$i_B=\frac{V_{Th}-V_0}{R_{Th}+(1+)R_E}$

ubstitute Eqs. $4\mathrm{.}69$ and $4\mathrm{.}70$ into Eq$.4\mathrm{.}72,$ we get the same obtained in Eq$.2\mathrm{.}25.$ Note that when we have incorporated the quivalent into the original circuit, we can obtain the solution ting a single equation.

$4\mathrm{.}20$ Find the Th$$venin equivalent circuit with respect to the terminals $a,b$ for the circuit shown. $($Hint: Define the voltage at the leftmost node as $v,$ and write two nodal equations with $V_{Th}$ as the right node voltage.$)$

Answer: $,V_{Th}=v_{ah}=30V,R_{Th}=10.$