Given i$1($t$)=204$cos$($t$)$i$1($t$)=204$cos$($t$)$ A and i$2($t$)=134$cos$($t$+30)$i$2($t$)=134$cos$($t$+30)$ A$,$ use superposition to compute the current II$.$

Use $2$ digits after the decimal point if necessary in your answers below.

a$.$ What are the phasors for i$1($t$)$i$1($t$)$and i$2($t$)$i$2($t$)$in polar form $($r$$$)($r$$$)$ with rr in Amps and in degrees?

I$1=($r$1)($$1)=($I$1=($r$1)($$1)=($Answer $1$ Question $8$A$)($A$)($Answer $2$ Question $8))$

$($note: r$1$r$1$ is positive and$$$1$$1$ can be positive or negative$)$

I$2=($r$2)($$2)=($I$2=($r$2)($$2)=($Answer $3$ Question $8$A$)($A$)($Answer $4$ Question $8))$

$($note: r$2$r$2$ is positive and$$$2$$2$ can be positive or negative$)$

b$.$ What values $($in Amps$)$ must be summed to produce the final answer?

Let$$I$$I$$ be the contribution of current source I$1$I$1$:

I$=($x$1)+$j$($y$1)=($I$=($x$1)+$j$($y$1)=($Answer $5$ Question $8)+$j$()+$j$($Answer $6$ Question $8)$A$)$A

$($note: both x$1$x$1$ and$$y$1$y$1$ can be positive or negative$)$

Let$$I$$I$$ be the contribution of current source$$I$2$I$2$:

I$=($x$2)+$j$($y$2)=($I$=($x$2)+$j$($y$2)=($Answer $7$ Question $8)+$j$()+$j$($Answer $8$ Question $8)$A$)$A

$($note: both x$2$x$2$ and$$y$2$y$2$ can be positive or negative$)$

c$.$ What is the final answer $($in Amps$)?$

I$=$I$+$I$=($x$1+$x$2)+$j$($y$1+$y$2)=($I$=$I$+$I$=($x$1+$x$2)+$j$($y$1+$y$2)=($Answer $9$ Question $8)+$j$()+$j$($Answer $10$ Question $8)$A$)$A