Alternating currents playa central role in systems for distributing, converting, and using electrical energy, so it’s important to look at power relationships in ac circuits. For an ac circuit with instantaneous current i and current amplitude /, we’ll consider an element of that circuit across which the instantaneous potential difference is v with voltage amplitude V. The instantaneous power p delivered to this circuit element is
p = vi.
Let’s first see what this means for individual circuit elements. We’ll assume in each case that i= / cos tot.
Suppose first that the circuit element is a pure resistance R, as in Fig. 32-3; then v = vRand i are in phase. We obtain the graph representing p by multiplying the heights
of the graphs of v and i in Fig. 32-3b at each instant. This graph is shown by the black curve in Fig. 32-lOa. The product vi is always positive because v and i are always either both positive or both negative. Hence energy is supplied to the resistor at every instant for both directions of i, although the power is not constant.
The power curve for a pure resistance is symmetrical about a value equal to one half ‘its maximum value VI, so the average power Pav is
An equivalent expression is
Also, V ms= Ims R, so we can express Pa• by any of the equivalent forms
Note that expressions in Eq. (32-28) have the same form as the corresponding relations for de circuit, Eq. (26-18). Also note that they are valid only for pure resistors, not for more complicated combinations of circuit elements.
Next we connect the source to a pure induct-or L, as in Fig. 32-4. The voltage v = vL leads the current i by 90°. When we multiply the curves ofv and i, the product vi is negative during the half of the cycle when v and i have opposite signs. The power curve, shown in Fig. 32-1 Db, is symmetrical about the horizontal axis; it is positive half the time and negative the other half, and the average power is zero. When p is positive, energy is being supplied to set up the magnetic field in the induct-or; when p is negative,
the field is collapsing and the induct-or is returning energy to the source. The net energy transfer over one cycle is zero.
Finally, we connect the source to a pure capacitor C, as in Fig. 32-5. The voltage v = Vc lags the current i by 90°. Figure 32-1Oc shows the power curve; the average power is again zero. Energy is supplied to charge the capacitor and is returned to the source when the capacitor discharges. The net energy transfer over one cycle is.again zero.
In any ac circuit, with any combination of resistors, capacitors, and instructors, the voltage v across the entire circuit has some phase angle; with respect to the current i. Then the instantaneous power p is given by
p = vi = [V cos(w + o)][1 cos wt].
The instantaneous power curve has the form shown in Fig. 32-IOd. The area between the positive loops and the horizontal axis is greater than that between the negative loops and the horizontal axis, and the average power is positive.
We can derive from Eq. (32-29) an expression for the average power Pay by using the identity for the cosine of the sum of two angles:
p = [V(cos tat cos; – sin wt sin ;)][1 cos wt]
= VI cos; cos? an – VI sin; cos tot sin an.
From the discussion in Section 32-2 that led to Eq. (32-4), we see that the average value of cos’ wt (over one cycle) is 112. The average value of cos on sin ox is zero because this product is equal to hin 2wt, whose average over a cycle is zero. So the average
power p is
When v and i are in phase, so ; = 0, the average power equals tVI = V”..I””.; when v and i are 90° out of phase, the average power is zero. In the general case, when v has a phase angle; with respect to i, the average power equals tI multiplied by V cos ;, the component of the voltage phosphor that is in phase with the current phosphor. Figure 32-11 shows the general relationship of the current and voltage phosphors. For the L-R-C series circuit, Figs. 32-8b and 32-8c show that V cos ; equals the voltage amplitude VR for the resistor; hence Eq. (32-30) is the average power dissipated in the resistor. On average there is no energy flow into or out of the induct-or or capacitor, so none of Pay goes into either of these circuit elements.
The factor cos ; is called the power factor of the circuit. For a pure resistance, ; = 0, cos; = I, and p••= V”..Irma’For a pure capacitor or induct-or, ; = ±900 , cos ; = 0, and p••= O.For an L-R-C series circuit the power factor is equal to RlZ; we leave the proof of this statement as a problem.
A low power factor (large angle; of lag or lead) is usually undesirable in power circuits. The reason is that for a given potential difference, a large current is needed to supply a given amount of power. This results in large lR losses in the transmission lines. Your electric power company may charge a higher rate to a client with a low power factor. Many types of ac machinery draw a lagging current; that is, the current drawn by the machinery lags the applied voltage; hence the voltage leads the current, so ; > 0 and cos ; < 1. The power factor can be corrected toward the ideal value of 1 by connecting a capacitor in parallel with the load. The current drawn by the capacitor leads the voltage (that is, the voltage across the capacitor lags the current), which compensates for the lagging current in the other branch of the circuit. The capacitor itself absorbs no net power from the line.