A motor attached to a 120 V/60 Hz power line draws an 7.90 A current. Its average energy dissipation is 810 W. How much series capacitance needs to be added to increase the power factor to 1.0?i found all of the below values but couldne't find the last onerms resistor voltage=103power factor= 0.854motor's resistance=13

Question

A motor attached to a 120 V/60 Hz power line draws an 7.90 A current. Its average energy dissipation is 810 W. How much series capacitance needs to be added to increase the power factor to 1.0?i found all of the below values but couldne&#039;t find the last onerms resistor voltage=103power factor= 0.854motor&#039;s resistance=13

Lilia Nolan · Accepted Answer

If you add series capacitance, the voltage drop across the capacitor may decrease/increase the voltage across the motor to a level that the motor may not work. Power factor correction is typically done with the capacitor in parallel with the motor.Apparent Power  S  =      V    S    I  =  120  V  ⋅  7.90  A  =  948  V  ⋅  APower Factor  p  f  =  P      /    S  =  810  W      /    948  V  ⋅  A  =  0.854   laggingReactive Power      Q    L    =      (          S      2        −          P      2        )    =      (    (    948    V    ⋅    A          )      2        −    (    810    W          )      2        )    =  492.54  V  A  RNOWInductive Reactance      Q    L    =      I    2        X    L        X    L    =      Q    L        /        I    2    =  492.54  V  A  R      /    (  7.90  A      )    2    =  7.89  ΩTo bring the pf = 1, leading capacitive reactive power must equal lagging inductive reactive power:      X    C    =      X    L    =  7.89  ΩCapacitance+  C  =  1      /    (  2  π  f      X    C    )  =  1      /    (  2  ⋅  π  ⋅  60  H  z  ⋅  7.89  Ω  )  =  336  μ  F

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