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- たつや すみだ
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3 e jωt e jωt e jωt
4 OFF ( 1) ( 2) ( 1) RC ( 2) RC ( 3) RL ( 4) RL R, L, C
5 ( Q ) Q (Quality Factor) RLC Q R RLC Q R Q LC RLC () () () (Dot convention) k () ()
6 () Y, Z, K, H, G Y Z K ( 1) H ( 2) G
7 RL RC RLC A 227 A A A A.4 j A A A.7 e jθ A.8 e x A.9 e jθ
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9 9 1 () R V I V = RI. (1.1) 1.3 R 1, R 2, R R S R S = R 1 + R 2 + R 3. (1.3) 1 I = V 1 R 1, I = V 2 R 2, I = V 3 R 3. (1.4) 1.2 I R 1 V 1 = R 1 I Ω, R V R 2 V 2 = R 2 I S, G R 3 V 3 = R 3 I G = 1 R. (1.2) I I V = RI V I = V/R V R S V = R S I
10 10 1 V I I I I 1 V = R 1 I 1 E V R L J V R L I 2 R 1 V = R 2 I 2 DC voltage source Load Resistance DC current source Load Resistance I 3 R 2 V = R 3 I V R 3 R P V = R P I (1.6) R P = I R P = R 2R 3 + R 1 R 3 + R 1 R 2 R 1 + R 2 + R 3. (1.9) V = R P I R P 1.3 V = V 1 + V 2 + V 3. (1.5) R S V = R S I (1.3) ( ) 1.4 R 1, R 2, R R P 1 R P = 1 R R R 3. (1.6) V = R 1 I 1, V = R 2 I 2, V = R 3 I 3. (1.7) I = I 1 + I 2 + I 3. (1.8) ( ) 1.5 G 1, G 2, G 3, G P (1) (2) I 1 = G 1 V, I 2 = G 2 V, I 3 = G 3 V. (1.10) I = I 1 + I 2 + I 3 I = (G 1 +G 2 +G 3 )V. (1.11) G P = G 1 +G 2 +G 3. (1.12)
11 () I V R L E R i V i I V R L R L V ( ) R L I = V /R L (R L = 0 () ) R L I ( ) R L V = R L J (R L = () ) R i I V i = R i I E V R L R i DC voltage source Load Resistance 1.5 R L R i E R i R L E = V i + V (1.13) R i R L V i = R i I, V = R L I. (1.14) E = (R i + R L )I, I = I V = R L I V = R L R i + R L E = E R i + R L. (1.15) R i R L E (1.16) V R L R L R i R i /R L 1 V E (1.17) V R L
12 V E 0.1 Ω R i I V R L (a) Ω 1.5 V Voltage (V) Current (A) E = 1.5 V R i = 0.1 Ω V I R i = 0.1 Ω E = 1.5 V 1.6 V V = I R L R i + R L E. (1.18) I = V R L (1.19) Voltage (V) R L (Ω) (b) R L (Ω) Current (A) R L V I V 2.8 Ω V 1.49 V R i V I i = V /R i J I Ω 1.5 V V I R (a) (b) R L R i R L R i
13 I J I i R i V R L DC current source Load Resistance 1.8 J R i I ( R L ) J = I i + I (1.20) R i R L I i = V R i, I = V R L. (1.21) ( 1 J = + 1 ) V, V = R i R L J R i R L (1.22) V I = V /R L I = 1 R L J R i R L = R L R i J (1.23) I R L R L R i R L /R i 1 I J (1.24) I R L
14 (a) c d i i > 0 c d d c 1.9 (b) i i > (a) c d v v > 0 c d v d c v 1.10 (b) v + c + d v v > 0 c d v v < 0 d c v c d v i i c d c d (a) (b) 1.9 v v c d c d (a) (b) 1.10 (=) (=) ( ) v +
15 15 i source i i i (a) Normal combination of the directions for current and voltage in the case of voltage source i e drop v i i i (b) Normal combination of the directions for current and voltage in the case of passive circuit elements e example v example 1.11 (a) (b) + v e
16 16 1 Imag. z [1] R V (t) 1 I(t) 45 o V (t) = RI(t). O Real 1.12 z = cos(45 ) + jsin(45 ) [2] R 1 R 2 R S R P R S 1/R P R 1 R 2 R S = R 1 + R 2, 1 R P = 1 R R 2. 1/3 cm *2 [3] j *1 z 1 = j2.0, z 2 = j1.0 z 1 z 2 z 1 /z 2 z 1 z 1 x+jy 2 z1 z 1 z 1 () z 1 z 2 = j3.0, z 1 z 2 = j0.50, z 1 = 2.2, ( 2 ) 0.33 cm cm cm [4] z = cos(45 ) + jsin(45 ) 1.12 z 1 = 1.0 j2.0. [5] f (t) = cos(ωt) + jsin(ωt) f (t) ω( 0) t *1 j + i i +i + j *2 1/4
17 17 d f (t) = ωsin(ωt) + jωcos(ωt) dt { } = jω cos(ωt) + jsin(ωt) = jωf (t), f (t) dt = 1 ω sin(ωt) j ω cos(ωt) = 1 { } cos(ωt) + jsin(ωt) jω = 1 jω f (t). [6] f (t) = e at g(t) = e bt a b f (t)g(t), f (t) g(t), d dt f (t), f (t)g(t) = e at e bt = e (a+b)t, f (t) g(t) = eat e bt = e(a b)t, d dt f (t) = d dt eat = ae at, f (t) dt = e at dt = 1 a eat. f (t) dt. Amplitude (arb. units) 1 f(t) 0 g(t) ωt (degree) 1.13 [7] f (t) = sin(ωt) g(t) = sin(ωt + 90 ) f (t) g(t) f (t) g(t) g(t) f (t)
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19 19 2 ω *1 V m I m R V m = R I m ( ) L V m = ωl I m 90 C V m = I m ωc 90 *1 ω i(t) v(t) = Ri(t) 2.1 [1] (resistor) (resistor) 2.1 [1] v(t) i(t) v(t) = Ri(t). (2.1) R (resistance) Ω, Ohm (conductance) ( S (Siemens)) (inductor) (inductor) 2.1 [2]
20 第 2 章 交流回路素子とその性質 抵抗 コイル コンデンサ 20 i(t) i(t) di(t) v(t) = L dt v(t) = 1 i(t) dt C 図 2.2 コイル (インダクタ) [2] 図 2.3 コンデンサ (キャパシタ) [3] はインダクタである コイルの両端に印加された電圧 費電力が負である とは 電力がその回路素子から供給 v(t) と抵抗に流れる電流 i(t) の間には 以下の関係があ されることを意味する 無から電力が供給されることは り ファラデーの電磁誘導の法則から導き出されるもの 無いので この状況は 回路素子に投入した電力がその である 回路素子で反射されてしまうことを意味する 本節で di(t) v(t) = L. dt (2.2) は 抵抗 コイル コンデンサの各素子に対してこのこ とを検証する ここで L をインダクタンス (inductance) という 単位 は H (ヘンリー, Henry) である なお 交流回路では この反射を抑制し 効率良く電 力を負荷に供給するための方策をとることになる この 電磁誘導による電圧は 電磁気学的には 誘導起電 力 即ち 起電力 である 従って 電磁気学的に見れ 方策を理解するためには 本講義で学ぶ交流回路理論の 学習が必要なのである ば コイルは電源のような能動素子として扱うべき素子 である しかし 電気回路では コイルを抵抗と同じ範 疇の受動素子として扱い そこに発生する誘導起電力を 受動素子の両端の電圧 即ち 電圧降下 として扱う 抵抗 抵抗 R に流れる電流を i(t) とするとき 抵抗での消 費電力 p R (t) は次式で与えられる このように扱う理由については 第 8 章の相互インダク p R (t) = R i(t)2. タンスの豆知識を参照されたし 従って 抵抗での消費電力は常に正であることがわかる コンデンサ (capacitor) コンデンサ (capacitor) は 図 2.3 の写真に示すよう な回路素子であり [3] 電流の積分に比例した電圧が端 子間に現れる素子である 日本語ではコンデンサである が 英語ではキャパシタである コンデンサの両端の電 圧 v(t) とそこに流れる電流 i(t) の間には 以下の関係が コイルとコンデンサ コイル L に流れる電流を i(t) とするとき コイルで の消費電力 p L (t) は 天下り的であるが 次式で与えら れる p L (t) = ある いわゆるコンデンサの充電の式である v(t) = 1 C i(t) dt. (2.4) (2.3) ここで C をキャパシタンス (capacitance) という 単 位は F (ファラッド, Farad) である 2.2 回路素子における電力とエネルギー ) ( d 1 Li(t)2. dt 2 (2.5) また コンデンサ C の電圧を v(t) とするとき コンデ ンサでの消費電力 p C (t) は 天下り的であるが 次式で 与えられる ) ( d 1 2 Cv(t). p C (t) = dt 2 (2.6) これらの式より 具体的な i(t) や v(t) の波形がわって いなくても コイルとコンデンサについては 抵抗と異 抵抗の場合には 電力は消費されるだけ 即ち電力は なり i(t) や v(t) の時間的変化の仕方によっては消費電 常に正であるが コイルとコンデンサの場合には 電力 力が負になり得る ということが読み取れると思う 即 が消費されるだけとは限らず 負になることもある 消 ち 抵抗では交流の場合も電力は消費だけであるが コ
21 v(t) = V m sin ωt i(t) = v(t) R v(t) = V m sin ωt i(t) = 1 L v(t) dt Current (A) Voltage (V) Voltage Freq. = 60 Hz R = 1 kohms Phase (degree) Power Current Power (W) Current (A) Voltage (V) Voltage Freq. = 60 Hz L = 100 mh Current Phase (degree) 270 Power Power (W) 2.5 V m = 100 V f = ω/(2π) = 60 Hz R = 1 kω 2.7 V m = 100 V f = ω/(2π) = 60 Hz L = 100 mh v(t) v(t) = V m sinωt (2.7) i(t) i(t) = v(t) R (2.8) = V m sinωt R (2.9) = I m sinωt (2.10) ω I m = V m R θ = v(t) v(t) = V m sinωt (2.11) i(t) i(t) = 1 v(t) dt (2.12) L = V m cosωt (2.13) ωl ( = I m sin ωt π ) (2.14) 2 cos sin ω
22 22 2 v(t) = V m sin ωt i(t) = C dv(t) dt v(t) 2.8 v(t) = V m sinωt (2.15) i(t) Voltage Current 40 i(t) = C d v(t) (2.16) dt Current (A) Voltage (V) 50 0 Power 20 0 Power (W) = ωcv m cosωt (2.17) ( = I m sin ωt + π ) (2.18) 2 sin cos Freq. = 60 Hz C = 1000 uf Phase (degree) ω I m = ωc V m 2.9 V m = 100 V f = ω/(2π) = 60 Hz C = 1000 µf I m = V m ωl θ = π 2 = θ = + π 2 = * 2 *2 () () = 90 (t = 0 )
23 i(t) = I m sin ωt v(t) = R i(t) i(t) = I m sin ωt v(t) = L di(t) dt Current (A) Voltage (V) Freq. = 60 Hz R = 10 Ohms 90 Current 180 Phase (degree) Power Voltage Power (W) Current (A) Voltage (V) Freq. = 60 Hz L = 10 mh 90 Current 180 Power Phase (degree) Voltage Power (W) 2.11 I m = 1 A f = ω/(2π) = 60 Hz R = 10 Ω 2.13 I m = 1 A f = ω/(2π) = 60 Hz L = 10 mh 2.4 * R 2.10 i(t) = I m sinωt (2.19) v(t) v(t) = R i(t) (2.20) = R I m sinωt (2.21) = V m sinωt (2.22) *3 ω V m = R I m θ = 0 = L 2.12 i(t) = I m sinωt (2.23) v(t) v(t) = L d i(t) (2.24) dt = ωl I m cosωt (2.25) ( = V m sin ωt + π ) (2.26) 2 cos sin
24 i(t) = I m sin ωt v(t) = i(t) dt C ω V m = I m ωc θ = π 2 = Current (A) Voltage (V) Current Freq. = 60 Hz C = 1000 uf Voltage Phase (degree) 270 Power I m = 1 A f = ω/(2π) = 60 Hz C = µf ω V m = ωl I m θ = + π 2 = C 2.14 Power (W) i(t) = I m sinωt (2.27) v(t) v(t) = 1 i(t) dt (2.28) C = 1 ωc I m cosωt (2.29) ( = V m sin ωt π ) (2.30) i(t) v(t) v(t) = Ri(t) + L d dt i(t) + 1 i(t) dt. (2.31) C i(t) = I m sinωt v(t) v(t) = v f + v s. (2.32) cos sin v f = A 1 e s 1t + A 2 e s 2t (2.33) v s = V m sin(ωt + θ) (2.34)
25 v(t) v R (t) v L (t) v C (t) R L C i(t) 2.16 v f t 0 v s t v(t) = V m sin(ωt + θ) V m θ i(t) (2.31) ( v(t) = I m [R sinωt + ωl 1 ) ] cosωt. (2.35) ωc V m θ ( V m = I m R 2 + ωl 1 ) 2, (2.36) ωc ( ωl 1 ) θ = tan 1 ωc (2.37) R
26 26 2 sin θ cos θ θ definition of sin and cos L [4] The term inductance was coined by Oliver Heaviside in February It is customary to use the symbol L for inductance, in honor of the physicist Heinrich Lenz. In the SI system the measurement unit for inductance is the henry, H, named in honor of the scientist who discovered inductance, Joseph Henry. (current) I i [5] The conventional symbol for current is I, which originates from the French phrase intensite de courant, or in English current intensity. This phrase is frequently used when discussing the value of an electric current, but modern practice often shortens this to simply current. The I symbol was used by Andre-Marie Ampere, after whom the unit of electric current is named, in formulating the eponymous Ampere s force law which he discovered in The notation travelled from France to Britain, where it became standard, although at least one journal did not change from using C to I until sin cos sin, cos x y sin cos d dθ sinθ = cosθ = sin (θ + π 2 ). (2.38) -1 θ -cos θ π sin(- + θ ) 2 π 2 θ θ π sin( + θ ) 2-1 cos θ π 2 θ 1 π -cos θ = sin(θ - ) 2 π cos θ = sin(θ + ) cos sin 90 sin cos ( sinθ dθ = cosθ = sin θ π ). (2.39) 2 90 inductor capacitor
27 inductor 2 capacitor condensare ( ) condenser potassium kalium ( ) sodium natrium ( ) titanium 3 aluminum 3 magnesium 3 germanium neon 3 xenon 3 uranium 3 ion anion cation (kation)
28 28 2 [1] L i(t) v(t) [3] sinωt t cos sin cos cos sin ( ) v(t) = L d i(t). (2.40) dt 8 d (ωt dt sinωt = ωcosωt = ωsin + π ). 2 (2.44) sinωt dt = 1 ω cosωt = 1 ( ω sin ωt π ). 2 (2.45) ±90 [2] C i(t) v(t) v(t) q(t) q(t) = Cv(t). (2.41) q(t) = i(t) dt. (2.42) v(t) = 1 i(t) dt. (2.43) C
29 29 1. () R L C v(t) i(t) R v(t) = Ri(t) 2. () R L C i(t) = I m sinωt L R v(t) = RI sinωt v(t) = L d dt i(t) C v(t) = 1 C i(t) dt L ( v(t) = ωl I m sin ωt + π ) 2 90 C v(t) = 1 ( ωc I m sin ωt π ) 2 90
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31 31 [1] [2] [3] [4] [5]
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33 33 3 a(t) = A m sin(ωt + θ) a(t) = A m e j(ωt+θ) a(t) = A m e j(ωt+θ) jωt A m A e = A m / 2 a(t) A *1 A = A e e jθ V I v(t) = R i(t) V = R I v(t) = L di dt V = jωl I v(t) = 1 C i dt V = 1 jωc I e jωt i(t) = I m sinωt i(t) = I m cosωt (3.1) i(t) = I m e jωt = I m exp(jωt) (3.2) *2 e jωt 3.2 e jωt e jωt = exp(jωt) = cosωt + jsinωt. (3.3) sin cos exp sin exp exp sin cos exp exp cos *1 A = A e θ θ *2 exp() e ()
34 34 3 : R : sin cos exp sin cos cos sin 90 cos sin i(t) = I m e jωt v(t) = L d i(t) (3.8) dt exp ( cos(ωt) = sin ωt + π ) 2 e j(ωt+ π 2 ) = e jωt e j π 2 = je jωt. (3.4) v(t) = jωl I m e jωt (3.9) v(t) = jωl i(t) (3.10) cos sin sin cos 90 sin cos exp ( sin(ωt) = cos ωt π ) 2 e j(ωt π 2 ) = e jωt e j π 2 = je jωt. (3.5) 3.3 e jωt e jωt sin cos (3.10) : ωl : 90 * v(t) = 1 C i(t) = I m e jωt i(t) dt (3.11) v(t) = 1 jωc I me jωt (3.12) exp v(t) = [ ]i(t) (3.6) v(t) = 1 jωc i(t) (3.13) (3.13) : 1 ωc : 90 * 4 v(t) = Ri(t) (3.7) R exp *3 j *4 j
35 exp e jωt e jωt e jωt i(t) = I m e j(ωt+θ) (3.14) I = I m e jθ (3.15) ω 3.5 V = R I, (3.16) V = jωl I, (3.17) V = 1 I. jωc (3.18) i(t) = I m e jωt I = I m (3.19) v(t) = V m e j(ωt+θ) V = V m e jθ (3.20) v(t) = R i(t) (3.21) exp V m e jθ e jωt = R I m e jωt (3.22) e jωt V m e jθ = R I m (3.23) V = R I. (3.24) v(t) = jωl i(t) (3.25) exp e jωt V m e jθ e jωt = jωl I m e jωt (3.26) V m e jθ = jωl I m (3.27) V = jωl I. (3.28) v(t) = 1 i(t) (3.29) jωc exp V m e jθ e jωt = 1 jωc I me jωt (3.30)
36 36 3 e jωt V m e jθ = 1 jωc I m (3.31) V = 1 jωc I. (3.32) 3.6 () A e A m 2 A e = A m 2. (3.33) i(t) = I m e j(ωt+θ) i(t) = I m e j(ωt+θ) I = I e e jθ, I e = I m 2 (3.34) V I v(t) = R i(t) V = R I, (3.37) v(t) = L di V = jωl I, (3.38) dt v(t) = 1 i dt V = 1 I. (3.39) C jωc 3 3 v(t) = V m sin(ωt + θ) v(t) = V m e j(ωt+θ). V = V e e jθ, V e = V m 2. (3.35) i(t) = I m sin(ωt + ϕ) i(t) = I m e j(ωt+ϕ). I = I e e jϕ, I e = I m 2. (3.36)
37 ωl ωl /ωC /ωC i(t) v(t) I V R R i(t) = I m sin(ωt + θ) v(t) = RI m sin(ωt + θ) Im V R I I = I e e jθ 3.1 i(t) v(t) I V L L i(t) = I m sin(ωt + θ) V jωl I Re v(t) = ωl I m sin(ωt + θ + 90 o ) Im I I e e jθ 3.2 i(t) i(t) = I m sin(ωt + θ) v(t) I V C Im Re C v(t) I = m sin(ωt + θ 90 o ) ωc I I e e jθ V I jωc Re 3.3
38 R V I P P = V I = RI 2 (3.40) 2 1/2 (= ) *5 (=) R v(t) = V m sinωt i(t) = I m sinωt p(t) p(t) = v(t)i(t) (3.41) ( T = 2π/ω) P ( ) P = 1 T T 0 1 = V m I m T 1 = V m I m 2 v(t)i(t) dt T 0 sin 2 ωt dt (3.42) 1/2 *5 1/2
39 39 cos v(t) v(t) = ωla sin(ωt + θ). (3.49) sin exp 1: sin exp i(t) = A sin(ωt + θ) v(t) sin exp v(t) v(t) = L di dt (3.43) sin v(t) exp v(t) = ωla cos(ωt + θ). (3.44) i(t) = Ae j(ωt+θ) (3.45) v(t) = jωlae j(ωt+θ) (3.46) sin exp exp v(t) = jωla cos(ωt + θ) ωla sin(ωt + θ) (3.47) exp sin sin exp 2: cos exp i(t) = A cos(ωt + θ) v(t) cos exp v(t) exp i(t) = Ae j(ωt+θ) (3.50) v(t) = jωlae j(ωt+θ) (3.51) cos exp exp v(t) = jωla cos(ωt + θ) ωla sin(ωt + θ) (3.52) exp cos sin exp 3: i(t) = A sin(ωt + θ) i(t) ( w(t) = K L di ) 2 (3.53) dt sin w(t) = K(ωLA) 2 cos 2 (ωt + θ) (3.54) 2 1 cos[2(ωt + θ)] = K(ωLA) 2 (3.55) exp w(t) = K(jωLA) 2 exp[2j(ωt + θ)] (3.56) = K(ωLA) 2 cos[2(ωt + θ)] (3.57) jk(ωla) 2 sin[2(ωt + θ)] (3.58) sin sin v(t) = L di dt (3.48)
40 40 3 j j j j j π/2 (90 ) j π/2 (90 ) jωl j j exp ( j = e j π 2 = exp j π ) 2 j (3.59) v(t) = jωl i(t) (3.60) = e j π 2 ωl I m e jωt (3.61) = ωl I m e j(ωt+ π 2 ) (3.62) π/2 1/(jωC) j 1 ( j = π e j 2 = exp j π ) 2 (3.63) π/2 sin(ωt+π/2) sin(ωt π/2) j π/2 (90 ) 1/ 2 2 P = (1/2)V m I m 1/2 1/2 P = V I () P = V I V m I m V m = [ ]I m V = [ ]I V = V m /2 I = I m P = V I V = [ ]I V = [ ]I (root-mean-square: rms) 2 OK v(t) = V m f (t) i(t) = I m g(t) f (t) g(t) T P = 1 T T 0 = V m I m 1 T v(t)i(t) dt T 0 V e = V m / 2 I e = I m / 2 f (t)g(t) dt. (3.64)
41 41 1 V e = V m T 1 I e = I m T T 0 T 0 f (t)g(t) dt, (3.65) f (t)g(t) dt (3.66) R f (t) = g(t) 1 V e = V m T 1 I e = I m T T 0 T 0 f (t) 2 dt, (3.67) g(t) 2 dt (3.68) RMS 1 V e = T 1 I e = T T 0 T 0 v(t) 2 dt, (3.69) i(t) 2 dt (3.70) () (root-mean-square: rms) RMS RMS RMS RMS
42 42 3 [1] ( 1) i(t) = I m sinωt L d dt i(t) i(t) 90 1 i(t) dt i(t) 90 C 1 C L d dt i(t) = ωl I m cosωt ( = ωl I m sin ωt + π ), 2 i(t) dt = 1 ωc I m cosωt = 1 ωc I m sin [2] ( 2) ( ωt π 2 i(t) = I m cosωt L d dt i(t) i(t) 90 1 i(t) dt i(t) 90 C 1 C ). L d dt i(t) = ωl I m sinωt ( = ωl I m cos ωt + π ), 2 i(t) dt = 1 ωc I m sinωt [3] j = 1 ωc I m cos ( ωt π 2 () j () j ). j = e j π 2 π 2 90 j = e j π 2 π 2 90 [4] e jωt i(t) = I m e jωt 1 C L d i(t) = jωl i(t), dt 1 i(t) dt = 1 C jωc i(t) L d ( I m e jωt) = jωl I m e jωt = jωl i(t). dt (3.71) ( I m e jωt) dt = 1 jωc I me jωt = 1 i(t). jωc (3.72) [5] [1] [2] [4] i(t) i(t) = I m e jωt = I m cosωt + ji m sinωt (3.73) L d i(t) = jωl i(t) dt = e +j π 2 ωl I m e jωt = ωl I m e j(ωt+ π 2 ) ( = ωl I m cos ωt + π ) 2 ( +jωl I m sin ωt + π ), (3.74) 2 1 C i(t) dt = 1 jωc i(t) = e j π 1 2 ωc I me jωt = 1 ωc I me j(ωt π 2 ) = 1 ( ωc I m cos ωt π ) 2 1 ( +j ωc I m sin ωt π ) (3.75) 2
43 43 V = V e θ, I = I e ϕ. 1. v(t) = V m sin(ωt + θ), i(t) = I m sin(ωt + ϕ) v(t) = V m e j(ωt+θ), i(t) = I m e j(ωt+ϕ) 2. R L v(t) = Ri(t) = V = R I v(t) = d i(t) = V = jωl I dt 2. e jωt C v(t) = i(t) dt = V = 1 jωc I V = V e e jθ, I = I e e jϕ V e = V m 2, I e = I m 2.
44
45 45 4 V I V = ZI Z Z Y I = Y V : V = RI R V I () Z sin cos 4.1 V = R I (4.1) V = Z I (4.2) Z Z Z (impedance) Z Ω (, Ohm) 1 R, ωl, Ω ωc ω [rad s 1 ] [s 1 ] L [H]
46 46 4 I V 1 = Z 1 I V 2 = Z 2 I V 3 = Z 3 I V V = Z I Z 1 Z 2 Z I V = V 1 + V 2 + V 3 ( ) [H] v = Ldi/dt [V] = [H][A][s] 1 = [H] = [V][A] 1 [s]. (4.3) V = Z S I Z S ωl [s 1 ][L] = [V][A] 1 = [Ω] (4.4) I V C [F] () [F] v = 1 i dt C [V] = [F] 1 [A][s] = [F] = [V] 1 [A][s]. (4.5) 1 ωc = (ωc) 1 ([s] 1 [F]) 1 = [V][A] 1 = [Ω] (4.6) 4.2 R jωl 1 jωc R Z (1) 1 I = V 1 Z 1, I = V 2 Z 2, I = V 3 Z 3. (4.7) (2) V = V 1 + V 2 + V 3. (4.8) V = Z S I Z S Z S = Z 1 + Z 2 + Z 3. (4.9) R Z (1) V = Z 1 I 1, V = Z 2 I 2, V = Z 3 I 3. (4.10)
47 Z = R + jωl I Z 1 = R 1 Z = R + jωc V Z 2 = jωl 1 Z = R + j ( ωl ) ωc Z 3 = 1 jωc V 4.5 I I I 1 I 2 I 3 V = Z 1 I 1 Z 1 V = Z 2 I 2 Z 2 V = Z 3 I 3 Z 3 V V = Z P I Z P Z = R + jx. (4.13) R ( (resistance)) X ( (reactance)) X > 0: X < 0: () () (2) I = I 1 + I 2 + I 3. (4.11) V = Z P I Z P 1 Z P = 1 Z Z Z 3. (4.12) 4.5 V I V = RI R
48 48 4 V = ZI Z( ) V 1 V 2 V 3 I = Y 1 V 1 I = Y 2 V 2 I = Y 3 V 3 Y 1 Y 2 Y 3 I V = V 1 + V 2 + V (admittance) I = Y V. (4.14) I = Y S V I V Y S V Y Z Y = 1 Z. (4.15) S (, Siemens) Y S = 1 Y Y Y 3 (4.17) Y P = Y 1 + Y 2 + Y 3 (4.18) 4.8 Y = G + jb (4.16) G (conductance) B (susceptance) B B > 0: B < 0: z = x + jy
49 4.11. 極座標形式の計算例 49 V ಐᑙṾ ٳ ǽ ಏǽɪʀɑǛ ǃಏ FGITGG DŽǺǹ DzǵǓȚǢǷȡ ǽ I I 1 = Y1 V ǢǷ V Y1 I 2 = Y2 V V I 3 = Y3 V ಐᑙṾǷ ᏩಐᑙṾǽ ၁ ɤɇɻ ᏩಐᑙṾ Y2 V Y3 కકȡકȅȑǽǾ ǃሱ ᄋઆDŽȡ ᕧǺǨȚǢǷ V I 図 4.8 関数電卓による直角座標系と極座標系の変換例 I = YPV 4.11 V YP 極座標形式の計算例 極座標形式の計算例を以下に示す 例えば かけ算の 場合には 図 4.7 アドミタンスの並列合成 (4 15 ) (2 30 ) = 8 45 (4.19) となる 割り算の場合には 指数関数形式 4 15 = z = rejθ = r exp[jθ ] これは極座標形式の一種であるが 指数関数を使う ので 本講義では 指数関数形式 と呼ぶことにす る この場合の θ の単位はラジアンが用いられる 数値を扱わない理論計算の時には この形式が良く 用いられるが 実際の工学的問題の場合には 偏角 が π 何倍などという形では出てこないので 次に示 すような角度を度で表した方式が良く用いられる 例えば 偏角が 1 ラジアン と言われても どれく (4.20) となる 関数電卓によっては こうした極座標表記と直角座標 表記の変換をしてくれるものがある 電気回路を扱う場 合には そのような電卓を良く使う 最近では スマホ アプリで無料のものがあるので利用するとよい なお 関数電卓で角度を扱うときには 電卓の角度の単位の設 定に気をつけること らいの角度なのかがすぐにわからないハズである 4.12 極座標形式 z = r θ この場合の θ の単位は度 が用いられる 度を用 いるのは数値を扱う工学的問題を対象とするからで ある 先ほどの 1 ラジアンは 度で表すと 約 57 である これならば だいたい 60 なので 三角定 規を思い浮かべれば どれくらいの角度なのかがす ぐにピンとくるハズである 交流電源の内部インピーダンスと内部 アドミタンス 現実の直流電源には内部抵抗なるものが潜んでいるこ とを以前に述べた 交流の電源の場合にも 現実の交流 電源には 内部インピーダンスが潜んでいる 図 4.9 に示すように 交流電源を交流電圧源としてみ た場合 電源端子から電源側を見たときに V = Z I を満 たす Z を電源の内部インピーダンスという なお 内部 インピーダンスがいくらか ということを計算したりす
50 50 4 Power Source V I E Z V I E Z V I OFF Z V I (a) OFF in the case of a voltage source 4.9 ( ) J Y V I OFF Y V I I I Power Source V J Y V (a) OFF in the case of a current source 4.10 ( ) "OFF" 4.10 I = Y V Y "OFF" 4.13 OFF OFF E = (a) J = (b) 4.14 V = ZI OFF (a) ( ) (b) ( ) I I V? V Z = 1 Y 4.12 E = ZJ, Y = 1 (4.21) Z 4.15
51 V = ZI, I = Y V (4.22) Z Y
52 Z 2 Z 2 Z 2 = 0 Z 2 Z 2 Z 2 = 0 Z 2 = Z 1 Z 2 Z 3 Remove Z 2 Z 1 Z 3 Z 2 = 0 Z 1 Z 2 Z 3 Remove Z 2 Z 1 Z 2 = Z
53 53 [3] R 1 R 2 R 3 R S [1] i(t) = I m sin(ωt + θ) R S = R 1 + R 2 + R 3 i(t) [4] R 1 R 2 R 3 R P I = I e e jθ 1 (/ 2) I e = I m [2] v(t) = V m sin(ωt + θ) v(t) V = V e e jθ = R S R 1 R 2 R 3 [5] z = re jθ z = r θ r = 2.0 θ = π 4 = 45 2 (/ 2) V e = V m [6] z = Imag. Imag. I m 2 θ I V m 2 θ V O Real O Real 4.14 i(t) = I m sin(ωt + θ) 4.15 v(t) = V m sin(ωt + θ)
54 π
55 55 j2 A. j 1. R = 1 Ω L = 1 mh C = 500 µf ω = 1000 rad/s 2 R R = 1.0 Ω L jωl = j (1000) ( ) = j 1.0 Ω C 1 jωc = j 1 (1000) ( ) = j 2.0 Ω 2. R, L, C Z 2 Z Z = (1.0 j1.0) Ω Z = 2 45 = ( ) Ω j -j R = 1 Ω L = 1 mh C = 500 µf ω = 1000 rad/s Z Z Z V m = 2.0 V ω = 1000 rad/s Z () () = = 2, = = V V e = V m 2 = 2 2 = 2 = 1.4 V V = 2 0 = ( ) V Z = ( 2 45 ) Ω I = V Z = = ( ) A Z
56 56 4 j2 j I V j -j R = 1 Ω L = 1 mh C = 500 µf ω = 1000 rad/s V m = 2 V V = ( ) V, I = ( ) A, V e = V = 1.4 V, V m = 2V e = 2.0 V, θ = 0, I e = I = 1.0 A, I m = 2I e = 1.4 A, ϕ = Voltage (V) and Current (A) 3 2 v(t) 1 i(t) Phase (degree) 4.18 R = 1 Ω L = 1 mh C = 500 µf ω = 1000 rad/s V m = 2 V V = V e θ v(t) = V m sin(ωt + θ), I = I e ϕ i(t) = I m sin(ωt + ϕ) 4.18
57 57 B. 1. R = 1 Ω L = 0.5 mh C = 1000 µf ω = 1000 rad/s 2 R 1 R = 1.0 S L 1 jωl = j 1 (1000) ( ) = j 2.0 S C jωc = j (1000) ( ) = j 1.0 S 2. R L C Y Y Y = (1.0 j1.0) S Y = 2 45 = ( ) S Y 4.19 j2 j j -j R = 1 Ω L = 0.5 mh C = 1000 µf ω = 1000 rad/s Y 3. Y V m = 2.0 V ω = 1000 rad/s Y () () = = 2, = = V V e = V m 2 = 2 2 = 2 = 1.4 V V = 2 0 = ( ) V Y = ( 2 45 ) S I = V Z = V Y = ( 2 0 ) ( 2 45 ) = ( ) A Y
58 58 4 j2 j V j I -j R = 1 Ω L = 0.5 mh C = 1000 µf ω = 1000 rad/s V m = 2.0 V V = ( ) V, I = ( ) A, V e = V = 1.4 V, V m = 2V e = 2.0 V, θ = 0, I e = I = 2.0 A, I m = 2I e = 2.8 A, ϕ = Voltage (V) and Current (A) 3 2 i(t) 1 v(t) Phase (degree) 4.21 R = 1 Ω L = 0.5 mh C = 1000 µf ω = 1000 rad/s V m = 2 V V = V e θ v(t) = V m sin(ωt + θ), I = I e ϕ i(t) = I m sin(ωt + ϕ) 4.21
59 Z Y Z Y 4 E V p I C R V R θ E a b p I V C = jωc V R =RI V b V a V C I Z = R + 1 jωc 2 1 Z = R + ( ωc ) 2 arg Z = tan 1 1 ( ωcr) 1 jωc θ R Z RC 5.1 R C Z Z = R + 1 jωc = R j 1 ωc. (5.1) Z Z arg Z θ ( ) 1 2 Z = R 2 +, (5.2) ωc ( arg Z = tan 1 1 ). (5.3) ωcr 5.1 RC RC 5.2 R C Y Y = 1 + jωc. (5.4) R Y Y argy θ ( ) 1 2 Y = + (ωc) 2, (5.5) R argy = tan 1 (ωcr). (5.6)
60 60 5 E I I R R C I R = E/R I C = jωce I Y = 1 C R + jωc Y = ( 2 1 R ) + (ωc) 2 arg Y = tan 1 (ωcr) E I I R R L I R = E/R I L = E/( jωl ) 1 1 Y = I L R + jωl Y = ( + R ) arg Y = tan 1 R ( ωl ) ( ωl ) 2 I θ I R I C E jωc θ Y 1/R θ I I R I L E 1 jωl θ Y 1/R 5.2 RC 5.4 RL E θ I E R L V R V R = RI Z = R 2 + (ωl) 2 V L = jωli arg Z = tan 1 ωl ( R ) V L I Z = R + jωl jωl 5.3 RL RL 5.3 R L Z θ Z R Z = R + jωl. (5.7) Z Z arg Z θ Z = arg Z = tan 1 ( ωl R R 2 + (ωl) 2, (5.8) ). (5.9) RL 5.4 R L Y Y = 1 R + 1 jωl = 1 R j 1 ωl. (5.10) Y Y argy θ ( ) 1 2 ( ) 1 2 Y = +, (5.11) R ωl ( argy = tan 1 R ). (5.12) ωl R C RC I E R R V R R I C C V C C I
61 E I C R a b p I V C = jωc V R =RI E I a R V R =RI b I C V C = jωc 5.5 CR RC p (= 0 V) b b p C R R C a b p a b p I V C = jωc V R =RI V R =RI I V C = jωc V p (a) V p V b (b) V R θ E V R θ E V b I V C V a I V C Va p a b b *1 p b V bp p b V b p V bp V b p V bp V b p V bp p b b p *2 5.6(a) 5.6(b) R C I (a) (b) R C V R = RI, (5.13) V C = 1 I. jωc (5.14) *1 b b b b ) *2 5.6 R C R C b b C R (a) 5.6(b) p a a V ap ( ) 5.6(a) p R 5.6(b) C R (5.13) C (5.14) b b
62 (a) p b V bp = V R I V bp 5.6(b) p b V b p = V C 90 I V b p 90 a C R E b V b R C p 5.8 (Phase-Shifter) V b V R V C V p E V a V C V V R V b Vb Vb Vp Va Vp Va Vb Vb Vb V b Vp Va Vp Va 5.3 ( 1) Vb Vb (phase-shifter) 5.8 R ωc E V 5.9 V R V C p b V bp p b V b V = V bp V b p = E V E V E V 5.9 R C 1 V E o b 2 R R ( 2) 5.4 ( 2) 5.10 V bo () E E /2 V bo E R 1 V bo = R E E 2 jωc 1 R 1 1 jωc 1 E = R jωc 1 = jωc 1R 1 1 E jωc 1 R (5.15)
63 Z 1 Z 3 I 5 Z 5 V 5 Z 2 Z 4 E A R 1 R 3 v 2 v 1 B v BC v 3 v 4 R 2 R 4 C D 5.11 E 5.5 V bo = E 2 (5.16) 5.11 Z 5 Z 1 Z 2 Z 3 Z 4 Z 1 Z 2 = Z 3 Z 4. (5.17) (Wheatstone bridge) (Maxwell bridge) (Wien bridge) R 1 R 2 = R 3 R 4 (5.18) 5.12 R 1 R 2 R 3 R 4 R 1 = R 3 R 4 R 2 (5.19) R v 3 v 4 R 3 v 3 = E, R 1 + R 3 (5.20) R 4 v 4 = E. R 2 + R 4 (5.21) v BC v BC = v 3 v 4. (5.22) v BC = 0 R 3 R 1 + R 3 R 4 R 2 + R 4 = 0. (5.23) R 1 R 2 = R 3 R 4. (5.24) R 1 R 4 = R 2 R 3
64 64 5 R 1 A R 3 C 3 R 2 C 4 R 4 E ( & ) [1] 1 Ω 10 MΩ ( 4 ) R i R i 5.13 [1] R 1 = A.BCD 10 E Ω R 1 R 2 = A.BCD 1 A 2 B 3 C 4 D R 3 /R 4 R 1 E R 1 4 R 1 = A.BCD 10 E Ω B C ( ) ω = 1 R 3 C 3 (5.25) R 1 = 2R 2 R 3 = R 4 C 3 = C ( R )( ) 1 + jωc 4 = R 1. (5.26) jωc 3 R 4 R 2 R 3 + C ( 4 + j ωr 3 C 4 R 4 C 3 1 ωr 4 C 3 ) = R 1 R 2. (5.27) R 3 R 4 + C 4 C 3 = R 1 R 2, (5.28) ωr 3 C 4 = 1 ωr 4 C 3 (5.29) R 1 = 2R 2 R 3 = R 4 C 3 = C 4 ω = 1 R 3 C 3 (5.30) R 3 C 3 ω [2] R 2 R 3 R 4 C 4 Z = R 1 + jωl 1 R 1 L 1 ω
65 L 1 R 1 A R 3 Z s L 1 C 1 1 Z s = jωl 1 + jωc1 1 = j ( ωl 1 ωc1 ) R 2 C 4 E R LC 5.15 Z p C 2 L 2 1 Z p = 1 jωc 2 + jωl2 ωl = j ( 2 1 ω ) 2 L 2 C LC [2] R 1 + jωl 1 = R 1 2 R 3, (5.31) + jωc 4 R 4 R 1 R 4 + jωl 1 R 4 = R 2 R 3. (5.32) 1 + jωc 4 R 4 R 1 R 4 + jωl 1 R 4 = R 2 R 3 (1 + jωc 4 R 4 ). (5.33) R 1 = R 2R 3, R 4 (5.34) L 1 = C 4 R 2 R 3 (5.35) R 1 L 1 R 2 R 3 R 4 C LC LC L C 5.17 LC Z s Z s = jωl jωc 1 ( = j ωl 1 1 ) ωc 1 (5.36) () 5.18 LC Z p 1 Z p = jωc jωl 2 ( ) ωl 2 = j 1 ω 2 L 2 C 2 (5.37) () 5.17 LC 5.18 LC
66 L 2 = 100 mh C 2 = 10 uf Series Zs L1 C 1 Reactance X ( ) Frequency (Hz) 5.19 LC L 1 = 100 mh C 1 = 10 uf Parallel Reactance X ( ) Zp C 2 L Frequency (Hz) LC L 1 = 100 mh C 1 = 10 µf L 2 = 100 mh C 2 = 10 µ F Z S Z P f = 1 khz
67 5.7. ( 1) RC ( 1) RC 5.21 E e(t) = E m sinωt E m = 10 2 V ω = 5000 rad/s C = 10 µf R = 10 Ω 3 *3 E I C R V C V R 5.21 RC 1. e(t) E E r θ 2. C R Z r θ 3. I r θ 4. C ()V C r θ 5. R ()V R r θ 6. E V C V R 7. I i(t) 8. e(t) i(t) 9. Z E I 1 e(t) = E m sin(ωt + θ) E = Em 2 θ E * E = = 10 0 = ( ) V. Z Z = R + 1 jωc Z 3 1 Z = 10 + j(5000) ( ) 1 = 10 j = 10 j = 10 j = 10 j20 = = ( ) Ω. I = E/Z E = (10 0 ) V Z = ( ) Ω I 4 I = E Z = = = = ( ) A. (5.38) C Z C C V C = Z C I Z C Z C = 1 jωc 1 = = j20 j = ( ) Ω V C V C = Z C I = (20 90 ) ( ) = = ( ) V.
68 V V R E V V C V (a) V R V C E (b) Voltage (V) Voltage (V) Current Current (A) 5.22 RC ( ) (a) o (b) Phase (degree) 5 R V R = RI V R 6 V R = Z R I = = = ( ) V. E V C V R E = ( ) V, V C = ( ) V, V R = ( ) V. E = V C + V R ; I = I m 2 ϕ i(t) i(t) = I m sin(ωt + ϕ). I = ( ) A I m ϕ I m = = A, ϕ = RC 10.0 Ω A I j20.0 Ω V Z Ω = (10.0 j20.0) Ω 5.24 RC Z E I 8 e(t) i(t) e(t) = 10 2sinωt = 14.1sinωt V, i(t) = 0.632sin(ωt ) A Z E I Z = 10.0 j20.0 = ( ) Ω, E = ( ) V, I = ( ) A E i(t) i(t) = 0.632sin(ωt ) A.
69 5.8. ( 2) RC ( 2) RC 5.25 E e(t) = E m sinωt E m = 10 2 V ω = 5000 rad/s C = 100 µf R = 10 Ω 3 I Y Y = j(5000) ( ) = j = j0.5 = = ( ) S. E I R R C I C 3 I = Y E E = (10 0 ) V Y = ( ) S I 5.25 RC 1. e(t) E E r θ 2. C R Y r θ 3. I r θ 4. C ()I C r θ 5. R ()I R r θ 6. I I C I R 7. I i(t) 8. e(t) i(t) 9. Y E I 1 e(t) = E m sin(ωt + θ) E = Em 2 θ E 2 E = = ( ) V. Y Y = 1 R + jωc 4 I = Y E = ( ) (10 0 ) = = ( ) A. C Y C C I C = Y C E Y C Y C = jωc = j = j0.5 = ( ) S. I C 5 I C = Y C E = ( ) (10 0 ) = ( ) A. (5.39) R I R = E/R I R 6 I R = E R = = ( ) A. (5.40) I I C I R I = ( j5.00) = ( ) A, I C = j5.00 = ( ) A, I R = 1.00 = ( ) A.
70 70 5 I A I C A I R A Voltage (V) 5.26 RC ( ) I = I R + I C Voltage (V) 10 0 Current 5 0 Current (A) o 7 ; Phase (degree) I = I m 2 ϕ i(t) 5.27 RC i(t) = I m sin(ωt + ϕ). I = ( ) A I m ϕ I m = = 7.21 A, ϕ = i(t) i(t) = 7.21sin(ωt ) A. 8 e(t) i(t) e(t) = 10 2sinωt = 14.1sinωt V, i(t) = 7.21sin(ωt ) A Y E I j0.500 S S = ( j0.500) S Y S A I 78.7 E V 5.28 RC Y E I Y = ( j0.500) = ( ) S, E = ( ) V, I = ( ) A. 5.28
71 5.9. ( 3) RL ( 3) RL 5.29 E e(t) = E m sinωt E m = 10 2 V ω = 5000 rad/s L = 10 mh R = 10 Ω 3 I Z Z = R + jωl. Z = 10 + j(5000) ( ) = 10 + j50 = = ( ) Ω. E L R V L V R 3 I = E/Z E = (10 0 ) V Z = ( ) Ω I 5.29 RL 1. e(t) E E r θ 2. L R Z r θ 3. I r θ 4. L ()V L r θ 5. R ()V R r θ 6. E V L V R 7. I i(t) 8. e(t) i(t) 9. Z E I 4 I = E Z = = = = ( ) A. (5.41) L Z L L V L = Z L I Z L Z L = jωl = j = j50.0 = ( ) Ω. V L V L = Z L I = (50 90 ) ( ) = = ( ) V. (5.42) 1 e(t) = E m sin(ωt + θ) E = Em 2 θ E E = = ( ) V 5 R V R = RI V R V R = Z R I = 10 ( ) = = ( ) V. 2 6
72 V V L 20 Voltage (V) 1.0 E V R V V 5.30 RL ( ) Voltage (V) 10 0 Current Current (A) E V L V R o Phase (degree) E = ( ) V, V L = ( ) V, V R = ( ) V. (5.43) E = V L + V R ; 5.31 RL j50.0 Ω 10.0 Ω Ω = ( j50.0) Ω Z V E 78.7 I A I = I m 2 ϕ 5.32 RL Z E I i(t) i(t) = I m sin(ωt + ϕ). I = ( ) A I m ϕ I m = = A, ϕ = Z E I Z = ( j50.0) = ( ) Ω, E = ( ) V, I = ( ) A i(t) i(t) = 0.277sin(ωt 78.7 ) A. 8 e(t) i(t) e(t) = 10 2sinωt = 14.1sinωt V, i(t) = 0.277sin(ωt 78.7 ) A. 5.31
73 5.10. ( 4) RL ( 4) RL 5.33 E e(t) = E m sinωt E m = 10 2 V ω = 5000 rad/s L = 1 mh R = 10 Ω 3 Y Y = j(5000) ( ) = 0.1 j0.2 = = ( ) S. E I I R R L I L 3 I = Y E E = 10 0 V Y = S I 5.33 RL I = Y E = ( ) (10 0 ) = = ( ) A. (5.44) 1. e(t) E E r θ 2. L R Y r θ 3. I r θ 4. L ()I L r θ 5. R ()I R r θ 6. I I L I R 7. I i(t) 8. e(t) i(t) 9. Y E I 1 e(t) = E m sin(ωt + θ) E = Em 2 θ E 2 E = = ( ) V Y Y = 1 R + 1 jωl. 4 L Y L L I L = Y L E Y L Y L = 1 jωl = = j0.2 j = ( ) S. I L 5 I L = Y L E = ( ) (10 0 ) = 2 90 = ( ) A. R I R = E/R I R 6 I R = E R = = ( ) A. I I L I R I = ( ) A, I L = ( ) A, I R = ( ) A.
74 74 5 I R A 20 Voltage (V) 10 I L 10 Current A I A Voltage (V) 0 0 Current (A) 5.34 RL ( ) o I = I L + I R ; I = I m 2 ϕ i(t) i(t) = I m sin(ωt + ϕ). I = ( ) A I m ϕ I m = = 3.16 A, ϕ = i(t) Phase (degree) 5.35 RL S V E j0.200 S I A Y S = (0.100 j0.200) S 5.36 RL Y E I 5.36 i(t) = 3.16sin(ωt 63.4 ) A. 8 e(t) i(t) e(t) = 10 2sinωt = 14.1sinωt V, i(t) = 3.16sin(ωt 63.4 ) A Y E I Y = ( j0.200) = ( ) S, E = ( ) V, I = ( ) A.
75 75 Z () ( 5.23 ) () I V P k B f (x)... I m = 1 A V e = 1 V sinθ dx... I m m I m V e e V e j( i) j i e jθ e e jθ k B B Botlzmann k B di dt di dt R... R... R... R... R (...) R R R R L L L L C C C C () 5.37
76 (BOSS PH-3) [3].
77 77 [1] i(t) = I m sin(ωt + θ) I = I m 2 e jθ I = I m 2 θ [2] v(t) = V m sin(ωt + θ) [3] R, L, C Z Z V I V I Z [4] Z = R + jωl + 1 jωc V = ZI R, L, C Y Y V I V I Y V = V m 2 e jθ V = V m 2 θ Y = 1 R + 1 jωl + jωc I = Y V Imag. I m 2 θ I O Real 5.38 i(t) = I m sin(ωt + θ) Imag. V m 2 θ V O Real 5.39 v(t) = V m sin(ωt + θ)
78 78 5 A. v(t) = V m sinωt V m = 10 2 V ω = 5000 rad/s R = 1 Ω C = 400 µf RC 3 1. v(t) V V (r θ ) Z = R + 1 jωc = (1.00 j0.500) Ω Z = = ( ) Ω Z I I V I V m V e V e = V m 2 = 10.0V V = ( ) V I = V Z = = ( ) = = ( ) A RC Z Z 4. I i(t) j10 j2 j5 I -2-1 j j0.5 Z V j5 -j -j10 -j ω = 5000 rad/s R = 1 Ω C = 400 µf Z 5.41 ω = 5000 rad/s R = 1 Ω C = 400 µf Z V I
79 79 Voltage v(t) and current i(t) v(t) i(t) 26.6 o -2-1 j2 j j0.5 Y -j Phase (degree) 360 -j ω = 5000 rad/s R = 1 Ω C = 400 µf Z v(t) i(t) 5.43 ω = 5000 rad/s R = 1 Ω L = 0.4 mh Y 2. RL Y Y I m = I e 2 = = = 12.7 A i(t) = 12.7sin(ωt ) A Y = 1 R + 1 jωl = (1.00 j0.500) S 5.41 B. i(t) = I m sinωt I m = 10 2 A ω = 5000 rad/s R = 1 Ω L = 0.4 mh RL 3 Y = = ( ) S Y V V I V 1. i(t) I I (r θ ) I m I e I e = I m 2 = 10.0A V = I Y = = ( ) = = ( ) V I = ( ) A
80 80 5 j10 j5 V I j5 -j ω = 5000 rad/s R = 1 Ω L = 0.4 mh Y I V 20 Current i(t) and voltage v(t) i(t) v(t) 26.6 o Phase (degree) ω = 5000 rad/s R = 1 Ω L = 0.4 mh Y i(t) v(t) V v(t) V m = V e 2 = = = 12.7 V v(t) = 12.7sin(ωt ) V 5.45
81 81 [1] [2] [3]
82
83 Power S E I I : V(V), I(A), P(W) θ = 0 o Current S = EI Voltage 1.0 θ = + 60 o E V S = V I S V(V), I(A), P(W) Current Power P Voltage Voltage Current Q * 1 cosθ V(V), I(A), P(W) Power cos 100 % θ = + 90 o Time (ms) , 60, 90, *1 Reactive Power 6.1
84 84 6 I i(t) E Z jx Z θ R p(t) = e(t)i(t) (6.4) = 2 E I sinωt sin(ωt θ) (6.5) = EI cosθ EI cos(2ωt θ) (6.6) ω E Z = R + jx = Z exp(jθ) e(t) i(t) p(t) = e(t)i(t) I = E Z = E E = Z ejθ Z e jθ = I e jθ (6.1) e(t) i(t) e(t) = E m sinωt, (6.2) i(t) = I m sin(ωt θ) (6.3) E m = 2 E, I m = 2 I e(t) EI cosθ EI cos(2ωt θ) Z R, L, C 6.2 R, L, C e(t)i(t) p(t) = EI cosθ EI cos(2ωt θ) (6.7) = EI cosθ R EI cosθ cos2ωt EI sinθ sin2ωt (6.8) θ = 0, cosθ = 1, sinθ = 0 p(t) = EI EI cos2ωt (6.9)
85 (a) R E(V), I(A), P(W) θ = 0 o Power Current p(t) = 0 (6.12) 6.3(b) -1.0 Voltage (b) L E(V), I(A), P(W) Voltage Current Power C θ = π, cosθ = 0, sinθ = 1 2 θ = - 90 o Voltage Current p(t) = + EI sin2ωt (6.13) (c) C E(V), I(A), P(W) Power p(t) = 0 (6.14) θ = + 90 o Time (ms) (c) L 6.3 R, L, C p(t) = EI (6.10) L C 6.3(a) L θ = + π, cosθ = 0, sinθ = +1 2 p(t) = EI sin2ωt (6.11) 6.3 p(t) = EI cosθ EI cosθ cos2ωt EI sinθ sin2ωt (6.15)
86 86 6 EI cosθ EI cosθ cos2ωt = EI cosθ (1 cos2ωt) = EI cosθ (2sin 2 ωt) (6.16) cosθ % θ EI sinθ sin2ωt EI EI cosθ EI sinθ 6.4 S S = EI = ZI 2 = E 2 /Z (6.17) [W] [VA] Q = S sinθ (6.20) [var] *2 sinθ (6.21) 6.7 S P Q θ 6.4(a) 6.4(b) P = EI cosθ = S cosθ (6.18) [W] cosθ (6.19) 6.8 *2 v Var
87 (a) (b) Q S θ P S θ P Q S S P Q cos θ 6.4 (a) S P Q θ (b) S = EĪ S = EI (6.22) Ī I I E = E e j0 = E Z = Z e jθ I = I e jθ E I EI = E I e jθ (6.23) EI EI = E I e jθ (6.24)
88 (a) ( ) 6.5(b) (I m sinωt) 2 = 1 2 I2 m (1 cos2ωt) cos2ωt I 2 m 6.5(c) v v/r () i v v i v i R ± v + N i ± S (a) moving coil meter i (c) wattmeter (b) moving iron meter ± Current coil ± Voltage coil i + v 6.5 [1,2] (a) (b) (c) Z L
89 89 [A] [B] [A] ( ) R E I R I R = E R 6.6(a) () ; 2 E 2 IR = R ( ) (b) R jx R jx I L I L = I R + ji X I L 2 2 IL = 2( I R 2 + I X 2 ) EI = E I L E I L E I L ( ) ( 1 )
90 90 6 I R = E/R 2 E 2 IR e(t) E I R = E/R R R jx i R (t) (a) I = E/R je/x 2 E R jx jx 2 I 2 I R e(t) i E X (t) I R = E/R R I X = je/x jx i R (t) i(t) (b) (a) R (b) R jx [3] 6 () 85 ( (4) ) ( ) ( 1) ( 2) 1 6.9(a)(b) 1 R ωl R 2ωL 1
91 91 L L I S = I L ΔV I S = I L (a) R (b) R C E E V I R I L R L jx L ji X Source Transmission line Z S = R S + jx S Load Z L = R L // jx L Z 1 Z = jωl + 1/R + jωc = jωl + R 1 + jωcr ( R = 1 + ω 2 C 2 R 2 + jω CR 2 ) L 1 + ω 2 C 2 R 2. (6.25) 1 Z 1 C CR 2 L 1 + ω 2 C 2 = 0. (6.26) R2 C (ω 2 R 2 L) C 2 R 2 C + L = 0. (6.27) C C = R2 ± (R 2 ) 2 4(ω 2 R 2 L)L 2(ω 2 R 2 L) = R2 ± R R 2 (2ωL) 2 2(ω 2 R 2 L) = R2 ± R (R + 2ωL)(R 2ωL) 2(ω 2 R 2. (6.28) L) (R 2ωL) C R ωl R 2ωL. (6.29) [B] [4] 6.10 *3 ( ) E Z S R S jx S 6.10 R L jx I S E V *3
92 92 6 V = E V V = (R S + jx S ) I S E ΔV ΔV X I L 6.10 I S = I L O θ I L = I S V R S I S φ θ jx S I S ΔV R E V V I 6.11 V = (R S + jx S ) I L R L I R jx ji X I I L = I R ji X V = (R S + jx S ) (I R ji X ) = (R S I R + X S I X ) + j (X S I R R S I X ) = V R + j V X V R = R S I R + X S I X, V X = X S I R R S I X P jq P = V I R, Q = V I X V R V X P Q V R = R SP + X S Q, V V X = X SP R S Q. V V P Q 6.11 V V E V E V E V E V V = E V = V R + j V X V V = V E = V + V R + j V X E 2 = ( V + V R ) 2 + V 2 X V E A = R S P + X S Q, (6.30) B = X S P R S Q (6.31) { E 2 = V + A } 2 { } B 2 + (6.32) V V W = V 2 W 2 + (2A E 2 )W + (A 2 + B 2 ) = 0.
93 93 I S = I L + I C ΔV I S = I L + I C I L I C E jx S I S E Source E Transmission line Z S = R S + jx S V I R R L jx L Load Z L = R L // jx L ji X Compensator O I C I S θ V ΔV φ R S I S I L 6.12 E = V W V 6.13 a = 1, b = 2A E 2, c = A 2 + B 2 V 2 = W = b ± b 2 4ac 2a E 2 = V 2 Q (6.33) E 2 = V 2 Q aq 2 + bq + c = 0. V = W = b ± b 2 4ac ± 0 E 2a a b c a = R 2 S + X 2 S, (6.34) b = 2 V 2 X S, (6.35) c = ( V 2 + R S P) 2 + X 2 S P2 V 4. (6.36) Q E = V Q = b ± b 2 4ac. (6.37) 2a ± E = V Q C *4 Q = Q +Q C E 2 = { V + R SP + X S Q } 2 { XS P R S Q } 2 + (6.33) V V *4 Q Q C 6.13 V Q C ( I C ) I S E = V
94 94 6 *5 1 E = V V 2 = kv 2 V = kv (+ ) V 2 = kv 2 V = kv ( ) 0 50 Mvar 0 E 10 kv ( 1) [4] E = 10 kv P = 25 MW Q = +50 MVar 25/ = Q > 0 R S = Ω X S = Ω ( ) *6 V V A B (6.30) (6.31) A = R S P + X S Q = kv 2, B = X S P R S Q = kv 2 E = 10 kv (6.32) V + 10 kv 3.2 kv V = 6.78 kv A B V V = A V + j B = ( j0.8678) kv V E E = 10 kv V V V E E E = V + V = ( j0.8678) kv = ( ) kv V 5 E V 5 I L I L = P jq V = (3.686 j7.373) ka = ( ) ka PF ( ) PF = cos( ) = () 6.14(a) *5 *6 () 4 3 V E 10 kv Q C
95 95 (6.37) Q (6.34) (6.36) a = R 2 S + X 2 S = Ω 2, b = 2 V 2 X S = 2 ( ) = V 2 Ω c = ( V 2 + R S P) 2 + X 2 S P2 V 4 = (( ) ( )) 2 +( ) 2 ( ) 4 = V 4. (6.37) + Q = Mvar (+ ) = Mvar ( ). Q C = Q Q = = Mvar (+ ) = = Mvar ( ) Q ( ) Q (+ ) Q C = 56.4 Mvar I L I C I C I C = jq C V Mvar = j = j5.635 ka 10 kv I L V = 10 kv I L = P jq V = (2.500 j5.000) ka = ( ) ka I L I C ( I S ) I S = I L + I C = ( j0.635) ka = ( ) ka PF = cos14.25 = () P 25 PF = = = P 2 +Q ( 6.354) V A = R S P + X S Q 0 = kv 2 B = X S P R S Q 0 = kv 2 V = A V + j B V = = ( j1.030) kv = ( ) kv + j E E = V + V = ( j1.030) kv = ( ) kv
96 96 6 V E 6 E (b) Q Q Q = Mvar Q C = Q Q = Mvar I C I C = j Q C Mvar = j = j53.40 ka V 10 kv I C 10 I C I L V Q C I L = (2.500 j5.000) ka = ( ) ka I S I S = I L + I C = ( j48.40) ka = ( ) ka PF = cos87.04 = E V V A = R S P + X S Q 0 = kv 2 B = X S P R S Q 0 = kv 2 V = A V + j B V = = ( j4.775) kv = ( ) kv + j E E = V + V = ( j4.775) kv = ( ) kv E V ( 2) [4] 1 Q C = Q Q 0 = 0 1 V (6.30) (6.31) A = R S P = kv 2 B = X S P = kv 2 E = 10 kv (6.32) V 2 = kv 2 V = kv E = 10 kv kv 10 kv 2.5%
97 97 2.5% 1 V V R = kv, V X = kv. V = ( j1.006) kv = ( ) kv E E = V + V = ( j1.006) kv = ( ) kv I L V = kv I L = P jq V = (2.565 j5.129) ka = ( ) ka Q C = Q I C = +j5.129 ka 6.14(c) (1 m) [5] L S(1 m) = µ { 0 log 2π ( 2h a ) + µ } S. (6.38) 4 h a µ S (µ S = 1) µ 0 (µ 0 = 4π 10 7 H/m) h = 10 m a = 5.64 mm L S(1m) = H/m 1 km R S(1 km) = Ω/km, L S(1 km) = H/km 60 Hz ω = 2πf = 377 rad/s R S(1 km) = Ω/km, X S(1 km) = Ω/km (6.39) ( 2) ( j0.3922) Ω [4] 100 mm m ( ρ = Ωm) S = 100 mm 2 ( a = 5.64 mm) (1 m) R S R S(1 m) = ρ 1 S = Ω/m
98 98 6 (a) 63.4 o E = 10.0 kv 4.98 o V = 6.78 kv I R = 3.69 ka V V R = 3.18 kv V X = kv I X = 7.37 ka I L = I S = 8.24 ka I C = 5.64 ka (b) 14.3 o I S 5.91 o 63.4 o I R = 2.50 ka E = 10.0 kv V = 10.0 kv V I X = 5.00 ka I = 5.59 ka I C = 5.13 ka (c) 63.4 o 5.77 o E = 10.0 kv I S = I R = 2.57 ka V = 9.75 kv V I X = 5.13 ka I = 5.74 ka 6.14 (a) (b) (b) 1
99 99 [1]
100 V = 10 0 V I = A S () S S = V I = (10 0 ) (50 60 ) = ( ) VA S = 500 VA VA 2. % cos( 60 ) = 0.5 = 50.0% 3. P P = S cos( 60 ) = = 250 W W
101 101 [1] C. K. Alexander and M. N. O. Sadiku: Fundamentals of Electric Circuits 5th Ed. (McGraw-Hill, New York, NY, 2013) pp [2] S. Tumanski: Principles of Electrical Measurement (Taylor & Francis, New York, NY, 2006) Chapter 3 Classic electrical measurement, pp [3] : ( ) (,, 2015) p [4] T. J. E. Miller: in Reactive Power Control In Electric Systems, Ed. Timothy J. E. Miller (John Wiley & Sons, New York, NY, 1982) Chapter 1 The theory of load compensation, pp [5], : (,, 1970) p. 276.
102
103 103 7 ω ω ω 7.1 0, () Q, 7.2 ( Q ) 7.1 ω Z = R +jx Z 1/ Z 7.2 ω 7.2 Abs. admittance (x10-3 S) RLC Series Circuit Y R = 20 Ω L = 100 mh C = 10 uf Q = Angular frequency (rad/s) 2000 V(ω) Z = R + jx 7.1 ω Z 7.2 () R = 20 Ω L = 100 mh C = 10 µf ()
104 104 7 V(ω) R L C Z s = R + j ( ωl 1 ) ωc 7.3 RLC L C ω 0 = 1 LC. (7.4) ( ) 1 f 0 = 2π LC. (7.5) L C 7.3 V () Z s I = V Z s (7.1) Y s = 1/Z s *1 Z s Z s Y s Y s Z s ( Z s = R + j ωl 1 ) ωc (7.2) Z s j( ) Z s Z s (=R) Z s ω 0 ω 0 L 1 = 0. (7.3) ω 0 C * ω 0 = 1/ LC Z s Z s R Y s 1/R I V /R R, L, C L = 100 mh C = 10 µf R L C ω 0 = 1 1 = (7.6) LC = 1000 rad/s (7.7) R R 0 Ω 10 Ω 20 Ω 50 Ω R ω 0 = 1000 rad/s Y s R
第 章 交流回路素子とその性質 抵抗 コイル コンデンサ it) it) dit) vt) = L vt) = 1 it) 図. コイル インダクタ) [] 図.3 コンデンサ キャパシタ) [3] はインダクタである コイルの両端に印加された電圧 費電力が負である とは 電力がその回路素子から供給
1 ω *1 V m I m R V m = R I m ) L V m = ωl I m 9 V m = I m ω 9 *1 ω it) vt) = Rit).1 [1].1.1.1 resistor) resistor).1 [1] vt) it) vt) = Rit)..1) R resistance) Ω, Ohm conductance) S Siemens)).1. inductor)
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