Jan,

I can give you a little help along your way, but since you've already

formulated an answer then my explanation can help you validate your

work.

Expressing sinusoidal signals (such as AC RMS) with the same frequency

(in this case 50Hz) as phasors is useful for consolidating by addition

the signals into a signal formula. The key to phasors is that this can

be done when the frequency is the same, but the amplitudes and phases

differ. For example,

N

Sum of A(k) cos(W(0) * t + Phi(k)) = A cos(W * t + Phi)

k=1

Note that on the left hand of the equation, I have written in ASCII

text as follows:

"W(0)" is short hand for writing W subscript 0 and represents the

radian measure of angle,

"Phi(k)" is short for Phi subscript k,

"t" is used as the classic symbol representing time.

This equation states that the sum of N cosine signals of differing

amplitudes and phase shifts, but with the same frequency, can always be

reduced to a single cosine signal of the same frequency. [Excerpted

from text "DSP First, A multimedia approach", by McClellan, et al. pp

31]. We know that this is only one half of the fun. Euler provided us

with complete fun by showing us that we could express cosine signals as

the real part of e^(W*t + Phi) * j. His brilliance helps us in one way

because it saves us from having to write annoying trigonometric

identities all over the place, and reduces the math to addition in the

exponents. Thus,

e^(Omega * j) = cos(Omega) + j * sin (Omega)

Real{ e^(Omega * j) } = cos(Omega)

Imaginary{ e^(Omega * j) } = sin(Omega)

Notes:

Omega is the English literal translation of the Greek symbol O,

j represents the imaginary symbol representing square root of -1.

Hence,

N

Sum of A(k) cos(W(0) * t + Phi(k))

k=1

= Sum of A(k) Re{ e^(W(0) * t + Phi(k)) * j }

= Sum of A(k) Re{ e^(W(0) * t) * j + e^(Phi(k)) * j }

= Re { e^(W(0) * t) * j } * Sum of A(k) * Re { e^(Phi(k)) * j }

So much for phasors. From your problem statement we have,

W(0) = 50 Hz

V = 240 Vrms

We know that, "Peak Voltage: Peak voltage tells you how far the voltage

swings, either positive or negative, from the point of reference... The

RMS voltage of a pure‡ sine wave is approximately .707*peak voltage."

[

http://www.bcae1.com/voltages.htm]

V = I*R [Ohms law]

Thus,

I = 240 Vrms/ 1 Ohm.

= 240rms cos(2*pi*50*t + Phi)

= 240/0.707 Peak Current * cos(2*pi*50*t + Phi)

a) Forty 50 watts incandescent lamps with unity power factor

Power = Watts = V * I

40 * 50 W = 240/0.707 Peak Voltage * cos(2*pi*50*t + 0) * I,

I = (40 * 50 W) / (240 / 0.707) Peak Current * cos(2*pi*50*t + 0)

= (40 * 50 W) / (240 / 0.707) * Re { e^(2*pi*50*t)*j }

b) Thirty-five 40 watts fluorescent lamps with a lagging power factor

of 0.9

35 * 40 W = 240/0.707 Peak Voltage * cos(2*pi*50*t + 2*pi*0.9) * I

I = (35 * 40 W) / (240 / 0.707) Peak Current * cos(2*pi*50*t +

2*pi*0.9)

= (35 * 40 W) / (240 / 0.707) Peak Current * Re { e^(2*pi*50*t +

2*pi*0.9)*j }

= (35 * 40 W) / (240 / 0.707) Peak Current * Re { e^(2*pi*50*t)*j *

e^(2*pi*0.9)*j }

c) One 2400 watts air conditioning system with a lagging power factor

of

0.65.

1 * 2400 W = 240/0.707 Peak Voltage * cos(2*pi*50*t + 2*pi*0.65) * I

I = (1 * 2400 W) / (240 / 0.707) Peak Current * cos(2*pi*50*t +

2*pi*0.65)

= ...

Etcetera, etcetera.... Now see if you can do the rest.

You're Welcome,

-Beagle