Question 1.5: Find the load impedance Z2 in the bridge circuit, shown in F...
Find the load impedance Find the load impedance Z_{2} in the bridge circuit, shown in Fig. 3.39a, for maximum power transfer from the source to the load.
The source voltage is 0.8660 + j0.5. We have to find the Thévenin’s equivalent voltage and impedance across the load impedance terminals. Replacing the bottom half of the Δ circuit by Y – Δ circuit, we get, with
Z_{a}=Z_{4}=0.9901 − j0.0990, Z_{ac}=Z_{3}=j1, Z_{ab}=Z_{5}=1.Z_{a}=\frac{j1 × 1}{(0.9901 − j0.0990) + j1 +1 } = 0.1888 + j0.4170
Z_{b}=\frac{(0.9901 − j0.0990) × 1}{(0.9901 − j0.0990) + j1 + 1} = 0.3942 − j0.2282
Z_{c}=\frac{(0.9901 − j0.0990) × j1}{(0.9901 − j0.0990) + j1 + 1} = 0.2282 + j0.3942.
The transformed circuit, with the voltage source short-circuited, is shown in Fig. 3.39b. Now,
Z_{eq}=(0.3942−j0.2282)+\frac{(−j10 + (0.1888 + j0.4170))(0.2282 + j0.3942)}{((−j10 + (0.1888 + j0.4170)) + (0.2282 + j0.3942))} = 0.6425+j0.1763The maximum average power transfer occurs with Z_{L}=Z^{*}_{eq} = 0.6425 − j0.1763. The V_{oc} across the load terminals of the transformed circuit, with the voltage source inserted, is
V_{oc}=\frac{((−j10 + (0.1888 + j0.4170)))(0.866 + j0.5)}{−j10 + (0.1888 + j0.4170) + (0.2282 + j0.3942)} = 0.9155+j0.4977 = 1.0420\angle (0.4979
Therefore, the maximum power transferred is
P_{m}=\frac{\left|V_{oc} \right| ^{2} }{R_{L} } =\frac{(1.0420^{2} )}{(8 × 0.6425)} = 0.2113 W.
Figure 3.40 show the power transferred for a range of values of the load resistor. The peak value occurs when the load resistance is equal to the Thévenin equivalent resistance.
