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CH5 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 Engineering Circuit Analysis

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Given instantaneous current ＆ voltage, the instantaneous power delivered to any device is : If it is a resistor If it is an inductor If it is a capacitor 5.1 Instantaneous Power Ch5 AC circuit power analysis

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The total amount of power supplied by the source = total amount of power delivered to the circuit elements Known Power supplied by the source Power delivered to R: Power delivered to L: P L (t) = v L (t) ∙ i L (t) +-+- R V0V0 +-+- 5.1 Instantaneous Power Ch5 AC circuit power analysis

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Since Ch5 AC circuit power analysis 5.1 Instantaneous Power

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How if both ＆ are sinusoidal signals? is the phase different between v(t) and i(t). E.g., in the 1 st order RL circuit,, leading to possible to be the averaged power? Time invariant twice of Ch5 AC circuit power analysis 5.1 Instantaneous Power

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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 = 1.2ms, the powers being absorbed by the resistor and capacitor respectively. If P S = P C + P R ? When t < 0,V s = 40V, When t = 0, V s = 100V, and Hence In this 1 st order RC circuit Hence, + ) - ~ Ch5 AC circuit power analysis 5.1 Instantaneous Power

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The average power (over time) for the instantaneous power The average power over time intervals to If is periodic as, where is the period Validate the previous suggestion ! Ch5 AC circuit power analysis 5.2 Average Power

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Example : With and, the average power for sinusoidal signals It is a periodic signal with a period of Phase different between ＆, or Validate the previous suggestion ! Ch5 AC circuit power analysis 5.2 Average Power

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Two observations: -- power absorbed by an ideal resistor,, -- power absorbed by a purely reactive element, or, Example 8.2 (p.213) Given and, Determine P and p(t). 5.2 Average Power Ch5 AC circuit power analysis

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P 5.2 Average Power

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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 meshFor the source on the right mesh The power of the left supplies 50W power and it is absorbed by the resistor and the source on the right + ) - ~ + - ~ Ch5 AC circuit power analysis 5.2 Average Power

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Remark: superposition is applicable for evaluating the average power of a non periodic function which can be decomposed into two periodic functions with different periods, i.e., Ch5 AC circuit power analysis 5.2 Average Power

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Example 8.6 (p218) Determine the average power delivered to the 4 Ω resistor by the current i 1 = 2 cos10t – 3 cos20t A. The two different frequency components of the current can be considered separately. i 1 = 2 cos10t – 3 cos20t A Ch5 AC circuit power analysis 5.2 Average Power

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- 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 the mean of the square (RMS) 5.3 Effectives values of Current ＆ Voltage Ch5 AC circuit power analysis For a resistor,

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RMS of the Sinusoidal wave. Given, For a sinusoidal function, its RMS value is of its magnitude Ch5 AC circuit power analysis 5.3 Effectives values of Current ＆ Voltage

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Effective value of multiple-frequency circuit Power can be determined using superposition frequencies,In general, given a circuit with of 100Hz of 50Hz Ch5 AC circuit power analysis 5.3 Effectives values of Current ＆ Voltage

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Example Given two sinusoidal currents with the frequencies of 60Hz and 120Hz, respectively, Ch5 AC circuit power analysis 5.3 Effectives values of Current ＆ Voltage

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Given One can determine the average 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 that the average power can take. Note that the unit of the P app is VA. 5.4 Apparent Power and Power Factor Ch5 AC circuit power analysis

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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 lags behind, inductive load, since leads, a leading PF of “—current is leading” a lagging PF of “—current is lagging” Ch5 AC circuit power analysis 5.4 Apparent Power and Power Factor

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Determine the apparent power supplied by the source3, and the power factor of the combined loads. Example 8.8(P222) +-+- Note that for the impedance 3+j4 Ω, only the resistive component will absorb power delivered by the source. ad +-+- Ch5 AC circuit power analysis 5.4 Apparent Power and Power Factor (a lagging PF)

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