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A coil with inductive resistance X_(L)=30 Omega and impedance Z=50 Omega is connected to the mains with effective oltage value V=100V. Find the phase differencebetween the current and the voltage, as well as the heat power generated in the coil. |
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Answer» Solution :`cancel(Z)=SQRT(R^(2)+ X_(L)^(2))` or ` R_(0)=sqrt(cancel(Z^(2))-X_(L)^(2))` The `TAN theta=(X_(L))/( sqrt(cancel(Z^(2))-X_(L)^(2)))` So ` cos varphi=(sqrt(cancel(Z^(2))-X_(L)^(2)))/(cancel(Z))=sqrt(1-((X_(L))/(cancel(Z)))^(2))` `varphi=cos ^(-1) sqrt(1- ((X_(L))/( cancel(Z)))^(2))=37^(@)`. The current lags by `varphi` BEHIND the voltage. also `P=VI cos varphi=(V^(2))/( cancel(Z^(2)))sqrt(cancel(Z^(2))-X_(L)^(2))=.160kW.` |
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