UNIWERSYTET TECHNOLOGICZNO-PRZYRODNICZY

w BYDGOSZCZY

WYDZIAŁ TELEKOMUNIKACJI I ELEKTROTECHNIKI

ZAKŁAD PODSTAW PROJEKT Z

ELEKTROTECHNIKI TEORII OBWODÓW

_______________________________________

PROJEKT

Temat projektu: Obliczanie stanów nieustalonych w obwodach liniowych.

Autor projektu:

Data przyjęcia projektu:

Ocena za projekt:

Stany nieustalone w układach prądu stałego.

1. Korzystając z metody klasycznej obliczyć prąd i₁(t), i₂(t) po zamknięciu wyłącznika pierwszego i drugiego. Drugi wyłącznik zamyka się po upływie czasu τ po zamknięciu pierwszego wyłącznika. Zachodzi najpierw stan nieustalony w₁ ͢τ w₂

2. Obliczyć prąd i₂(t) metodą operatorową.

3. Narysować wykres i₁(t), na którym należy pokazać stan nieustalony spowodowany zamknięciem w₁ i w₂.

	Dane: E =100[V] L =125 [mH] = 0,125[H] R1 = R2 = R3 = 25[Ω] C = 190[µF] = 190·10^-6[F]

Zamknięcie włącznika w₁

E = U_R3(t) - U_L(t) - U_R1(t) = 0

E- R₃·I(t) - U_L(t) - R₁·I(t) = 0

U_L(t) = L$\frac{dI(t)}{\text{dt}}$

E = R₃·I(t)+ L$\frac{dI(t)}{\text{dt}}\ $+ R₁·I(t)

E = (R₁+R₃)·I(t) + L$\frac{dI(t)}{\text{dt}}$

I₁(t) = I_U + I_P(t)

I₁(t) = $\frac{E}{R1 + R3} + A e^{- \frac{t}{\tau}}$

t →  ∞ IU = ? L = $\frac{d(I_{U} = \text{const}.)}{\text{dt}} + I_{U} (R_{1} R_{3})$ = E IU·(R1+R3) = E IU= $\frac{E}{R_{1} + R_{3}}\ $=$\ \frac{100}{25 + 25} = 2\ \lbrack A\rbrack$ IU = 2 [A]

IP(t) = A·e^λt = A$e^{- \ \frac{R}{L} \bullet t}$ = A·$e^{- \frac{t}{\tau}}$ $\frac{d I_{P}(t)}{\text{dt}}$ + IP·(R1+R3) = 0 L · λ + R = 0 λ=$- \frac{R1 + R3}{L} = - \frac{25 + 25}{0,125} = - 400$ $$\tau = \left| \frac{1}{\lambda} \right| = \left| - \frac{1}{400} \right| = 0,0025\ \lbrack s\rbrack$$ I(t=0) = ? I(t=0) = A$e^{0} + \frac{E}{R_{1} R_{3}}$ 0 = A + $\frac{E}{R1 + R3}$ A = 0 - $\frac{E}{\ R1 + R3}$=$- \frac{100}{25 + 25} = - 2\left\lbrack A \right\rbrack$ IP(t) = -2·$e^{- \frac{t}{0,0025}}$ [A]

I₁(t)= -2 ·e^−400t + 2 = 2(1 − e^−400t) [A]

I₁(τ)=-2·e^λτ + 2 = −2(1 − e^{−4000, 0025}) = 1,26 [A]

Zamknięcie włącznika w2 I1^*(t) = I1P^*(t) + I1u^* I2^*(t) = I2P^*(t) + I2u^* I1u^* =  ? I2u^* =  ?
` dla t = ∞ I1u^*(t) = 0 I1u^*(t) = ? L=$\frac{d {(I}_{U} = \text{const}.)}{\text{dt}} + I_{1U} (R_{1} + R_{3})$ I1u^*(R1 + R3) = E $\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ I}_{1u}^{*} = \frac{E}{R_{1} + R_{3}}$=$\frac{100}{25 + 25} = 2\lbrack A\rbrack$
I1p^*(t) = ? I2p^*(t) = ? λ1 = ? λ2 = ? $$= {(R}_{1} + jw L) + \frac{R_{3} (R_{2} + \frac{1}{jw C})}{R_{3} + (R_{2} + \frac{1}{jw C})}$$ $${\ \ \ \ \ \ \ \ \ \ \ (R}_{1} + jw L) + \frac{R_{3} (R_{2} + \frac{1}{\lambda C})}{R_{3} + (R_{2} + \frac{1}{\lambda C})} = 0$$
25+λ·0,125+$\frac{25 (25 + \frac{1}{\lambda 190 10^{- 6}})}{25 (25 + \frac{1}{\lambda 190 10^{- 6}})}$=0 25+λ·0,125+$\frac{625 + \frac{416666,667}{\lambda})}{50 + \frac{16666,667}{\lambda})}$=0 25+λ·0,125+$\frac{650 \lambda + 416666,667}{50 \lambda + 16666,667}$=0 /·(50λ + 16666, 667) 1250·λ+416666,675+6,25·λ^2+2083,333·λ+625·λ+416666,667=0 λ1,2 =−202, 6 ± j32, 3 I1p^*(t) = [A1cos(32,3t)+A2sin(32,3t)]e^−202, 6t[A] I2p^*(t) = [B1cos(32,3t)+B2sin(32,3t)]e^−202, 6t[A] I1^*(t) = 2 + [A1cos(32,3t)+A2sin(32,3t)]e^−202, 6[A] I2^*(t) = 0 + [B1cos(32,3t)+B2sin(32,3t)]e^−202, 6t[A] t^* = 0 I1^*(t=0) = 2 + [A1cos(32,30) + A2sin(32,30)]e^−202, 60 I2^*(t=0) = [B1cos(32,30) + B2sin(32,30)]e^−202, 60 I1^*(t=0) = 2 + A1 $$\left\{ \begin{matrix} I_{1}^{*}\left( t = 0 \right) - 2{= \ A}_{1}\text{\ \ \ \ \ \ \ \ \ \ \ }(1) \\ I_{2}^{*}\left( t = 0 \right) = B_{1}\text{\ \ \ \ \ \ \ \ \ \ }\ (2)\ \text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ $$ $${\frac{d I_{1}^{*}\left( t \right)}{\text{dt}} = 2 - 202,6 \bullet e}^{- 202,6}\left\lbrack A_{1}\cos\left( 32,3t \right) + A_{2}\sin\left( 32,3t \right) \right\rbrack + e^{- 202,6}$$ [−A132,3sin(32,3t)+A232,3cos(32,3t)] $${\frac{d I_{2}^{*}\left( t \right)}{\text{dt}} = 32,3 \bullet e}^{- 202,6}\left\lbrack B_{1}\cos\left( 32,3t \right) + B_{2}\sin\left( 32,3t \right) \right\rbrack + e^{- 202,6}\text{\ \ }$$ [−B132,3sin(32,3t)+B232,3cos(32,3t)] $\left. \ \frac{d I_{1}^{*}\left( t \right)}{\text{dt}} \right|$t=0= - 202,6·A1+32, 3·A2 (3) $\left. \ \frac{d I_{2}^{*}\left( t \right)}{\text{dt}} \right|$t=0= - 202,6·B1+32, 3·B2 (4) $$\left\{ \begin{matrix} I_{1}^{*}\left( t \right) - I_{2}^{*}\left( t \right) - I_{3}^{*}\left( t \right) = 0 \\ E - I_{3}^{*}\left( t \right) R_{3} - L \frac{d I_{1}^{*}\left( t \right)}{\text{dt}} - I_{1}^{*}\left( t \right) R_{1} = 0 \\ I_{3}^{*}\left( t \right) R_{3} - U_{C}\left( t \right) - I_{2}^{*}\left( t \right) R_{2} = 0 \\ \end{matrix} \right.\ $$ t* = 0 $$\left\{ \begin{matrix} I_{1}^{*}\left( 0 \right) - I_{2}^{*}\left( 0 \right) - I_{3}^{*}\left( 0 \right) = 0 \\ I_{3}^{*}\left( 0 \right) R_{3} - L \frac{d I_{1}^{*}\left( 0 \right)}{\text{dt}} - I_{1}^{*}\left( 0 \right) R_{1} = E \\ I_{3}^{*}\left( 0 \right) R_{3} - U_{C}\left( 0 \right) - I_{2}^{*}\left( 0 \right) R_{2} = 0 \\ \end{matrix} \right.\ $$ I1^*(0) = I1^*(τ) = 1, 26 [A] UC(0) = 0 $$\left\{ \begin{matrix} 2 I_{2}^{*}\left( 0 \right) = 1,26 \\ I_{3}^{*}\left( 0 \right) 25 + 0,125 \\ I_{3}^{*}\left( 0 \right) = I_{2}^{*}\left( 0 \right) \\ \end{matrix} \right.\ \frac{d I_{1}^{*}\left( 0 \right)}{\text{dt}} + 1,26 25 = 100$$ $$\left\{ \begin{matrix} I_{2}^{*}\left( 0 \right) = 0,63 \\ \frac{dI_{1}^{*}\left( 0 \right)}{\text{dt}} = \frac{100 - 0,63 25 - 1,26 25}{0,125} \\ I_{3}^{*}\left( 0 \right) = I_{2}^{*}\left( 0 \right) = 0,63 \\ \end{matrix} \right.\ = 422$$ A1 = 1,26-2 = -0,74 (1) B1 = 0,63 (2) 422=202,6·A1+32,3·A2 (3) $\left. \ \frac{d I_{2}^{*}\left( t \right)}{\text{dt}} \right|$t=0=? (4) $$\left\{ \begin{matrix} \left. \ \frac{dI_{1}^{*}\left( t \right)}{\text{dt}} \right|_{t = 0}\ - \left. \ \frac{dI_{2}^{*}\left( t \right)}{\text{dt}} \right|_{t = 0}\ - \left. \ \frac{dI_{3}^{*}\left( t \right)}{\text{dt}} \right|_{t = 0}\ = 0 \\ R_{3} \left. \ \frac{dI_{3}^{*}\left( t \right)}{\text{dt}} \right|_{t = 0} + L \left. \ \frac{d^{2}I_{1}^{*}\left( t \right)}{dt^{2}} \right|_{t = 0} + R_{1} \left. \ \frac{dI_{2}^{*}\left( t \right)}{\text{dt}} \right|_{t = 0} = 0 \\ R_{3} \left. \ \frac{dI_{3}^{*}\left( t \right)}{\text{dt}} \right|_{t = 0} - \left. \ \frac{dU_{C}\left( t \right)}{\text{dt}} \right|_{t = 0} - R_{2} \left. \ \frac{dI_{2}^{*}\left( t \right)}{\text{dt}} \right|_{t = 0} = 0 \\ \end{matrix} \right.\ $$ $$\frac{dU_{C}\left( t \right)}{\text{dt}} \bullet C = I_{2}^{*}\left( 0 \right)$$ $$\frac{dU_{C}\left( t \right)}{\text{dt}} = \frac{I_{2}^{*}(0)}{C} = \frac{0,63}{190 10^{- 6}} = 3315,78 = 3,3 10^{3}$$ $$\left\{ \begin{matrix} 422 - y_{1} - y_{2} = 0 \\ 25 y_{2} + 0,125 y_{3} + 25 422 = 0 \\ 25 y_{2} - 3,3 10^{3} - 25 y_{1} = 0 \\ \end{matrix} \right.\ $$ $$\left\{ \begin{matrix} - y_{1} - y_{2} = - 293,97 \\ 25 y_{2} + 0,125 y_{3} = - 10550 \\ - 25 y_{1} + 25 y_{2} = 3,3 10^{3} \\ \end{matrix} \right.\ $$ $$\left\{ \begin{matrix} y_{1} = - 421 = \left. \ \frac{dI_{2}^{*}\left( t \right)}{\text{dt}} \right|_{t = 0} \\ y_{2} = - 1 = \left. \ \frac{dI_{3}^{*}\left( t \right)}{\text{dt}} \right|_{t = 0} \\ y_{3} = - 84200 = \left. \ \frac{d^{2}I_{1}^{*}\left( t \right)}{dt^{2}} \right|_{t = 0} \\ \end{matrix} \right.\ $$ $$\left\{ \begin{matrix} 422 = - 202,6 A_{1} + 32,3 A_{2} \\ - 421 = - 202,6 B_{1} + 32,3 B_{2} \\ \end{matrix} \right.\ $$ A2 = 1, 032 B2 = −1, 218 I1^*(t)=[2+(−0, 74cos(32, 3t)+1, 032sin(32, 3t))]e^−202, 6t [A] I2^*(t)=[0, 63cos(32, 3t)+(−1, 218)sin(32, 3t)]e^−202, 6t [A]

Metoda operatorowa obliczenia I₂(t)

I₂(t) =  ?

I₁^*(0) = I₁^*(τ) = 1, 26 [A]

U_C(0) = 0

$$V_{1}\left( s \right) = \frac{\sum_{}^{}{(E\left( s \right) \bullet Y\left( s \right))}}{\sum_{}^{}{Y(s)}} = \frac{\left( \frac{E}{s} + L I_{1}^{*}(0) \right) (\frac{1}{R_{1} + sL})}{\frac{1}{R_{1} + sL} + \frac{1}{R_{3}} + \frac{1}{R_{2} + \frac{1}{\text{sC}}}}$$

$\frac{E}{s} + L I_{1}^{*}\left( 0 \right) = \frac{100}{s} + 0,125 1,26 = \frac{100 + 0,1575s}{s}$

$$V_{1}\left( s \right) = \frac{\frac{100 + 0,1575s}{s\left( R_{1} + sL \right)}}{\frac{1}{R_{1} + sL} + \frac{1}{R_{3}} + \frac{\text{sC}}{R_{2} Cs + 1}} = \frac{\frac{100 + 0,0790665}{s \left( R_{1} + sL \right)}}{\frac{R_{3}\left( R_{2} Cs + 1 \right) + \left( R_{1} + sL \right) \left( R_{2} Cs + 1 \right) + sC\left( R_{1} + sL \right) R_{3}}{\left( R_{1} + sL \right) R_{3} \left( R_{2} C_{s} + 1 \right)}} =$$

$$= \frac{100 + 0,1575s}{s} \frac{R_{3}\left( R_{2} Cs + 1 \right)}{R_{3} \left( R_{2} Cs + 1 \right) + \left( R_{1} + sL \right) \left( R_{2} Cs + 1 \right) + sC\left( R_{1} + sL \right) R_{3}}$$

$$I_{2}\left( s \right) = \frac{V_{1}(s)}{R_{2} + \frac{1}{\text{sC}}} = V_{1}\left( s \right) \frac{1}{R_{2} sC} = V_{1}\left( s \right) \frac{\text{sC}}{R_{2} Cs + 1} = \frac{(100 + 0,1575s) C R_{3}}{R_{3} \left( R_{2} Cs + 1 \right) + \left( R_{1} + sL \right) \left( R_{2} Cs + 1 \right) + sC\left( R_{1} + sL \right) R_{3}} =$$

$$= \frac{\left( 100 + 0,1575s \right) 190 10^{- 6} 25}{25 \left( 25 190 10^{- 6}s + 1 \right) + \left( 25 + 0,125s \right) \left( 25 190 10^{- 6}s + 1 \right) + s190 10^{- 6} \left( 25 + 0,125s \right) 25} =$$

$$= \frac{0,15 + 0,00023625s}{0,0375s + 25 + 0,0375s + 25 + 0,0001875s^{2} + 0,125s + 0,0375s + 0,0001875s^{2}} = \frac{0,15 + 0,00023625s}{0,000375s^{2} + 0,2375s + 50}\overset{\Rightarrow}{}\frac{M\left( s \right)}{N\left( s \right)}$$

0, 000375s² + 0, 2375s + 50 = 0

s_1, 2 = −202, 6 ± j32, 3

$$I_{2}^{*}\left( t \right) = \ 2 \bullet Re\left\lbrack \frac{M\left( s_{1} \right)}{N'\left( s_{1} \right)} \bullet e^{s_{1}t} \right\rbrack$$

M(s₁)=0, 15 + 0, 00023625(−202,6±j32,3) = 0, 075187 + j0, 042953

N^′(s) = 2 • 0, 000375s + 0, 2375 = 0, 00075s + 0, 2375

N^′(s₁) = 0, 00075(−202,6±j32,3) + 0, 2375 = −0, 0000025 + j0, 136368

$$I_{2}^{*}\left( t \right) = \ 2 \bullet Re\left\lbrack \frac{0,075187 + j0,042953}{- 0,0000025 + j0,136368} \bullet e^{\left( - 202,6 \pm j32,3 \right)t} \right\rbrack = 2 \bullet Re\left\lbrack 0,31497 - j0,55136 \bullet e^{- = - 202,6t} \bullet e^{j32,3t} \right\rbrack = \ 2 \bullet Re\left\lbrack 0,63498 \bullet e^{- j60,26} \bullet e^{- 202,6t} \bullet e^{j32,3t} \right\rbrack = 1,26996 \bullet e^{- 202,6t} \bullet \cos\left( 202,6t - 60,26 \right) = \left\lbrack 1,26996 \bullet \cos( - 60,26) \bullet \cos\left( 32,3t \right) + 1,26996 \bullet sin\left( - 60,26 \right) \bullet \sin\left( 32,3t \right) \right\rbrack \bullet e^{- 202,6t} = \left\lbrack \mathbf{0,63 \bullet}\cos\left( \mathbf{32,3}\mathbf{t} \right)\mathbf{+ ( - 1,103) \bullet}\sin\left( \mathbf{32,3}\mathbf{t} \right) \right\rbrack\mathbf{\bullet}\mathbf{e}^{\mathbf{- 202,6}\mathbf{t}}$$