Question

Find the phasor form of i(t) = 10 cos (10t + 63°) + 15 cos (10t − 42°) A . (Round the final answers to two decimal places and insert ± in the answer field(s), if required.) The phasor form of i(t) = 10 cos (10t + 63°) + 15 cos (10t − 42°) A is ∠ ° A.

Answer 1

Answer:

$11.12\angle -4.11^o$

Step-by-step explanation:

Phasor Form of Sinusoids

A phasor is a complex number in polar form that can be applied to circuit analysis.

When two or more sinusoidal functions are added or subtracted, we can simplify the calculations by using this notation

$R\angle \theta$

Where R is the rms value of the sinusoidal function:

$R=Amplitude/\sqrt{2}$

The total current is split into

$i_1(t) = 10 cos (10t + 63^o)$

$i_2(t) = 15 cos (10t-42^o)$

The phasor of the first current is

$\displaystyle P_1=\frac{10}{\sqrt{2}}\ \angle 63^o=7.071\angle 63^o$

With components

$P_{1x}=7.071cos63^o=3.21\\P_{1y}=7.071sin63^o=6.30$

The phasor of the second current is

$\displaystyle P_2=\frac{15}{\sqrt{2}}\ \angle -42^o=10.61\angle -42^o$

With components

$P_{2x}=10.61cos(-42^o)=7.88\\P_{1y}=10.61sin(-42^o)=-7.10$

The sum of both phasors is

$P_{tx}=11.09\\P_{ty}=-0.80$

The magnitude is

$P=\sqrt{11.09^2+(-0.80)^2}=11.12$

The angle is

$\displaystyle tan\theta=\frac{-0.80}{11.09}$

$\theta=-4.11^o$

Corresponding to a resulting phasor

$11.12\angle -4.11^o$