Question 8.20: Convert the wye circuit in Figure 8-63 to a delta circuit.

Convert the wye circuit in Figure 8-63 to a delta circuit.

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Use Equations 8-4 : RA=R1R2+R1R3+R2R3R2R_{A}= \frac{R_{1}R_{2}+ R_{1} R_{3}+ R_{2}R_{3}}{R_{2}}, 8-5 : RB=R1R2+R1R3+R2R3R1R_{B}= \frac{R_{1}R_{2}+ R_{1}R_{3}+ R_{2} R_{3}}{R_{1}}, and 8-6 : RC=R1R2+R1R3+R2R3R3R_{C}= \frac{R_{1}R_{2}+ R_{1}R_{3}+ R_{2}R_{3}}{R_{3}}.

RA=R1R2+R1R3+R2R3R2R_{A}= \frac{R_{1}R_{2}+ R_{1}R_{3}+ R_{2}R_{3}}{R_{2}}

 

      =(1.0 kΩ)(2.2 kΩ)+(1.0 kΩ)+(5.6 kΩ)+(2.2 kΩ)(5.6 kΩ)2.2 kΩ=9.15 kΩ\ \ \ \ \ \ = \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

 

RB=R1R2+R1R3+R2R3R1R_{B}= \frac{R_{1}R_{2}+ R_{1}R_{3}+ R_{2} R_{3}}{R_{1}}

 

      =(1.0 kΩ)(2.2 kΩ)+(1.0 kΩ)+(5.6 kΩ)+(2.2 kΩ)(5.6 kΩ)1.0 kΩ=20.1 kΩ\ \ \ \ \ \ = \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

 

RC=R1R2+R1R3+R2R3R3R_{C}= \frac{R_{1}R_{2}+ R_{1}R_{3}+ R_{2}R_{3}}{R_{3}}

 

      =(1.0 kΩ)(2.2 kΩ)+(1.0 kΩ)+(5.6 kΩ)+(2.2 kΩ)(5.6 kΩ)5.6 kΩ=3.59 kΩ\ \ \ \ \ \ = \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.

Screenshot (552)

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