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Find the voltage across the 15Ω resistor in the given network using mesh analysis.
1 Answer
written 2.5 years ago by |
Solution:
Apply KVL to loop (1) 120⌊a−20(a+b)−j20(b)−j15(a)=0−(20+j15)a−(20+j20)b=−120⌊0......(1) Apply KVL to second loop −j10(a)−j15(b)−15(a)+j20(b)+j15(a)=0−(15−j5)a+(j5)b=0Δ=|−(20+j15)−(20+j20)−(15−j5)(j5)|=−(20+j15)(j5)−(20+j20)(15−j5)=−325−300j=442.29−137.29
Δa=|−120⌊0−(20+j20)0(j5)|=−(120)(j5)=−600j=600 ⌊ - 90 ∴a=ΔaΔ=600L−90442.29⌊−137.29=1.356447.29 Amp Hence voltage across 15Ω resistor is V15=15(a)=15(1.356⌊47.29)V15=20.3447.29 Volts