*Make sure to read what is ac circuit first.*

## Conservation of AC Power

To see this, consider the circuit in Figure.(1a), where two load impedances **Z**_{1} and **Z**_{2} are connected in parallel across an ac source **V**.

Figure 1. An ac voltage source supplied loads connected in : (a) parallel, (b) series. |

KCL gives

(1) |

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)

(2) |

where **S**_{1} and **S**_{2} denote the complex powers delivered to loads **Z**_{1} and **Z**_{2}, respectively.

If the loads are connected in series with the voltage source, as shown in Figure.(1b), KVL yields

(3) |

The complex power supplied by the source is

(4) |

where **S**_{1} and **S**_{2} denote the complex powers delivered to loads **Z**_{1} and **Z**_{2}, respectively.

We conclude from Equations.(2) and (4) that whether the loads are connected in series or in parallel (or in general), the total power *supplied* by the source equals the total power *delivered* to the load. Thus, in general, for a source connected to *N* loads.

(5) |

This means that the total complex 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.

## Conservation of AC Power Examples

For better understanding let us review the examples below :

1. Figure.(2) shows a load being fed by a voltage source through a transmission line. The impedance of the line is represented by the (4 + *j*2) 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.

Figure 2 |

*Solution :*

The total impedance is

The current through the circuit is

(a) For the source, the complex power is

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

The complex power absorbed by the line is

or

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

The complex power absorbed by the load is

The real power is 1708 W and the reactive power is 1139 VAR (leading). Note that **S**_{s} = **S**_{line} + **S**_{L}, as expected. We have used the rms values of voltages and currents.

2. In the circuit of Figure.(3), **Z**_{1} = 60∠-30^{o} Ω and **Z**_{2} = 40∠45^{o} Ω. Calculate the total : (a) apparent power, (b) real power, (c) reactive power, and (d) pf, supplied by the source and seen by the source.

Figure 3 |

*Solution :*

The current through **Z**_{1} is

while the current through **Z**_{2} is

The complex powers absorbed by the impedances are

The total complex power is

(a) The total apparent power is

(b) The total real power is

(c) The total reactive power is

(d) The pf = P_{t}/|**S**_{t}| = 462.4/481.6 = 0.96 (lagging).

We may cross check the result by finding the complex power **S**_{s} supplied by the source.

which is the same as before.

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*Reference: Fundamentals of electric circuits by Charles K. Alexander and Matthew N. O. Sadiku*.

Untuk Bahasa Indonesia baca Konservasi Daya di Rangkaian AC.