Conversion of sin and cosine Terms
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Q1 Alternating currents i1 and i2 flowing into a circuit node are given by
i1 = 0.02 sin wt
i2 = 0.032 cos ( wt - π/3)
Determine an expression for the output current i = i1 + i2 the form R sin (wt + α) and thus state its amplitude and phase angle
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Solution Summary
This analysis shows hows to use trigonometrical expressions and reduction to form a single magnitude phase current signal for the output current of the form i = i1 + i2 the form R sin (wt + α) = 0.0503sin(wt + 0.324) and thus, states its amplitude and phase angle. Step by step calculations are given.
Solution Preview
Use the general expression for cos(A - B) = cos(A).cos(B) + sin(A).sin(B) to convert 0.032 cos (wt - n/3) to
0.032 cos (wt - n/3) = 0.032*{cos(wt).cos(n/3) + sin(wt).sin(n/3)} = 0.032*{0.5cos(wt) + 0.866sin(wt)}, since cos(n/3) = 0.5 & sin(n/3) = 0.866
Multiplying out fully by the 0.032 factor we obtain
i2 = 0.032 cos (wt - n/3) = 0.016cos(wt) + 0.0277sin(wt)
Now we make the sume
i = i1 + i2 = 0.02sin(wt) + 0.016cos(wt) + 0.0277sin(wt) = ...
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