Power in Alternating-Current Circuits — Question 5

Question 5

Excess shunt capacitance can increase current again

A fixed-voltage load has real power P and positive inductive reactive power Q0. An ideal shunt capacitor C is variable.

1. Find source rms current as a function of C and the capacitance minimizing it.

2. Find the other nonnegative capacitance giving the same source current as no capacitor, and identify its power-factor sign.

Question 5 - Solution

Excess shunt capacitance can increase current again

(a) Method and calculation

$$Q_{\mathrm{net}}=Q_0-\omega CV^2,\quad I_s^2=\frac{P^2+(Q_0-\omega CV^2)^2}{V^2}.$$$$\boxed{C_*=Q_0/(\omega V^2),\quad I_{s,\min}=P/V}.$$

(b) Interpretation and further calculation

$$I_s(C)=I_s(0)\Rightarrow(Q_0-\omega CV^2)^2=Q_0^2.$$$$\boxed{C=2C_*\ \text{is the nonzero solution; }Q_{\mathrm{net}}=-Q_0\ \text{is leading}}.$$

Physical check

Current reduction is not monotonic with added capacitance beyond unity power factor.