• 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:

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A sinusoidal alternating voltage is to be rectified. (a) Suggest one advantage of full-wave rectification as compared with half-wave rectification.

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:

(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 I²(R₁ + R₂)

(ii) (P_R =) I²R₁

(c)

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 = I²R_A / I²R_B = R_A / R_B = {4ρl / [πd²]} / {ρl / 2[πd²]} = 8 or 8:1

(d)

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 / R_A} / {V² / 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.