# Exercice phis

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par Sbiro Abdelkrim                série d'exercices dipole RC   ‫ا‬        ‫ل‬   ‫ر‬                    ‫نا‬   ‫ا ا‬

t
du c U −τ
.           ‫أ‬     ،          ‫د ا‬       ‫ا‬   ‫ض‬        = e                  ‫قا‬    2- 1‫ا رس‬    ‫:1-1 ا‬     ‫ا‬
dt    τ
.         ξ e o = 1,35.10−4 J      ‫ا‬     ‫و‬       ‫ا‬   ‫ا‬     ‫د نا‬       ‫لا‬       ، -5-1 5,4J -4-1 300V -3-1

r = 1Ω - 3-2       τ = 1,2ms 2-2                1-2-
*********************************************************************************************
‫ا‬
du c                                   du       du
R.c.       + uc = E :  ‫ ⇐ ا د ا‬u R = R.i = R. c = R.c. c :  u R + u c = E : ‫ات‬   ‫ا‬      ‫ن‬                -1-1-1
dt                                     dt       dt
τ      1
. c = = 4 = 10−4 F = 100µF τ = 1s 4-1 .‫ د ا د ا ا رس‬τ = R.c 1-3 . ‫ا‬                          ‫ق وا‬        : 2-1
R 10 Ω
1-2 . ‫و‬          ‫ذات أ‬  ‫ا‬   ‫ا‬     ‫ة ، وه‬         ‫ةا‬    ‫ةآ آ‬    τ     ‫2-2 آ آ‬     2-   ξ e = 7,2.10 −3 J : 5-1

********************************************************************************************

:          ‫ا‬        ‫ا‬
‫ار .‬          ‫ه أ‬                  ‫‪ R = 0,5kΩ‬ا‬    ‫و‬      ‫أو‬               ‫‪، c = 100 µF‬‬       ‫ن‬          ‫غ‬

‫.‬       ‫ا‬            ‫‪uc‬‬    ‫ات ا‬    ‫2‬      ‫ا‬
‫.‬            ‫‪ R.c‬و رن‬
‫ا‬          ‫ا‬      ‫‪τ‬‬            ‫1-ا‬
‫ل د ا د د و ة ‪.τ‬‬                  ‫2-‬
‫0 = ‪.t‬‬      ‫ا‬        ‫ا‬                ‫ا‬      ‫3- د‬
‫.‬     ‫ا‬         ‫و‬     ‫ا‬       ‫ا‬       ‫ا‬        ‫4-‬
‫.‬             ‫ا‬       ‫ا‬       ‫ا‬        ‫5-‬
‫ده .‬           ‫ا دة ا ارة و د‬            ‫ا‬      ‫ا‬          ‫6- ا‬
‫. 2 – ا ا رس 3- ‪-5 1,8.10 −3 J - 4 6V‬‬ ‫ا‬               ‫: 1 - ‪ τ = R.c = 0,5.10 3 Ω × 100.10 −6 F = 0,05s‬و‬        ‫أ‬
‫ل ل.‬      ‫ار‬        ‫دت ا ارة‬           ‫ا‬       ‫و‬       ‫ا آ‬      ‫ا‬        ‫ا‬       ‫‪ - 6 0J‬ا‬
‫*********************************************************************************************‬

‫1‬
‫⇐ ‪. h' = 11,2cm‬‬        ‫: 1- ‪c(u co − u c ) = m.g .h' - 3 mgh = 25.10 −3 kg.10 N .kg −1 .0,4m = 0,1J -2 2,88.10 −2 J‬‬       ‫أ‬
‫2‬     ‫2‬

‫2‬
‫:‬       ‫ا‬          ‫ا‬
‫دس:‬          ‫ا‬            ‫ا‬
‫ا ر‪ K‬ةآ‬                            ‫ا‬         ‫ا‬        ‫ا آ‬
‫: ‪. E = 5V ، g = 10 N / kg ، c = 400 µF‬‬        ‫.‬                     ‫ا‬
‫ك؟ ذا؟‬        ‫ا‬             ‫1 (ه‬
‫.‬     ‫ا‬      ‫و‬     ‫ا‬          ‫ا‬          ‫ا‬      ‫2(ا‬
‫فا‬             ‫ا‬    ‫ا‬     ‫ك،‬      ‫ا‬        ‫ا ر ‪K‬‬                    ‫3(‬
‫‪ m = 5g‬ر ع ‪ . h‬أ‬          ‫ك ذات ا‬    ‫ود ا‬              ‫ف‬        ‫ا‬
‫ا ر ع ‪.h‬‬
‫ه ‪. h' = 7cm‬‬       ‫4-( ا ا ار ع ا‬
‫ذا ؟‬        ‫4-1-‬
‫ك.‬     ‫دود ا‬          ‫4-2 – ا‬
‫:‬        ‫أ‬
‫ك.‬   ‫ا‬         ‫ور ا ر ا‬          ‫ا آ‬       ‫ا‬     ‫آ‬      ‫ما‬    ‫نا‬       ‫ن ‪K‬‬                     ‫ك‬       ‫ا‬          ‫1-‬
‫‪ξ‬‬     ‫01.5‬  ‫3−‬
‫1‬
‫ا‬            ‫= ‪ 1-4 -4 . h = e‬ن‬      ‫3−‬
‫2 - ‪= 0,1m = 10cm ⇐ ξ e = m.g .h -3 ξ e = c.E 2 = 5.10 −3 J‬‬
‫01 × 01.5 ‪m.g‬‬                                        ‫2‬
‫ا‬       ‫ا‬     ‫ا‬      ‫رج ا‬    ‫ك=‬   ‫دود ا‬     ‫ل 4-2‬   ‫ل‬         ‫ا و‬        ‫ىا‬           ‫د‬              ‫ا‬
‫'‪mgh‬‬     ‫‪5.10 −3 kg .10 N .kg −1 .7.10 −2 m‬‬
‫= ‪.r‬‬          ‫=‬                                    ‫: %07 = 7,0 =‬
‫‪ξe‬‬                   ‫‪5.10 −3 J‬‬
.     ‫ وو ة آ‬b‫ و‬a ‫ ، د‬u c (t ) = a.t + b :   ‫آ‬  ‫د‬                     ‫ا م‬          ‫ ا‬uc   ‫ا‬
.i(t) : ‫ل‬      ‫ا رة ا‬‫ . آ‬i(t) ‫ة ا ر‬              ‫، وا‬      ‫ا‬          q (t)   ‫ا‬ ‫2– د‬
. t=0       ‫ا‬           ‫ا‬   ‫و‬     ‫ ا‬Eo         ‫3-1 – ا‬     : ‫ا ا‬   ‫3- أ‬
.t=15s         ‫ا‬        ‫ ا‬E1       ‫3-2 – ا‬
.   ‫فا‬               ‫ ا‬E 3 = mgh      ‫ا‬        ‫ا‬       - 3-3
. E2        ‫فا‬               ‫ا‬      ‫3-4 – ا‬
. g = 10 N / kg       . ‫ك‬       ‫3-5 – دود ا‬
*********************************************************************************************

u c (t ) = a.t + 5     ‫ و‬b=5V ⇐ 5 = a. × 0 + b ⇐ uc=5V ، t=0        ‫ و‬u c (t ) = a.t + b : -1 - II
2−5
u c (t ) = −0,2.t + 5 : ‫ و‬a =      = −0,2V / s ⇐ 2 = a. × 15 + 5    ⇐ uc=2V ، t=15s    :     ‫و‬
15
q = c.u c = c.(−0,2.t + 5) = 1 × (−0,2.t + 5) = −0,2.t + 5:    : q (t)     ‫ا‬    –2
dq d
.      ‫را‬                  ‫أن ر ا‬      ‫ ا رة - ل‬i (t ) =     = (−0,2.t + 5) = −0,2 A
dt dt
1         1                         1
E3 = mgh = 0,1kg ×10N.kg −1 × 3m = 3J -3-3 E1 = cu c = × 1 × 2 2 = 2 J -2-3 E o = cE 2 = 12,5 J -1-3 -3
2

2         2                         2
E3      3
r=      =       = 0,28. = 28% -5-3 E 2 = E o − E1 = 12,5 − 2 = 10,5 J    ‫فا‬             ‫ا‬     -4-3
E 2 10,5

SBIRO Abdelkrim
(pour toute observation contactez moi par mail)
sbiabdou@yahoo.fr

.      ‫ن وا‬   ‫ا‬     ‫وأ ل ا‬       ‫أد‬

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