Question B.7: Finding Thevenin’s equivalent circuit Represent the network ...

The open-circuit voltage, which is Thevenin’s voltage between terminals a and b, can be found from the voltage divider rule in Eq. (B.10); that is,

V_{2} =R_2I_S=\frac{R_{2}}{R_{1}+R_{2}} V_{ S }                                       (B.10)

\begin{aligned}V_{ Th } &=\frac{R_{2}}{R_{1}+R_{2}} V_{ CC } \\&=\frac{7.5 k }{15 k +7.5 k } \times 12=4 V\end{aligned}                                      (B.14)

If source V_{ CC } is set to zero and test voltage V_{ X } is applied across terminals a and b, the circuit for determining R_{\text {Th }}  is shown in Fig. B.12. R_{\text {Th }} becomes the parallel combination of R_{1} \text { and } R_{2} . That is,

\begin{aligned}R_{ Th }=\frac{V_{ X }}{I_{ X }} &=R_{1} \| R_{2} \\&=15 k \| 7.5 k =5 k \Omega\end{aligned}                                          (B.15)