This circuit (without C) is similar to the circuit We now consider the R-L-C fragment in...

 

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This circuit (without C) is similar to the circuit We now consider the R-L-C fragment in the dashed-line box to be the load (which is inductive), and cannot be changed. For the most economical operation, it would be helpful if the current (relabeled iTnew) and voltage were in phase (phase angle for both voltage and current to be the same). The equivalent phrase is to say that we desire the power factor¹ to be 1 or the total load to be purely restive. Capacitor C is to be added to the circuit as shown to bring the power factor to unity. Your job is to calculate the capacitance of Cc that eliminates the imaginary term from the original i. Assume that the original i is 0.184-68º. IOW, ITnew must be the real part of i=0.184-68°. v₂ = 10 20° (0 = 1000 rad/s iTnew ~ Cc LOAD mell 2092 0.1 H 20 μF HH This circuit (without C) is similar to the circuit We now consider the R-L-C fragment in the dashed-line box to be the load (which is inductive), and cannot be changed. For the most economical operation, it would be helpful if the current (relabeled iTnew) and voltage were in phase (phase angle for both voltage and current to be the same). The equivalent phrase is to say that we desire the power factor¹ to be 1 or the total load to be purely restive. Capacitor C is to be added to the circuit as shown to bring the power factor to unity. Your job is to calculate the capacitance of Cc that eliminates the imaginary term from the original i. Assume that the original i is 0.184-68º. IOW, ITnew must be the real part of i=0.184-68°. v₂ = 10 20° (0 = 1000 rad/s iTnew ~ Cc LOAD mell 2092 0.1 H 20 μF HH