Question

Find the phasor form of:

1. 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.)

Answer 1

`I=24.598\angle 50.377`

Step-by-step explanation:

A tension or current expressed in cosine form and with a positive sign can be converted directly into a phasor. This is done by indicating the tension and the offset angle:

`Acos(10\omega t +\phi)=A\angle \phi`

So:

`i(t)=10cos(10t+63)+15cos(10t-42)=10\angle 63 + 15\angle42=I`

You can sum the phasors simply using a calculator, however, let's do it manually:

Let's find the rectangular form of each phasor using the next formulas:

`A=√(a^2+b^2) \\\phi=arctan((b)/(a))`

For
`10\angle 63`

`63=arctan((b)/(a) )\\\\tan(63)=(b)/(a) \\\\b=a*tan(63)`

`10=√(a^2+(a*tan(63))^2) \\\\10^2=a^2+a^2*(tan(63))^2\\\\Solving\hspace{3}for\hspace{3}a\\\\a=\sqrt{(100)/(1+tan(63)^2) } =4.539904997\\\\and\hspace{3}b\\b=√(100-a^2) =8.910065242`

So:

`Z_1=a+bj=4.539904997+8.910065242j`

For
`15\angle 42`

`42=arctan((b_2)/(a_2) )\\\\tan(42)=(b_2)/(a_2) \\\\b_2=a_2*tan(42)`

`15=√(a_2^2+(a_2*tan(42))^2) \\\\15^2=a_2^2+a_2^2*(tan(42))^2\\\\Solving\hspace{3}for\hspace{3}a_2\\\\a_2=\sqrt{(225)/(1+tan(42)^2) } =11.14717238\\\\and\hspace{3}b_2\\\\b_2=√(225-a^2) =10.0369591`

So:

`Z_2=a_2+b_2j=11.14717238+10.0369591j`

Hence:

`Z_T=Z_1+Z_2=(4.539904997+11.14717238)+(8.910065242+10.0369591)j\\Z_T=15.68707738+18.94702434j`

Finally:

`I=√(15.68707738^2+18.94702434^2) \angle arctan ((18.94702434)/(15.68707738) )=24.598\angle 50.377`