A series circuit with R=10 Ω, L=50 mH and C=100 μF consumes power P from a 200 V,50 Hz supply. Change only L so that power remains P but power factor becomes lagging. Find the new L in mH to two decimals.
For fixed R and supply, equal power requires equal |impedance| and hence equal squared net reactance. Initially X=ω(.05)−1/(ωC)≈−16.123 Ω, capacitive. The new reactance must be +16.123 Ω. Thus Lnew=2/(ω²C)−.05≈.152642 H=152.64 mH, using ω=2π×50.
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