Time-averaged power

Several mathematical relationships have been developed using the vector electric field intensity induced inside a sphere as a function of the incident electric field intensity (8,9). The square of the average internal electric field intensity, lEl 2, was derived and related to the time-average power density incident upon the sample within the stripline. 

Thus, with the usual notation, if the voltage is described by C/osin(instantaneous power drawn from the voltage source is Ic U, the time average power dissipated is P where... 

If there is to be a net extraction of power from the source, there must be a component I of /in phase with U, as shown in Fig. 2.31 I leads to power loss, whereas the capacitative component Ic does not. Therefore the time average power loss is... 

The time-averaged power radiated per unit solid angle is... 

Daily Production time average Power Consumption hrs 18 18 18 18... 

The angular differential cross-section is calculated by taking dP/dQ, where dP is the time-averaged power scattered into solid angle dQ, and dividing by the intensity of the incident light. This is equivalent to writing da/dQ = r )/ Eq so that... 

In the mvestigation of the response of a system subject to an external field, we are usually mterested in the response (r, i)> of the property directly coupled to the field, i.e., (SA(r, t)) = (SB(r, t)). For such mvestigations, the time-averaged power Q(co) absorbed by the system from a CW monochromatic field over the period 2n jco is given by... 

We define to be the total, time-averaged power propagating along an optical waveguide in the increasing z-direction of Fig. 11-1 (a). In terms of the... 

The total, time-averaged power flow of the radiation field parallel to the waveguide axis has magnitude This follows from Eq. (11-25) as... 

Poynting s vector theorem can be derived from Maxwell s equations as a consequence of power conservation. It shows that the time-averaged power produced by a distribution of currents with density J within volume is given by[l]... 

Taking into account the boundary conditions nxE =nxE and nxH = nx H on S, we express the time-averaged power absorbed by the particle as... 

The time-averaged power absorbed by the particle then becomes... 

---