Lesson 7 Flashcards — Reactive and Apparent Power

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1. What do capacitors and inductors do to the phase between voltage and current?

A capacitor makes current lead voltage by 90°; an inductor makes current lag by 90°. Both store and return energy each cycle rather than consuming it.

2. Do reactive elements consume real power?

No. They consume no real power — but they still draw current, and that current is real, as are the conductors that must carry it.

3. Define real, reactive, and apparent power with their units.

All three use the same phase angle \(\phi = \phi_V - \phi_I\). Real \(P = V_\text{rms}I_\text{rms}\cos\phi\) in watts (W). Reactive \(Q = V_\text{rms}I_\text{rms}\sin\phi\) in volt-amperes reactive (VAR) — and \(Q\) is signed: a lagging, inductive load has \(\phi > 0\) and \(Q > 0\), a leading, capacitive load has \(\phi < 0\) and \(Q < 0\). Apparent \(S = V_\text{rms}I_\text{rms}\) in volt-amperes (VA).

4. How do P, Q, and S combine?

As a right triangle: \(S^2 = P^2 + Q^2\). They are not added arithmetically — 3 kW with 4 kVAR is 5 kVA, not 7.

5. A load draws 2.3 kW at pf 0.92 lagging from a 115 V bus. Find the current and \(Q\).

Run the triangle backward. The power factor gives \(S\) from \(P\): \(S = P/\text{pf} = 2300/0.92 = 2500\ \text{VA}\). Then \(S\) gives the current: \(I = S/V_\text{rms} = 2500/115 = 21.7\ \text{A}\). Then the triangle gives \(Q\) with its sign: \(Q = +\sqrt{S^2 - P^2} = +\sqrt{2500^2 - 2300^2} = +\sqrt{960{,}000} = +980\ \text{VAR}\) — positive, because the load is lagging.

6. Several loads share one bus. Which powers add?

\(P\) and \(Q\) add; \(S\) does not. Sum the real powers and the signed reactive powers — \(P_\text{bus} = \sum P\) and \(Q_\text{bus} = \sum Q\) — then rebuild the hypotenuse, \(S_\text{bus} = \sqrt{P_\text{bus}^2 + Q_\text{bus}^2}\). Because the signs are kept, a leading load's \(-Q\) cancels a lagging load's \(+Q\): two 4 kW loads at pf 0.80, one lagging and one leading, need 8 kVA of bus, not \(5 + 5 = 10\) kVA. That cancellation is power-factor correction.

7. Define power factor.

\(\text{pf} = \cos\phi = P/S\), running from 0 to 1. A pf of 1 means every volt-ampere is doing useful work. Since \(\cos\phi > 0\) either way, always say leading or lagging alongside the number.

8. If reactive power does no work, why does a utility care about power factor?

Because the current associated with \(Q\) still flows through generators, transformers, and wire, and that current hurts three ways. It causes \(I^2R\) losses in every conductor it crosses. It ties up capacity — the hardware is thermally limited by current, so it must be sized for \(S\) rather than \(P\), which is why generators, transformers, and cables are rated in VA. And it degrades voltage regulation: reactive current through line reactance sags the voltage, the mechanism behind grid voltage collapse.

9. What does 'lagging' power factor tell you about the load?

That current lags voltage, so the load is inductive — which is what a motor should be.

10. What are the two most common errors on this objective?

Adding \(P\) and \(Q\) arithmetically instead of as a right triangle, and mixing peak with RMS values. Every equation here takes RMS.

11. A motor draws 15 A at 240 V with pf = 0.80 lagging. Find S, P, and Q.

\(S = (240)(15) = 3600\ \text{VA}\); \(P = S\cos\phi = 3600(0.80) = 2880\ \text{W}\); and \(\phi = +\cos^{-1}(0.80) = +36.9^\circ\) (lagging). Get the sine exactly rather than rounding through the angle: \(\sin\phi = \sqrt{1 - 0.80^2} = 0.60\), so \(Q = 3600(0.60) = +2160\ \text{VAR}\) — positive, because the load is inductive. Check: \(\sqrt{2880^2 + 2160^2} = 3600\) ✓

12. What does correcting power factor to 1.0 achieve?

The same real power is delivered with less current, so conductors run cooler, \(I^2R\) losses drop, and existing hardware can serve more load.

13. So what do customers actually pay for — the VARs themselves?

Nobody pays for the VARs. Reactive power consumes zero net energy: \(Q\) flows into the load's fields and straight back out each cycle, and a house meter reads kW·h, which is real energy — so a homeowner pays nothing extra for a reactive load. What everybody pays for is the amperes the VARs require. At fixed \(P\) and \(V\), \(I = P/(V\,\text{pf})\), and those extra amperes cause \(I^2R\) losses upstream, tie up VA-rated equipment capacity, and sag the voltage. Large industrial and commercial customers see the bill directly, as kVA demand charges or power-factor penalty clauses that commonly kick in below a pf of about 0.90–0.95; residential customers do not.