سلسلة تمارين في الفيزياء حول دراسة الظواهر الكهربائية من اعداد الاستاذ فرقاني ف

‫اا‬ ‫را ا ا‬ ‫ا‬ ‫ﻥ ﻡد‬‫– ا وب‬ ‫ﻥﺏ‬ ‫وزارة ا ﺏ ا‬‫ا ذ ‪ :‬ﻥ رس‬ ‫دة ا م ا‬ ‫إن‬ ‫ا‬ ‫‪:‬ا ما وا‬ ‫ا‬ ‫ا ة‪ 3:‬ت‬ ‫م‪3:‬عت‪،‬ر‪،‬تر‬‫‪Sujet : 3AS 03 - 01‬‬ ‫‪ :‬درا اه آ ‪.‬‬ ‫ا يا‬‫‪2011/2010 :‬‬ ‫ا ا را‬ ‫‪2011/03/10 :‬‬ ‫ر‬ ‫)‪u (V‬‬ ‫ن א ول ‪(*) :‬‬ ‫א‬ ‫‪ C1 = 2µF‬ﺕ ﺕ ﺕ ‪ u = 100 V‬ﻥ ﺏ ﻡ ﻡ‬ ‫‪ -1‬ﺕ ﻡ‬ ‫‪ C2 = 0.5 µF‬آ ﺏ ‪:‬‬ ‫‪ C1‬رﺏ ﺏ ‪. C2‬‬ ‫أ‪ -‬ا ا ﺏ ا‬ ‫آ ﻡ ﺏ رﺏ ‪.‬‬ ‫ب‪ -‬أﺡ ا ﺕ ﺏ‬ ‫‪ C1 = 1µF‬و ا ﻥ‬ ‫‪ -2‬ﻡ ن ﻡ ن ا ا و‬ ‫ﺕ ﺕ ا ‪. u = 300 V‬‬ ‫‪ ، C2 = 2µF‬ﻥ ﺏ‬ ‫ﺏ‪:‬‬ ‫ا آ رﺕ‬ ‫أ‪ -‬أ أن ا ا‬ ‫ا ا ‪.C‬‬ ‫‪ ، 1 = 1 + 1‬أﺡ‬ ‫‪C C1 C2‬‬ ‫ب‪ -‬أﺡ ﺵ ا ا ‪. C‬‬ ‫اﺕ ‪.‬‬ ‫‪C2 ، C1‬‬ ‫ا‬ ‫ﺝـ‪ -‬أﺡ ‪ u1‬و ‪ u2‬ا ﺕ ا ﺏ ﺏ‬ ‫‪. 5mF‬‬ ‫آ ﻡ ‪. C1 = 0.2 mF‬‬ ‫‪ -3‬ﻡ ﻡ ت ﻡ‬ ‫دﻡ ه ا ت ل ﻡ ﻡ‬ ‫ﺕ‬ ‫أ‪-‬‬ ‫ب‪ -‬ﺡ د د ا ت ا ‪.‬‬ ‫ﺕ را ﺵ ﺕ ﺏ ‪. I = 0.2 mA‬‬ ‫‪ C = 3.2 mF‬ﺕ ﺏ‬ ‫‪ -4‬ﻡ‬ ‫إ اغ ا ﺕ ﻡ ؟ آ ؟‬ ‫أ‪ -‬ه‬ ‫‪.‬‬ ‫ﺏا ا‬ ‫ﺏ ﻡ ور ‪ 4‬د‬ ‫ب‪ -‬أﺡ ﺵ آ س و ا ﺕ ا ﺏ ‪ u‬ﺏ ا‬‫‪. ∆tmax‬‬ ‫وز ‪ . 40V‬أﺡ ا ﻡ ا‬ ‫ﺝـ‪ -‬إذا أن ا ﺕ ا ﺏ ا‬ ‫א ن א ‪(*) :‬‬ ‫‪ -1‬ﻥ ا ارة ا ﺏ ف س ﻡ ‪.‬‬‫ا ‪t=0‬‬ ‫ﺏ ا ﺏ ء‪،‬ﻥ ا‬ ‫ﻡ ي ا ارة ﺏ ر ﺏ ا ة ‪I = 20 µA‬‬ ‫ﻥ‬ ‫ا ﺕ ﺏ ا أزﻡ ﻡ ﻥ‬ ‫ﻥ‬ ‫ا ا وﻥ ا ول ا ‪:‬‬‫‪:.‬‬ ‫‪،،‬‬ ‫–‪:‬‬ ‫–‬

‫‪2:‬‬ ‫‪2011/2010 :‬‬ ‫)‪- (01‬‬ ‫–‬ ‫‪-‬‬ ‫أ‪ -‬اآ ا ا ﺕ ﺏ ﺏ ‪ u‬و ‪. t‬‬ ‫ب‪ -‬أر ا ن )‪ u = f(t‬و ا ﺏ ب ا ‪.‬‬‫‪ -2‬أرﺏ ﻡ ت ﺕ ‪ C4 = 4 µF ، C3 = 1.5 µF ، C2 = 0.5 µF ، C1 = 2 µF :‬ﺕ رﺏ ﺏ‬ ‫ا‪:‬‬ ‫‪ ، C3 ، C2‬أ أن ‪. C5 = C 2 + C3‬‬ ‫‪. 100V‬‬ ‫ﻥ ي ا ارة ﺏ ﺕ‬ ‫ت ‪ .‬أوﺝ ‪. C‬‬ ‫ا ذات ا‬ ‫أ‪ -‬ﻥ ‪ C5‬ه ا‬ ‫ارة ‪.‬‬ ‫اا ا‬ ‫ب‪ -‬إذا ا ﻥ ‪ C‬ه‬ ‫ﺝـ‪ -‬أوﺝ ﺵ ا‬ ‫ا‪.‬‬ ‫د‪ -‬أوﺝ ﺵ آ ﻡ‬ ‫ﻡ ﻡ ا تا‬ ‫א نא ‪:‬‬ ‫ﺕ ا ارة ا ﺏ ا ﺕ ﻡ ﺕ ا ﺏ ﺕ ا آ‬ ‫ا ﺏ ‪ E‬وﻡ وﻡ ا اﺥ ﻡ ‪ ،‬وﺵ ﻡ ‪، C‬‬ ‫ﻥ أوﻡ ﻡ وﻡ ‪ ، R‬را اه از ﻡ ذو ذاآ ة )ا ‪(1-‬‬ ‫‪ -1‬ﻥ ا د ا )‪ (1‬ا ‪ t = 0‬أ‬ ‫ا‪.‬‬ ‫أ‪ -‬ﺏ ﻡ ذا ث ا ى ا ي ‪.‬‬ ‫ﺏ ا ﺕ )‪ uC = f(t‬ﺏ‬ ‫ب‪ -‬أآ ا د ا‬ ‫ا‪.‬‬ ‫ا د‪.‬‬ ‫ﺡ‬ ‫ه‬ ‫‪u‬‬ ‫=‬ ‫‪E‬‬ ‫(‬ ‫‪1‬‬ ‫ـ‬ ‫ـ‬ ‫‪t‬‬ ‫)‬ ‫رة‬ ‫أن ا‬ ‫ﺝـ‪ -‬ﺏ‬ ‫‪RC‬‬ ‫‪e‬‬ ‫د‪ -‬ﻡ آ ﺕ ات ‪ u‬ﺏ ‪. t‬‬ ‫)‪. q = f(t‬‬ ‫و ه ا ةﺏ ﺵ ا‬ ‫هـ‪ -‬أ ﻡ ﺝ آ ﺏ ا د ا‬ ‫‪.u=E‬‬ ‫و‪ -‬ﻡ ه د ﻥ ﺕ ا س ﻡ ا ن )‪ u = f(t‬ا ‪ t = 0‬ﻡ ا‬ ‫د دون ﺏ ه ن ‪.‬‬ ‫ا تا ا‬ ‫‪ -2‬ﻥ ا د ا )‪: (2‬‬ ‫أ‪ -‬ﺏ ﻡ ذا ث ا ي ا ي ‪.‬‬ ‫ا ‪.‬ﻡ ﺡ ه ا‬ ‫ب‪ -‬أآ ا د ﺏ ا ﺕ )‪ u = f(t‬ﺏ‬‫ا د‪:‬‬ ‫ﺕ ا ﻡ وﺏ آ‬ ‫ﺝـ‪ -‬أوﺝ ا رات ا ﻡ د ا‬ ‫‪:‬‬ ‫• ﺵةا ر‪.‬‬ ‫‪. ،،‬‬ ‫–‪:‬‬ ‫–‬

‫‪3:‬‬ ‫‪2011/2010 :‬‬ ‫)‪- (01‬‬ ‫–‬ ‫‪-‬‬ ‫•ﺵ ا ‪.‬‬ ‫ا ا وﻡ ‪. R‬‬ ‫•ا ﺕﺏ‬ ‫•ا ا ﻥ ا ﻥ ا ‪.‬‬ ‫‪. t1/ 2‬‬ ‫=‬ ‫‪τ‬‬ ‫‪ln‬‬ ‫‪2‬‬ ‫‪:‬‬ ‫رة‬ ‫ﺏ‬ ‫ﻡ أ ءﺕ إ ا‬ ‫‪ -3‬أ أن زﻡ ﺕ‬ ‫‪2‬‬‫ﺡ ا أ أن ﻡ س ا ن ﻡ ر ا زﻡ ا‬ ‫‪ -4‬ﻡ ﻡ ا ا ﻥ ا‬ ‫‪. t=τ‬‬ ‫‪2‬‬ ‫א ن א א ‪ ) :‬ﺏ ر ‪ – 2008‬ر ت ( )**(‬‫)ا ‪ (2-‬را ا‬ ‫‪ ،‬ا ح ا ذ ﺕ ﻡ ﻡ ا ارة ا‬ ‫ﺡ ا لا‬ ‫ا‪:‬‬ ‫‪ . RC‬ﺕ ن ا ارة ﻡ ا‬ ‫‪ -‬ﻡ ﺕ ﺕ آ ﺏ ﺏ ‪. E = 12 V‬‬ ‫‪ -‬ﻡ ) ﻡ ﻥ ( ‪. C = 1.0 µF‬‬ ‫‪ -‬ﻥ أوﻡ ﻡ وﻡ ‪. R = 5 . 103 Ω‬‬ ‫‪-‬ﺏد ‪.‬‬ ‫‪ -1‬ﻥ ا د ا )‪ (t = 0‬ا )‪. (1‬‬ ‫أ‪ /‬ﻡ ذا ث ‪.‬‬ ‫ﻡ ه ة ا ر ا ﻡ ﺕ ا ﺏ ‪uAB‬‬ ‫ب‪ /‬آ‬ ‫ا ﺕ اﺵ ل ا ارة ا ﺏ‬ ‫ﺝ ـ‪ /‬ﺏ أن ا د ا‬ ‫‪.‬‬ ‫‪RC‬‬ ‫‪du AB‬‬ ‫‪+‬‬ ‫‪u AB‬‬ ‫=‬ ‫‪E‬‬ ‫رﺕ‬ ‫‪dt‬‬ ‫ارة ‪ ،‬و ﺏ ﺏ ل ا‬ ‫رة )‪ (τ‬ا ﺏ ا‬ ‫د‪ -‬أ‬ ‫ا ي أﻥ ر ﺏ ﻥ ا م ا و ﺡ ات )‪. (SI‬‬ ‫ا ﺏ )‪ -1‬ﺝـ( ﺕ ا رة ‪:‬‬ ‫هـ‪ /‬ﺏ أن ا د ا‬ ‫) ‪t / τ‬ـ‪ e‬ـ‪. uAB = E(1‬‬ ‫‪τ‬ﻡ ا ن‪.‬‬ ‫ﺕ ا ﺏ )‪ uAB = f(t‬و ﺏ آ ﺕ‬ ‫و‪ /‬أر ﺵ ا ا ﻥ ا‬ ‫ا ﺕ ‪ uAB‬ا ‪ t = 5τ‬و ‪ . E‬ﻡ ذا ﺕ ؟‬ ‫ي‪ /‬رن ﺏ‬ ‫‪ -2‬ﺏ ا ﻥ ء ﻡ ا را ا ﺏ ‪ ،‬ﻥ ا د ا )‪. (2‬‬ ‫أ‪ /‬ﻡ ذا ث ‪.‬‬ ‫ا ارة ا ﺏ ‪.‬‬ ‫ب‪ /‬أﺡ ا ا ا‬ ‫א ن א س ‪ ) :‬ﺏ ر ‪ – 2008‬م ﺕ ( )**(‬ ‫ﺵ ﻡ ﻡ ‪ ، (C) ،‬ﻥ ﺏ ا ﻡ‬ ‫ا ﺏا ‪:‬‬ ‫ا‬ ‫‪ -‬ﻡ آ ﺏ و ﺕ ﺕ ﺏ ‪ E = 3 V‬ﻡ وﻡ ا اﺥ‬ ‫ﻡ‪.‬‬ ‫‪ -‬ﻥ اوﻡ ﻡ وﻡ ‪. R = 104 Ω‬‬ ‫‪.K -‬‬ ‫ﺕ ا ﺏ )‪ uc(t‬ﺏ‬ ‫را را ﻡ‬ ‫‪:‬‬ ‫‪. ،،‬‬ ‫–‪:‬‬ ‫–‬

‫‪4:‬‬ ‫‪2011/2010 :‬‬ ‫)‪- (01‬‬ ‫–‬ ‫‪-‬‬ ‫)‪ uc(t‬ا‬ ‫ا ‪ .‬ﻥ ﺏ ا اه از ﻡ ذي ذاآ ة )ا ‪. (4-‬‬ ‫ا‬ ‫ﺵ ﺵ را ا ه از ا‬ ‫‪ t=0‬ه‬ ‫‪K‬ا‬ ‫ﻥا‬ ‫ا ‪5-‬‬ ‫‪ -1‬ﻡ ه ﺵ ة ا ر ا ﺏ ا ر ا ارة ﺏ ﻡ ة ‪ ∆t = 15 s‬ﻡ ؟‬ ‫ﺏ ا ﻡ ‪ ، τ‬و ﺏ أن ﻥ وﺡ ة س ا ﻡ ‪.‬‬ ‫‪ -2‬أ ا رة ا‬ ‫‪ -3‬ﺏ ﻥ ‪ τ‬و ا ا )‪. (C‬‬ ‫ا ) ا ‪:(t=0‬‬ ‫‪ -4‬ﺏ‬ ‫أ‪ /‬اآ رة ﺵ ة ا ر ا ﺏ )‪ i(t‬ا ر ا ارة ﺏ )‪ q(t‬ﺵ ا ‪.‬‬ ‫ا ﺏ ا )‪. q(t‬‬ ‫ب‪ /‬اآ رﺕ ا ﺕ ا ﺏ )‪ uc(t‬ﺏ‬ ‫‪.‬‬ ‫‪uc‬‬ ‫‪+‬‬ ‫‪RC‬‬ ‫‪du c‬‬ ‫=‬ ‫‪E‬‬ ‫‪:‬‬ ‫رة‬ ‫ﺏ‬ ‫)‪ uc(t‬ﺕ‬ ‫اﺕ‬ ‫ﺝـ‪ /‬ﺏ أن ا د ا‬ ‫‪dt‬‬‫ﺏ ‪،A‬‬ ‫ا رة ا‬ ‫ا ﺏ ﺏ رة ) ‪ . uc (t) = E (1- e-t/A‬ا‬ ‫‪ -5‬ﺡ ا د ا‬ ‫وﻡ ه ﻡ ا ؟‬ ‫** ا ذ ‪ :‬ﻥ رس **‬ ‫ﻥ ﺝ إﺏ‬ ‫ﻥ ﻡد ﻥ ﺏ‬ ‫ا وب ‪-‬‬ ‫‪[email protected]‬‬ ‫‪Tel : 0771998109‬‬ ‫ا ا وﻥ ﺏ ي ﺥ ا روس أو ا ر و ﺡ ‪.‬‬ ‫وﺵ ا ﻡ‬ ‫ﻥ ﻡ ه ا ا ع و ‪ .‬أدﺥ ﻡ ا ذ ‪:‬‬ ‫‪sites.google.com/site/faresfergani‬‬ ‫‪:‬‬ ‫‪. ،،‬‬ ‫–‪:‬‬ ‫–‬

‫‪5 : 2011/2010 :‬‬ ‫)‪- (01‬‬ ‫–‬ ‫‪-‬‬‫‪Sujet : 3AS 03 - 01‬‬ ‫أﺝ ﺏ‬‫‪Q = UC1‬‬ ‫‪ :‬درا اه آ ‪.‬‬ ‫ا يا‬‫‪Q = 100 . 2 .10-6 = 2 . 10-4 C‬‬ ‫א ن א ول ‪:‬‬ ‫‪ -1‬أ‪ -‬ا ا ﺏ ا ‪:‬‬ ‫آ ﻡ ﺏ رﺏ ‪:‬‬ ‫ب‪ -‬ا ﺕ ﺏ‬ ‫‪ C2 ، C1‬ﺏ رﺏ ن ‪:‬‬ ‫ذات ا‬ ‫ا‬ ‫ﺏ ض أن ‪ Q2 ، Q1‬ه ﺵ‬‫)‪Q = Q1 + Q2 ....................... (1‬‬‫نا ﺕ ﺏ‬ ‫ا ‪ ،‬و آ ن أن ا ارة ا ﺕ ن ﻡ ا‬ ‫‪ C2 ، C1‬ﻡ ﺏ‬ ‫‪-‬ا‬ ‫ه ﻥ أي ‪:‬‬‫‪U2 = U1‬‬‫= ‪▪ U1‬‬ ‫‪Q1‬‬ ‫→‬ ‫‪Q1 = U1C1‬‬ ‫‪C1‬‬‫= ‪▪ U2‬‬ ‫‪Q2‬‬ ‫→‬ ‫‪Q2 = U2C2 = U1C2‬‬ ‫‪C2‬‬ ‫)‪ (1‬ﻥ ‪:‬‬ ‫ﺏ‬‫‪Q = U1C1 + U1C2‬‬‫)‪Q = U1(C1 + C2‬‬ ‫→‬ ‫‪Q‬‬ ‫‪U1 = C1 + C2‬‬ ‫‪2 .10-4‬‬‫‪U1 = 2 .10-6 + 0.5 .10-6 = 80 V = U2‬‬ ‫‪:‬‬ ‫‪. ،،‬‬ ‫–‪:‬‬ ‫–‬

6: 2011/2010 : - (01) – - : 1 = 1 + 1 ‫ إ ت أن‬-2 C C1 C2U = U1 + U2 ............................. (1) : ‫ﻥ ن ﺝ ا ﺕ ات‬ ‫ﺡ‬ : ‫و‬I = I1 = I2 → : ‫وﻡ‬Q = Q1 = Q2▪ U = Q C▪ U1 = Q1 = Q C1 C1▪ U2 = Q2 = Q C2 C2 : (1) ‫ﺏ‬QQ Q : ‫ ﺵ ا ا‬-‫ب‬ =+C C1 C2Q 11 = Q( + )C C1 C211 1 =+C C1 C21 1 1 2+1 3C = 10-6 + 2 .10-6 = 2 .10-6 = 2 .10-6 2 .10-6 2C = = µF 33 Q → Q = UCU= C .( ‫ا‬ ‫ا ا هﻥ ا ﺕﺏ‬ ‫)ا ﺕ ﺏ‬Q = 300 . 2 . 10-6 = 2 . 10-4 C . 3 : ،، :– –

‫‪7 : 2011/2010 :‬‬ ‫)‪- (01‬‬ ‫–‬ ‫‪-‬‬ ‫‪ C2 ، C1‬ﻡ و آ ن أن ه ﺕ ا ﻡ ﺏ‬ ‫ا ذات ا‬ ‫ﺝـ‪ -‬ﺡ ب ‪: U2 ، U1‬‬ ‫ا ا أي ‪:‬‬ ‫ﺡىﻥ ﺵ‬ ‫ﺵ ا ا هﻥ ﺵ‬ ‫ا ﺕ نﺵ آ ﻡ‬‫‪Q = Q1 = Q2 = 2 . 10-4 C‬‬‫▪‬ ‫‪U1‬‬ ‫=‬ ‫‪Q1‬‬ ‫=‬ ‫‪Q‬‬ ‫=‬ ‫‪2 .10-4‬‬ ‫=‬ ‫‪200‬‬ ‫‪V‬‬ ‫‪C1‬‬ ‫‪C1‬‬ ‫‪6‬ـ‪10‬‬‫▪‬ ‫‪U1‬‬ ‫=‬ ‫‪Q2‬‬ ‫=‬ ‫‪Q‬‬ ‫=‬ ‫‪2 .10-4‬‬ ‫‪= 100‬‬ ‫‪V‬‬ ‫‪C2‬‬ ‫‪C2‬‬ ‫‪6‬ـ‪2 .10‬‬ ‫اﺏ‬ ‫)‪ (C > C1‬و ه ا‬ ‫تا‬ ‫واﺡ ة ﻡ ا‬ ‫و ‪. U1 + U2 = 300 V :‬‬ ‫وا‬ ‫‪ -3‬أ‪ -‬ﺕ ا ت ‪:‬‬ ‫ا ع ن‪:‬‬ ‫ا ا أآ ﻡ ﻡ‬ ‫ا ع‪.‬‬ ‫ب‪ -‬د ا ت ا ‪:‬‬ ‫إذا ا ﻥ ‪ n‬ه د ا ت ا‬‫‪C = C1 + C1 + ........ C1‬‬ ‫‪n‬ﻡ ة‬‫‪C = nC1‬‬ ‫→‬ ‫‪C‬‬ ‫=‪n‬‬ ‫‪C1‬‬ ‫‪5 .10-3‬‬ ‫غ‬ ‫رﺏ ﻥ أوﻡ ﻡ ه ا‬ ‫إ اغ ا ﺕ ﻡ و ذ‬ ‫‪ -4‬أ‪ -‬ﻥ‬‫‪n = 0.2 .10-3 = 25‬‬ ‫ه ا ا ا وﻡ ‪.‬‬ ‫ب‪ -‬ﺵ آ س ﺏ ﻡ ور ‪: 4 min‬‬ ‫ﺏ أن ﺵ ة ا ر ﺏ ن ‪:‬‬‫=‪I‬‬ ‫‪Q‬‬ ‫→‬ ‫=‪Q‬‬ ‫‪I ∆t‬‬ ‫‪∆t‬‬‫‪Q = 0.2 .10-3 . 4 . 60 = 4.8 .10-2 C‬‬ ‫‪ -‬ا ﺕ ﺏ ﻡ ور ‪: 4 min‬‬ ‫‪Q‬‬ ‫‪: ∆tmax‬‬ ‫ﺝـ‪ -‬ا ﻡ ا‬‫=‪U‬‬ ‫‪C‬‬ ‫‪4.8 .10-2‬‬‫‪U = 3.2 .10-3 = 15 V‬‬ ‫‪I‬‬ ‫=‬ ‫‪Q‬‬ ‫→‬ ‫‪Q = I ∆t‬‬ ‫‪∆t‬‬ ‫‪U‬‬ ‫=‬ ‫‪Q‬‬ ‫→‬ ‫‪Q = UC‬‬ ‫‪C‬‬ ‫‪:‬‬ ‫‪. ،،‬‬ ‫–‪:‬‬ ‫–‬

8 : 2011/2010 : - (01) – - : ‫وﻡ‬I ∆t = UC → ∆t = UC : I∆tmax → U = Umax : ‫آﺏ ا‬ ‫وﻡ‬∆t max = UmaxC I∆t max = 40 . 3.2 .10-3 = 640 s = 10 min ,40s 0.2 .10-3U = Q → Q = UC ............... (1) : ‫א نא‬ C : t ‫ و‬U ‫ ا ا ﺕ ﺏ ﺏ‬-‫ أ‬-1I= Q → Q = It .................. (2) t : ‫و آ ن أن ﺵ ة ا ر ﺏ ن‬UC = It → U = I t : ‫( ن‬2) ، (1) ‫ﻡ‬ C : u = f(t) ‫ا ن‬ 10 t(s) U(V) 20 30 40 50 . 9 ،، :– – 8 7 6 5 4 3 2 1 0 0 10 :

9 : 2011/2010 : - (01) – -U = at ................................... (1) : ‫ﺡ ب ا‬- :‫ﻡ ا ن‬ .‫ه ﻡ ا ن‬a: ‫ﺡ‬ :‫و ﺏ‬ I : ‫( ﻥ‬2) ، (1) ‫ﺏ ﺏا‬U = t ................................ (2) CI → I =a C=Ca 9-0a = = 0.2 45 - 0C = 20 .10-6 = 10-4 F = 100 µF : C5 = C2 + C3 ‫ إ ت أن‬-‫ أ‬-2 0.2 C2 IA I2 B I3 C3 I B A U : ‫ﺡ ﻥ نا‬ C5 : ‫وﻡ‬ :‫و‬I = I2 + I3Q = Q2 + Q3 ......................... (1)U2 = U3 = U▪ Q → Q = UC5 U= C5▪ U2 = Q2 → Q2 = U2C2 = UC2 C2 : . ،، :– –

‫‪10 : 2011/2010 :‬‬ ‫)‪- (01‬‬ ‫‪–-‬‬‫▪‬ ‫‪U3‬‬ ‫=‬ ‫‪Q3‬‬ ‫→‬ ‫‪Q3 = U3C3 = UC3‬‬ ‫ا )‪: (1‬‬ ‫‪C3‬‬ ‫ﺏ‬‫‪UC5 = UC2 + UC3‬‬ ‫ب‪ C -‬ا ‪:‬‬ ‫ا ارة ا ﺏ ﺕ‬‫‪UC5 = U(C2 + C3 ) → C5 = C2 + C3‬‬‫‪C5 = 0.5 . 10-6 + 1.5 . 10-6 = 2 . 10-6 F = 2 µF‬‬ ‫آ‪:‬‬ ‫‪C1‬‬ ‫‪V C5‬‬ ‫‪C4‬‬ ‫ا ن‪:‬‬ ‫ا‬ ‫ا ت ذات ا ت ‪ C5 ، C2 ، C1‬ﻡ‬‫‪11 1 1‬‬ ‫‪=+ +‬‬‫‪C C1 C4 C5‬‬‫‪1 1 1 1 2+1+ 2 5‬‬‫‪C = 2 .10-6 + 4 .10-6 + 2 .10-6 = 4 .10-6 = 4 .10-6‬‬‫‪C = 4 .10-6 = 8 .10-7 F = 0.8 µF‬‬ ‫ﺝـ‪ -‬ﺵ ا ا ‪:‬‬ ‫‪5‬‬ ‫ا نﺵ ﺕ ﻡ و وﻡ و ا‬ ‫ﺏ أن ا ت ذات ا ت ‪ C ، C2 ، C1‬ﻡ‬ ‫ا ذات ا ‪ C‬أي ‪:‬‬‫‪Q1 = Q4 = Q5 = 8 . 10-5 C‬‬‫ا ذات‬ ‫ا ع نﻡ عﺵ ﻡ و‬ ‫‪ C3 ، C2‬ﻡ‬ ‫‪ -‬ﺏ أن ا ذات ا‬ ‫ا ‪ C5‬أي ‪:‬‬‫‪Q2 + Q3 = Q5‬‬ ‫ذات ا‬ ‫‪ -‬ا ﺕ ان ﺏ ا‬‫)‪Q2 + Q3 = 8 . 10-5 C ........................................(1‬‬ ‫‪ C3 ، C2‬ﻡ و أي ‪:‬‬‫‪U2 = U3‬‬‫‪Q2 = Q3‬‬ ‫→‬ ‫‪Q3‬‬ ‫=‬ ‫‪C3‬‬ ‫‪Q2‬‬‫‪C2 C3‬‬ ‫‪C2‬‬ ‫‪:‬‬ ‫‪. ،،‬‬ ‫–‪:‬‬ ‫–‬

‫‪11 :‬‬ ‫‪2011/2010 :‬‬ ‫)‪- (01‬‬ ‫–‬ ‫‪-‬‬‫‪Q3‬‬ ‫‪1.5 .10-6‬‬ ‫‪Q2‬‬ ‫→‬ ‫‪Q3‬‬ ‫‪= 3Q2‬‬ ‫)‪..................... (2‬‬ ‫‪= 0.5 .10-6‬‬ ‫ا )‪ (1‬ﻥ ‪:‬‬ ‫ﺏ‬‫‪Q2 + 3Q2 = 8 . 10-5‬‬ ‫و ﻡ ا )‪ (2‬ن ‪:‬‬‫‪4Q2 = 8 . 10-5 → Q2 = 2 . 10-5 C‬‬ ‫א نא ‪:‬‬‫‪Q3 = 3 . 2 . 10-5 = 6 . 10-5 C‬‬ ‫‪ -1‬أ‪ -‬ﻡ ث ا ى ا ي ‪:‬‬‫ﻥ ا وﻥ ت ﻡ‬ ‫و ا ىا ي ا‬ ‫ﻡ ﺕ ا د ا )‪ (1‬ﺕ ا‬ ‫ﺕ اآ ا وﻥ ت ه ا ا س ‪.‬‬ ‫ا س ‪ M‬إ ا س ‪ B‬دارة ا ‪ ،‬ﺡ‬ ‫ب‪ -‬ا د ا ‪:‬‬ ‫ﺡ ﻥ ن ﺝ ا ﺕ ات ‪:‬‬‫‪uAB = uAM + uMB‬‬‫‪E = R i + uC‬‬ ‫‪dq‬‬‫‪R dt + uC = E‬‬‫‪R‬‬ ‫) ‪d(C uC‬‬ ‫‪+‬‬ ‫‪uC‬‬ ‫=‬ ‫‪E‬‬ ‫‪dt‬‬‫‪RC‬‬ ‫‪duC‬‬ ‫‪+‬‬ ‫‪uC‬‬ ‫=‬ ‫‪E‬‬ ‫‪dt‬‬‫‪duC‬‬ ‫‪+‬‬ ‫‪1‬‬ ‫‪uC‬‬ ‫=‬ ‫‪E‬‬ ‫‪dt‬‬ ‫‪RC‬‬ ‫‪RC‬‬ ‫و ه ﻡ د ﺕ ﻡ ا رﺝ ا و ‪.‬‬ ‫‪1‬‬ ‫‪-t‬‬ ‫‪:‬‬ ‫‪u‬‬ ‫=‬ ‫أن‬ ‫ت‬ ‫ﺝـ‪ -‬إ‬ ‫) ‪E(1- e RC‬‬ ‫‪:‬‬ ‫‪1‬‬ ‫‪. ،،‬‬ ‫–‪:‬‬ ‫–‬ ‫‪-t‬‬‫) ‪uC = E(1- e RC‬‬ ‫‪:‬‬

12 : 2011/2010 : - (01) – -du = E(0 - d 1 = E(- (- 1 1 E 1 : ‫ا دا‬ -t -t -tdt dt ‫دا‬ (e RC )) e RC )) = e RC RC RC ‫ﺏ‬E 1 + E 1 E -t -t RC e RC (1 - e RC ) =R RCE 1 + E - E 1 E -t -t e RC e RC =R RC RC RC EE . ‫إذن ا ا ه ﺡ‬ = : u = f(t) ‫ ا ن‬-‫ب‬RC RC E u C (V) t(s) τ :q ‫ﺏ‬ ‫ ا د ا‬-‫هـ‬ : ‫ﺡ ﻥ ن ﺝ ا ﺕ ات‬uAB = uAM + uMB . ‫و ه ﻡ د ﺕ ﻡ ا رﺝ ا و‬E = R i + uC : u = E ‫ ﻡ ا‬t = 0 ‫ د ﺕ ا س ن ا‬-‫و‬ dq qE = R dt + C .t=0 ‫ﻥ ﻡ د ا س‬-dq 1 E : ‫ﻡ ا أو ﻡ د ﻡ ا‬ ‫ا س ه ا‬- + q=dt RC R : ‫ أي‬t = 0 ‫ا‬ ‫ ه ﻡ ا س و ا ي وي ا‬a ‫ﺡ‬uC = at . ،، :– :‫ﺏ‬ du –a = ( dt )t=0 :

13 : 2011/2010 : - (01) – -du E 1 = -tdt RC e RCt=0 → du E → a= E = dt RC RC . u = E t : ‫ ﺕ ن ﻡ ا‬t = 0 ‫إذن ﻡ د ﻡ س ا ن ا‬ RC :‫ﻡ د ا س ن‬ ‫ ﺏ‬u =E ‫ ن‬u = E ‫ ﺕ ا س ﻡ ا‬-E = E → 1 → t = RC = τ t t =1 : ‫ ﻡ ذا ث ا ى ا ي‬-‫ أ‬-2 RC RC ‫( ﺕ غ ا و ا ى ا ي ﺕ د ا وﻥ ت ا اآ‬2) ‫و ا د ا‬. ‫ دارة ا وﻡ‬M ‫ا س‬ ‫ إ و ا‬، ‫ أ ء ا‬M ‫ و ا أﺕ ﻡ ا س‬B ‫ا س‬ :u ‫ﺏ‬ ‫ ا د ا‬-‫ب‬ : ‫ﺡ ﻥ ن ﺝ ا ﺕ ات‬uAB = uAM + uMB : ‫ ن ا أ ﺥ رج ا ارة و ﻡ‬uAB = 00 = Ri + uCR d(C uC ) + uC = 0 dtRC duC + uC = 0 dtduC + 1 uC = 0 dt RC 1 -t . u ‫ﻡ ا رﺝ ا و ﺡ‬ ‫وه ﻡ د ﺕ‬ = Ee RC 1 du E - 1 t : ‫ ا رات ا ﻡ و ا ﻥ ت‬-‫ﺝـ‬ -t =- :‫ﻡ‬ → e RCu = Ee RC :‫*ﺵةا ر‬ RC RC dq d(C u) du E 1i= = = C = C (- -t dt dt dt RC e RC ) E - 1 t - 1 t R i = - e RC = - I0e RC : . ،، :– –

14 : 2011/2010 : - (01) – - . ‫را ﺏ‬ ‫ةا‬ ‫ا‬ ‫ه‬ I0 = E ‫ﺡ‬ R i(A) τ t(s) -I0 : ‫*ﺵ ا‬ ‫ـ‬tq = CuC = C(Ee RC ) ‫ـ‬1t ‫ـ‬1t q = CEe RC = Q0e RC : ‫ا‬ ‫ ه ﺵ ا‬Q0 = CE ‫ﺡ‬ q(C) Q0 = EC t(ms) τ : ‫ا ا وﻡ‬ ‫*ا ﺕ ﺏ‬uR = Ri :‫ن‬ ‫وﻡ‬ i = - E 1 ‫ﺏ‬ ‫وﺝ ﻥ‬ -t R e RCuR = R( - E 1 ) R -t e RC : . ،، :– –

15 : 2011/2010 : - (01) – - . uR0 = E : ‫ أي‬E 1 ‫ا ﺕا ﺏ‬ -t uR = - Ee RC ‫ا ا وﻡ ﻡ و ة ا آ ا ﺏ‬ uR(V) t(s) τ -E : ‫*ا ا ﻥ‬E(C) = 1 Cu2 = 1 C(Ee ‫ـ‬ 1 t )2 2 RC 2 E(C) = 1 CE2e ‫ـ‬ 2 t = ‫ـ‬2 t RC E0(c)e RC 2 ‫اا‬ ‫ه‬ E0(C) = 1 ‫ﺡ‬ CE 2 E(c)(J) E(c)0 t(ms) τ/2 : . ،، :– –

16 : 2011/2010 : - (01) – - : ‫ زﻡ ﺕ ا إ ا‬-3 :‫ﺏ‬E(C) = 1 CE 2e ‫ـ‬ 2 t = ‫ـ‬ 2 RC τt E0(c)e 2t = t1/2 → E(C) = E0(C) 2 :‫ﺏ ﻥ‬E0(C) = ‫ـ‬ 2 t1/ 2 → 1 = e ‫ـ‬ 2 t1/ 2 2 τ 2 τ E0(c)e- 2 = 1 → - 2 = - ln 2 → t1/ 2 = τ ln 2 τ t1/2 ln 2 τ t1/ 2 2 : t= τ ‫ﻡ ر ا زﻡ‬ t=0 ‫ ا‬E(C) = f(t) ‫ إ ت أن ﻡ س ا ن‬-4 2 : ‫ ا ي ن رة ﻡ ﻡ د ﻡ ا‬t = 0 ‫ ا‬E(C) = f(t) ‫ﻥ ﻡ د ﻡ س ا‬E(C) = a t + b . a = ( dE(C) )t=0 ‫ أي‬t = 0 ‫ا‬ ‫ ﻡ ه ا ا ن و وي ا‬a ‫ﺡ‬ dt : ‫رة ا‬ ‫و ذ ﻥ‬a ‫ﻥ‬ : - 2 tE(C) = E0(C) e τ : ‫وﻡ‬dE(C) 2 - 2 t dt τ =- E0(C)e τ ( dE(C) 2 - 2 (0) 2 dt τ a = - τ E0(C)t=0 → )t=0 = - E0(C)e τ → : ‫وﻡ ﻡ د ا س ﺕ‬ 2 :b ‫ﻥ ا ن‬E(C) = - τ E0(C) t + b : E(C) = f(t)‫ﻡ ا ن ى‬t = 0 → E(C) = E0(C) : ‫ﻡ د ا سا ﺥ ةﻥ‬ ‫ﺏ‬E0(C) 2 → b = E0(C) = - τ E0(C) (0) + b : ‫وﻡ ﺕ ﻡ د ا سآ‬ : . ،، :– –

‫‪17 :‬‬ ‫‪2011/2010 :‬‬ ‫)‪- (01‬‬ ‫–‬ ‫‪-‬‬ ‫‪2‬‬ ‫ﻡد ا‬ ‫ن ‪ E(C) = 0‬ﺏ‬ ‫ﺕ ا س ﻡ ﻡ ر ا زﻡ‬‫)‪E(C) = - τ E0(C) t + E0(C‬‬ ‫ا ‪ t=0‬ﻡ‬ ‫إذن أن ﻡ س ا ن )‪E(C) = f(t‬‬ ‫سﻥ ا ا ا ‪:‬‬ ‫‪2‬‬‫)‪0 = - τ E0(C) t + E0(C‬‬‫‪2‬‬ ‫)‪E0(C‬‬ ‫‪t‬‬ ‫=‬ ‫)‪E0(C‬‬ ‫→‬ ‫‪2‬‬ ‫‪t‬‬ ‫‪=1‬‬ ‫→‬ ‫‪τ‬‬‫‪τ‬‬ ‫‪τ‬‬ ‫=‪t‬‬ ‫‪2‬‬ ‫‪τ‬‬ ‫ر ا زﻡ‬ ‫=‪t‬‬ ‫‪2‬‬ ‫‪E(C) J‬‬ ‫وا ا ‪:‬‬ ‫‪τ/2‬‬ ‫)‪t (s‬‬ ‫ا ﺏ ا ا ه از ا‬ ‫א نא א ‪:‬‬ ‫‪τ‬‬ ‫‪ -1‬و ا د ا )‪ (1‬ﺕ ا ‪.‬‬ ‫ب‪ -‬ه ة ا ر ا ﻡ ﺕ ا ﺏ رﺏ‬ ‫‪B‬‬ ‫‪Y‬‬ ‫ب‪ -‬إﺏ از ا د ا ‪:‬‬ ‫ﺡ ﻥ ن ﺝ ا ﺕ ات‬‫‪uAM = uAB + uBM‬‬ ‫‪. ،،‬‬ ‫–‪:‬‬ ‫–‬‫‪E = uAB + R i‬‬ ‫‪dq‬‬‫‪E = uAB + R dt‬‬ ‫‪:‬‬

18 : 2011/2010 : - (01) – -E = u AB + RC du AB : τ ‫ رة‬-‫ب‬ dt : ‫ ر ﺏ ﻥ‬τ ‫ إ ت أن‬-RC du AB + u AB = E . ‫ ر ﺏ ﻥ‬τ ‫إذن‬ dt τ = RC[τ] = [R][C] [Q] [Q][τ] = [U] [I][∆t] → [τ] = [∆t] [I] [U] = [I] = [I] : ‫ ه ﺡ د ا‬uAB = E(1- e-t/τ ) ‫ إ ت أن‬-‫هـ‬uAB = E(1- e-t/τ )du AB = - E e-t/τ =- E e-t/τ dt τ RC : ‫ا دا‬ ‫ﺏ‬RC( E e-t/τ ) + E(1- e-t/τ ) = E . ‫إذن ا ا ه ﺡ د ا‬ RCEe-t/τ + E - Ee-t/τ = E → E = E : ‫ا ا ﻥ‬- E uAB(V) τ t(s) : E ‫ و‬t = 5τ uAB ‫ ا رﻥ ﺏ‬-‫ي‬uAB = E(1- e-t/τ )t = 5τ → uAB = E(1- e-5τ/τ ) = uAB = E(1 - e-5 ) ≈ E ، :– – : .،

‫‪19 :‬‬ ‫‪2011/2010 :‬‬ ‫)‪- (01‬‬ ‫–‬ ‫‪-‬‬ ‫ﺕ‬ ‫ا‬ ‫‪ ، E‬و ﻥ ﻡ ذ أن‬ ‫‪ t = 5τ‬ﺕ وي ﺕ‬ ‫‪ uAB‬ا‬ ‫إذن‬ ‫ا ‪. t = 5τ‬‬ ‫‪-2‬أ‪ -‬ث ﺕ ‪.‬‬ ‫ب‪ -‬ا ا ا ارة ‪:‬‬‫)‪E(C‬‬ ‫=‬ ‫‪1‬‬ ‫‪C‬‬ ‫‪u‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪AB‬‬ ‫‪ uAB = E‬و ﻡ ‪:‬‬ ‫أ‬ ‫ﻡ نا ﺕ أ‬ ‫ﺕ نا أ‬‫)‪E0(C‬‬ ‫=‬ ‫‪1‬‬ ‫‪C‬‬ ‫‪E2‬‬ ‫‪2‬‬‫)‪E0(C‬‬ ‫=‬ ‫‪1 10-6‬‬ ‫‪(12)2‬‬ ‫=‬ ‫‪7.2‬‬ ‫‪.10-5‬‬ ‫‪J‬‬ ‫‪2‬‬ ‫א نא س‪:‬‬ ‫‪ -1‬ﺵ ة ا ر ا ر ا ارة ﺏ ﻡ ة ‪: ∆t = 15 s‬‬ ‫ﺏ ‪ 15s‬ﺕ ن ا ا ﺏ )‪ (RC‬ﺡ ﻥ م دا و ه ن ‪:‬‬‫‪uC = E‬‬ ‫و ﺡ ﻥ ن ﺝ ا ﺕ ات ‪:‬‬‫‪E = uC + uR‬‬‫‪E = E + Ri → Ri = 0 → i = 0‬‬ ‫ﺏ ا ﻡ ‪:τ‬‬ ‫ب‪ -‬ا رة ا‬ ‫‪τ = RC‬‬ ‫‪ -‬إ ت أن ﺏ ا ﻡ ‪ τ‬ﻥ وﺡ ة س ا ﻡ ‪:‬‬‫]‪[τ] = [R][C‬‬ ‫]‪[Q‬‬ ‫]‪[Q‬‬‫= ]‪[τ‬‬ ‫]‪[U‬‬ ‫= ]‪[I‬‬ ‫‪[I][∆t‬‬ ‫]‬ ‫→‬ ‫]‪[τ] = [∆t‬‬ ‫]‪[I‬‬ ‫= ]‪[U‬‬ ‫]‪[I‬‬ ‫إذن ﺏ ا ﻡ ‪ τ‬ﻥ وﺡ ة س ا ﻡ ‪.‬‬ ‫‪ -4‬أ‪ -‬رة )‪ i(t‬ﺏ )‪: q(t‬‬ ‫)‪dq(t‬‬ ‫ب‪ -‬رة )‪ uC(t‬ﺏ )‪: q(t‬‬ ‫= )‪i(t‬‬ ‫ﺝـ‪ -‬إﺏ از ا د ا ‪:‬‬ ‫‪dt‬‬ ‫ﺡ ﻥ ن ﺝ ا ﺕ ات ‪:‬‬ ‫)‪q(t‬‬ ‫‪uC(t) = C‬‬‫‪uE = uR + uC‬‬‫‪E = R i + uC‬‬ ‫‪:‬‬ ‫‪. ،،‬‬ ‫–‪:‬‬ ‫–‬

20 : 2011/2010 : - (01) – - dqE=R + uC dtRC duC + uC = E dtduC + 1 E dt RC uC = RC . ‫ﻡ ا رﺝ ا و‬ ‫وه ﻡ د ﺕ‬ :A‫ـ‬ ‫ ا رة ا‬-5 :uC = (1- e-t/A )duC = E( 0 - ( - 1 e-t/A ) = E e-t/A : ‫ا دا‬ ‫ﺏ‬ dt A AE e-t/A + 1 E (1- e-t/A ) = EA RC RCE e-t/A + E - E e-t/A = EA RC RC RCE e-t/A - E e-t/A = 0A RCE e-t/A = E e-t/A . A = RC = τ ‫ﺏ ﺏ ﻥ‬A RC ‫ أو ا ﻡ ا زم إ ﺕ‬63% ‫ ) ﺏ ا ﻡ ( ا ﻡ ا زم ا ﺏ‬A : ‫ا لا‬- . ‫ ﻡ زﻡ ا‬20% ‫أو ه‬ ‫ا‬ ‫ ﻡ‬37% ‫ﻡ و ـ‬ ‫ﺵا إ‬ ** ‫ ﻥ رس‬: ‫** ا ذ‬ ‫ﻥ ﻡد ﻥ ﺏ‬ - ‫ا وب‬ [email protected] Tel : 0771998109 . ‫ا روس أو ا ر و ﺡ‬ ‫ا ا وﻥ ﺏ ي ﺥ‬ ‫ﻥ ﺝ إﺏ‬ ‫وﺵ ا ﻡ‬ : ‫ أدﺥ ﻡ ا ذ‬. ‫ﻥ ﻡ ه ا ا ع و‬ sites.google.com/site/faresfergani : . ،، :– –

‫اا‬ ‫را ا ا‬ ‫ا‬ ‫ﻥ ﻡد‬‫– ا وب‬ ‫ﻥﺏ‬ ‫وزارة ا ﺏ ا‬‫ا ذ ‪ :‬ﻥ رس‬ ‫دة ا م ا‬ ‫إن‬ ‫ا‬ ‫‪:‬ا ما وا‬ ‫ا‬ ‫ا ة‪ 3:‬ت‬ ‫م‪3:‬عت‪،‬ر‪،‬تر‬‫‪Sujet : 3AS 03 - 02‬‬ ‫‪ :‬درا اه آ ‪.‬‬ ‫ا يا‬‫ا ا را ‪2011/2010 :‬‬ ‫‪2010/12/31 :‬‬ ‫ر‬ ‫ﺏ و ﺕ وي ‪. 1.5 A‬‬ ‫א ن א ول ‪(*) :‬‬ ‫‪ -1‬وﺵ ﻡ وﻡ ‪ r = 6 Ω‬و ذاﺕ ‪ L = 1 H‬زه ﺕ ر آ ﺏ ﺵ ﺕ‬ ‫ا ﺵ ا ما ا ‪.‬‬ ‫‪-‬أ ا ﺕ ﺏ‬ ‫ﺕﺕ آ ﺏ‬ ‫‪ -2‬وﺵ ) ‪ ( L = 0.1 H , r‬ﻡ ﻥ ﺏ‬ ‫‪ 6V‬زه ﺕ ر آ ﺏ ﺵ ﺕ ‪. 1.5A‬‬ ‫ﻡ‬ ‫أ‪ -‬أ ا وﻡ ا اﺥ ﺵ ‪.‬‬ ‫ب‪ -‬ﻥ ر ا ﺵ ﺕ را آ ﺏ ﺕ ﺵ ﺕ ﺏ ا ﻡ و‬ ‫ا نا ﺏ ‪:‬‬ ‫ا ‪. t = 0.5 s‬‬ ‫اﺵ‬ ‫‪-‬أ ا ﺕ ﺏ‬ ‫‪ -3‬وﺵ ذاﺕ ‪ L‬وﻡ وﻡ ‪ r = 8 Ω‬ﻥ ر ﺏ ﺕ ر آ ﺏ ﺵ ﺕ‬ ‫ﻡ ة و ﺕ ﺏ ‪i(t) = 10 t - 3 :‬‬ ‫ا =‪t‬‬ ‫اﺵ‬ ‫ما ﺕ ﺏ‬ ‫‪L‬‬ ‫‪-‬‬ ‫‪. 0.15 s‬‬ ‫א ن א ‪(*) :‬‬ ‫ا وﺵ )‪ ( L , r‬و ﻥ أوﻡ ﻡ وﻡ‬ ‫دارة آ ﺏ ﺕ‬ ‫‪ ، R = r = 12 Ω‬ﻡ ﺕ ﺕ ﻡ ﻡ وﻡ ا اﺥ ﻡ و ﺕ‬ ‫ا آ ا ﺏ ‪ . E‬ﻥ ا ارة إ را اه از ﻡ آ ﺏ‬ ‫ا‪:‬‬ ‫ﺵ ﺵ را ا ه از ا ﻥ ا ‪:‬‬ ‫ا ا ‪. 3 V/div‬‬ ‫‪ -1‬ﻡ ذا آ ﺏ ن ؟ ‪.‬‬ ‫‪ -2‬آ ﺕ ا ﺵ ‪.‬‬ ‫‪ -3‬أ ﺵ ة ا ر ا ر ا ارة ‪.‬‬ ‫‪ -4‬أ ا ة ا آ ا ﺏ ‪.‬‬‫‪:‬‬ ‫‪. ،،‬‬ ‫–‪:‬‬ ‫–‬

‫‪2:‬‬ ‫‪2011/2010 :‬‬ ‫)‪- (02‬‬ ‫–‬ ‫‪-‬‬ ‫ﺵ )‪. (r ≠ 0‬‬ ‫א ن א ‪(**) :‬‬ ‫ا ا وا ﺕ ن ا‬ ‫ﻥ ا ارة ا ﺏ ا‬ ‫ﻡ ﻡ ﺕ ﺕ ا آ ا ﺏ ‪ ، ، E‬ﻥ أوﻡ ﻡ وﻡ‬ ‫‪ ، R‬وﺵ ذاﺕ ‪ L‬و ﻡ وﻡ ا اﺥ ‪. r‬‬ ‫إه ل‬ ‫او‬ ‫اا‬ ‫‪ -1‬أﺝ‬ ‫ا وﻡ ا اﺥ ﺵ )‪. (r = 0‬‬ ‫ﻡا د‬ ‫ﺏ ﺵ ة ا ر )‪. τ ، I0 ، i(t‬‬ ‫أ‪ -‬أآ ا د ا‬ ‫ﻡ ر ا نﺏ آ‬ ‫ب‪ -‬أآ ا رات ا‬ ‫ا‪:‬‬ ‫ا ا وﻡ ‪.‬‬ ‫•ا ﺕ ﺏ‬ ‫اﺵ ‪.‬‬ ‫•ا ﺕ ﺏ‬ ‫ا ﺕ ‪ uR‬ﺏ‬ ‫•ا ا ﻥ ا ﺵ ‪.‬‬ ‫ا ا وﻡ ‪.‬‬ ‫و‬ ‫‪ -2‬أ آ ﺏ ا د ا ﺏ‬ ‫م إه ل ا وﻡ ا اﺥ‬ ‫‪ -3‬أ ﻥ ا ا‬ ‫א ن א א ‪ ) :‬إﻡ ن ا ا ﻥ – ‪(**) ( 2008/2007‬‬‫ﺕ ا ﺏ ﺕ ا آ ا ﺏ ‪ ، E‬وﺵ ذاﺕ ‪ L‬و ﻡ وﻡ ا اﺥ ‪، r‬‬ ‫ﺕ دارة آ ﺏ ﻡ ﻡ‬ ‫ﻥ أوﻡ ﻡ وﻡ ‪ ، R = 90 Ω‬را اه از ﻡ ذو ذاآ ة )ا ‪. (1-‬‬‫ا ن )‪ (1‬ﺕ ات ا ﺕ ﺏ‬ ‫ﺵ ﺵ را ا ه از ا ﻥ )‪، (2) ، (1‬‬ ‫‪ -1‬ﻥ ا‬ ‫ا اﻡ ‪.‬‬ ‫ا ‪ ، uE‬و ا ن )‪ (2‬ﺕ ات ا ﺕ ‪ uR‬ﺏ‬‫ﺏ ارة ﺕ ﻡ ا ل ا ﻥ )‪(2) ، (1‬‬ ‫أ‪ -‬ﺏ ﺏ ا ر آ ﺕ رﺏ را ا ه از ا‬ ‫ب‪ -‬ا دا ه ا ﻥ أوﺝ ‪:‬‬ ‫• ا ةا آ ا ﺏ ‪.‬‬ ‫• ﺵةا را ﺏ ا ما ا ‪.‬‬ ‫• ا وﻡ ا اﺥ ﺵ ‪.‬‬ ‫• ذاﺕ ا ﺵ ‪.‬‬ ‫ا ‪1-‬‬ ‫)‪u(V‬‬ ‫ا ‪1 2-‬‬ ‫‪KE‬‬ ‫‪-+‬‬ ‫‪R L,r‬‬ ‫‪2‬‬ ‫)‪t(ms‬‬ ‫‪2‬‬ ‫‪1‬‬‫‪:‬‬ ‫‪. ،،‬‬ ‫–‪:‬‬ ‫–‬

‫‪3:‬‬ ‫‪2011/2010 :‬‬ ‫)‪- (02‬‬ ‫–‬ ‫‪-‬‬ ‫ﺵ ة ا ر )‪ i = f(t‬ا ر ﺏ ارة ‪.‬‬ ‫ا نا ‪.‬‬ ‫‪ -2‬ﻥ‬ ‫ا د‪.‬‬ ‫ا دا ا ﺕ‬ ‫أ‪ -‬اآ‬ ‫ه‬ ‫‪i‬‬ ‫=‬ ‫‪I0e‬‬ ‫ـ‬ ‫‪t‬‬ ‫رة‬ ‫أن ا‬ ‫ب‪ -‬ﺏ‬ ‫‪τ‬‬ ‫א ن א س ‪ ) :‬ﺏ ر ‪ – 2008‬ر ت ( )**(‬ ‫ك و ﻡ ات وﺵ ﻡ وﻡ )‪ (r‬و ذاﺕ )‪(L‬‬ ‫ﺏ ضﻡ‬ ‫ﺏ ذي ﺕ ﺕ آ ﺏ ﺏ ‪ E = 4.5V‬و‬ ‫ا‬ ‫ﻥﺏ‬ ‫‪) K‬ا ‪. (1-‬‬ ‫ﺝ ﻡ ور ا ر‬ ‫‪ -1‬اﻥ ﻡ ا ارة ور إﺝ ﺏ و ﺏ‬ ‫نا ﺕ ا ﺏ ﺏ‬ ‫ا ﺏ وﺝ ا ا‬ ‫ا‪.‬‬ ‫ا ﺵ وﺏ‬ ‫‪ -2‬ا ‪ t = 0‬ﻥ ا ‪. (K) :‬‬ ‫)‪ i(t‬ر ا ﺏ ا ر‬ ‫ا ﺕ ا ةا‬ ‫ﻥ ن ﺝ ا ﺕ ات ‪ ،‬أوﺝ ا د ا‬ ‫أ‪ -‬ﺏ‬ ‫ا ارة ‪.‬‬ ‫‪r‬‬ ‫‪-t‬‬ ‫ﻡ ا ) ‪ I0 i(t) = I0 (1- e L‬ه ا ة ا‬‫ر‬ ‫ا ﺏﺕ‬ ‫ب‪ -‬ﺏ أن ا د ا‬ ‫ا ﺏ ا ر ا ارة ‪.‬‬ ‫ر ا ﺏ ﺏ رة ) ‪ (r) i(t) = 0.45 (1- e-10t‬ﺏ ﻥ و )‪ (i‬ﺏ ﻡ ‪ .‬أ‬ ‫‪ -3‬ﺕ ا ة ا‬ ‫ا دا ‪:‬‬ ‫)‪ (I0‬ر ا ﺏ ا ر ا ارة ‪.‬‬ ‫أ‪ /‬ا ة ا‬ ‫ب‪ /‬ا وﻡ )‪ (r‬ﺵ ‪.‬‬ ‫ﺝـ‪ /‬ا اﺕ )‪ (L‬ﺵ ‪.‬‬ ‫د‪ /‬ﺏ ا ﻡ )‪ (τ‬ا ارة ‪.‬‬ ‫ا ما ا ؟‬ ‫‪-4‬أ‪ /‬ﻡ ا ا ﻥ ا ﺵ‬ ‫اﺵ ‪.‬‬ ‫ب‪ /‬اآ رة ا ﺕ ا ﺏ ا ﺏ‬ ‫ا ﺵ ا )‪. (t = 0.3s‬‬ ‫ا ﺕا ﺏ ﺏ‬ ‫ﺝـ‪ /‬أ‬ ‫** ا ذ ‪ :‬ﻥ رس **‬ ‫ﻥ ﺝ إﺏ‬ ‫ﻥ ﻡد ﻥ ﺏ‬ ‫ا وب ‪-‬‬ ‫‪[email protected]‬‬ ‫‪Tel : 0771998109‬‬ ‫ا ا وﻥ ﺏ ي ﺥ ا روس أو ا ر و ‪.‬‬ ‫وﺵ ا ﻡ‬ ‫ﻥ ﻡ ه ا ا ع و ‪ .‬أدﺥ ﻡ ا ذ ‪:‬‬ ‫‪sites.google.com/site/faresfergani‬‬ ‫‪:‬‬ ‫‪. ،،‬‬ ‫–‪:‬‬ ‫–‬

‫‪4:‬‬ ‫‪2011/2010 :‬‬ ‫)‪- (02‬‬ ‫–‬ ‫‪-‬‬ ‫أﺝ ﺏ‬‫‪Sujet : 3AS 03 - 02‬‬ ‫اه آ ‪.‬‬ ‫ا ي ا ‪ :‬درا‬ ‫‪di‬‬ ‫ا ما ا ‪:‬‬ ‫א ن א ول ‪:‬‬‫‪uL = L dt + ri‬‬‫‪di‬‬ ‫‪ -1‬ا ﺕ ﺏ ا ﺵ‬ ‫‪=0‬‬ ‫ا ما ا ﺕ نﺵةا ر ﺏ و ‪:‬‬‫‪dt‬‬‫‪uL = ri‬‬ ‫وﻡ ‪:‬‬‫‪uL = 6 .1.5 = 9V‬‬ ‫‪ -2‬ا وﻡ ا اﺥ ﺵ ‪:‬‬ ‫‪di‬‬ ‫ا ما ا ﺕ نﺵةا ر ﺏ و ‪:‬‬‫‪uL = L dt + ri‬‬‫‪di‬‬ ‫وﻡ ‪:‬‬ ‫‪:‬‬ ‫‪=0‬‬‫‪dt‬‬ ‫ﺏ ﻥ‪:‬‬‫‪uL = ri‬‬‫‪uL = 6 V → i = 1.5 A‬‬ ‫ا ‪: t = 0.5 s‬‬ ‫اﺵ‬ ‫ب‪ -‬ا ﺕ ﺏ‬‫‪6 = r . 1.5 → r = 6 = 4Ω‬‬ ‫و‪:‬‬ ‫‪ di‬ا ن ﻡ ا‬ ‫ﺕا‬ ‫‪1.5‬‬ ‫‪dt‬‬ ‫‪di‬‬‫)‪uL = L dt + r i ....................... (1‬‬ ‫‪. ،،‬‬ ‫–‪:‬‬ ‫ﻡ ا ن‪:‬‬‫‪di = tan α = 0 - 3 = - 3‬‬ ‫–‬‫‪dt 1.5 - 0.5‬‬ ‫‪:‬‬

5: 2011/2010 : - (02) – -t = 0.5 s → i = 1A : (1) ‫ا رة‬ ‫ﺏ‬ -3uL = (0.1(-3)) + (4 . 2) = 7.7 V : ‫اﺵ‬ ‫ما ﺕ ﺏ‬ L di :‫ﺏ‬uL = L dt + rii = 10 t – 3di = 10dtuL = 10L + 8( 10t – 3)uL = 10L + 80t – 24t = 0.15 s → u = 00 = (80 . 0.15) + 10L – 2410L = 24 – (80 . 0.15)10L = 12 → L = 1.2 H : ‫א نא‬ : (r = 0) i(t) ‫ﺏ‬ ‫ ا د ا‬-1 :‫ا‬ ‫ﻥ ن ﺝ ا ﺕ ات‬uE = uR + uL di → di +Ri=E → di R E → di R i= E R → di + R i = E RE=Ri+L L + i= + dt dt dt L L dt L L R dt L R L : ‫آﺏ‬ τ= R ، E : ‫أن‬ ‫و‬ L I0 = Rdi 1 1dt + τ i = I0 τ di + 1 i = I0 dt τ τ i (t) = I0 (1- e-t/τ ) : ‫و ه ﻡ د ﺕ ﻡ ا رﺝ ا و‬ :‫ا‬ ‫ﻥ ن ﺝ ا ﺕ ات‬uE = uR + uL0=R di → di +R i =0 → di + R i =0 i+L L dt dt dt L : . ،، :– –

6: 2011/2010 : - (02) –- di 1 ‫و ه ﻡ د ﺕ ﻡ ا رﺝ ا و‬ dt + τ i = 0 i (t) = I0e-t/τ : : ‫ ا رات ا‬-‫ب‬ : ‫ا ا وﻡ‬ ‫•ا ﺕ ﺏ‬ :‫ا‬uR = R ii (t) = I0 (1- e-t/τ )uR = R I0 ( 1- e-t/τ )→ uR = E (1- e-t/τ ) R R uR = E (1- e-t/τ ) :‫ه‬t = 0 → uR = 0t = ∞ → uR = E uR(V) Eτ t(s) :‫ا‬uR = R ii (t) = I0 e-t/τuR = R I0 e-t/τ → uR = R E e-t/τ R : uR = E e-t/τ ، :– – .،

7: 2011/2010 : - (02) – -t = 0 → uR = E uR(V) :‫ه‬t = ∞ → uR = 0 E t(ms) τ : ‫اﺵ‬ ‫ﺕﺏ‬ ‫•ا‬ : ‫ا‬ di di = I0 ( 0 - (- 1 e-t/τ ) ) = I0 e-t/τ →uL = L . I0 e-t/τ E R e-t/τuL = L dt dt τ τ τ = L. .i = I0 (1 - e-t/τ ) → R L uL = E e-t/τ : ‫وه‬t = 0 → uL = Et = ∞ → uL = 0 uL(V) E t(ms) τ: . ،، :– –

8: 2011/2010 : - (02) – - di :‫ا‬uL = L dt : ‫وه‬i = I0 e-t/τ → di I0 e-t/τ →uL = L( - I0 e-t/τ ) = L(- E . R e-t/τ ) =- τ τ R L dt uL = - E e-t/τt = 0 → uL = - Et = ∞ → uL = 0 uL(V) t(s) τ -E τ : ‫•ا ا ﻥ ا ﺵ‬ :‫ا‬E(L) = 1 Li2 2i = I0 ( 1 - e-t/τ ) E(L) = 1 LI02 (1- e-t/τ )2 2t = 0 → E(L) = 0 :‫ه‬ ‫ ه ا ا‬E(L)0 :t =∞→ E(L) = 1 LI02 = E(L)0 2 . ‫ا ﻥ اﺵ‬ : . ،، :– –

9: 2011/2010 : - (02) – - E(L)(J) E(L)0 τ/2 t(s) :‫ا‬E(L) = 1 Li2 2i = I0 e-t/τ → E(L) = 1 LI02 ( e-t/τ )2 2 E(L) = 1 LI02e-2t/τ 2t = 0 → E(L) = 1 LI02 = E(L)0 :‫ه‬ 2 ‫ ه ا ا‬E(L)0 :t = ∞ → E(L) = 0 . ‫ا ﻥ اﺵ‬ E(L)(J) E(L)0 t(ms) τ/2 : . ،، :– –

10 : 2011/2010 : - (02) – - : uR ‫ﺏ‬ ‫ ا د ا‬-2 :‫ا‬ ‫ﻥ ن ﺝ ا ﺕ ات‬uE = uR + uL diE = uR + L dt : uR =Ri→ i = uR → di = 1 duR ‫أن‬ ‫و‬ R dt R dtE = uR + L 1 du R → L duR + uR = E R dt R dt du R + R uR = ER dt L L . ‫و ه ﻡ د ﺕ ﻡ ا رﺝ ا و‬ :‫ا‬ ‫ﻥ ن ﺝ ا ﺕ ات‬uE = uR + uL di0 = uR + L dt : uR =Ri→ i = uR → di = 1 duR ‫أن‬ ‫و‬ R dt R dt0 = uR + L 1 du R → L duR + uR = 0 R dt R dt du R + R uR =0 dt L . ‫ﻡ ا رﺝ ا و‬ ‫وه ﻡ د ﺕ‬ : r ≠ 0 ‫( ﻡ أﺝ‬1) ‫أ‬ ‫ ﻥ‬-3 ‫ ا‬-1 : i(t) ‫ﺏ‬ ‫دا‬ :‫ا‬ ‫ﻥ ن ﺝ ا ﺕ ات‬uE = uR + uL . ،، :– – diE = R i + L + ri dt diL + (R + r) i = E dt :

11 : 2011/2010 : - (02) – -L di ER + r dt + i = R + rτ di + i = I0 dt : ‫ﻥ‬τ ‫ﺏا‬ di + 1 i = I0 dt τ τ i (t) = I0 (1- e-t/τ ) : ‫و ه ﻡ د ﺕ ﻡ ا رﺝ ا و‬ :‫ا‬ ‫ﻥ ن ﺝ ا ﺕ ات‬uE = uR + uL di0 = Ri + L dt + ri diL + (R + r) i = 0 dtdi R + rdt + L i = 0 di 1 dt + τ i = 0 i (t) = I0e-t/τ : ‫ﻡ ا رﺝ ا و‬ ‫وه ﻡ د ﺕ‬ : ‫ ا رات ا‬-‫ب‬ : ‫ا ا وﻡ‬ ‫•ا ﺕ ﺏ‬ ‫ا‬ :uR = R ii (t) = I0 (1- e-t/τ ) → uR = R I0 ( 1- e-t/τ ) → uR = R E (1- e-t/τ ) R+r uR = ER (1 - e-t/τ ) R+r :‫ه‬t = 0 → uR = 0t=∞ → uR = ER <E R+r : . ،، :– –

12 : 2011/2010 : - (02) – - uR(V) E ER/R+r t(s)uR = R i :‫ا‬i (t) = I0 e-t/τ :‫ه‬uR = R I0 e-t/τ → uR = R E e-t/τ R+r uR = ER e-t/τ R+rt=0 → uR = ER <E R+rt = ∞ → uR = 0 uR(V) E ER/R+r t(ms) : . ،، :– –

13 : 2011/2010 : - (02) – - : ‫اﺵ‬ ‫•ا ﺕ ﺏ‬ : ‫ا‬ di : ‫وه‬uL = L dt + rii = I0 ( 1 - e-t/τ ) → di = I0 ( 0- 1 e-t/τ ) ) = I0 e-t/τ dt (- τ τuL = L. I0 e-t/τ + rI0 (1 - e-t/τ ) → uL = L E L e-t/τ + E.r (1 - e-t/τ ) τ R+r R+r R+ruL = Ee-t/τ + E.r - E.r e-t/τ → uL = E.r + (E - E.r )e-t/τ R+r R+r R+r R+ruL = E.r E.R + E.r - E.r )e-t/τ R+r +( R+r uL = E.r + E.R e-t/τ R+r R+rt=0 → E.r E.R E (R + r) uL = R +r + R +r = R + r = Et=∞ → E.r uL = R + r uL(V) E E.r/R+r t(ms) :‫ا‬ diuL = L dt + rii = I0 e-t/τ → di = I0 1 e-t/τ ) = - I0 e-t/τ dt (- τ τ :. ،، :– –

14 : 2011/2010 : - (02) – -uL = L (- I0 e-t/τ ) + rI0 e-t/τ → uL = L( - E L ) e-t/τ + E.r e-t/τ τ R+r R+r R+ruL = Ee-t/τ + E.r e-t/τ → uL = E.r - E)e-t/τ R+r ( R+ruL E.r - E.R - E.r )e-t/τ =( R+r uL = - E.R e-t/τ R+r : ‫وه‬t=0 → E.R uL = - R + rt = ∞ → uL = 0 uL(V) t(s) -E.R/R+r : ‫א نא‬ : ‫ ﻡ آ ﻡ ﺏ ن‬-1‫ا وﻡ‬ ‫ ﺏ‬UAM ‫ﺕ ات ا ﺕ‬ ‫اﻥ‬ ‫أن أ‬ ‫ﻡ ﺥ ل رﺏ را ا ه از ا‬ :‫ن‬ . ‫اﺵ‬ ‫ار‬ ‫ ﺏ‬UBM ‫ﺕ ات ا ﺕ‬ ‫وا ﺥ‬ (2)‫ن‬ ‫نا ﺕ‬ ‫ا وﻡ و ا‬ ‫ﻡ ﺥ لﺕ ا نا ي ن ﺝ‬- ‫أوﻡ ا ي‬UA > UM > UB ‫ ﺏ‬UBM ‫ات‬ ‫ﺕ‬ ‫ ﺏ‬UAM ‫( ﺕ ات‬1) ‫إذن ا ن‬UAM = UA – UM > 0 : ‫ ﺕ ف ا ﺵ‬-2UBM = UB – UM < 0 . ‫اﺵ‬‫ﻡز‬ ‫ﺏآ‬ ‫ﺏ‬ ‫ﺕ فﻥ‬ ‫اﺵ ه ا ﺏﺕ‬ .( ‫ﺕ رآ ﺏ ﻡ ) ﺏ‬ : . ،، :– –

‫‪15 : 2011/2010 :‬‬ ‫)‪- (02‬‬ ‫‪–-‬‬‫ﻡا ‪،‬‬ ‫ا‬ ‫ﻡﺹ‬ ‫ا وﻡ و ا ﺵ‬ ‫را رﺏ‬ ‫‪ -3‬ﺵ ة ا ر ا ر ا ارة‪:‬‬ ‫ا وﻡ ‪.‬‬ ‫ﺏ ا رﺏ‬ ‫ﺵةا را‬ ‫ﺵ ة ا ر ا ر ﺏ ارة ه ﻥ ﺵ ة ا‬‫‪UR = RI‬‬ ‫→‬ ‫‪I = UR‬‬ ‫ب ﺵ ة ا ر ا ر ﺏ ارة ﻥ‬ ‫‪R‬‬ ‫ﻡ ا ن ‪ UR = 2 . 3 = 6V :‬و ﻡ ‪:‬‬ ‫‪6‬‬‫‪I = = 0.5 A‬‬ ‫‪ -4‬ا ة ا آ ا ﺏ ‪: E‬‬ ‫ﻥ ن ﺝ ا ﺕ ات ‪:‬‬ ‫‪12‬‬‫‪UE = UL + UR‬‬ ‫‪di‬‬‫)‪E = rI + RI ( = 0‬‬ ‫‪dt‬‬‫‪E = (R + r)I‬‬‫‪E = (12 +12) 0.5 = 12 V‬‬ ‫א نא א ‪:‬‬ ‫‪-1‬أ‪ -‬آ رﺏ را ا ه از ا ‪:‬‬ ‫‪KE‬‬ ‫‪-+‬‬ ‫‪R‬‬ ‫‪L‬‬ ‫‪BM‬‬ ‫‪A‬‬ ‫‪Y2 Y1‬‬ ‫ب‪ -‬ا ة ا آ ا ﺏ ‪:‬‬ ‫أي ‪uE = E :‬‬ ‫ا ﺏ و ﻡ وي ة ا آ ا ﺏ‬ ‫‪-‬ا ﺕ ﺏ‬ ‫ﻡ ا ن)‪. uE = 10V : (1‬‬ ‫إذن ‪E = 10 V :‬‬ ‫▪ﺵةا را ﺏ ا ما ا ‪:‬‬ ‫ﻡ ا ن)‪. uR0 = 9 V : (2‬‬ ‫‪:‬‬ ‫‪:‬‬ ‫‪. ،،‬‬ ‫–‪:‬‬ ‫–‬

16 : 2011/2010 : - (02) – -uR0 = RI0 → I0 = uR0 R 9 : ‫▪ ا وﻡ ا اﺥ ﺵ‬I0 = 90 = 0.1AI0 = E r → R +r = E → r= E -R R+ I0 I0r = 10 - 90 = 10 Ω : ‫▪ ذاﺕ ا ﺵ‬ 0.1 . τ = 1.5 . 10-3 : ‫ﻡ ا ن‬τ = L → L = (R + r)τ R+rL = (90 +10) .1.5 .10-3 = 0.15 H :‫ﺏ ﺵةا ر‬ ‫ ا د ا‬-‫أ‬-2 : ‫ﻥ ن ﺝ ا ﺕ ات‬UAB = uAM + uMB di0 = L + ri + Ri dt diL + (r + R)i = 0 dtdi R + r + i=0dt Ldi 1 (τ = Ldt + τ i = 0 ) R + r) . ‫ﻡ ا رﺝ ا و‬ ‫وه ﻡ د ﺕ‬ :‫ا د‬ ‫ ا ﻡ‬-‫ﺝـ‬ - ti = I0e τdi - I0 - tdt τ = e τ : ‫ا دا ﻥ‬ ‫ﺏ‬-I - t 1 - tτ τ e τ + (I0e τ ) = 0 : . ،، :– –

17 : 2011/2010 : - (02) – -- I0 - t I0 - t τ τ e τ + e τ =0 → 0 = 0 . ‫دا‬ ‫إذن ا ا آ ر ه‬ :‫ا‬ ‫ا ﺵ وﺏ‬ :‫א نא س‬ ‫ ﺝ ﻡ ور ا ر و ﺝ ﺵ ا ﺕ ﺏ‬-1 I ‫ــ‬ + : ‫ إ د ا د ا‬-‫ أ‬-2 : ‫ﻥ ن ﺝ ا ﺕ ات‬uE = uL . ‫و ه ﻡ د ﺕ ﻡ ا رﺝ ا و‬ diE = L + ri dt diL + ri = E dt r -t : i(t) = I0 (1- ‫ﺕ ﻡا‬ ‫ إ ت أن ا د ا‬-‫ب‬ eL ) r -ti(t) = I0 (1- e L )di(t) = I0 r r = E r r = E - r t dt L -t r L -t L e eL eL L : ‫ا دا ﻥ‬ ‫ﺏ‬L E r + r E (1- r = E -t -t eL eL ) Lr rr -t -tEe L + E(1- e L ) = E rr - t - t L + E - L = E → E=EEe Ee . ‫دا‬ ‫إذن ا ا ه‬ : ‫ ﺵ ة ا ر ا‬-3 : ‫ﻡﺝ‬ : . ،، :– –

18 : 2011/2010 : - (02) – - r : ‫وﻡ ﺝ أﺥ ى‬ -ti = I0 (1- e L ) ....................... (1)i = 0.45(1 - e-10t ) ....................... (1) . I0 = 0.45 A : ‫( ﻥ‬2) ، (1) ‫ﺏ ﺏا‬ : ‫ ﺵ‬r ‫ ا وﻡ‬-‫ب‬ E → EI0 = r r= I0r = 4.5 = 10 Ω : ‫ ﺵ‬L ‫ ا اﺕ‬-‫ﺝـ‬ 0.45 : (2) ، (1) ‫ﺏ ﺏا‬r → 10 10 = 100 L= = =1HL r 10 : τ ‫ ﺏ ا ﻡ‬-‫د‬τ= L rτ = 1 = 0.1 s 10 : ‫ا ما ا‬ ‫ ا ا ﻥ ا ﺵ‬-‫ أ‬-4E(L) = 1 Li2 2 : ‫ وﻡ‬i = I0 : ‫ا م ا ا‬E(L) = 1 LI02 2E(L) = 1 .1. (0.45)2 = 0.10 J 2 : ‫اﺵ‬ ‫ رة ا ﺕ ا ﺏ‬-‫د‬ : diuL = L dt + rii = I0 ( 1 - r = E (1- r t ) -t r - eL ) eLdi E r = -tdt L eL : . ،، :– –

19 : 2011/2010 : - (02) – - : ‫ ﻥ‬uL ‫رة‬ ‫ﺏ‬uL = L. E r E r -t +r -t L eL r (1-e L ) rr -t -tuL = E e L + E(1- e L ) rr → uL = E -t -tuL = E e L + E - E e L . ( ‫دارة ﺕ ي ﻥ أوﻡ‬ ‫ )ﻥ آ‬E ‫ن ﺏ و ﻡ وي ـ‬ ‫أي أن ا ﺕ ﺏ ا ﺵ‬ ** ‫ ﻥ رس‬: ‫** ا ذ‬ ‫ﻥ ﺝ إﺏ‬ ‫ﻥ ﻡد ﻥ ﺏ‬ - ‫ا وب‬ [email protected] Tel : 0771998109 . ‫ا ا وﻥ ﺏ ي ﺥ ا روس أو ا ر و‬ ‫وﺵ ا ﻡ‬ : ‫ أدﺥ ﻡ ا ذ‬. ‫ﻥ ﻡ ه ا ا ع و‬ sites.google.com/site/faresfergani : . ،، :– –

‫اا‬ ‫را ا ا‬ ‫ا‬ ‫ﻥ ﻡد‬‫– ا وب‬ ‫ﻥﺏ‬ ‫وزارة ا ﺏ ا‬‫ا ذ ‪ :‬ﻥ رس‬ ‫دة ا م ا‬ ‫إن‬ ‫ا‬ ‫‪:‬ا ما وا‬ ‫ا‬ ‫ا ة‪ 3:‬ت‬ ‫م‪3:‬عت‪،‬ر‪،‬تر‬‫‪Sujet : 3AS 03 - 03‬‬ ‫‪ :‬درا اه آ ‪.‬‬ ‫ا يا‬‫ا ا را ‪2011/2010 :‬‬ ‫‪2011/03/10 :‬‬ ‫ر‬ ‫א ن א ول ‪ ) :‬ﺏ ر ‪ – 2009‬م ﺕ ( )**(‬ ‫ا‪:‬‬ ‫اﻡ‬ ‫ن ا ارة ا ا ا ‪ 1-‬ﻡ ا‬ ‫ا‬ ‫‪.E=6V‬‬ ‫‪-‬ﻡ آ‬ ‫‪ -‬ﻡ ﺱ ‪. C = 1.2 µF‬‬ ‫‪ -‬ﻥ أوﻡ ﻡ وﻡ ‪. R = 5 kΩ‬‬ ‫‪.K -‬‬ ‫ﻥا ‪:‬‬ ‫ﻥ ن ا ات أو ا د ا‬ ‫‪-1‬‬ ‫‪R،E،‬و‪.C‬‬ ‫)‪duC (t‬‬ ‫‪،‬‬ ‫)‪uC(t‬‬ ‫‪dt‬‬ ‫‪1‬‬ ‫‪-t‬‬ ‫‪.‬‬ ‫آ‬ ‫= )‪uC (t‬‬ ‫‪E‬‬ ‫)‬ ‫‪:‬‬ ‫رة‬ ‫ا‬ ‫‪ -2‬ﻡ أن ا د ا ا‬ ‫‪( 1 - e RC‬‬ ‫‪ -3‬ﺡ د وﺡ ة ا ار ‪ ، RC‬ﻡ ﻡ ا‬ ‫؟ اذآ اﺱ ‪.‬‬ ‫ارة ا‬ ‫)‪ uC(t‬ا ت ا وﻥ ا ول ا ‪:‬‬ ‫اا‬ ‫‪ -4‬أﺡ‬ ‫)‪t(ms‬‬ ‫‪0‬‬ ‫‪6 12 18 24‬‬ ‫)‪uC(t) (V‬‬ ‫ا ﻥ )‪. uC(t) = f(t‬‬ ‫‪ -5‬أرﺱ ا‬ ‫ا‬ ‫‪ ، C ، R ، E‬أو‬ ‫‪ -6‬أو ا رة ا ة ا ر ا )‪i(t‬‬ ‫ﻡ )∞ = ‪. (t‬‬‫‪+‬‬ ‫)‪ ( t = 0‬و ) ∞ = ‪. ( t‬‬‫‪-‬‬ ‫‪ -7‬أآ رة ا ا ﺏ ا ﻥ ا ‪ ،‬أ‬ ‫‪:‬‬ ‫‪ ) :‬ﺏ ر ‪ – 2009‬ر ت ( )**(‬ ‫א نא‬ ‫ﻥ ا آ ا ﺏ ا ا ا ا ﺏ ﺏ لا‬ ‫)‪ (C‬ﻡ ﻥ ‪.‬‬ ‫▪ﻡ‬ ‫▪ ﻥ أوﻡ ﻡ وﻡ )‪( R = R’ = 470 Ω‬‬ ‫▪ ﻡ ذي ﺕ ﺕ ﺏ )‪. (E‬‬ ‫▪ ﺏ د )‪ ، (k‬أ ك ﺕ ‪.‬‬ ‫‪ /1‬ﻥ ا د ا )‪ (1‬ا ) ‪: ( t = 0‬‬ ‫‪. ،،‬‬ ‫–‪:‬‬ ‫–‬

‫‪2:‬‬ ‫‪2011/2010 :‬‬ ‫)‪- (03‬‬ ‫–‬ ‫‪-‬‬ ‫ﺏ ا ر ا ارة ﻡ ﺏ ا ﺕ ‪. uR ، uC‬‬ ‫ا ﺝ ا را‬ ‫أ‪ /‬ﺏ‬ ‫ا ﺕ ا ‪.q‬‬ ‫ا ‪ q = qA‬أوﺝ ا د ا‬ ‫‪ uC‬و ‪ uR‬ﺏ ﺵ‬ ‫ب‪/‬‬ ‫‪A‬و‪α‬ﺏ ‪.E،R،C‬‬ ‫‪. q(t) = A (1- e-αt ) :‬‬ ‫ها د ﻡا‬ ‫ﺝـ‪ /‬ﺕ‬ ‫)‪. (E‬‬ ‫ا )‪ ، (5V‬ا‬ ‫ﻥا ﺏ‬ ‫د‪ /‬إذا آ ﻥ ا ﺕ ا ﺏ‬ ‫ا )‪. (C‬‬ ‫هـ‪ /‬ﻡ ﺕ ا آ ﺕ ن )‪ . (EC = 5 mJ‬ا‬ ‫‪ /2‬ﻥ ا د ا ن ا )‪: (2‬‬ ‫أ‪ /‬ﻡ ذا ث ؟‬ ‫)‪ (2) (1‬د )‪. (k‬‬ ‫ﺏ اﻡا ا‬ ‫ب‪ /‬رن ﺏ‬ ‫א ن א ‪ ) :‬ﺏ ر ‪ – 2008‬م ﺕ ( )**(‬ ‫)ا ‪: (2-‬‬ ‫ﺕ ي ا ارة ا ﺏ ا‬ ‫‪ -‬ﻡ ﺕ ﺕ ا ﺏ ﺏ ‪. E = 12V‬‬ ‫‪ -‬ﻥ أوﻡ ﻡ وﻡ ‪. R = 10 Ω‬‬ ‫‪ -‬وﺵ ذاﺕ ‪ L‬و ﻡ وﻡ ‪. r‬‬ ‫‪.K -‬‬ ‫را اه از ﻡ ذي ذاآ ة ‪ ،‬ر ا ﺕ‬ ‫‪ -1‬ﻥ‬ ‫ﻡ ا ارة ا ﺏ ‪ ،‬آ‬ ‫ا ﺏ )‪ (uAB‬و )‪ . (ucB‬ﺏ‬ ‫رﺏ ا ارة ا ﺏ ﺏ ﺥ ه ا ا ز ‪.‬‬‫‪ -2‬ﻥ ا ‪ K‬ا ‪) t = 0‬ا ‪ (3-‬ا )‪ uBA=f(t‬ا ه را ا ه از ا ‪.‬‬ ‫ا م ا ا أوﺝ ‪:‬‬ ‫ﻡ ﺕ ا ارة‬ ‫أ‪ /‬ا ﺕ ا ﺏ )‪. (uBA‬‬ ‫ب‪ /‬ا ﺕ ا ﺏ )‪. (uCB‬‬ ‫ا ارة ‪.‬‬ ‫را ر‬ ‫ﺝـ‪ /‬ا ة ا‬ ‫‪ -3‬ﺏ د ا ن )ا ‪ . (3-‬ا ‪:‬‬ ‫أ‪ (τ) /‬ﺏ ا ﻡ ا ارة ‪.‬‬ ‫‪:‬‬ ‫‪. ،،‬‬ ‫–‪:‬‬ ‫–‬

‫‪3:‬‬ ‫‪2011/2010 :‬‬ ‫)‪- (03‬‬ ‫–‬ ‫‪-‬‬ ‫ب‪ /‬ﻡ وﻡ و ذاﺕ ا ﺵ ‪.‬‬ ‫ا ﻥ اﺵ ‪.‬‬ ‫‪ -4‬أ ا ا‬ ‫א ن א א ‪ ) :‬ﺏ ر ‪ – 2009‬ر ت( )**(‬ ‫ا ﺏا ‪:‬‬ ‫ﻥﺏ ا ا‬ ‫▪ ﻡ ذي ﺕ ﺕ ﺏ )‪. (E = 12V‬‬ ‫▪ وﺵ ذاﺕ )‪ (L = 300 mH‬وﻡ وﻡ )‪. (r = 10Ω‬‬ ‫▪ ﻥ أوﻡ ﻡ وﻡ )‪. (R = 110Ω‬‬ ‫▪ )‪) . (k‬ا ‪. (1-‬‬‫ا ارة ‪.‬‬ ‫ا ﺕ ﺵةا را ﺏ‬ ‫‪ -1‬ا ) ‪ (t = 0 s‬ﻥ ا )‪ : (k‬أوﺝ ا د ا‬‫ز ا ارة ؟‬ ‫رة ﺵ ة ا ر ا ﺏ ‪ I0‬ا ي‬ ‫‪ -2‬آ ن ك ا ﺵ ا م ا ا ؟ و ﻡ ه‬ ‫ﺏ ا ال‪. 1-‬‬ ‫‪-‬‬ ‫‪t‬‬ ‫دا ا‬ ‫‪i‬‬ ‫=‬ ‫‪A‬‬ ‫(‬ ‫‪1‬‬ ‫‪-‬‬ ‫‪e‬‬ ‫‪τ‬‬ ‫)‬ ‫‪ -3‬ﺏ ر ا‬ ‫ﻡ ‪A‬و‪.τ‬‬ ‫أ‪ /‬أوﺝ ا رة ا‬ ‫اﺵ ‪.‬‬ ‫رة ا ﺕ ا ﺏ ‪ uBC‬ﺏ‬ ‫ب‪ /‬ا‬ ‫ما ا ‪.‬‬ ‫ا ﺕ ا ﺏ ‪ uBC‬ا‬ ‫‪./4‬أ‪ /‬أ‬ ‫ب‪/‬ارﺱ آ ﺵ ا ن )‪. uBC = f(t‬‬ ‫א ن א س ‪ 3 ) :‬ع ف ‪(**) ( 025/12 – 01‬‬ ‫أوﻡ ﻡ وﻡ‬ ‫ﺕ ا آ ا ﺏ ‪،E‬ﻡ ر ‪،C‬ﻥ‬ ‫ﻡﻡ ﺕاﺏ‬ ‫ﺕ دارة آ ﺏ‬‫ان‬ ‫‪ ،‬ﻥ ﺏ ﺕ ر ﺵ ة ا ر ا ر ﺏ ارة ﺥ ل ا ﻡ‬ ‫‪. (1-‬‬ ‫‪) R = 50 Ω‬ا‬ ‫‪ -1‬ﻥ ا د‬ ‫ا )‪ (1‬ا‬ ‫ا )ا‬ ‫‪. (2-‬‬ ‫‪E‬‬ ‫ا ‪1-‬‬ ‫ا ‪2-‬‬ ‫‪-+‬‬ ‫)‪i(A‬‬ ‫‪0.2‬‬ ‫‪C‬‬ ‫‪K‬‬ ‫‪1‬‬ ‫‪R‬‬ ‫‪2‬‬ ‫‪A‬‬‫‪BM‬‬ ‫)‪t(ms‬‬ ‫‪:‬‬ ‫‪5‬‬ ‫ا دا ا ن أوﺝ ‪:‬‬ ‫• ﺵةا را ﺏ ا ما ا ‪.‬‬ ‫• ا ةا آ ا ﺏ ‪.‬‬ ‫• ا‪.‬‬ ‫‪.‬‬ ‫‪،،‬‬ ‫–‪:‬‬ ‫–‬

‫‪4:‬‬ ‫‪2011/2010 :‬‬ ‫)‪- (03‬‬ ‫–‬ ‫‪-‬‬ ‫‪Ln u‬‬ ‫‪ -2‬ﻥ ا د ا )‪. (2‬‬‫‪2.3‬‬ ‫ا ىا ي‪.‬‬ ‫أ‪ -‬ﺏ ﻡ ذا ث ا‬ ‫‪2.3‬‬ ‫ﺏ )‪ u u = f(t‬ا ﺕ ﺏ‬ ‫ب‪ -‬أوﺝ ا د ا‬ ‫ا‪.‬‬ ‫‪.‬‬ ‫‪u‬‬ ‫=‬ ‫‪Ee‬‬ ‫ـ‬ ‫‪t‬‬ ‫ﺝـ‪ -‬ﺕ ﻡ أن ه ا د ﻡ ا‬ ‫‪τ‬‬ ‫)‪t(ms‬‬ ‫‪ -3‬ﻥ ل ا ا ﺏ ﺏ أﺥ ى '‪ C‬ﻥ ا و‬ ‫اﻥ ء ا ﻥ ا ﻡ ﺝ ‪ ،‬ا ن ا ﺏ‬ ‫ا ﺏ اﻡ‬ ‫ﺕ ات ا ر ا ي ﺕ ﺏ‬ ‫)‪. ln u = f(t‬‬ ‫أ‪ -‬أوﺝ ا رة ا ﺏ ‪. E ، τ' ، t ، ln u‬‬ ‫أوﺝ ﺏ ا ﻡ‬ ‫ب‪ -‬ا دا ه ا رة و آ ا ا ن ا‬ ‫ا '‪. C‬‬ ‫ارة ا‬ ‫** ا ذ ‪ :‬ﻥ رس **‬ ‫ﻥ ﺝ إﺏ‬ ‫ﻥ ﻡد ﻥ ﺏ‬ ‫ا وب ‪-‬‬ ‫‪[email protected]‬‬ ‫‪Tel : 0771998109‬‬ ‫ا ا وﻥ ﺏ ي ﺥ ا روس أو ا ر و ‪.‬‬ ‫وﺵ ا ﻡ‬ ‫ﻥ ﻡ ه ا ا ع و ‪ .‬أدﺥ ﻡ ا ذ ‪:‬‬ ‫‪sites.google.com/site/faresfergani‬‬‫‪:‬‬ ‫‪. ،،‬‬ ‫–‪:‬‬ ‫–‬

5: 2011/2010 : - (03) – - ‫أﺝ ﺏ‬Sujet : 3AS 03 - 03 . ‫ درا اه آ‬: ‫ا يا‬ : ‫א ن א ول‬ : ‫ إ د ا د ا‬-1 : ‫ﻥ ن ﺝ ا ﺕ ات‬uE = uR + uCE = Ri + uC di + uCE=R dtdi 1 Edt + RC uC = RC . ‫و ه ﻡ د ﺕ ﻡ ا رﺝ ا و‬ : ‫دا‬ ‫ ه‬uC = E(1- e-t/RC ) ‫ إ ت أن‬-2 :uC = E(1- e-t/RC ) : ‫ا دا‬ ‫ﺏ‬duC = E e-t/RC dt RCE e-t/RC + 1 ( 1 - e-t/RC ) = ERC RC RCE e-t/RC + E - E e-t/RC = E → EE =RC RC RC RC RC RC . ‫دا‬ ‫إذن ا ا ه‬ : τ = RC ‫ ﺕ و ة‬-3[τ] = [R][C] [Q] [Q][τ] = [U] [I][∆t] → [τ] = [∆t] [I] [U] = [I] = [I] . ‫ ر ﺏ ﻥ‬τ ‫إذن‬ . 63% ‫ا ﺏ‬ ‫ ه ا ة ا زﻡ‬: ‫ ا ول ا‬- ‫ ﺏ ا ﻡ‬: ‫إ‬- : . ،، :– –

6: 2011/2010 : - (03) – - : ‫ ا ول‬-4 t (ms) 0 6 12 18 24 uC (V) 0 3.79 5.19 5.70 5.89 : uC = f(t) ‫ ا ن‬-5 7 uC (V) 6 5 4 3 2 1 t(s) 0 0 5 10 15 20 25 30 : i = f(t) ‫ـ‬ ‫ ا رة ا‬-6i = dq = d(C uC ) = C duCdt dt dt : ‫وﺝ ﻥ ﺏ‬duC = E e-t/RC : dt RCi = C E e-t/RC → i = E e-t/RC : ‫ رة ا ا ﻥ ا‬-7 RC RE(C) = 1 uC2 C 2 : . ،، :– –

7 : 2011/2010 : - (03) – -E(C) = 1 E2(1- e-t/RC ) : ‫א نא‬ C ‫ ﺝ ا ر و ﺕ ا ﺕ‬-‫ أ‬-1 2t→∞ → E(C) = 1 E2 C 2E(C) = 1 .1.2 .10-6 (6)2 = 2.16 .10-5 F = 2.16 F 2 : ‫ ﺏ‬uR ، uC +‫ــ‬ E uC C5 uR I : q ‫ ﺏ‬uR ‫ و‬uC ‫ ا‬-‫ب‬ q :q ‫ﺏ‬ ‫ا د ا‬-uC = C dquR = R i = R dt : ‫ﻥ ن ﺝ ا ﺕ ات‬uE = uR + uC . ‫و ه ﻡ د ﻡ ا رﺝ ا و‬ dq qE=R + dt Cdq 1 + q=Edt RC :E،R،C ‫ﺏ‬α‫و‬A ‫ ا‬-‫ﺝـ‬q = A (1- e-αt )dq = A e-αtdt α : ‫ا دا‬ ‫ﺏ‬A e-αt + A (1- e-αt ) = Eα RC R : . ،، :– –

‫‪8:‬‬ ‫‪2011/2010 :‬‬ ‫)‪- (03‬‬ ‫–‬ ‫‪-‬‬‫‪A‬‬ ‫‪e-αt‬‬ ‫‪+‬‬ ‫‪A‬‬ ‫‪-‬‬ ‫‪A‬‬ ‫‪e-αt‬‬ ‫=‬ ‫‪E‬‬‫‪α‬‬ ‫‪RC‬‬ ‫‪RC‬‬ ‫‪R‬‬ ‫أن ن ‪:‬‬ ‫ذ‬ ‫دا و‬ ‫اا ه‬‫‪▪ α = RC‬‬‫▪‬ ‫‪A‬‬ ‫=‬ ‫‪E‬‬ ‫→‬ ‫‪A = EC‬‬ ‫‪RC‬‬ ‫‪R‬‬ ‫د‪: E -‬‬ ‫ﺏ ﻥ ن ﺝ ا ﺕ ات آ ﺏ ‪:‬‬ ‫ﻡا ا‬‫‪E = R + uC‬‬ ‫ﻥ ا ﺕ مﺵةا رو ‪:‬‬‫‪E = uC → E = 5V‬‬ ‫هـ‪ -‬ا ‪:‬‬ ‫‪:‬‬‫)‪E(C‬‬ ‫=‬ ‫‪1‬‬ ‫‪C‬‬ ‫‪uC2‬‬ ‫‪2‬‬ ‫ن ‪: uC = E‬‬ ‫و ﻡﺕ ا آ‬‫)‪E(C‬‬ ‫=‬ ‫‪1‬‬ ‫‪C‬‬ ‫‪E2‬‬ ‫→‬ ‫‪C‬‬ ‫=‬ ‫)‪2E(C‬‬ ‫‪ -2‬أ‪ -‬ﺝ ا د ا‬ ‫‪2‬‬ ‫‪E2‬‬ ‫ب‪ -‬ا رﻥ ﺏ ‪ τ‬ا‬‫‪C‬‬ ‫=‬ ‫‪2‬‬ ‫(‬ ‫)‪5 .10-3‬‬ ‫=‬ ‫‪4‬‬ ‫‪.10-4‬‬ ‫‪F‬‬ ‫=‬ ‫‪400‬‬ ‫‪µF‬‬ ‫ا )‪ (1‬ﺕ ﺝ ا‬ ‫‪(5)2‬‬ ‫ا )‪ (2‬ﺕ ﺝ ا‬ ‫ا ا وﻡ ‪.‬‬ ‫)‪ (2‬ﺕ غ ا‬ ‫)‪: (2) ، (1‬‬ ‫ﻡ وﻡ وا ة ا ‪:‬‬‫‪τ1 = RC‬‬ ‫ﻡ وﻡ ’‪ R = R‬ا ‪:‬‬‫‪τ2 = (R + R’)C = 2RC → τ2 = 2τ1 → τ2 > τ1‬‬ ‫א نא ‪:‬‬ ‫‪ -1‬آ رﺏ ا ارة ﺏ ا ا ه از ا ‪:‬‬ ‫‪y1 y2‬‬ ‫‪ +‬ــ‬ ‫‪ -2‬أ‪ -‬ا ﺕ ‪ uBA‬ا م ا ا ‪:‬‬ ‫ﻡ ا نو ا ما ا ‪:‬‬‫‪uBA = uBA0 = 10 V‬‬ ‫‪. ،،‬‬ ‫–‪:‬‬ ‫–‬ ‫‪:‬‬

9: 2011/2010 : - (03) – -uCA = uCB + uBA : uCB -‫ب‬ : ‫ﻥ ن ﺝ ا ﺕ ات‬uCA0 = uCB0 + uBA0E = uCB0 + uBA0 : ‫و ا ما ا‬uCB0 = E – uBA0uCB0 = 12 – 10 = 2V : ‫ ﺵ ة ا ر ا‬-‫ﺝـ‬ : ‫و ا ما ا‬uBA = R i : τ -‫ أ‬-3uBA0 = RI0 → I0 = u BA0 ‫ﻡ ا‬t=0 ‫ﻡ ﺥ لﺕ ا س‬ R : ‫ ﻡ وﻡ و ذاﺕ ا ﺵ‬-‫ب‬ 10I0 = 10 = 1 A . τ = 2 . 10-3 s : ‫ ن‬uBA = E▪ I0 = E→ R+r= E → r= E -R R+r I0 I0r = 12 - 10 = 2 Ω 1▪ τ = L → L = τ ( R + r) R+rL = 2 .10-3 (10 + 2) = 2.4 .10-2 H : ‫ا ﻥ اﺵ‬ ‫ ا ا‬-4E(L) = 1 L I02 2E(L) = 1 . 2.4 .10-2 (1)2 = 1.2 .10-2 J 2 : ‫א نא א‬ : uC ‫ﺏ‬ ‫ ا د ا‬-1 : ‫ﻥ ن ﺝ ا ﺕ ات‬uAC = uAM + uMC . ،، :– – :

10 : 2011/2010 : - (03) – - diE = Ri + L + ri ................. (1) dt diL + (R + r) i = E dtdi R + r E . ‫و ه ﻡ د ﺕ ﻡ ا رﺝ ا و‬ + i=dt L L . ‫ك ﻥ أوﻡ‬ ‫ وﻡ ا ﺵ ﺕ ه ا‬di = 0 ‫ ا م ا ا ﺕ ن ﺵ ة ا ر ﺏ ا ن‬-2 dt : ‫ رة ﺵ ة ا ر‬- :‫ﻥ‬ ‫ا دا‬ di ‫ﺏ‬ =0 dtR+r E → i = E r i= R+ : τ ، A ‫ رة‬-‫ أ‬-3 LLi = A (1- e-t/τ )di = A e-t/τdt τ : ‫ا دا‬ ‫ﺏ‬A e-t/τ + R+ r . A ( 1 - e-t/τ ) = Eτ L LA e-t/τ + (R + r)A - (R + r)A e-t/τ = Eτ L L L : ‫أن ن‬ ‫ا واة‬ ‫و‬ ‫دا‬ ‫اا ه‬▪ A = (R + r)A → τ= L τ L R+r▪ (R + r)A = E → E L L A= R +r : ‫اﺵ‬ ‫ ﺏ‬uBC ‫ رة ا ﺕ‬-‫ب‬ diuBC = L dt + ri▪ i = E (1- e-t/τ ) R+r▪ di = E . 1 e- t/τ = E R+r r-t/τ = E e-t/τ dt R+ τ R+r L L r : uBC ‫رة‬ ‫ﺏ‬ : . ،، :– –

11 : 2011/2010 : - (03) – -u BC = E e-t/τ + r E r (1- e-t/τ ) L. R+ Lu BC = E e-t/τ + Er r (1- e-t/τ ) R+ : ‫ ا م ا ا‬uBC -‫ أ‬-4 : ‫ ن‬t = ∞ → e‫ـ‬t / τ = 0 ‫ا م ا ا أ ن‬ Er E ruBC = E (0) + R + r (1 - (0)) = R + r E r 10 .12uBC = R + r = 110 +10 = 1V : uBC = f(t) ‫ ر ا ن ﺏ آ‬-‫ب‬ uBC(V) 14 12 10 8 6 4 2 2 4 6 8 10 1 t (s) 0 . ‫ﺕ ن ﻡ وﻡ‬ :‫א نא س‬ 0 : ‫ ﺵ ة ا ر ا م ا ا‬-1t = 0 → i = I0 = 0.2 A ‫ا ةا آ ا ﺏ‬- :‫ﻡ ا ن‬ : . ،، :– –