Engineering Circuit Analysis CH 5 AC circuit power
- Slides: 20
Engineering Circuit Analysis CH 5 AC circuit power analysis 5. 1 Instantaneous Power 5. 2 Average Power 5. 3 Effectives values of Current & Voltage 5. 4 Apparent Power and Power Factor
Ch 5 AC circuit power analysis 5. 1 Instantaneous Power Given instantaneous power current delivered to any devices is : If it is a resistor If it is a inductor If it is a capacitor & voltage, the instantaneous power
Ch 5 AC circuit power analysis 5. 1 Instantaneous Power The total amount of power supplied by the source = total amount of power delivered to the circuit elements Known R V 0 + - + Power supplied by the source - Power delivered to R: Power delivered to L: PL(t) = VL(t) ∙ i. L(t)
Ch 5 AC circuit power analysis 5. 1 Instantaneous Power Since
Ch 5 AC circuit power analysis 5. 1 Instantaneous Power How if both & sinusoidal signals? It is known twice of Time invariant , leading to averaged power? possible to be the
Ch 5 AC circuit power analysis 5. 1 Instantaneous Power • Example 8. 1 Given a voltage source of 40 + 60μ(t) V, a 5 μF capacitor and a 200 Ω resistor being connected in series. Find at t = 1. 2 ms, the powers being absorbed by the resistor and capacitor respectively. If PS = PC + PR? When t<0 , Vs=40 V ) + ~- When t=0 , Vs=100 V , and Hence In this 1 st order RC circuit Hence ,
Ch 5 AC circuit power analysis 5. 1 Instantaneous Power The average power (over time) for the instantaneous power The average power over time intervals --If is periodic as to , where Validate the previous suggestion ! is the period
Ch 5 AC circuit power analysis 5. 1 Instantaneous Power • Example : the averaged power for It is a periodic signal with a period of Phase different between Validate the previous suggestion ! & , or
Ch 5 AC circuit power analysis 5. 2 Average Power Two observations: -- power absorbed by an ideal resistor, , -- power absorbed by a purely reactive element, Example 8. 2 (p. 213) Given and Determine P and p(t).
Ch 5 AC circuit power analysis 5. 2 Average Power • Example 8. 4 Determine the average power absorbed by each of the passive elements and supplied by each of the sources. Apply KVL, we can determine the currents are: The current that passes through the resistor is: ) + ~- Hence , Both L & C are reactive elements , and For the source on the left mesh For the source on the right mesh The power of the left supplies 50 W power and it is absorbed by the resistor and the source on the right
Ch 5 AC circuit power analysis 5. 2 Average Power P
Ch 5 AC circuit power analysis 5. 2 Average Power Remark : superposition is applicable for evaluating the averaged power of a non periodic function which can be decomposed into two periodic functions with different periods,
Ch 5 AC circuit power analysis 5. 2 Average Power • Example 8. 6 (p 218) Determine the average power delivered to the 4 Ω resistor by the current i 1 = 2 cos 10 t – 3 cos 20 t A. The two different frequency components of the current can be considered separately. i 1 = 2 cos 10 t – 3 cos 20 t A
Ch 5 AC circuit power analysis 5. 3 Effectives values of Current & -- represent the actual effect of a periodic function (current voltage ) in term of delivering / absorbing power -- transforms an AC circuit into an equivalent DC circuit -- square root of a mean square (RMS)
Ch 5 AC circuit power analysis 5. 3 Effectives values of Current & RMS of the Sinusoidal wave. Given , For a sinusoidal function , its RMS is of its magnitude
Ch 5 AC circuit power analysis 5. 3 Effectives values of Current & Effective value of multiple-frequency circuit of 50 Hz of 100 Hz Power can be determined using super position In general, given a circuit with frequencies,
Ch 5 AC circuit power analysis 5. 3 Effectives values of Current & Example …
Ch 5 AC circuit power analysis 5. 4 Apparent Power and Power Factor Given One can determine the averaged power delivered to the network is or However, the effective power delivered to the network would be is defined as the apparent power – The maximal value of the average power can be. Note that the unit of the Papp is VA.
Ch 5 AC circuit power analysis 5. 4 Apparent Power and Power Factor Power factor ( ) In the above case, If passive elements are all resistive, If passive elements are all reactive, , If passive elements are neither purely resistive, nor purely reactive , capacitive load, since , inductive load, since , a leading PF of “—current is leading” a lagging PF of “—current is lagging” lags behind leads
Ch 5 AC circuit power analysis 5. 4 Apparent Power and Power Factor Example 8. 8(P 222) Determine the apparent power supplied by the source 3, and the power factor of the combined loads. + - Note that for the impedance 3+j 4Ω, only the resistive component will absorb power delivered by the source. ad
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