What am i doing wrong? $(100$ts$)$ Refer to Fig. P$4.$ Calculate the equivalent resistance seen from the DC power supply $($

ig$.$ P$4$

$R_{\text{R}\text{e}\text{o}}=\frac{1}{\frac{1}{1}+\frac{1}{1}+\frac{1}{1}}=\frac{1}{3}$ Req $=0\mathrm{.}33$

$R_S=2\mathrm{.}2+3\mathrm{.}3=5\mathrm{.}5$

$R_{\text{f}\text{o}\text{t}}=RSTRR\frac{4}{\frac{?}{?}}R\frac{3}{R}S=5.S+0\mathrm{.}33=5\mathrm{.}83$

$2$

Given $v_x(t)=5sin(t-\frac{}{6})$;$v_y(t)=3sin(t-\frac{}{4})dots$ answer the following

a$)$ the ac$-$Phasors, $\frac{?}{\text{b}\text{a}\text{r}}(V{)}_x\frac{,}{b}ar(V{)}_y$

$x-\frac{}{6}$

$($:$5cos(2100t-60)+3cos12cost-45\}$

$\sqrt[\phantom{2}]{5^2+3^2}=5\mathrm{.}83$

b

$sin=\frac{3}{5\mathrm{.}83}=si{n}^{-1}(\frac{3}{5\mathrm{.}83})=30$

b$)\frac{?}{\text{b}\text{a}\text{r}}(V{)}_z\frac{?}{b}=ar(V{)}_x\frac{+}{b}ar(V{)}_y$

$x$

$-4?$

Given $v_x(t)=5sin(t-\frac{}{6})$;$v_y(t)=3sin(t-\frac{}{4})dots$ answer the following

a$)$ the ac$-$Phasors, $\frac{?}{\text{b}\text{a}\text{r}}(V{)}_x\frac{,}{b}ar(V{)}_y$

$x-\frac{}{6}$

$($:$5cos(2100t-60)+3cos12cost-45\}$

$\sqrt[\phantom{2}]{5^2+3^2}=5\mathrm{.}83$

b

$sin=\frac{3}{5\mathrm{.}83}=si{n}^{-1}(\frac{3}{5\mathrm{.}83})=30$

b$)\frac{?}{\text{b}\text{a}\text{r}}(V{)}_z\frac{?}{b}=ar(V{)}_x\frac{+}{b}ar(V{)}_y$

$x$

$-4$