^{2}/ R or E^{2}/ R
There is the same p.d. is across
resistors A and B when they are in parallel. Hence,

Ratio = {V

^{2}/ R_{A}} / {V^{2 }/ R_{B}} = R_{B}/ R_{A}= 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.

Question 675 part b(i) and b(ii) are wrong

ReplyDeleteThe 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.

Yeah, it was wrong initially. Sorry.

DeleteAnd 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.

No worries

DeleteThe 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! :)