Average power formula is the important calculation in electrical circuit. The

Consider the general case of instantaneous power absorbed by an arbitrary combination of circuit elements under sinusoidal excitation, as shown in Figure.(1).

Let the voltage and current at the terminals of the circuit be

where

We apply the trigonometric identity

and express Equation.(3) as

This shows us that the instantaneous power has two parts. The first is constant or time independent. Its value depends on the phase difference between the voltage and the current. The second is a sinusoidal function whose frequency is 2ω, which is twice the angular frequency of the voltage or current.

The sketch of

When

The instantaneous power changes with time and is therefore difficult to measure. The

Although Equation.(6) shows the averaging done over

Substituting

The first integrand is constant, and the average of a constant is the same constant. The second integrand is a sinusoid. We know that the average of a sinusoid over its period is zero because the area under the sinusoid during a positive half-cycle is canceled by the area under it during the following negative half-cycle. Thus, the second term in Equation.(7) vanishes and the average power becomes

Since cos(θ

Note that

The phasor form of

We recognize the real part of this expression as the average power

Consider two special cases of Equation.(10). When θ

where |

showing that a purely reactive circuit absorbs no average power. In summary,

The instantaneous power is given by

Applying the trigonometric identity

gives

or

The average power is

which is the constant part of

The current through the impedance is

The average power is

The current

The average power supplied by the voltage source is

The current through the resistor is

and the voltage across it is

The average power absorbed by the resistor is

which is the same as the average power supplied. Zero average power is absorbed by the capacitor.

We apply mesh analysis as shown in Figure.(4b). For mesh 1,

For mesh 2,

or

For the voltage source, the current flowing from it is

Following the passive sign convention, this average power is absorbed by the source, in view of the direction of

For the current source, the current through it is

The average power supplied by the current source is

It is negative according to the passive sign convention, meaning that the current source is supplying power to the circuit.

For the resistor, the current through it is

For the capacitor, the current through it is

The average power absorbed by the capacitor is

For the inductor, the current through it is

Notice that the inductor and the capacitor absorb zero average power and that the total power supplied by the current source equals the power absorbed by the resistor and the voltage source, or

indicating that power is conserved.

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*instantaneous power p(t)*absorbed by an element is the product of the*instantaneous voltage v(t)*across the element and the*instantaneous current i(t)*through it.## Instantaneous and Average Power Formula

Assuming the passive sign convention,(1) |

TheIt is the rate at which an element absorbs energy.instantaneous power(in watts) is the power at any instant of time.

Consider the general case of instantaneous power absorbed by an arbitrary combination of circuit elements under sinusoidal excitation, as shown in Figure.(1).

Figure 1. Sinusoidal source and passive linear circuit |

(2a) |

(2b) |

*V*and_{m}*I*are the amplitudes (or peak values), and θ_{m}_{v}and θ_{i}are the phase angles of the voltage and current, respectively. The instantaneous power absorbed by the circuit is(3) |

(4) |

(5) |

The sketch of

*p(t)*in Equation.(5) is shown in Figure.(2), where T = 2π/ω is the period of voltage or current. We observe that*p(t)*is periodic,*p(t)*=*p*(*t*+*T*), and has a period of_{0}*T*=_{0}*T*/2, since its frequency is twice that of voltage or current. We also observe that*p(t)*is positive for some part of each cycle and negative for the rest.When

*p(t)*is positive, power is absorbed by the circuit. When*p(t)*is negative, power is absorbed by the source; that is, power is transferred from the circuit to the source. This is possible because of the storage elements (capacitots and inductors) in the circuit.Figure 2. The instantaneous power p(t) entering a circuit |

*average power*is more convenient to measure. In fact, the wattmeter, the instrument for measuring power, responds to average power.TheThus, the average power is given byaverage power, in watts, is the average of the instantaneous power over one period.

(6) |

*T*, we would get the same result if we performed the integration over the actual period of*p(t)*which is*T*=_{0}*T*/2.Substituting

*p(t)*in Equation.(5) to (6) results(7) |

(8) |

_{v}- θ_{i}) = cos(θ_{v}- θ_{i}), what is important is the difference in the phases of the voltage and current.Note that

*p(t)*is time-varying while P does not depend on time. To find the instantaneous power, we must necessarily have v(t) and i(t) in the time domain. But we can find the average power when voltage and current are expressed in the time domain, as in Equation.(8), or when they are expressed in the frequency domain.The phasor form of

*v(t)*and*i(t)*in Equation.(2) are**V**=*V*∠θ_{m}_{v}and**I**=*I*∠θ_{m}_{i}, respectively.*P*is calculated using Equation.(8) or using phasors**V**and**I**. To use phasors, we notice that(9) |

*P*according to Equation.(8). Hence,(10) |

_{v}=θ_{i}, the voltage and current are in phase. This implies a purely resistive circuit or resistive load R, and(11) |

**I**|^{2}=**I**x**I***. Equation.(11) shows that a purely resistive circuit absorbs power at all times. When θ_{v}- θ_{i}= ± 90^{o}, we have purely reactive circuit, and(12) |

A resistive load (R) absorbs power at all times, while a reactive load (LorC) absorbs zero average power.

## Instantaneous and Average Power Formula Examples

For better understanding let us review the examples below**1. Given that***v(t)*= 120 cos(377*t*+ 45^{o}) V and*i(t)*= 10 cos(377*t*- 10^{o}) A**find the instantaneous power and the average power absorbed by the passive linear network of Figure.(1)**__Solution :__The instantaneous power is given by

Applying the trigonometric identity

gives

or

The average power is

which is the constant part of

*p(t)*above.**2. Calculate the average power absorbed by an impedance Z = 30 -***j*70 Ω when a voltage V = 120 ∠0^{o}is applied across it*Solution :*The current through the impedance is

The average power is

**3. For the circuit in Figure.(3), find the average power is supplied by the source and the average power absorbed by the resistor.**Figure 3 |

*Solution :*The current

**I**is given byThe average power supplied by the voltage source is

The current through the resistor is

and the voltage across it is

The average power absorbed by the resistor is

which is the same as the average power supplied. Zero average power is absorbed by the capacitor.

**4. Determine the average power generated by each source and the average power absorbed by each passive element in the circuit of Figure.(4a).**Figure 4 |

*Solution :*We apply mesh analysis as shown in Figure.(4b). For mesh 1,

For mesh 2,

or

For the voltage source, the current flowing from it is

**I**_{2}= 10.58∠79.1^{o}A and the voltage across it is 60∠30^{o}V, so that the average power isFollowing the passive sign convention, this average power is absorbed by the source, in view of the direction of

**I**_{2}and the polarity of the voltage source. That is, the circuit is delivering average power to the voltage source.For the current source, the current through it is

**I**_{1}= 4∠0^{o}and the voltage across it isThe average power supplied by the current source is

It is negative according to the passive sign convention, meaning that the current source is supplying power to the circuit.

For the resistor, the current through it is

**I**_{1}= 4∠0^{o}and the voltage across it is 20**I**_{1}= 80∠0^{o}, so that the power absorbed by the resistor isFor the capacitor, the current through it is

**I**_{2}= 10.58∠79.1^{o}and the voltage across it is -*j*5**I**_{2}= (5∠-90^{o})(10.58∠79.1^{o}) = 52.9∠79.1 - 90^{o }.The average power absorbed by the capacitor is

For the inductor, the current through it is

**I**_{1}-**I**_{2}= 2 -*j*10.39 = 10.58∠-79.1^{o}. The voltage across it is*j*10(**I**_{1}-**I**_{2}) = 10.58∠-79.1^{o}+ 90^{o}. Thus, the average power absorbed by the inductor isNotice that the inductor and the capacitor absorb zero average power and that the total power supplied by the current source equals the power absorbed by the resistor and the voltage source, or

indicating that power is conserved.

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