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In a series LR circuit, the voltage drop across inductor is 8 volt and across resistor is 6 volt. Then voltage applied and power factor of circuit respectively are:A. 14V,0.8B. 10V,0.8C. 10 V,0.6D. 14 V,0.6

Answer» `V_(L)=8 V,V_(R)=6V,V=sqrt((V_(L)^(2)+V_(R)^(2))=10 V`
Power factor `=cos phi=(V_(R))/V=6/10 =0.6`


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