Answer:
[tex]I=24.598\angle 50.377[/tex]
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:
[tex]Acos(10\omega t +\phi)=A\angle \phi[/tex]
So:
[tex]i(t)=10cos(10t+63)+15cos(10t-42)=10\angle 63 + 15\angle42=I[/tex]
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:
[tex]A=\sqrt{a^2+b^2} \\\phi=arctan(\frac{b}{a})[/tex]
For [tex]10\angle 63[/tex]
[tex]63=arctan(\frac{b}{a} )\\\\tan(63)=\frac{b}{a} \\\\b=a*tan(63)[/tex]
[tex]10=\sqrt{a^2+(a*tan(63))^2} \\\\10^2=a^2+a^2*(tan(63))^2\\\\Solving\hspace{3}for\hspace{3}a\\\\a=\sqrt{\frac{100}{1+tan(63)^2} } =4.539904997\\\\and\hspace{3}b\\b=\sqrt{100-a^2} =8.910065242[/tex]
So:
[tex]Z_1=a+bj=4.539904997+8.910065242j[/tex]
For [tex]15\angle 42[/tex]
[tex]42=arctan(\frac{b_2}{a_2} )\\\\tan(42)=\frac{b_2}{a_2} \\\\b_2=a_2*tan(42)[/tex]
[tex]15=\sqrt{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{\frac{225}{1+tan(42)^2} } =11.14717238\\\\and\hspace{3}b_2\\\\b_2=\sqrt{225-a^2} =10.0369591[/tex]
So:
[tex]Z_2=a_2+b_2j=11.14717238+10.0369591j[/tex]
Hence:
[tex]Z_T=Z_1+Z_2=(4.539904997+11.14717238)+(8.910065242+10.0369591)j\\Z_T=15.68707738+18.94702434j[/tex]
Finally:
[tex]I=\sqrt{15.68707738^2+18.94702434^2} \angle arctan (\frac{18.94702434}{15.68707738} )=24.598\angle 50.377[/tex]
