Convert the wye circuit in Figure 8-63 to a delta circuit.
Question 8.20: Convert the wye circuit in Figure 8-63 to a delta circuit.
Use Equations 8-4 : R_{A}= \frac{R_{1}R_{2}+ R_{1} R_{3}+ R_{2}R_{3}}{R_{2}}, 8-5 : R_{B}= \frac{R_{1}R_{2}+ R_{1}R_{3}+ R_{2} R_{3}}{R_{1}}, and 8-6 : R_{C}= \frac{R_{1}R_{2}+ R_{1}R_{3}+ R_{2}R_{3}}{R_{3}}.
R_{A}= \frac{R_{1}R_{2}+ R_{1}R_{3}+ R_{2}R_{3}}{R_{2}}\ \ \ \ \ \ = \frac{(1.0 \ k\Omega ) (2.2 \ k\Omega )+ (1.0 \ k\Omega )+ (5.6 \ k\Omega )+(2.2 \ k\Omega ) (5.6 \ k\Omega )}{2.2 \ k\Omega }= 9.15 \ k\Omega
R_{B}= \frac{R_{1}R_{2}+ R_{1}R_{3}+ R_{2} R_{3}}{R_{1}}
\ \ \ \ \ \ = \frac{(1.0 \ k\Omega ) (2.2 \ k\Omega ) + (1.0 \ k\Omega )+ (5.6 \ k\Omega )+(2.2 \ k\Omega ) (5.6 \ k\Omega )}{1.0 \ k\Omega }= 20.1 \ k\Omega
R_{C}= \frac{R_{1}R_{2}+ R_{1}R_{3}+ R_{2}R_{3}}{R_{3}}
\ \ \ \ \ \ = \frac{(1.0 \ k\Omega ) (2.2 \ k\Omega ) + (1.0 \ k\Omega )+ (5.6 \ k\Omega )+(2.2 \ k\Omega ) (5.6 \ k\Omega )}{5.6 \ k\Omega }= 3.59 \ k\Omega
The resulting delta circuit is shown in Figure 8-64.
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