14: Power in AC Circuits
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- Jeffrey Holmes
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1 E1.1 Analysis of Circuits ( ) AC Power: 14 1 / 11
2 Average Power E1.1 Analysis of Circuits ( ) AC Power: 14 2 / 11
3 Average Power Intantaneous Power dissipated inr: p(t) = v2 (t) R E1.1 Analysis of Circuits ( ) AC Power: 14 2 / 11
4 Average Power Intantaneous Power dissipated inr: p(t) = v2 (t) R E1.1 Analysis of Circuits ( ) AC Power: 14 2 / 11
5 Average Power Intantaneous Power dissipated inr: p(t) = v2 (t) R Average Power dissipated in R: P = 1 T T 0 p(t)dt E1.1 Analysis of Circuits ( ) AC Power: 14 2 / 11
6 Average Power Intantaneous Power dissipated inr: p(t) = v2 (t) R Average Power dissipated in R: P = 1 T T 0 p(t)dt= 1 R 1 T T 0 v2 (t)dt E1.1 Analysis of Circuits ( ) AC Power: 14 2 / 11
7 Average Power Intantaneous Power dissipated inr: p(t) = v2 (t) R Average Power dissipated in R: P = 1 T T 0 p(t)dt= 1 R 1 T T 0 v2 (t)dt= v 2 (t) R E1.1 Analysis of Circuits ( ) AC Power: 14 2 / 11
8 Average Power Intantaneous Power dissipated inr: p(t) = v2 (t) R Average Power dissipated in R: P = 1 T T 0 p(t)dt= 1 R 1 T T 0 v2 (t)dt= v 2 (t) R v 2 (t) is the value ofv 2 (t) averaged over time E1.1 Analysis of Circuits ( ) AC Power: 14 2 / 11
9 Average Power Intantaneous Power dissipated inr: p(t) = v2 (t) R Average Power dissipated in R: P = 1 T T 0 p(t)dt= 1 R 1 T T 0 v2 (t)dt= v 2 (t) R v 2 (t) is the value ofv 2 (t) averaged over time We define the RMS Voltage (Root Mean Square): V rms v 2 (t) E1.1 Analysis of Circuits ( ) AC Power: 14 2 / 11
10 Average Power Intantaneous Power dissipated inr: p(t) = v2 (t) R Average Power dissipated in R: P = 1 T T 0 p(t)dt= 1 R 1 T T 0 v2 (t)dt= v 2 (t) R v 2 (t) is the value ofv 2 (t) averaged over time We define the RMS Voltage (Root Mean Square): V rms v 2 (t) The average power dissipated inrisp = v 2 (t) R = (V rms) 2 R V rms is the DC voltage that would cause R to dissipate the same power. E1.1 Analysis of Circuits ( ) AC Power: 14 2 / 11
11 Average Power Intantaneous Power dissipated inr: p(t) = v2 (t) R Average Power dissipated in R: P = 1 T T 0 p(t)dt= 1 R 1 T T 0 v2 (t)dt= v 2 (t) R v 2 (t) is the value ofv 2 (t) averaged over time We define the RMS Voltage (Root Mean Square): V rms v 2 (t) The average power dissipated inrisp = v 2 (t) R = (V rms) 2 R V rms is the DC voltage that would cause R to dissipate the same power. We use small letters for time-varying voltages and capital letters for time-invariant values. E1.1 Analysis of Circuits ( ) AC Power: 14 2 / 11
12 Cosine Wave RMS Cosine Wave: v(t) = 5cosωt. Amplitude isv = 5 V. E1.1 Analysis of Circuits ( ) AC Power: 14 3 / 11
13 Cosine Wave RMS Cosine Wave: v(t) = 5cosωt. Amplitude isv = 5 V. Squared Voltage: v 2 (t) = V 2 cos 2 ωt = V 2( cos2ωt) E1.1 Analysis of Circuits ( ) AC Power: 14 3 / 11
14 Cosine Wave RMS Cosine Wave: v(t) = 5cosωt. Amplitude isv = 5 V. Squared Voltage: v 2 (t) = V 2 cos 2 ωt = V 2( cos2ωt) Mean Square Voltage: v 2 = V 2 2 sincecos2ωt averages to zero. E1.1 Analysis of Circuits ( ) AC Power: 14 3 / 11
15 Cosine Wave RMS Cosine Wave: v(t) = 5cosωt. Amplitude isv = 5 V. Squared Voltage: v 2 (t) = V 2 cos 2 ωt = V 2( cos2ωt) Mean Square Voltage: v 2 = V 2 2 sincecos2ωt averages to zero. E1.1 Analysis of Circuits ( ) AC Power: 14 3 / 11
16 Cosine Wave RMS Cosine Wave: v(t) = 5cosωt. Amplitude isv = 5 V. Squared Voltage: v 2 (t) = V 2 cos 2 ωt = V 2( cos2ωt) Mean Square Voltage: v 2 = V 2 RMS Voltage: V rms = v 2 = 1 2 V = 3.54 V 2 sincecos2ωt averages to zero. E1.1 Analysis of Circuits ( ) AC Power: 14 3 / 11
17 Cosine Wave RMS Cosine Wave: v(t) = 5cosωt. Amplitude isv = 5 V. Squared Voltage: v 2 (t) = V 2 cos 2 ωt = V 2( cos2ωt) Mean Square Voltage: v 2 = V 2 RMS Voltage: V rms = v 2 = 1 2 V = 3.54 V 2 sincecos2ωt averages to zero. Note: Power engineers always use RMS voltages and currents exclusively and omit the rms subscript. For example UK Mains voltage = 230 V rms = 325 V peak. E1.1 Analysis of Circuits ( ) AC Power: 14 3 / 11
18 Cosine Wave RMS Cosine Wave: v(t) = 5cosωt. Amplitude isv = 5 V. Squared Voltage: v 2 (t) = V 2 cos 2 ωt = V 2( cos2ωt) Mean Square Voltage: v 2 = V 2 2 sincecos2ωt averages to zero. RMS Voltage: V rms = v 2 = 1 2 V = 3.54 V= Ṽ Note: Power engineers always use RMS voltages and currents exclusively and omit the rms subscript. For example UK Mains voltage = 230 V rms = 325 V peak. In this lecture course only, a ~ overbar means 2: thusṽ = 1 2 V. E1.1 Analysis of Circuits ( ) AC Power: 14 3 / 11
19 Power Factor Suppose voltage and current phasors are: V = V e jθ V I = I e jθ I E1.1 Analysis of Circuits ( ) AC Power: 14 4 / 11
20 Power Factor Suppose voltage and current phasors are: V = V e jθ V v(t) = V cos(ωt+θ V ) I = I e jθ I i(t) = I cos(ωt+θ I ) E1.1 Analysis of Circuits ( ) AC Power: 14 4 / 11
21 Power Factor Suppose voltage and current phasors are: V = V e jθ V v(t) = V cos(ωt+θ V ) I = I e jθ I i(t) = I cos(ωt+θ I ) E1.1 Analysis of Circuits ( ) AC Power: 14 4 / 11
22 Power Factor Suppose voltage and current phasors are: V = V e jθ V v(t) = V cos(ωt+θ V ) I = I e jθ I i(t) = I cos(ωt+θ I ) Power dissipated in loadz is p(t) = v(t)i(t) = V I cos(ωt+θ V )cos(ωt+θ I ) E1.1 Analysis of Circuits ( ) AC Power: 14 4 / 11
23 Power Factor Suppose voltage and current phasors are: V = V e jθ V v(t) = V cos(ωt+θ V ) I = I e jθ I i(t) = I cos(ωt+θ I ) Power dissipated in loadz is p(t) = v(t)i(t) = V I cos(ωt+θ V )cos(ωt+θ I ) = V I ( 1 2 cos(2ωt+θ V +θ I )+ 1 2 cos(θ V θ I ) ) E1.1 Analysis of Circuits ( ) AC Power: 14 4 / 11
24 Power Factor Suppose voltage and current phasors are: V = V e jθ V v(t) = V cos(ωt+θ V ) I = I e jθ I i(t) = I cos(ωt+θ I ) Power dissipated in loadz is p(t) = v(t)i(t) = V I cos(ωt+θ V )cos(ωt+θ I ) = V I ( 1 2 cos(2ωt+θ V +θ I )+ 1 2 cos(θ V θ I ) ) = 1 2 V I cos(θ V θ I )+ 1 2 V I cos(2ωt+θ V +θ I ) E1.1 Analysis of Circuits ( ) AC Power: 14 4 / 11
25 Power Factor Suppose voltage and current phasors are: V = V e jθ V v(t) = V cos(ωt+θ V ) I = I e jθ I i(t) = I cos(ωt+θ I ) Power dissipated in loadz is p(t) = v(t)i(t) = V I cos(ωt+θ V )cos(ωt+θ I ) = V I ( 1 2 cos(2ωt+θ V +θ I )+ 1 2 cos(θ V θ I ) ) = 1 2 V I cos(θ V θ I )+ 1 2 V I cos(2ωt+θ V +θ I ) E1.1 Analysis of Circuits ( ) AC Power: 14 4 / 11
26 Power Factor Suppose voltage and current phasors are: V = V e jθ V v(t) = V cos(ωt+θ V ) I = I e jθ I i(t) = I cos(ωt+θ I ) Power dissipated in loadz is p(t) = v(t)i(t) = V I cos(ωt+θ V )cos(ωt+θ I ) = V I ( 1 2 cos(2ωt+θ V +θ I )+ 1 2 cos(θ V θ I ) ) = 1 2 V I cos(θ V θ I )+ 1 2 V I cos(2ωt+θ V +θ I ) Average power: P = 1 2 V I cos(φ) where φ = θ V θ I E1.1 Analysis of Circuits ( ) AC Power: 14 4 / 11
27 Power Factor Suppose voltage and current phasors are: V = V e jθ V v(t) = V cos(ωt+θ V ) I = I e jθ I i(t) = I cos(ωt+θ I ) Power dissipated in loadz is p(t) = v(t)i(t) = V I cos(ωt+θ V )cos(ωt+θ I ) = V I ( 1 2 cos(2ωt+θ V +θ I )+ 1 2 cos(θ V θ I ) ) = 1 2 V I cos(θ V θ I )+ 1 2 V I cos(2ωt+θ V +θ I ) Average power: P = 1 2 V I cos(φ) where φ = Ṽ Ĩ cos(φ) = θ V θ I E1.1 Analysis of Circuits ( ) AC Power: 14 4 / 11
28 Power Factor Suppose voltage and current phasors are: V = V e jθ V v(t) = V cos(ωt+θ V ) I = I e jθ I i(t) = I cos(ωt+θ I ) Power dissipated in loadz is p(t) = v(t)i(t) = V I cos(ωt+θ V )cos(ωt+θ I ) = V I ( 1 2 cos(2ωt+θ V +θ I )+ 1 2 cos(θ V θ I ) ) = 1 2 V I cos(θ V θ I )+ 1 2 V I cos(2ωt+θ V +θ I ) Average power: P = 1 2 V I cos(φ) where φ = Ṽ Ĩ cos(φ) = θ V θ I cosφ is the power factor E1.1 Analysis of Circuits ( ) AC Power: 14 4 / 11
29 Power Factor Suppose voltage and current phasors are: V = V e jθ V v(t) = V cos(ωt+θ V ) I = I e jθ I i(t) = I cos(ωt+θ I ) Power dissipated in loadz is p(t) = v(t)i(t) = V I cos(ωt+θ V )cos(ωt+θ I ) = V I ( 1 2 cos(2ωt+θ V +θ I )+ 1 2 cos(θ V θ I ) ) = 1 2 V I cos(θ V θ I )+ 1 2 V I cos(2ωt+θ V +θ I ) Average power: P = 1 2 V I cos(φ) where φ = Ṽ Ĩ cos(φ) = θ V θ I cosφ is the power factor φ > 0 a lagging power factor (normal case: Current lags Voltage) φ < 0 a leading power factor (rare case: Current leads Voltage) E1.1 Analysis of Circuits ( ) AC Power: 14 4 / 11
30 Complex Power If Ṽ = 1 2 V e jθ V and Ĩ = 1 2 I e jθ I The complex power absorbed byz iss Ṽ Ĩ where * means complex conjugate. E1.1 Analysis of Circuits ( ) AC Power: 14 5 / 11
31 Complex Power If Ṽ = 1 2 V e jθ V and Ĩ = 1 2 I e jθ I The complex power absorbed byz iss Ṽ Ĩ where * means complex conjugate. Ṽ Ĩ = Ṽ e jθ V Ĩ e jθ I E1.1 Analysis of Circuits ( ) AC Power: 14 5 / 11
32 Complex Power If Ṽ = 1 2 V e jθ V and Ĩ = 1 2 I e jθ I The complex power absorbed byz iss Ṽ Ĩ where * means complex conjugate. Ṽ Ĩ = Ṽ e jθ V Ĩ e jθ I = Ṽ Ĩ e j(θ V θ I ) E1.1 Analysis of Circuits ( ) AC Power: 14 5 / 11
33 Complex Power If Ṽ = 1 2 V e jθ V and Ĩ = 1 2 I e jθ I The complex power absorbed byz iss Ṽ Ĩ where * means complex conjugate. Ṽ Ĩ = = Ṽ Ṽ Ĩ e jθ V e jφ Ĩ e jθ I = Ṽ Ĩ e j(θ V θ I ) E1.1 Analysis of Circuits ( ) AC Power: 14 5 / 11
34 Complex Power If Ṽ = 1 2 V e jθ V and Ĩ = 1 2 I e jθ I The complex power absorbed byz iss Ṽ Ĩ where * means complex conjugate. Ṽ Ĩ = = Ṽ Ṽ Ĩ e jθ V e jφ = Ṽ Ĩ e jθ I = Ṽ Ĩ e j(θ V θ I ) Ĩ cosφ+jṽ Ĩ sinφ E1.1 Analysis of Circuits ( ) AC Power: 14 5 / 11
35 Complex Power If Ṽ = 1 2 V e jθ V and Ĩ = 1 2 I e jθ I The complex power absorbed byz iss Ṽ Ĩ where * means complex conjugate. Ṽ Ĩ = = Ṽ Ṽ Ĩ e jθ V e jφ = Ṽ Ĩ e jθ I = Ṽ Ĩ e j(θ V θ I ) Ĩ cosφ+jṽ Ĩ sinφ = P +jq E1.1 Analysis of Circuits ( ) AC Power: 14 5 / 11
36 Complex Power If Ṽ = 1 2 V e jθ V and Ĩ = 1 2 I e jθ I The complex power absorbed byz iss Ṽ Ĩ where * means complex conjugate. Ṽ Ĩ = = Ṽ Ṽ Ĩ e jθ V e jφ = Ṽ Ĩ e jθ I = Ṽ Ĩ e j(θ V θ I ) Ĩ cosφ+jṽ Ĩ sinφ = P +jq E1.1 Analysis of Circuits ( ) AC Power: 14 5 / 11
37 Complex Power If Ṽ = 1 2 V e jθ V and Ĩ = 1 2 I e jθ I The complex power absorbed byz iss Ṽ Ĩ where * means complex conjugate. Ṽ Ĩ = = Ṽ Ṽ Ĩ e jθ V e jφ = Ṽ Ĩ e jθ I = Ṽ Ĩ e j(θ V θ I ) Ĩ cosφ+jṽ Ĩ sinφ = P +jq Complex Power: S ṼĨ = P +jq measured in Volt-Amps (VA) E1.1 Analysis of Circuits ( ) AC Power: 14 5 / 11
38 Complex Power If Ṽ = 1 2 V e jθ V and Ĩ = 1 2 I e jθ I The complex power absorbed byz iss Ṽ Ĩ where * means complex conjugate. Ṽ Ĩ = = Ṽ Ṽ Ĩ e jθ V e jφ = Ṽ Ĩ e jθ I = Ṽ Ĩ e j(θ V θ I ) Ĩ cosφ+jṽ Ĩ sinφ = P +jq Complex Power: S ṼĨ = P +jq measured in Volt-Amps (VA) Apparent Power: S Ṽ Ĩ measured in Volt-Amps (VA) E1.1 Analysis of Circuits ( ) AC Power: 14 5 / 11
39 Complex Power If Ṽ = 1 2 V e jθ V and Ĩ = 1 2 I e jθ I The complex power absorbed byz iss Ṽ Ĩ where * means complex conjugate. Ṽ Ĩ = = Ṽ Ṽ Ĩ e jθ V e jφ = Ṽ Ĩ e jθ I = Ṽ Ĩ e j(θ V θ I ) Ĩ cosφ+jṽ Ĩ sinφ = P +jq Complex Power: S ṼĨ = P +jq measured in Volt-Amps (VA) Apparent Power: S Ṽ Ĩ Average Power: measured in Volt-Amps (VA) P R(S) measured in Watts (W) E1.1 Analysis of Circuits ( ) AC Power: 14 5 / 11
40 Complex Power If Ṽ = 1 2 V e jθ V and Ĩ = 1 2 I e jθ I The complex power absorbed byz iss Ṽ Ĩ where * means complex conjugate. Ṽ Ĩ = = Ṽ Ṽ Ĩ e jθ V e jφ = Ṽ Ĩ e jθ I = Ṽ Ĩ e j(θ V θ I ) Ĩ cosφ+jṽ Ĩ sinφ = P +jq Complex Power: S ṼĨ = P +jq measured in Volt-Amps (VA) Apparent Power: S Ṽ Ĩ Average Power: Reactive Power: measured in Volt-Amps (VA) P R(S) measured in Watts (W) Q I(S) Measured in Volt-Amps Reactive (VAR) E1.1 Analysis of Circuits ( ) AC Power: 14 5 / 11
41 Complex Power If Ṽ = 1 2 V e jθ V and Ĩ = 1 2 I e jθ I The complex power absorbed byz iss Ṽ Ĩ where * means complex conjugate. Ṽ Ĩ = = Ṽ Ṽ Ĩ e jθ V e jφ = Ṽ Ĩ e jθ I = Ṽ Ĩ e j(θ V θ I ) Ĩ cosφ+jṽ Ĩ sinφ = P +jq Complex Power: S ṼĨ = P +jq measured in Volt-Amps (VA) Apparent Power: S Ṽ Ĩ measured in Volt-Amps (VA) Average Power: P R(S) measured in Watts (W) Reactive Power: Q I(S) ( Measured) in Volt-Amps Reactive (VAR) Power Factor: cosφ cos Ṽ Ĩ = P S E1.1 Analysis of Circuits ( ) AC Power: 14 5 / 11
42 Complex Power If Ṽ = 1 2 V e jθ V and Ĩ = 1 2 I e jθ I The complex power absorbed byz iss Ṽ Ĩ where * means complex conjugate. Ṽ Ĩ = = Ṽ Ṽ Ĩ e jθ V e jφ = Ṽ Ĩ e jθ I = Ṽ Ĩ e j(θ V θ I ) Ĩ cosφ+jṽ Ĩ sinφ = P +jq Complex Power: S ṼĨ = P +jq measured in Volt-Amps (VA) Apparent Power: S Ṽ Ĩ measured in Volt-Amps (VA) Average Power: P R(S) measured in Watts (W) Reactive Power: Q I(S) ( Measured) in Volt-Amps Reactive (VAR) Power Factor: cosφ cos Ṽ Ĩ = P S Machines and transformers have capacity limits and power losses that are independent of cos φ; their ratings are always given in apparent power. E1.1 Analysis of Circuits ( ) AC Power: 14 5 / 11
43 Complex Power If Ṽ = 1 2 V e jθ V and Ĩ = 1 2 I e jθ I The complex power absorbed byz iss Ṽ Ĩ where * means complex conjugate. Ṽ Ĩ = = Ṽ Ṽ Ĩ e jθ V e jφ = Ṽ Ĩ e jθ I = Ṽ Ĩ e j(θ V θ I ) Ĩ cosφ+jṽ Ĩ sinφ = P +jq Complex Power: S ṼĨ = P +jq measured in Volt-Amps (VA) Apparent Power: S Ṽ Ĩ measured in Volt-Amps (VA) Average Power: P R(S) measured in Watts (W) Reactive Power: Q I(S) ( Measured) in Volt-Amps Reactive (VAR) Power Factor: cosφ cos Ṽ Ĩ = P S Machines and transformers have capacity limits and power losses that are independent of cos φ; their ratings are always given in apparent power. Power Company: Costs apparent power, Revenue average power. E1.1 Analysis of Circuits ( ) AC Power: 14 5 / 11
44 Power in R, L, C For any impedance, Z, complex power absorbed: S = ṼĨ = P +jq E1.1 Analysis of Circuits ( ) AC Power: 14 6 / 11
45 Power in R, L, C For any impedance, Z, complex power absorbed: S = ṼĨ = P +jq Using (a)ṽ = ĨZ (b)ĩ Ĩ = Ĩ 2 we gets = Ĩ 2 Z = Ṽ 2 Z E1.1 Analysis of Circuits ( ) AC Power: 14 6 / 11
46 Power in R, L, C For any impedance, Z, complex power absorbed: S = ṼĨ = P +jq Using (a)ṽ = ĨZ (b)ĩ Ĩ = Ĩ 2 we gets = Ĩ 2 Z = Ṽ 2 Resistor: S = Ĩ 2 R = Ṽ 2 R φ = 0 Absorbs average power, no VARs (Q = 0) Z E1.1 Analysis of Circuits ( ) AC Power: 14 6 / 11
47 Power in R, L, C For any impedance, Z, complex power absorbed: S = ṼĨ = P +jq Using (a)ṽ = ĨZ (b)ĩ Ĩ = Ĩ 2 we gets = Ĩ 2 Z = Ṽ 2 Resistor: S = Ĩ 2 R = Ṽ 2 R φ = 0 Absorbs average power, no VARs (Q = 0) Z Inductor: S = jĩ 2 ωl = j Ṽ 2 ωl φ = +90 No average power, Absorbs VARs (Q > 0) E1.1 Analysis of Circuits ( ) AC Power: 14 6 / 11
48 Power in R, L, C For any impedance, Z, complex power absorbed: S = ṼĨ = P +jq Using (a)ṽ = ĨZ (b)ĩ Ĩ = Ĩ 2 we gets = Ĩ 2 Z = Ṽ 2 Resistor: S = Ĩ 2 R = Ṽ 2 R φ = 0 Absorbs average power, no VARs (Q = 0) Z Inductor: S = jĩ 2 ωl = j Ṽ 2 ωl φ = +90 No average power, Absorbs VARs (Q > 0) Capacitor: S = j Ĩ 2 ωc = j Ṽ 2 ωc φ = 90 No average power, Generates VARs (Q < 0) E1.1 Analysis of Circuits ( ) AC Power: 14 6 / 11
49 Power in R, L, C For any impedance, Z, complex power absorbed: S = ṼĨ = P +jq Using (a)ṽ = ĨZ (b)ĩ Ĩ = Ĩ 2 we gets = Ĩ 2 Z = Ṽ 2 Resistor: S = Ĩ 2 R = Ṽ 2 R φ = 0 Absorbs average power, no VARs (Q = 0) Z Inductor: S = jĩ 2 ωl = j Ṽ 2 ωl φ = +90 No average power, Absorbs VARs (Q > 0) Capacitor: S = j Ĩ 2 ωc = j Ṽ 2 ωc φ = 90 No average power, Generates VARs (Q < 0) VARs are generated by capacitors and absorbed by inductors E1.1 Analysis of Circuits ( ) AC Power: 14 6 / 11
50 Power in R, L, C For any impedance, Z, complex power absorbed: S = ṼĨ = P +jq Using (a)ṽ = ĨZ (b)ĩ Ĩ = Ĩ 2 we gets = Ĩ 2 Z = Ṽ 2 Resistor: S = Ĩ 2 R = Ṽ 2 R φ = 0 Absorbs average power, no VARs (Q = 0) Z Inductor: S = jĩ 2 ωl = j Ṽ 2 ωl φ = +90 No average power, Absorbs VARs (Q > 0) Capacitor: S = j Ĩ 2 ωc = j Ṽ 2 ωc φ = 90 No average power, Generates VARs (Q < 0) VARs are generated by capacitors and absorbed by inductors The phase,φ, of the absorbed power,s, equals the phase ofz E1.1 Analysis of Circuits ( ) AC Power: 14 6 / 11
51 Tellegen s Theorem Tellegen s Theorem: The complex power, S, dissipated in any circuit s components sums to zero. x n = voltage at node n V b, I b = voltage/current in branchb (obeying passive sign convention) E1.1 Analysis of Circuits ( ) AC Power: 14 7 / 11
52 Tellegen s Theorem Tellegen s Theorem: The complex power, S, dissipated in any circuit s components sums to zero. x n = voltage at node n V b, I b = voltage/current in branchb a bn (obeying passive sign convention) +1 ifv b ends at node n 1 ifv b starts from noden 0 else E1.1 Analysis of Circuits ( ) AC Power: 14 7 / 11
53 Tellegen s Theorem Tellegen s Theorem: The complex power, S, dissipated in any circuit s components sums to zero. x n = voltage at node n V b, I b = voltage/current in branchb a bn (obeying passive sign convention) +1 ifv b ends at node n 1 ifv b starts from noden 0 else e.g. branch 4 goes from 2 to 3 a 4 = [0, 1, 1] E1.1 Analysis of Circuits ( ) AC Power: 14 7 / 11
54 Tellegen s Theorem Tellegen s Theorem: The complex power, S, dissipated in any circuit s components sums to zero. x n = voltage at node n V b, I b = voltage/current in branchb a bn (obeying passive sign convention) +1 ifv b ends at node n 1 ifv b starts from noden 0 else e.g. branch 4 goes from 2 to 3 a 4 = [0, 1, 1] Branch voltages: V b = n a bnx n (e.g. V 4 = x 3 x 2 ) E1.1 Analysis of Circuits ( ) AC Power: 14 7 / 11
55 Tellegen s Theorem Tellegen s Theorem: The complex power, S, dissipated in any circuit s components sums to zero. x n = voltage at node n V b, I b = voltage/current in branchb a bn (obeying passive sign convention) +1 ifv b ends at node n 1 ifv b starts from noden 0 else e.g. branch 4 goes from 2 to 3 a 4 = [0, 1, 1] Branch voltages: V b = n a bnx n (e.g. V 4 = x 3 x 2 ) noden: b a bni b = 0 b a bni b = 0 E1.1 Analysis of Circuits ( ) AC Power: 14 7 / 11
56 Tellegen s Theorem Tellegen s Theorem: The complex power, S, dissipated in any circuit s components sums to zero. x n = voltage at node n V b, I b = voltage/current in branchb a bn (obeying passive sign convention) +1 ifv b ends at node n 1 ifv b starts from noden 0 else e.g. branch 4 goes from 2 to 3 a 4 = [0, 1, 1] Branch voltages: V b = n a bnx n (e.g. V 4 = x 3 x 2 ) noden: b a bni b = 0 b a bnib = 0 Tellegen: b V bib = b n a bnx n Ib E1.1 Analysis of Circuits ( ) AC Power: 14 7 / 11
57 Tellegen s Theorem Tellegen s Theorem: The complex power, S, dissipated in any circuit s components sums to zero. x n = voltage at node n V b, I b = voltage/current in branchb a bn (obeying passive sign convention) +1 ifv b ends at node n 1 ifv b starts from noden 0 else e.g. branch 4 goes from 2 to 3 a 4 = [0, 1, 1] Branch voltages: V b = n a bnx n (e.g. V 4 = x 3 x 2 ) noden: b a bni b = 0 b a bnib = 0 Tellegen: b V bib = b n a bnx n Ib = n b a bnib x n E1.1 Analysis of Circuits ( ) AC Power: 14 7 / 11
58 Tellegen s Theorem Tellegen s Theorem: The complex power, S, dissipated in any circuit s components sums to zero. x n = voltage at node n V b, I b = voltage/current in branchb a bn (obeying passive sign convention) +1 ifv b ends at node n 1 ifv b starts from noden 0 else e.g. branch 4 goes from 2 to 3 a 4 = [0, 1, 1] Branch voltages: V b = n a bnx n (e.g. V 4 = x 3 x 2 ) noden: b a bni b = 0 b a bnib = 0 Tellegen: b V bib = b n a bnx n Ib = n b a bni b x n= n x n b a bni b E1.1 Analysis of Circuits ( ) AC Power: 14 7 / 11
59 Tellegen s Theorem Tellegen s Theorem: The complex power, S, dissipated in any circuit s components sums to zero. x n = voltage at node n V b, I b = voltage/current in branchb a bn (obeying passive sign convention) +1 ifv b ends at node n 1 ifv b starts from noden 0 else e.g. branch 4 goes from 2 to 3 a 4 = [0, 1, 1] Branch voltages: V b = n a bnx n (e.g. V 4 = x 3 x 2 ) noden: b a bni b = 0 b a bnib = 0 Tellegen: b V bib = b n a bnx n Ib = n b a bni b x n= n x n b a bni b = n x n 0 E1.1 Analysis of Circuits ( ) AC Power: 14 7 / 11
60 Tellegen s Theorem Tellegen s Theorem: The complex power, S, dissipated in any circuit s components sums to zero. x n = voltage at node n V b, I b = voltage/current in branchb a bn (obeying passive sign convention) +1 ifv b ends at node n 1 ifv b starts from noden 0 else e.g. branch 4 goes from 2 to 3 a 4 = [0, 1, 1] Branch voltages: V b = n a bnx n (e.g. V 4 = x 3 x 2 ) noden: b a bni b = 0 b a bnib = 0 Tellegen: b V bib = b n a bnx n Ib = n b a bni b x n= n x n Note: b S b = 0 b P b = 0 and also b a bnib = n x n 0 b Q b = 0. E1.1 Analysis of Circuits ( ) AC Power: 14 7 / 11
61 Power Factor Correction Ṽ = 230. Motor modelled as5 7jΩ. Ĩ = Ṽ R + Ṽ Z L E1.1 Analysis of Circuits ( ) AC Power: 14 8 / 11
62 Power Factor Correction Ṽ = 230. Motor modelled as5 7jΩ. Ĩ = Ṽ R + Ṽ Z L = 46 j32.9 A E1.1 Analysis of Circuits ( ) AC Power: 14 8 / 11
63 Power Factor Correction Ṽ = 230. Motor modelled as5 7jΩ. Ĩ = Ṽ R + Ṽ Z L = 46 j32.9 A= E1.1 Analysis of Circuits ( ) AC Power: 14 8 / 11
64 Power Factor Correction Ṽ = 230. Motor modelled as5 7jΩ. Ĩ = Ṽ R + Ṽ Z L = 46 j32.9 A= S = ṼĨ = 10.6+j7.6 kva E1.1 Analysis of Circuits ( ) AC Power: 14 8 / 11
65 Power Factor Correction Ṽ = 230. Motor modelled as5 7jΩ. Ĩ = Ṽ R + Ṽ Z L = 46 j32.9 A= S = ṼĨ = 10.6+j7.6 kva= kva E1.1 Analysis of Circuits ( ) AC Power: 14 8 / 11
66 Power Factor Correction Ṽ = 230. Motor modelled as5 7jΩ. Ĩ = Ṽ R + Ṽ Z L = 46 j32.9 A= S = ṼĨ = 10.6+j7.6 kva= kva cosφ = P S = cos36 = 0.81 E1.1 Analysis of Circuits ( ) AC Power: 14 8 / 11
67 Power Factor Correction Ṽ = 230. Motor modelled as5 7jΩ. Ĩ = Ṽ R + Ṽ Z L = 46 j32.9 A= S = ṼĨ = 10.6+j7.6 kva= kva cosφ = P S = cos36 = 0.81 Add parallel capacitor of 300 µf: E1.1 Analysis of Circuits ( ) AC Power: 14 8 / 11
68 Power Factor Correction Ṽ = 230. Motor modelled as5 7jΩ. Ĩ = Ṽ R + Ṽ Z L = 46 j32.9 A= S = ṼĨ = 10.6+j7.6 kva= kva cosφ = P S = cos36 = 0.81 Add parallel capacitor of 300 µf: Z C = 1 jωc = 10.6jΩ E1.1 Analysis of Circuits ( ) AC Power: 14 8 / 11
69 Power Factor Correction Ṽ = 230. Motor modelled as5 7jΩ. Ĩ = Ṽ R + Ṽ Z L = 46 j32.9 A= S = ṼĨ = 10.6+j7.6 kva= kva cosφ = P S = cos36 = 0.81 Add parallel capacitor of 300 µf: Z C = 1 jωc = 10.6jΩ ĨC = 21.7j A E1.1 Analysis of Circuits ( ) AC Power: 14 8 / 11
70 Power Factor Correction Ṽ = 230. Motor modelled as5 7jΩ. Ĩ = Ṽ R + Ṽ Z L = 46 j32.9 A= S = ṼĨ = 10.6+j7.6 kva= kva cosφ = P S = cos36 = 0.81 Add parallel capacitor of 300 µf: Z C = 1 jωc = 10.6jΩ ĨC = 21.7j A Ĩ = 46 j11.2 A = A E1.1 Analysis of Circuits ( ) AC Power: 14 8 / 11
71 Power Factor Correction Ṽ = 230. Motor modelled as5 7jΩ. Ĩ = Ṽ R + Ṽ Z L = 46 j32.9 A= S = ṼĨ = 10.6+j7.6 kva= kva cosφ = P S = cos36 = 0.81 Add parallel capacitor of 300 µf: Z C = 1 jωc = 10.6jΩ ĨC = 21.7j A Ĩ = 46 j11.2 A = A S C = ṼĨ C = j5 kva E1.1 Analysis of Circuits ( ) AC Power: 14 8 / 11
72 Power Factor Correction Ṽ = 230. Motor modelled as5 7jΩ. Ĩ = Ṽ R + Ṽ Z L = 46 j32.9 A= S = ṼĨ = 10.6+j7.6 kva= kva cosφ = P S = cos36 = 0.81 Add parallel capacitor of 300 µf: Z C = 1 jωc = 10.6jΩ ĨC = 21.7j A Ĩ = 46 j11.2 A = A S C = ṼĨ C = j5 kva S = ṼĨ = 10.6+j2.6 kva E1.1 Analysis of Circuits ( ) AC Power: 14 8 / 11
73 Power Factor Correction Ṽ = 230. Motor modelled as5 7jΩ. Ĩ = Ṽ R + Ṽ Z L = 46 j32.9 A= S = ṼĨ = 10.6+j7.6 kva= kva cosφ = P S = cos36 = 0.81 Add parallel capacitor of 300 µf: Z C = 1 jωc = 10.6jΩ ĨC = 21.7j A Ĩ = 46 j11.2 A = A S C = ṼĨ C = j5 kva S = ṼĨ = 10.6+j2.6 kva= kva E1.1 Analysis of Circuits ( ) AC Power: 14 8 / 11
74 Power Factor Correction Ṽ = 230. Motor modelled as5 7jΩ. Ĩ = Ṽ R + Ṽ Z L = 46 j32.9 A= S = ṼĨ = 10.6+j7.6 kva= kva cosφ = P S = cos36 = 0.81 Add parallel capacitor of 300 µf: Z C = 1 jωc = 10.6jΩ ĨC = 21.7j A Ĩ = 46 j11.2 A = A S C = ṼĨ C = j5 kva S = ṼĨ = 10.6+j2.6 kva= kva cosφ = P S = cos14 = 0.97 E1.1 Analysis of Circuits ( ) AC Power: 14 8 / 11
75 Power Factor Correction Ṽ = 230. Motor modelled as5 7jΩ. Ĩ = Ṽ R + Ṽ Z L = 46 j32.9 A= S = ṼĨ = 10.6+j7.6 kva= kva cosφ = P S = cos36 = 0.81 Add parallel capacitor of 300 µf: Z C = 1 jωc = 10.6jΩ ĨC = 21.7j A Ĩ = 46 j11.2 A = A S C = ṼĨ C = j5 kva S = ṼĨ = 10.6+j2.6 kva= kva cosφ = P S = cos14 = 0.97 Average power to motor,p, is10.6 kw in both cases. E1.1 Analysis of Circuits ( ) AC Power: 14 8 / 11
76 Power Factor Correction Ṽ = 230. Motor modelled as5 7jΩ. Ĩ = Ṽ R + Ṽ Z L = 46 j32.9 A= S = ṼĨ = 10.6+j7.6 kva= kva cosφ = P S = cos36 = 0.81 Add parallel capacitor of 300 µf: Z C = 1 jωc = 10.6jΩ ĨC = 21.7j A Ĩ = 46 j11.2 A = A S C = ṼĨ C = j5 kva S = ṼĨ = 10.6+j2.6 kva= kva cosφ = P S = cos14 = 0.97 Average power to motor,p, is10.6 kw in both cases. Ĩ, reduced from56.5 ց 47 A ( 16%) lower losses. E1.1 Analysis of Circuits ( ) AC Power: 14 8 / 11
77 Power Factor Correction Ṽ = 230. Motor modelled as5 7jΩ. Ĩ = Ṽ R + Ṽ Z L = 46 j32.9 A= S = ṼĨ = 10.6+j7.6 kva= kva cosφ = P S = cos36 = 0.81 Add parallel capacitor of 300 µf: Z C = 1 jωc = 10.6jΩ ĨC = 21.7j A Ĩ = 46 j11.2 A = A S C = ṼĨ C = j5 kva S = ṼĨ = 10.6+j2.6 kva= kva cosφ = P S = cos14 = 0.97 Average power to motor,p, is10.6 kw in both cases. Ĩ, reduced from56.5 ց 47 A ( 16%) lower losses. Effect ofc: VARs= 7.6 ց 2.6 kvar E1.1 Analysis of Circuits ( ) AC Power: 14 8 / 11
78 Power Factor Correction Ṽ = 230. Motor modelled as5 7jΩ. Ĩ = Ṽ R + Ṽ Z L = 46 j32.9 A= S = ṼĨ = 10.6+j7.6 kva= kva cosφ = P S = cos36 = 0.81 Add parallel capacitor of 300 µf: Z C = 1 jωc = 10.6jΩ ĨC = 21.7j A Ĩ = 46 j11.2 A = A S C = ṼĨ C = j5 kva S = ṼĨ = 10.6+j2.6 kva= kva cosφ = P S = cos14 = 0.97 Average power to motor,p, is10.6 kw in both cases. Ĩ, reduced from56.5 ց 47 A ( 16%) lower losses. Effect ofc: VARs= 7.6 ց 2.6 kvar, power factor= 0.81 ր E1.1 Analysis of Circuits ( ) AC Power: 14 8 / 11
79 Ideal Transformer A transformer has 2 windings on the same magnetic core. E1.1 Analysis of Circuits ( ) AC Power: 14 9 / 11
80 Ideal Transformer A transformer has 2 windings on the same magnetic core. Ampère s law: N r I r = lφ µa ; Faraday s law: V r N r = dφ dt. E1.1 Analysis of Circuits ( ) AC Power: 14 9 / 11
81 Ideal Transformer A transformer has 2 windings on the same magnetic core. Ampère s law: N r I r = lφ µa ; Faraday s law: V r dt. N 1 : N 2 +N 3 shows the turns ratio between the windings. The indicates the voltage polarity of each winding. N r = dφ E1.1 Analysis of Circuits ( ) AC Power: 14 9 / 11
82 Ideal Transformer A transformer has 2 windings on the same magnetic core. Ampère s law: N r I r = lφ µa ; Faraday s law: V r dt. N 1 : N 2 +N 3 shows the turns ratio between the windings. The indicates the voltage polarity of each winding. N r = dφ SinceΦis the same for all windings, V 1 N 1 = V 2 N 2 = V 3 N 3. E1.1 Analysis of Circuits ( ) AC Power: 14 9 / 11
83 Ideal Transformer A transformer has 2 windings on the same magnetic core. Ampère s law: N r I r = lφ µa ; Faraday s law: V r dt. N 1 : N 2 +N 3 shows the turns ratio between the windings. The indicates the voltage polarity of each winding. N r = dφ SinceΦis the same for all windings, V 1 N 1 = V 2 N 2 = V 3 N 3. Assume µ N 1 I 1 +N 2 I 2 +N 3 I 3 = 0 E1.1 Analysis of Circuits ( ) AC Power: 14 9 / 11
84 Ideal Transformer A transformer has 2 windings on the same magnetic core. Ampère s law: N r I r = lφ µa ; Faraday s law: V r dt. N 1 : N 2 +N 3 shows the turns ratio between the windings. The indicates the voltage polarity of each winding. N r = dφ SinceΦis the same for all windings, V 1 N 1 = V 2 N 2 = V 3 N 3. Assume µ N 1 I 1 +N 2 I 2 +N 3 I 3 = 0 These two equations allow you to solve circuits and also imply that S i = 0. E1.1 Analysis of Circuits ( ) AC Power: 14 9 / 11
85 Ideal Transformer A transformer has 2 windings on the same magnetic core. Ampère s law: N r I r = lφ µa ; Faraday s law: V r dt. N 1 : N 2 +N 3 shows the turns ratio between the windings. The indicates the voltage polarity of each winding. N r = dφ SinceΦis the same for all windings, V 1 N 1 = V 2 N 2 = V 3 N 3. Assume µ N 1 I 1 +N 2 I 2 +N 3 I 3 = 0 These two equations allow you to solve circuits and also imply that S i = 0. Special Case: E1.1 Analysis of Circuits ( ) AC Power: 14 9 / 11
86 Ideal Transformer A transformer has 2 windings on the same magnetic core. Ampère s law: N r I r = lφ µa ; Faraday s law: V r dt. N 1 : N 2 +N 3 shows the turns ratio between the windings. The indicates the voltage polarity of each winding. N r = dφ SinceΦis the same for all windings, V 1 N 1 = V 2 N 2 = V 3 N 3. Assume µ N 1 I 1 +N 2 I 2 +N 3 I 3 = 0 These two equations allow you to solve circuits and also imply that S i = 0. Special Case: For a 2-winding transformer this simplifies to V 2 = N 2 N 1 V 1 andi L = I 2 = N 1 N 2 I 1 E1.1 Analysis of Circuits ( ) AC Power: 14 9 / 11
87 Ideal Transformer A transformer has 2 windings on the same magnetic core. Ampère s law: N r I r = lφ µa ; Faraday s law: V r dt. N 1 : N 2 +N 3 shows the turns ratio between the windings. The indicates the voltage polarity of each winding. N r = dφ SinceΦis the same for all windings, V 1 N 1 = V 2 N 2 = V 3 N 3. Assume µ N 1 I 1 +N 2 I 2 +N 3 I 3 = 0 These two equations allow you to solve circuits and also imply that S i = 0. Special Case: For a 2-winding transformer this simplifies to V 2 = N 2 N 1 V 1 andi L = I 2 = N 1 N 2 I 1 Hence V 1 I 1 = ( N1 N 2 ) 2 V2 I L = ( N1 N 2 ) 2Z E1.1 Analysis of Circuits ( ) AC Power: 14 9 / 11
88 Ideal Transformer A transformer has 2 windings on the same magnetic core. Ampère s law: N r I r = lφ µa ; Faraday s law: V r dt. N 1 : N 2 +N 3 shows the turns ratio between the windings. The indicates the voltage polarity of each winding. N r = dφ SinceΦis the same for all windings, V 1 N 1 = V 2 N 2 = V 3 N 3. Assume µ N 1 I 1 +N 2 I 2 +N 3 I 3 = 0 These two equations allow you to solve circuits and also imply that S i = 0. Special Case: For a 2-winding transformer this simplifies to V 2 = N 2 N 1 V 1 andi L = I 2 = N 1 N 2 I 1 Hence V 1 I 1 = ( N1 N 2 ) 2 V2 I L = ( N1 N 2 ) 2Z Equivalent to a reflected impedance of( N1 N 2 ) 2Z E1.1 Analysis of Circuits ( ) AC Power: 14 9 / 11
89 Transformer Applications E1.1 Analysis of Circuits ( ) AC Power: / 11
90 Transformer Applications Power Transmission Suppose a power transmission cable has 1 Ω resistance. 100 kva@1 kv =100 A Ĩ2 R = 10 kw losses. E1.1 Analysis of Circuits ( ) AC Power: / 11
91 Transformer Applications Power Transmission Suppose a power transmission cable has 1 Ω resistance. 100 kva@1 kv =100 A Ĩ2 R = 10 kw losses. 100 kva@100 kv =1 A Ĩ2 R = 1 W losses. E1.1 Analysis of Circuits ( ) AC Power: / 11
92 Transformer Applications Power Transmission Suppose a power transmission cable has 1 Ω resistance. 100 kva@1 kv =100 A Ĩ2 R = 10 kw losses. 100 kva@100 kv =1 A Ĩ2 R = 1 W losses. Voltage Conversion Electronic equipment requires 20 V but mains voltage is240 V. E1.1 Analysis of Circuits ( ) AC Power: / 11
93 Transformer Applications Power Transmission Suppose a power transmission cable has 1 Ω resistance. 100 kva@1 kv =100 A Ĩ2 R = 10 kw losses. 100 kva@100 kv =1 A Ĩ2 R = 1 W losses. Voltage Conversion Electronic equipment requires 20 V but mains voltage is240 V. Interference protection Microphone on long cable is susceptible to interference from nearby mains cables. An N : 1 transformer reduces the microphone voltage byn but reduces interference byn 2. E1.1 Analysis of Circuits ( ) AC Power: / 11
94 Transformer Applications Power Transmission Suppose a power transmission cable has 1 Ω resistance. 100 kva@1 kv =100 A Ĩ2 R = 10 kw losses. 100 kva@100 kv =1 A Ĩ2 R = 1 W losses. Voltage Conversion Electronic equipment requires 20 V but mains voltage is240 V. Interference protection Microphone on long cable is susceptible to interference from nearby mains cables. An N : 1 transformer reduces the microphone voltage byn but reduces interference byn 2. Isolation There is no electrical connection between the windings of a transformer so circuitry (or people) on one side will not be endangered by a failure that results in high voltages on the other side. E1.1 Analysis of Circuits ( ) AC Power: / 11
95 Summary : S = ṼĨ = P +jq whereṽ = V rms = 1 2 V. E1.1 Analysis of Circuits ( ) AC Power: / 11
96 Summary : S = ṼĨ = P +jq whereṽ = V rms = 1 For an impedance Z: S = Ĩ 2 Z = Ṽ 2 Z 2 V. E1.1 Analysis of Circuits ( ) AC Power: / 11
97 Summary : S = ṼĨ = P +jq whereṽ = V rms = 1 For an impedance Z: S = Ĩ 2 Z = Ṽ 2 Z Apparent Power: S = Ṽ Ĩ used for machine ratings. 2 V. E1.1 Analysis of Circuits ( ) AC Power: / 11
98 Summary : S = ṼĨ = P +jq whereṽ = V rms = 1 For an impedance Z: S = Ĩ 2 Z = Ṽ 2 Z Apparent Power: S = Ṽ Ĩ used for machine ratings. Average Power: P = R(S) = Ṽ Ĩ cosφ (in Watts) Reactive Power: Q = I(S) = Ṽ Ĩ sinφ (in VARs) 2 V. E1.1 Analysis of Circuits ( ) AC Power: / 11
99 Summary : S = ṼĨ = P +jq whereṽ = V rms = 1 For an impedance Z: S = Ĩ 2 Z = Ṽ 2 Z Apparent Power: S = Ṽ Ĩ used for machine ratings. Average Power: P = R(S) = Ṽ Ĩ cosφ (in Watts) Reactive Power: Q = I(S) = Ṽ Ĩ sinφ (in VARs) Power engineers always useṽ andĩ and omit the ~. 2 V. E1.1 Analysis of Circuits ( ) AC Power: / 11
100 Summary : S = ṼĨ = P +jq whereṽ = V rms = 1 For an impedance Z: S = Ĩ 2 Z = Ṽ 2 Z Apparent Power: S = Ṽ Ĩ used for machine ratings. Average Power: P = R(S) = Ṽ Ĩ cosφ (in Watts) Reactive Power: Q = I(S) = Ṽ Ĩ sinφ (in VARs) Power engineers always useṽ andĩ and omit the ~. 2 V. Tellegen: In any circuit b S b = 0 b P b = b Q b = 0 E1.1 Analysis of Circuits ( ) AC Power: / 11
101 Summary : S = ṼĨ = P +jq whereṽ = V rms = 1 For an impedance Z: S = Ĩ 2 Z = Ṽ 2 Z Apparent Power: S = Ṽ Ĩ used for machine ratings. Average Power: P = R(S) = Ṽ Ĩ cosφ (in Watts) Reactive Power: Q = I(S) = Ṽ Ĩ sinφ (in VARs) Power engineers always useṽ andĩ and omit the ~. 2 V. Tellegen: In any circuit b S b = 0 b P b = b Q b = 0 : add parallel C to generate extra VARs E1.1 Analysis of Circuits ( ) AC Power: / 11
102 Summary : S = ṼĨ = P +jq whereṽ = V rms = 1 For an impedance Z: S = Ĩ 2 Z = Ṽ 2 Z Apparent Power: S = Ṽ Ĩ used for machine ratings. Average Power: P = R(S) = Ṽ Ĩ cosφ (in Watts) Reactive Power: Q = I(S) = Ṽ Ĩ sinφ (in VARs) Power engineers always useṽ andĩ and omit the ~. 2 V. Tellegen: In any circuit b S b = 0 b P b = b Q b = 0 : add parallel C to generate extra VARs : V i N i and N i I i = 0 (implies S i = 0) E1.1 Analysis of Circuits ( ) AC Power: / 11
103 Summary : S = ṼĨ = P +jq whereṽ = V rms = 1 For an impedance Z: S = Ĩ 2 Z = Ṽ 2 Z Apparent Power: S = Ṽ Ĩ used for machine ratings. Average Power: P = R(S) = Ṽ Ĩ cosφ (in Watts) Reactive Power: Q = I(S) = Ṽ Ĩ sinφ (in VARs) Power engineers always useṽ andĩ and omit the ~. 2 V. Tellegen: In any circuit b S b = 0 b P b = b Q b = 0 : add parallel C to generate extra VARs : V i N i and N i I i = 0 (implies S i = 0) For further details see Hayt et al. Chapter 11. E1.1 Analysis of Circuits ( ) AC Power: / 11
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