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 Electronics Paper 2, May/June 2009
 Questions: 1 2 3 4 5, 6 &7 8 9 10 Main
Weakness/Remedies
Strength

Question 2
(a)Copy and complete Table 1 with respect to a.c. signals.
 CONVERT FROM TO MULTIPLE   FACTOR Peak r.m.s. Peak average Peak peak-to-peak 2.0
(b) A  sinusoidal alternating voltage has a peak value of 100v and a frequency of 1000Hz.

(i) Write an expression for the instantaneous value of the voltage.

(ii)Calculate the value at an instant of 0.2 millisecond after the start of the cycle.

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OBSERVATION

(a)

 CONVERT FROM TO MULTIPLER/   FACTOR Peak r.m.s. 0.707 Peak Average 0.637 Peak Peak-to-peak 2.000

(b)        In general, instantaneous values for sinusoidal voltage signals were given by the expression.

V(t) = V peak sin  (2πft)

In this case, frequency (f)  = 1000 and peak voltage Vpeak was given as 100v.

Therefore the desired expression was
v(t)  =  100 sin  (2000πt) volts

(c)        The instantaneous value at t = 0.2 milli-seconds was
v(0.2)  =  100sin (2000π x 0.2 x 10-3)
=   100sin (0.4π)
=   100 x 0.951
=   95.1v

This question was on circuit analysis (i.e. sinusoidal alternating current signals) characterized by the following features:
Vrms  =  Vmax     =   0.707Vmax
√2
Vaverage   =  2Vmax   =   0.637Vmax
π
Most of the candidates appeared not to be aware of these relationships.

In parts b and c, many candidates missed them by taking sin (0.4 π) as sin(1.25663°) instead of sin (1.25663radians).