Power in Alternating-Current Circuits — Question 3

Question 3

A shunt capacitor reduces source current while preserving ideal load power

A load consumes 1200 W at lagging power factor .600 from an ideal 200-V rms, 50-Hz source. Add an ideal parallel capacitor for unity source power factor. Assume the load voltage stays fixed.

1. Find load reactive power and required capacitance.

2. Find source current before and after correction and the ratio of losses in a small unchanged supply resistance, neglecting its voltage drop.

Question 3 - Solution

A shunt capacitor reduces source current while preserving ideal load power

(a) Method and calculation

$$Q_r=P\tan(\cos^{-1}.600)=1600\,var,\quad Q_C=-\omega CV^2.$$$$\boxed{C=1600/[(100\pi)(200)^2]=127.324\,\mu F}.$$

(b) Interpretation and further calculation

$$I_{\mathrm{before}}=P/(V\cdot.600)=10.0\,A,\quad I_{\mathrm{after}}=P/V=6.00\,A.$$$$\boxed{P_{{\mathrm{line}},\mathrm{after}}/P_{{\mathrm{line}},\mathrm{before}}=(6/10)^2=.360}.$$

Physical check

The capacitor supplies reactive current locally; ideal correction does not reduce the specified real load power.