Problem $2(26$pts$)$:

$\frac{36}{16}$

$\frac{166}{196}$

$\frac{166}{256}$

$\frac{16*16}{16+16}=\frac{256}{32}$

$=4+160$

$4+16=20$

$32*\frac{16}{20}=32*\frac{4}{5}=\frac{128}{5}$

$R_{th}=2002$

$v+h=\frac{3}{20}$

$v+h=25\mathrm{.}6=\frac{128}{5}$

b$)$ Determine the value of the Maximum power transfered to $R_b$

$\frac{(v+h{)}^2}{4+th}=\frac{(25\mathrm{.}6{)}^2}{4(20)}=\frac{(25,6{)}^2}{80}$

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