11.7 Conservation of AC Power

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11.7 Conservation of AC Power

The principle of conservation of power applies to ac circuits as well as to dc circuits (see Section 1.5).

To see this, consider the circuit in Fig. 11.23(a), where two load impedances Z1 and Z2 are connected in parallel across an ac source V. KCL gives

$$\mathbf{I} = \mathbf{I}_1 + \mathbf{I}_2 \tag{11.52}$$

The complex power supplied by the source is (from now on, unless otherwise specified, all values of voltages and currents will be assumed to be rms values)

$$S = VI^* = V(I_1^* + I_2^*) = VI_1^* + VI_2^* = S_1 + S_2 \qquad (11.53)$$

11.3 and 11.4 that average power is conserved in ac circuits.

In fact, we already saw in Examples

Practice Problem 11.12

An ac voltage source supplied loads connected in: (a) parallel, (b) series.

where S1 and S2 denote the comple x powers delivered to loads Z1 and Z2, respectively.

If the loads are connected in series with the voltage source, as shown in Fig. 11.23(b), KVL yields

$$\mathbf{V} = \mathbf{V}_1 + \mathbf{V}_2 \tag{11.54}$$

The complex power supplied by the source is

$$S = VI^* = (V_1 + V_2)I^* = V_1I^* + V_2I^* = S_1 + S_2 \quad (11.55)$$

where S1 and S2 denote the comple x powers delivered to loads Z1 and Z2, respectively.

We conclude from Eqs. (11.53) and (11.55) that whether the loads are connected in series or in parallel (or in general), the total po wer supplied by the source equals the total power delivered to the load. Thus, in general, for a source connected to N loads,

$$S = S_1 + S_2 + \dots + S_N \tag{11.56}$$

This means that the total comple x power in a network is the sum of the complex powers of the individual components. (This is also true of real power and reactive power, but not true of apparent power.) This expresses the principle of conservation of ac power:

The complex, real, and reactive powers of the sources equal the respective sums of the complex, real, and reactive powers of the individual loads.

From this we imply that the real (or reactive) power flow from sources in a network equals the real (or reactive) power flow into the other elements in the network.

Example 11.13 Figure 11.24 sho ws a load being fed by a v oltage source through a transmission line. The impedance of the line is represented by the (4 + j2) Ω impedance and a return path. Find the real power and reactive power absorbed by: (a) the source, (b) the line, and (c) the load.

Solution:

The total impedance is

$$\mathbf{Z} = (4+j2) + (15-j10) = 19 - j8 = 20.62 \underline{\smash{\big)}\,22.83^\circ}$$

$\Omega$

In fact, all forms of ac power are conserved: instantaneous, real, reactive, and complex.

The current through the circuit is

through the circuit is
\n

$$\mathbf{I} = \frac{\mathbf{V}_s}{\mathbf{Z}} = \frac{220/0^{\circ}}{20.62/-22.83^{\circ}} = 10.67/22.83^{\circ} \text{ A rms}$$

(a) For the source, the complex power is

$$S_s = V_s I^* = (220/0^\circ)(10.67/-22.83^\circ)$$

= 2347.4/-22.83° = (2163.5 - j910.8) VA

From this, we obtain the real power as 2163.5 W and the reactive power as 910.8 VAR (leading).

(b) For the line, the voltage is

$$\mathbf{V}_{\text{line}} = (4 + j2)\mathbf{I} = (4.472 \underline{/ 26.57^{\circ}})(10.67 \underline{/ 22.83^{\circ}})$$ $$= 47.72 \underline{/ 49.4^{\circ}} \text{ V rms}$$

The complex power absorbed by the line is

$$S_{line} = V_{line}I^* = (47.72/49.4^{\circ})(10.67/-22.83^{\circ})$$

= 509.2/26.57° = 455.4 + j227.7 VA

or

$$S_{\text{line}} = |I|^2 Z_{\text{line}} = (10.67)^2 (4 + j2) = 455.4 + j227.7 VA$$

That is, the real power is 455.4 W and the reactive power is 227.76 VAR (lagging).

(c) For the load, the voltage is

$$\mathbf{V}_L = (15 - j10)\mathbf{I} = (18.03 \text{/} - 33.7^{\circ})(10.67 \text{/} 22.83^{\circ})$$

= 192.38 \text{/} - 10.87° V rms

The complex power absorbed by the load is

$$\mathbf{S}_L = \mathbf{V}_L \mathbf{I}^* = (192.38 \text{/} - 10.87^\circ)(10.67 \text{/} - 22.83^\circ)$$

= 2053 \text{/} - 33.7^\circ = (1708 - j1139) VA

The real power is 1708 W and the reactive power is 1139 VAR (leading). Note that Ss = Sline + SL, as expected. We have used the rms values of voltages and currents.

In the circuit in Fig. 11.25, the 60- Ω resistor absorbs an average power of 240 W. Find V and the complex power of each branch of the circuit. What is the overall complex power of the circuit? (Assume the current through the 60-Ω resistor has no phase shift.)

Answer: 240.7 ^⧸ 21.45° V (rms); the 20- Ω resistor: 656 VA; the (30 − j10) Ω impedance: 480 − j160 VA; the (60 + j20) Ω impedance: 240 + j80 VA; overall: 1376 − j80 VA.

Practice Problem 11.13

For Practice Prob. 11.13.

For Example 11.14.

Solution:

The current through Z1 is

$$\mathbf{I}_1 = \frac{\mathbf{V}}{\mathbf{Z}_1} = \frac{120/10^{\circ}}{60/-30^{\circ}} = 2/40^{\circ} \text{ A rms}$$

while the current through Z2 is

$$I_2 = \frac{V}{Z_2} = \frac{120/10^{\circ}}{40/45^{\circ}} = 3/-35^{\circ}$$

A rms

The complex powers absorbed by the impedances are

$$\mathbf{S}_1 = \frac{V_{\text{rms}}^2}{\mathbf{Z}_1^*} = \frac{(120)^2}{60/30^\circ} = 240/-30^\circ = 207.85 - j120 \text{ VA}$$

\n

$$\mathbf{S}_2 = \frac{V_{\text{rms}}^2}{\mathbf{Z}_2^*} = \frac{(120)^2}{40/-45^\circ} = 360/45^\circ = 254.6 + j254.6 \text{ VA}$$

The total complex power is

$$S_t = S_1 + S_2 = 462.4 + j134.6 VA$$

(a) The total apparent power is

arent power is
\n

$$|\mathbf{S}_t| = \sqrt{462.4^2 + 134.6^2} = 481.6 \text{ VA}.$$

(b) The total real power is

$$P_t = \text{Re}(S_t) = 462.4 \text{ W or } P_t = P_1 + P_2.$$

(c) The total reactive power is

$$Q_t = \text{Im}(S_t) = 134.6 \text{ VAR or } Q_t = Q_1 + Q_2.$$

(d) The pf = Pt∕∣St∣ = 462.4∕481.6 = 0.96 (lagging).

We may cross check the result by finding the complex power Ss supplied by the source.

$$\mathbf{I}_t = \mathbf{I}_1 + \mathbf{I}_2 = (1.532 + j1.286) + (2.457 - j1.721)$$

= 4 - j0.435 = 4.024 \underline{/ -6.21° A rms

$$\mathbf{S}_s = \mathbf{V}\mathbf{I}_t^* = (120/10°)(4.024/6.21°)$$

= 482.88/16.21° = 463 + j135 VA

which is the same as before.

Two loads connected in parallel are respectively 3 kW at a pf of 0.75 leading and 6 kW at a pf of 0.95 lagging. Calculate the pf of the com bined two loads. Find the complex power supplied by the source. Practice Problem 11.14

Answer: 0.9972 (leading), 9 − j0.6742 kVA.