電気通信大学 情報理工学研究科 機械知能システム学専攻 2020年8月実施 選択問題6 電気回路学 Author 祭音Myyura (co-authored with GPT 6 Astra)
Description 図1の回路はスイッチ S S S L [ H ] L\,[\mathrm H] L [ H ] C [ F ] C\,[\mathrm F] C [ F ] R [ Ω ] R\,[\Omega] R [ Ω ] E [ V ] E\,[\mathrm V] E [ V ] 0 0 0 S S S t = 0 t=0 t = 0 0 0 0
(1) スイッチ S S S t ≥ 0 t\ge0 t ≥ 0 i 1 ( t ) i_1(t) i 1 ( t ) i 2 ( t ) i_2(t) i 2 ( t )
(2) スイッチ S S S R R R i 2 ( t = ∞ ) i_2(t=\infty) i 2 ( t = ∞ )
(3) R = 1 [ Ω ] R=1\,[\Omega] R = 1 [ Ω ] α = 1 / ( 2 C ) \alpha=1/(2C) α = 1/ ( 2 C ) β = 1 / L C \beta=1/\sqrt{LC} β = 1/ L C α = β \alpha=\beta α = β R R R i 2 ( t ) i_2(t) i 2 ( t ) α , E , t \alpha,E,t α , E , t
表1 ラプラス変換表：F ( s ) = ∫ 0 ∞ f ( t ) e − s t d t \displaystyle F(s)=\int_0^\infty f(t)e^{-st}\,dt F ( s ) = ∫ 0 ∞ f ( t ) e − s t d t
元関数 f ( t ) f(t) f ( t ) 像関数 F ( s ) F(s) F ( s ) 注 e − k t e^{-kt} e − k t 1 s + k \dfrac1{s+k} s + k 1 k > 0 k>0 k > 0 t n n ! e − k t \dfrac{t^n}{n!}e^{-kt} n ! t n e − k t 1 ( s + k ) n + 1 \dfrac1{(s+k)^{n+1}} ( s + k ) n + 1 1 k > 0 k>0 k > 0 n n n
電球が 10000 10000 10000 m m m n n n 1 [ Ω ] 1\,[\Omega] 1 [ Ω ] E [ V ] E\,[\mathrm V] E [ V ] m m m n n n 4 [ Ω ] 4\,[\Omega] 4 [ Ω ]
図2：m = 4 , n = 3 m=4,n=3 m = 4 , n = 3 10000 10000 10000 m = 4 , n = 3 m=4,n=3 m = 4 , n = 3
题目描述
第1题
图1的电路由开关 S S S L [ H ] L\,[\mathrm H] L [ H ] C [ F ] C\,[\mathrm F] C [ F ] R [ Ω ] R\,[\Omega] R [ Ω ] E [ V ] E\,[\mathrm V] E [ V ] 0 0 0 t = 0 t=0 t = 0 S S S 0 0 0
(1) 用图1的电流 i 1 ( t ) , i 2 ( t ) i_1(t),i_2(t) i 1 ( t ) , i 2 ( t ) t ≥ 0 t\ge0 t ≥ 0
(2) 闭合开关足够长时间后，电阻 R R R i 2 ( t = ∞ ) i_2(t=\infty) i 2 ( t = ∞ )
(3) 设 R = 1 [ Ω ] R=1\,[\Omega] R = 1 [ Ω ] α = 1 / ( 2 C ) \alpha=1/(2C) α = 1/ ( 2 C ) β = 1 / L C \beta=1/\sqrt{LC} β = 1/ L C α = β \alpha=\beta α = β α , E , t \alpha,E,t α , E , t R R R i 2 ( t ) i_2(t) i 2 ( t )
表1 拉普拉斯变换表：F ( s ) = ∫ 0 ∞ f ( t ) e − s t d t \displaystyle F(s)=\int_0^\infty f(t)e^{-st}\,dt F ( s ) = ∫ 0 ∞ f ( t ) e − s t d t
原函数 f ( t ) f(t) f ( t ) 像函数 F ( s ) F(s) F ( s ) 条件 e − k t e^{-kt} e − k t 1 s + k \dfrac1{s+k} s + k 1 k > 0 k>0 k > 0 t n n ! e − k t \dfrac{t^n}{n!}e^{-kt} n ! t n e − k t 1 ( s + k ) n + 1 \dfrac1{(s+k)^{n+1}} ( s + k ) n + 1 1 k > 0 k>0 k > 0 n n n
第2题
共有 10000 10000 10000 m m m n n n 1 [ Ω ] 1\,[\Omega] 1 [ Ω ] E [ V ] E\,[\mathrm V] E [ V ] m , n m,n m , n 4 [ Ω ] 4\,[\Omega] 4 [ Ω ]
图2画出了 m = 4 , n = 3 m=4,n=3 m = 4 , n = 3 10000 10000 10000 m = 4 , n = 3 m=4,n=3 m = 4 , n = 3
Kai 問1 (1), (2)
並列枝の電圧が等しく、キャパシタ電流は i 1 − i 2 i_1-i_2 i 1 − i 2
L d i 1 d t + R i 2 = E , R i 2 = 1 C ∫ 0 t ( i 1 − i 2 ) d τ . \begin{gathered}
\boxed{L\frac{di_1}{dt}+Ri_2=E},\\
\boxed{Ri_2=\frac1C\int_0^t(i_1-i_2)\,d\tau}.
\end{gathered} L d t d i 1 + R i 2 = E , R i 2 = C 1 ∫ 0 t ( i 1 − i 2 ) d τ . 初期条件は i 1 ( 0 ) = 0 , i 2 ( 0 ) = 0 i_1(0)=0,i_2(0)=0 i 1 ( 0 ) = 0 , i 2 ( 0 ) = 0 R C i ˙ 2 = i 1 − i 2 RC\dot i_2=i_1-i_2 RC i ˙ 2 = i 1 − i 2 i 1 = i 2 i_1=i_2 i 1 = i 2 i 2 ( ∞ ) = E / R \boxed{i_2(\infty)=E/R} i 2 ( ∞ ) = E / R
問1 (3)
零初期値でラプラス変換すると L s I 1 + R I 2 = E / s LsI_1+RI_2=E/s L s I 1 + R I 2 = E / s I 1 = ( 1 + R C s ) I 2 I_1=(1+RCs)I_2 I 1 = ( 1 + RC s ) I 2 R = 1 R=1 R = 1
I 2 ( s ) = E s ( L C s 2 + L s + 1 ) = E β 2 s ( s 2 + 2 α s + β 2 ) . \begin{gathered}
I_2(s)=\frac{E}{s(LCs^2+Ls+1)}\\
=\frac{E\beta^2}{s(s^2+2\alpha s+\beta^2)}.
\end{gathered} I 2 ( s ) = s ( L C s 2 + L s + 1 ) E = s ( s 2 + 2 α s + β 2 ) E β 2 . β = α \beta=\alpha β = α
I 2 ( s ) = E [ 1 s − 1 s + α − α ( s + α ) 2 ] , I_2(s)=E\left[\frac1s-\frac1{s+\alpha}-\frac\alpha{(s+\alpha)^2}\right], I 2 ( s ) = E [ s 1 − s + α 1 − ( s + α ) 2 α ] , したがって
i 2 ( t ) = E { 1 − ( 1 + α t ) e − α t } ( t ≥ 0 ) . \boxed{i_2(t)=E\{1-(1+\alpha t)e^{-\alpha t}\}\quad(t\ge0)}. i 2 ( t ) = E { 1 − ( 1 + α t ) e − α t } ( t ≥ 0 ) . 初期値0、最終値 E E E R = 1 R=1 R = 1
配列抵抗は R a = 4 m / n R_a=4m/n R a = 4 m / n m n = 10000 mn=10000 mn = 10000
P = E 2 R a ( 1 + R a ) 2 = E 2 R a + 2 + 1 / R a . P=\frac{E^2R_a}{(1+R_a)^2}
=\frac{E^2}{R_a+2+1/R_a}. P = ( 1 + R a ) 2 E 2 R a = R a + 2 + 1/ R a E 2 . R a + 1 / R a ≥ 2 R_a+1/R_a\ge2 R a + 1/ R a ≥ 2 R a = 1 R_a=1 R a = 1 n = 4 m n=4m n = 4 m
m = 50 , n = 200 . \boxed{m=50,\qquad n=200}. m = 50 , n = 200 . この組は自然数の制約を満たすので、連続変数としての最大値 P max = E 2 / 4 P_{\max}=E^2/4 P m a x = E 2 /4
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