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I have been doing a few mock multi choice questions in preparation for a College exam.
Doing okay but have come across this one which has stumped me.

Here goes,
Two ac voltages V1 and V2 have values of 20v and 30v respectively.

If V1 leads V2 by 45 degrees the resultant voltage will be?

Anyway, The answer apparently is 46v at 18 degree angle.

How would you arrive at this answer,because I am clueless!

Any help in simple terms would be much appreciated,

Thanks in advance

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Hey Gerry.

It's all to do with phasor diagrams and vectors and their magnitudes.

Off the top of my head I cant explain it, and I cant post drawings on to the forum (pc issues).

Someone else (fresh out of college) may be able to help.

Last edited by a moderator:


Hi, best way to explain it is with a phasor diagram.

I've drawn it out for you - see attachment.

Hope this helps.




  • Thread Starter Thread Starter
  • #4
Lenny and especially Wayne, superb answer,thanks for taking the time.
It has really helped.
Any football questions that need answering I will try and reciperocate.

Cheers Gerry



Theres no formula as such, it's all about where the lines intersect.

With the given angle being 45 degrees draw your first line at your chosen scale, then draw the second at the 45 degree angle from that. Then the further two (dotted) lines to complete the shape(makes it easier to understand) although all you need is the triangle (vector).

The resultant voltage will be the longest line in the triangle drawn between the two ends of the other two lines. measure this and compare to your scale and measure the angle between this longest line and the other original line.

Hope this helps.



Have a look at this.

The diagram below shows two phasors A and B. These are then combined to give a new phasor C.
In this case the magnitude of C will be given by (by Pythagoras' theorem):
and the phase difference between the reference angle (in this case phasor A) and C is given by:

This is a standard formula for the two parts of a phasor - the resultant, and the phase angle.

P.S this will only work if the phases are at 90 degrees to one another. other than that it's the cosine rule.
Last edited by a moderator:


Hi Tim77,

You might regret asking for this lol:D

As the voltages aren't at right angles you can't use pythagoras in this example - you have to use

the cosine rule

It goes like this:

For the voltage....

X² = ( a² + b² ) – 2abCosθ

Where ...

X = required voltage

a = V2 (30v)

b = V1 (20v)

θ = 180° - 45°(phase angle)

Which gives:

X² = (900 + 400) – (2 x 30 x 20 x Cos135°)

X² = 1300 - ( -848.52 )

X² = 2148.52

Therefore ..

Voltage = √2148.52

= 46.35 volts

For the angle...

CosB = c² + a² - b²
. . . . . . . . .2ca


B = required angle

c = new voltage 46v

a = V2 (30v)

b = V1 (20v)

Which gives:

CosB = (2116 + 900) – 400
. . . . . . . . .2 x 46 x 30

CosB = 2616
. . . . . . .2760

CosB = 0.9478

Therefore… Angle = 18.59°

So there you go - my brain's sore now:)

I told you it was easier to explain with a phasor diagram lol

See you tomorrow,

Last edited by a moderator:



Not sure where the question originates from - maybe Gerry.slater65

can let us know if he checks back in.


Hey Tim.

In old money it was the 2360 part 2, electrical science and principals.

I've just dug out all my old college stuff, it's amazing what they teach and how much of it you actually use in everyday works.

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