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I have the question " The instantaneous values of two alternating currents are given by i1 = 20sinwt amperes and i2 = 10sin(wt + pi/3) amperes. By plotting i1 and i2 on the same axes, using the same scale, over one cycle, and adding ordinates at intervals, obtain a sinusoidal expression for i1 + i2."

The graph for the question is:

enter image description here

The answer for the sinusoidal expression is:

enter image description here

I understand everything except for the 19 degrees. The solutions say that the resultant waveform leads the curve i1 = 20sinwt by 19 degrees which is 0.332 rads, however I do not understand where the 19 degrees came from.

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    \$\begingroup\$ One curve is a bit "in front" of the other, isn't it? the 19° is simply the amount of a full 360° circle that the first one is in front of the other. We call that property of a wave "phase". \$\endgroup\$ Feb 19, 2017 at 21:43
  • \$\begingroup\$ Could you please explain further because I understand that one curve is abit in front of the other but I still don't understand why it's by 19 degrees. Thanks. \$\endgroup\$
    – Dan Khan
    Feb 19, 2017 at 21:51
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    \$\begingroup\$ that's simply the advance it has, as read from the graph. There's nothing more to explain – one full period is always 360°, and the dashed line is simply \$\frac{19}{360}\$ of that behind the solid one. \$\endgroup\$ Feb 19, 2017 at 21:54
  • \$\begingroup\$ So because your using the graph to find this angle difference. Does this mean that it may vary as it is an estimate ? Thanks. \$\endgroup\$
    – Dan Khan
    Feb 19, 2017 at 23:12

3 Answers 3

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Convert them into phasor forms and THEN add them and then you will understand that that is where the 19 degrees comes from, you can determine it mathematically there is no need to estimate it.

Google "convert sinusoid to phasor form" and that should help. If you have questions please ask and I can try to help.

phasor example

Look at the green circled part of the example but do it backwards. Imagine you are starting with the $$1.12cos(2\pi ft-2.68)$$ term and want to get to the top phasor term $$1.12\angle -153^o$$ (ignore the middle step, it is not relevant here). Do you see how the conversion is done, assuming the 2.68 in the cosine term is in radians? You want to do the same conversion for your two sin terms and then add them together and you will get your result!

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  • \$\begingroup\$ Thank you but convert what into phasor form ? :) \$\endgroup\$
    – Dan Khan
    Feb 19, 2017 at 23:48
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    \$\begingroup\$ the i1 and i2 terms that are in time domain sinusoid form \$\endgroup\$
    – scuba
    Feb 20, 2017 at 0:04
  • \$\begingroup\$ Could you please elaborate I still don't understand. Maybe with an example ? Sorry it's just that it's really confusing me :) \$\endgroup\$
    – Dan Khan
    Feb 20, 2017 at 0:21
  • \$\begingroup\$ updated my answer with an example from here: ee.ic.ac.uk/hp/staff/dmb/courses/ccts1/01000_Phasors.pdf \$\endgroup\$
    – scuba
    Feb 20, 2017 at 0:29
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    \$\begingroup\$ for 20sin(wt) there is no phase offset so the angle is 0. For 10sin(wt+pi/3) the angle offset is +pi/3 or +60 degrees which would directly be the term to the right of the angle symbol in phasor notation, and obviously the term to the left of the angle symbol is just the magnitude. \$\endgroup\$
    – scuba
    Feb 20, 2017 at 0:51
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If you look carefully at the sum of the two graphs, the peak and the zero crossing is not at the same place as the other two.

The shift of the peak and the zero crossings is called the phase angle. The phase angle of the first graph is zero and the phase angel of the second graph: $${\pi \over 3} = 60^\circ .$$

The phase angle of the sum is 0.332 radians, to convert to degrees use: $$0.332 \cdot {180^\circ \over \pi } = 19^\circ$$

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  • \$\begingroup\$ No but where does the 19 degrees or 0.332 come from ? How do you measure the 19 degrees? \$\endgroup\$
    – Dan Khan
    Feb 20, 2017 at 11:38
  • \$\begingroup\$ @Dan Khan you did say: "I understand everything except for the 19 degrees." Do you want to edit your question? \$\endgroup\$
    – skvery
    Feb 20, 2017 at 16:31
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It is how two sine waves are added. Check this document http://dspguru.com/sites/dspguru/files/Sum_of_Two_Sinusoids.pdf

You can google for demonstration of the formulas. As you will see you will get those 19 degrees.

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    \$\begingroup\$ Could you please expand this answer? When the link goes down the answer won't be worth anything. \$\endgroup\$
    – Voltage Spike
    Jan 28, 2021 at 20:08

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