Find the Thevenin equivalent circuit between A and B of the circuit in Figure 8-33.
Find the Thevenin equivalent circuit between A and B of the circuit in Figure 8-33.
First, remove R_{L}. Then V_{TH} equals the voltage across R_{2} + R_{3} as shown in Figure 8-34(a), because V_{4} = 0 V since there is no current through it.
V_{TH}= \left(\frac{R_{2} + R_{3}}{R_{1}+ R_{2} + R_{3}} \right) V_{S} = \left(\frac{690 \ \Omega }{1690 \ \Omega } \right) 10 \ V = 4.08 \ VTo find R_{TH}, first replace the source with a short to simulate a zero internal resistance. Then R_{1} appears in parallel with R_{2} + R_{3}, and R_{4} is in series with the series-parallel combination of R_{1}, R_{2}, and R_{3}, as indicated in Figure 8-34(b).
R_{TH}= R_{4} + \frac{R_{1}(R_{2}+ R_{3})}{R_{1}+ R_{2}+ R_{3}}= 1000 \ \Omega + \frac{(1000 \ \Omega )(690 \ \Omega )}{1690 \ \Omega }= 1410 \ \OmegaThe resulting Thevenin equivalent circuit is shown in Figure 8-34(c),