Find the current I drawn by the bridge circuit, shown in Fig. 2.31a, from the source.

Question

Accepted Answer

The circuit can be considered as the combination of twoΔ circuits with a shared resistor. To make the analysis simpler, we have to convert one of them into an equivalent circuit. Replacing the top half of theΔ circuit, we get, with

[latex]Z_{bc} =3 ,Z_{bc} = 1, Z_{bc} =1[/latex]

[latex]Z_{a} = \frac{1 imes 1}{3+1+1}=0.2[/latex]

[latex]Z_{b} = \frac{3 imes 1}{3+1+1}=0.6[/latex]

[latex]Z_{c} = \frac{3 imes 1}{3+1+1}=0.6.[/latex]

The transformed circuit is shown in Fig. 2.31b. Now,

[latex]Z_{eq}= 0.6 +\frac{(0.6 + 1)(3 + 0.2)}{(0.6 + 1 + 3 + 0.2)} =\frac{5}{3}.[/latex]

Therefore,

[latex]I= \frac{V}{Z_{eq}} =(1)\frac{3}{5} = 0.6 A,[/latex]

which is the same as found by nodal and loop analyses.
Let us get back the Δ circuit from the Y circuit.

[latex]Z_{bc}= \frac{Z_{a} Z_{b} +Z_{b} Z_{c}+Z_{c} Z_{a} }{Z_{a}} = \frac{(0.2)(0.6) + (0.2)(0.6) + (0.6)(0.6)}{0.2} =3[/latex]

[latex]Z_{ac}= \frac{Z_{a} Z_{b} +Z_{b} Z_{c}+Z_{c} Z_{a} }{Z_{b}} = \frac{(0.2)(0.6) + (0.2)(0.6) + (0.6)(0.6)}{0.6} =1[/latex]

[latex]Z_{ab}= \frac{Z_{a} Z_{b} +Z_{b} Z_{c}+Z_{c} Z_{a} }{Z_{c}} = \frac{(0.2)(0.6) + (0.2)(0.6) + (0.6)(0.6)}{0.6} =1.[/latex]

Question 1.3: Find the current I drawn by the bridge circuit, shown in Fig...

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