( )C ( ) ( ) ( )C ( ) ( ) ( ) ( ) ( ) ( ) ( )C ( ) ( ) ( ) ( ) ( ) ( ) () () ( ) - laadj-lyes
ﺍﻟــﺴﻨﺔ ﺍﻟﺜﺎﻟﺜﺔ ﺛﺎﻧﻮﻱ ﺍﻟﻮﺣﺪﺓ ﺍﻟﺜـــﺎﻟﺜﺔ :ﺍﻟﻈـــﻮﺍﻫﺮ ﺍﻟﻜـــﻬﺮﺑﺎﺋﻴﺔ ﺷــﻌﺒﺔ ﲤﺎﺭﻳﻦ ﺍﻟﻮﺣﺪﺓ ﺍﻟﺜﺎﻟﺜﺔ ﻣﻦ ﻣﻮﺍﺿﻴﻊ ﺷﻬﺎﺩﺓ ﺍﻟﺘﻌﻠﻴﻢ ﺍﻟﺜﺎﻧﻮﻱ ﺍﻟﻌﻠﻮﻡ ﺍﻟﺘﺠﺮﻳﺒﻴﺔ – ﺭﻳﺎﺿﻴﺎﺕ – ﺗﻘﲏ ﺭﻳﺎﺿﻲ ﺍﻟﺘﻤﺮﻳﻦ ﺍﻷﻭﻝ : ﻋﻠﻮﻡ ﲡﺮﻳﺒﻴﺔ 2008 ﻗﺼﺪ ﺷﺤﻦ ﻣﻜﺜﻔﺔ ﻣﻔﺮﻏﺔ ،ﺳﻌﺘﻬﺎ ) ، ( Cﻧﺮﺑﻄﻬﺎ ﻋﻠﻰ ﺍﻟﺘﺴﻠﺴﻞ ﻣﻊ ﺍﻟﻌﻨﺎﺻﺮ ﺍﻟﻜﻬﺮﺑﺎﺋﻴﺔ ﺍﻟﺘﺎﻟﻴﺔ : ﻣﻮﻟﺪ ﻛﻬﺮﺑﺎﺋﻲ ﺫﻭ ﺗﻮﺗﺮ ﺛﺎﺑﺖ E = 3Vﻣﻘﺎﻭﻣﺘﻪ ﺍﻟﺪﺍﺧﻠﻴﺔ ﻣﻬﻤﻠﺔ .ﻧﺎﻗﻞ ﺃﻭﻣﻲ ﻣﻘﺎﻭﻣﺘﻪ . R = 104 W ﻗﺎﻃﻌﺔ . K ﻹﻇﻬﺎﺭ ﺍﻟﺘﻄﻮﺭ ﺍﻟﺰﻣﲏ ﻟﻠﺘﻮﺗﺮ ) uC ( tﺑﲔ ﻃﺮﰲ ﺍﳌﻜﺜﻔﺔ ،ﻧﺼﻠﻬﺎ ﺑﺮﺍﺳﻢ ﺍﻫﺘﺰﺍﺯ ﻣﻬﺒﻄﻲ ﺫﻱ ﺫﺍﻛﺮﺓ .ﺍﻟﺸﻜﻞ-4- ﺍﻟﺸــﻜﻞ-4- ﻧﻐﻠﻖ ﺍﻟﻘﺎﻃﻌﺔ Kﰲ ﺍﻟﻠﺤﻈﺔ t = 0ﻓﻨﺸﺎﻫﺪ ﻋﻠﻰ ﺷﺎﺷﺔ ﺭﺍﺳﻢ ﺍﻻﻫﺘﺰﺍﺯ ﺍﳌﻬﺒﻄﻲ ﺍﳌﻨﺤﲎ ) uC ( tﺍﳌﻤﺜﻞ ﰲ ﺍﻟﺸﻜﻞ-5- ) uC (V .1ﻣﺎ ﻫﻲ ﺷﺪﺓ ﺍﻟﺘﻴﺎﺭ ﺍﻟﻜﻬﺮﺑﺎﺋﻲ ﺍﳌﺎﺭ ﰲ ﺍﻟﺪﺍﺭﺓ ﺑﻌﺪ ﻣﺪﺓ Dt = 15 sﻣﻦ ﻏﻠﻘﻬﺎ ؟ .2ﺃﻋﻂ ﺍﻟﻌﺒﺎﺭﺓ ﺍﳊﺮﻓﻴﺔ ﻟﺜﺎﺑﺖ ﺍﻟﺰﻣﻦ ، tﻭﺑﲔ ﺃﻥ ﻟﻪ ﻧﻔﺲ ﻭﺣﺪﺓ ﻗﻴﺎﺱ ﺍﻟﺰﻣﻦ . .3ﻋﲔ ﺑﻴﺎﻧﻴﺎ ﻗﻴﻤﺔ tﻭﺍﺳﺘﻨﺘﺞ ﺍﻟﺴﻌﺔ ) ( Cﻟﻠﻤﻜﺜﻔﺔ . .4ﺑﻌﺪ ﻏﻠﻖ ﺍﻟﻘﺎﻃﻌﺔ ) ﰲ ﺍﻟﻠﺤﻈﺔ uC (V ) : ( t = 0 ﺃ -ﺃﻛﺘﺐ ﻋﺒﺎﺭﺓ ﺷﺪﺓ ﺍﻟﺘﻴﺎﺭ ﺍﻟﻜﻬﺮﺑﺎﺋﻲ ) i ( tﺍﳌﺎﺭ ﰲ ﺍﻟﺪﺍﺭﺓ ﺑﺪﻻﻟﺔ ) q ( tﺷﺤﻨﺔ ﺍﳌﻜﺜﻔﺔ . ﺏ-ﺍﻛﺘﺐ ﻋﺒﺎﺭﺓ ﺍﻟﺘﻮﺗﺮ ﺍﻟﻜﻬﺮﺑﺎﺋﻲ ) uC ( tﺑﲔ ﻟﺒﻮﺳﻲ ﺍﳌﻜﺜﻔﺔ ﺑﺪﻻﻟﺔ ﺍﻟﺸﺤﻨﺔ ) . q ( t duC ﺟ -ﺑﲔ ﺃﻥ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﺿﻠﻴﺔ ﺍﻟﱵ ﺗﻌﱪ ﻋﻦ ) uC ( tﺗﻌﻄﻰ ﺑﺎﻟﻌﺒﺎﺭﺓ = E : dt 0,5 ) t ( ms . uC + RC ﺍﻟﺸــﻜﻞ-5- 2 t - ö æ A .5ﻳﻌﻄﻰ ﺣﻞ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﺿﻠﻴﺔ ﺍﻟﺴﺎﺑﻘﺔ ﺑﺎﻟﻌﺒﺎﺭﺓ . uC ( t ) = E ç 1 - e ÷ :ﺍﺳﺘﻨﺘﺞ ﺍﻟﻌﺒﺎﺭﺓ ﺍﳊﺮﻓﻴﺔ ﻟﻠﺜﺎﺑﺖ . Aﻭﻣﺎ ﻫﻮ ﻣﺪﻟﻮﻟﻪ ﺍﻟﻔﻴﺰﻳﺎﺋﻲ ؟ è ø ﺍﳊﻞ ﺍﳌﻔﺼﻞ : . .1ﲢﺪﻳﺪ ﺷﺪﺓ ﺍﻟﺘﻴﺎﺭ ﺍﻟﻜﻬﺮﺑﺎﺋﻲ ﺍﳌﺎﺭ ﰲ ﺍﻟﺪﺍﺭﺓ ﺑﻌﺪ ﻣﺪﺓ Dt = 15 sﻣﻦ ﻏﻠﻘﻬﺎ . ﻣﻦ ﺍﻟﺒﻴﺎﻥ ﳌﺎ Dt = 15 sﳒﺪ uC = 3V :ﻭﺑﺎﻟﺘﺎﱄ u R = E - uC = 0 :؛ ﻟﻜﻦ u R = Ri :ﻭﻣﻨﻪ . i = 0 .2ﻛﺘﺎﺑﺔ ﻟﻌﺒﺎﺭﺓ ﺍﳊﺮﻓﻴﺔ ﻟﺜﺎﺑﺖ ﺍﻟﺰﻣﻦ ، tﻭﺗﺒﲔ ﺃﻥ ﻟﻪ ﻧﻔﺲ ﻭﺣﺪﺓ ﻗﻴﺎﺱ ﺍﻟﺰﻣﻦ . ﻋﺒﺎﺭﺓ ﺛﺎﺑﺖ ﺍﻟﺰﻣﻦ ﻫﻲ . t = RC : [U ] . [Q ] = [Q ] = [ I ][T ] = T ﻭﺑﺎﻟﺘﺎﱄ [ ] : ] [ I ] [U ] [ I ] [I .3ﺗﻌﲔ ﺑﻴﺎﻧﻴﺎ ﻗﻴﻤﺔ tﻭﺍﺳﺘﻨﺘﺎﺝ ﺍﻟﺴﻌﺔ ) ( Cﻟﻠﻤﻜﺜﻔﺔ . = ] ، [t ] = [ R ][ Cﻭﻣﻨﻪ ﺛﺎﺑﺖ ﺍﻟﺰﻣﻦ ﻣﺘﺠﺎﻧﺲ ﻣﻊ ﻭﺣﺪﺓ ﺍﻟﺰﻣﻦ . ﻣﻦ ﺍﻟﺒﻴﺎﻥ ﳌﱠﺎ uC = 0, 63 ´ E = 0, 63 ´ 3 = 1,89V :ﳒﺪ . t = 1, 25 ´ 2 = 2,5s : 2, 5 t ﻭﻟﺪﻳﻨﺎ t = RC :ﻭﺑﺎﻟﺘﺎﱄ : = Cﺇﺫﻥ = 2, 5 ´ 10 - 4 F = 250 m F : 4 10 R =.C ) dq ( t .4ﺃ -ﻋﺒﺎﺭﺓ ﺷﺪﺓ ﺍﻟﺘﻴﺎﺭ ﺍﻟﻜﻬﺮﺑﺎﺋﻲ ) i ( tﺍﳌﺎﺭ ﰲ ﺍﻟﺪﺍﺭﺓ ﺑﺪﻻﻟﺔ ) q ( tﺷﺤﻨﺔ ﺍﳌﻜﺜﻔﺔ : dt = ) i (t 1 ﺏ -ﻋﺒﺎﺭﺓ ﺍﻟﺘﻮﺗﺮ ﺍﻟﻜﻬﺮﺑﺎﺋﻲ ) uC ( tﺑﲔ ﻟﺒﻮﺳﻲ ﺍﳌﻜﺜﻔﺔ ﺑﺪﻻﻟﺔ ) q ( tﺷﺤﻨﺔ ﺍﳌﻜﺜﻔﺔ .q ( t ) : C ﺟ -ﻛﺘﺎﺑﺔ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﺿﻠﻴﺔ ﺍﻟﱵ ﺗﻌﱪ ﻋﻦ ) . uC ( t ﺣﺴﺐ ﻗﺎﻧﻮﻥ ﲨﻊ ﺍﻟﺘﻮﺗﺮﺍﺕ ﻟﺪﻳﻨﺎ . uC + u R = E : ﺻﻔﺤﺔ 1 = ) uC ( t ) dq ( t ) du ( t duC ، u R ( t ) = R.i ( t ) = Rﻭﻣﻨﻪ = E : ﻭﻟﺪﻳﻨﺎ : = RC C dt dt dt .5 uC + RC ﺍﺳﺘﻨﺘﺎﺝ ﺍﻟﻌﺒﺎﺭﺓ ﺍﳊﺮﻓﻴﺔ ﻟﻠﺜﺎﺑﺖ . Aﻭﲢﺪﻳﺪ ﻣﺪﻟﻮﻟﻪ ﺍﻟﻔﻴﺰﻳﺎﺋﻲ ؟ t - ö duC ( t ) E - At æ ﻟﺪﻳﻨﺎ ، uC ( t ) = E ç 1 - e A ÷ :ﻭﺑﺎﻟﺘﺎﱄ : = e dt A è ø . t - ö æ RC - At du E -t ﻭﻟﺪﻳﻨﺎ ، uC + RC C = E :ﺇﺫﻥ ، E ç 1 - e A ÷ + RC e A = E :ﺃﻱ e = 1 : A dt A è ø t RC æ RC ö -A ﻭﻣﻨﻪ A = RC :ﻭﻫﻮ ﺛﺎﺑﺖ ﺍﻟﺰﻣﻦ çﻭﺑﺎﻟﺘﺎﱄ - 1 = 0 : ﺇﺫﻥ - 1 ÷ e = 0 : A è A ø ﻣﺪﻟﻮﻟﻪ ﺍﻟﻔﻴﺰﻳﺎﺋﻲ :ﻫﻮ ﺍﳌﺪﺓ ﺍﻟﺰﻣﻨﻴﺔ ﻟﺒﻠﻮﻍ ﺷﺤﻨﺔ -ﻓﺮﻕ ﺍﻟﻜﻤﻮﻥ ﺑﲔ ﻃﺮﰲ -ﺍﳌﻜﺜﻔﺔ 63%ﻣﻦ ﻗﻴﻤﺘﻬﺎ ﺍﻷﻋﻈﻤﻴﺔ ،ﻭﻫﻮ ﻣﺆﺷﺮ ﳌﺪﺓ ﺷﺤﻦ ﺍﳌﻜﺜﻔﺔ . + t A - . 1- e ﺍﻟﺘﻤﺮﻳﻦ ﺍﻟﺜﺎﱐ : ﻋﻠﻮﻡ ﲡﺮﻳﺒﻴﺔ 2008 ﲢﺘﻮﻱ ﺍﻟﺪﺍﺭﺓ ﺍﻟﻜﻬﺮﺑﺎﺋﻴﺔ ﺍﳌﺒﻴﻨﺔ ﰲ ﺍﻟﺸﻜﻞ -2-ﻋﻠﻰ : · ﻣﻮﻟﺪ ﺗﻮﺗﺮﻩ ﺍﻟﻜﻬﺮﺑﺎﺋﻲ ﺛﺎﺑﺖ . E = 12V · ﻧﺎﻗﻞ ﺃﻭﻣﻲ ﻣﻘﺎﻭﻣﺘﻪ . R = 10W · ﻭﺷﻴﻌﺔ ﺫﺍﺗﻴﺘﻬﺎ Lﻭﻣﻘﺎﻭﻣﺘﻬﺎ . r · ﻗﺎﻃﻌﺔ . K .1ﻧﺴﺘﻌﻤﻞ ﺭﺍﺳﻢ ﺍﻫﺘﺰﺍﺯ ﻣﻬﺒﻄﻲ ﺫﻱ ﺫﺍﻛﺮﺓ ،ﻹﻇﻬﺎﺭ ﺍﻟﺘﻮﺗﺮﻳﻦ ﺍﻟﻜﻬﺮﺑﺎﺋﻴﲔ ) ( u BAﻭ ) . ( uCB ﺑﲔ ﻋﻠﻰ ﳐﻄﻂ ﺍﻟﺪﺍﺭﺓ ﺍﻟﻜﻬﺮﺑﺎﺋﻴﺔ ،ﻛﻴﻒ ﻳﺘﻢ ﺭﺑﻂ ﺍﻟﺪﺍﺭﺓ ﺍﻟﻜﻬﺮﺑﺎﺋﻴﺔ ﲟﺪﺧﻠﻲ ﻫﺬﺍ ﺍﳉﻬﺎﺯ ؟ ﺍﻟﺸﻜــﻞ -2- .2ﻧﻐﻠﻖ ﺍﻟﻘﺎﻃﻌﺔ Kﰲ ﺍﻟﻠﺤﻈﺔ . t = 0ﳝﺜﻞ ﺍﻟﺸﻜﻞ -3-ﺍﳌﻨﺤﲎ ) u BA = f ( t ﺍﳌﺸﺎﻫﺪ ﻋﻠﻰ ﺷﺎﺷﺔ ﺭﺍﺳﻢ ﺍﻻﻫﺘﺰﺍﺯ ﺍﳌﻬﺒﻄﻰ . ) u BA (V ﻋﻨﺪﻣﺎ ﺗﺼﺒﺢ ﺍﻟﺪﺍﺭﺓ ﰲ ﺣﺎﻟﺔ ﺍﻟﻨﻈﺎﻡ ﺍﻟﺪﺍﺋﻢ ،ﺃﻭﺟﺪ ﻗﻴﻤﺔ : ﺃ -ﺍﻟﺘﻮﺗﺮ ﺍﻟﻜﻬﺮﺑﺎﺋﻲ ) . ( u BA ﺏ-ﺍﻟﺘﻮﺗﺮ ﺍﻟﻜﻬﺮﺑﺎﺋﻲ ) . ( uCB ﺟ -ﺍﻟﺸﺪﺓ ﺍﻟﻌﻈﻤﻰ ﻟﻠﺘﻴﺎﺭ ﺍﳌﺎﺭ ﰲ ﺍﻟﺪﺍﺭﺓ . .3ﺑﺎﻻﻋﺘﻤﺎﺩ ﻋﻠﻰ ﺍﻟﺒﻴﺎﻥ )ﺍﻟﺸـﻜﻞ (-3-ﺍﺳﺘﻨﺘﺞ : 2 ﺃ -ﻗﻴﻤﺔ ) (tﺛﺎﺑﺖ ﺍﻟﺰﻣﻦ ﺍﳌﻤﻴﺰ ﻟﻠﺪﺍﺭﺓ . ﺏ-ﻣﻘﺎﻭﻣﺔ ﻭﺫﺍﺗﻴﺔ ﺍﻟﻮﺷﻴﻌﺔ . ) t ( ms .4ﺍﺣﺴﺐ ﺍﻟﻄﺎﻗﺔ ﺍﻷﻋﻈﻤﻴﺔ ﺍﳌﺨﺰﻧﺔ ﰲ ﺍﻟﻮﺷﻴﻌﺔ . ﺍﻟﺸــﻜﻞ -3- ﺍﳊﻞ ﺍﳌﻔﺼﻞ : . .1ﺗﻮﺿﻴﺢ ﻛﻴﻔﻴﺔ ﺭﺑﻂ ﺍﻟﺪﺍﺭﺓ ﺍﻟﻜﻬﺮﺑﺎﺋﻴﺔ ﲟﺪﺧﻠﻲ ﻫﺬﺍ ﺟﻬﺎﺯ ﺭﺍﺳﻢ ﺍﻹﻫﺘﺰﺍﺯ ﺍﳌﻬﺒﻄﻲ . ﺃﻧﻈﺮ ﺍﻟﺸﻜﻞ ﺍﳌﻘﺎﺑﻞ . .2ﻣﻦ ﺍﻟﺒﻴﺎﻥ ﰲ ﺍﻟﻨﻈﺎﻡ ﺍﻟﺪﺍﺋﻢ ﳒﺪ : ﺃ -ﺍﻟﺘﻮﺗﺮ ﺍﻟﻜﻬﺮﺑﺎﺋﻲ ) . u BA = u R = 10V : ( u BA ﺏ -ﺍﻟﺘﻮﺗﺮ ﺍﻟﻜﻬﺮﺑﺎﺋﻲ ) . uCB = ub = E - u R = 12 - 10 = 2V : ( uCB u R 10 = ﺟ -ﺍﻟﺸﺪﺓ ﺍﻟﻌﻈﻤﻰ ﻟﻠﺘﻴﺎﺭ ﺍﳌﺎﺭ ﰲ ﺍﻟﺪﺍﺭﺓ = 1 A : R 10 2 0 =I .3ﺃ -ﲢﺪﻳﺪ ﻗﻴﻤﺔ ) (tﺛﺎﺑﺖ ﺍﻟﺰﻣﻦ ﺍﳌﻤﻴﺰ :ﻣﻦ ﺍﻟﺒﻴﺎﻥ ﳌﺎ u R = 0, 63 ´ 10 = 6, 3Vﳒﺪ . t = 0,8 ´ 2 = 1, 6s : ﺏ -ﻣﻘﺎﻭﻣﺔ ﻭﺫﺍﺗﻴﺔ ﺍﻟﻮﺷﻴﻌﺔ . ﺻﻔﺤﺔ 2 ub 2 ﻟﺪﻳﻨﺎ :ﰲ ﺍﻟﻨﻈﺎﻡ ﺍﻟﺪﺍﺋﻢ ub = r.Iﻭﺑﺎﻟﺘﺎﱄ = = 2W : I 1 L ﻭﻟﺪﻳﻨﺎ : = L = t ( R + r ) = 1, 6 ´ 10 -3 ´ 12 = 1,92 ´ 10 -3 H ، t R+r =. r .4ﺣﺴﺎﺏ ﺍﻟﻄﺎﻗﺔ ﺍﻷﻋﻈﻤﻴﺔ ﺍﳌﺨﺰﻧﺔ ﰲ ﺍﻟﻮﺷﻴﻌﺔ . 1 2 1 2 ﻟﺪﻳﻨﺎ E ( L ) = L.I 2 :ﻭﺑﺎﻟﺘﺎﱄ . E ( L ) = ´ 1,92 ´ 10-3. (1) = 9, 6 ´ 10-4 J : 2 ﺍﻟﺘﻤﺮﻳﻦ ﺍﻟﺜﺎﻟﺚ : ﺭﻳﺎﺿﻴﺎﺕ ﻭ ﺗﻘﲏ ﺭﻳﺎﺿﻲ 2008 ﺑﻐﺮﺽ ﻣﻌﺮﻓﺔ ﺳﻠﻮﻙ ﻭﳑﻴﺰﺍﺕ ﻭﺷﻴﻌﺔ ﻣﻘﺎﻭﻣﺘﻬﺎ ) ( rﻭﺫﺍﺗﻴﺘﻬﺎ ) ، ( Lﻧﺮﺑﻄﻬﺎ ﻋﻠﻰ ﺍﻟﺘﺴﻠﺴﻞ ﲟﻮﻟﺪ ﺫﻱ ﺗﻮﺗﺮ ﻛﻬﺮﺑﺎﺋﻲ ﺛﺎﺑﺖ E = 4,5Vﻭ ﻗﺎﻃﻌﺔ . Kﺍﻟﺸﻜﻞ -1- .1ﺃﻧﻘﻞ ﳐﻄﻂ ﺍﻟﺪﺍﺭﺓ ﻋﻠﻰ ﻭﺭﻗﺔ ﺍﻹﺟﺎﺑﺔ ﻭﺑﲔ ﻋﻠﻴﻪ ﺟﻬﺔ ﻣﺮﻭﺭ ﺍﻟﺘﻴﺎﺭ ﺍﻟﻜﻬﺮﺑﺎﺋﻲ ﻭﺟﻬﱵ ﺍﻟﺴﻬﻤﲔ ﺍﻟﺬﻳﻦ ﳝﺜﻼﻥ ﺍﻟﺘﻮﺗﺮ ﺍﻟﻜﻬﺮﺑﺎﺋﻲ ﺑﲔ ﻃﺮﰲ ﺍﻟﻮﺷﻴﻌﺔ ﻭﺑﲔ ﻃﺮﰲ ﺍﳌﻮﻟﺪ . .2ﰲ ﺍﻟﻠﺤﻈﺔ t = 0ﻧﻐﻠﻖ ﺍﻟﻘﺎﻃﻌﺔ : K ﺃ -ﺑﺘﻄﺒﻴﻖ ﻗﺎﻧﻮﻥ ﲨﻊ ﺍﻟﺘﻮﺗﺮﺍﺕ ،ﺃﻭﺟﺪ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﺿﻠﻴﺔ ﺍﻟﱵ ﺗﻌﻄﻲ ﺍﻟﺸﺪﺓ ﺍﻟﻠﺤﻈﻴﺔ ) i ( tﻟﻠﺘﻴﺎﺭ ﺍﻟﺸـــﻜﻞ -1- ﺍﻟﻜﻬﺮﺑﺎﺋﻲ ﺍﳌﺎﺭ ﰲ ﺍﻟﺪﺍﺭﺓ . r - t ö æ ﺏ-ﺑﲔ ﺃﻥ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﺿﻠﻴﺔ ﺍﻟﺴﺎﺑﻘﺔ ﺗﻘﺒﻞ ﺣﻼ ﻣﻦ ﺍﻟﺸﻜﻞ ÷ i ( t ) = I 0 ç 1 - e Lﺣﻴﺚ I 0ﻫﻲ ﺍﻟﺸﺪﺓ ﺍﻟﻌﻈﻤﻰ ﻟﻠﺘﻴﺎﺭ ﺍﻟﻜﻬﺮﺑﺎﺋﻲ ﺍﳌﺎﺭ ﰲ ﺍﻟﺪﺍﺭﺓ . è ø .3ﺗﻌﻄﻰ ﺍﻟﺸﺪﺓ ﺍﻟﻠﺤﻈﻴﺔ ﻟﻠﺘﻴﺎﺭ ﺍﻟﻜﻬﺮﺑﺎﺋﻲ ﺑﺎﻟﻌﺒﺎﺭﺓ i ( t ) = 0, 45 (1 - e -10 t ) :ﺣﻴﺚ ) ( tﺑﺎﻟﺜﺎﻧﻴﺔ ﻭ ) ( iﺑﺎﻵﻣﺒﲑ . ﺃﺣﺴﺐ ﻗﻴﻢ ﺍﳌﻘﺎﺩﻳﺮ ﺍﻟﻜﻬﺮﺑﺎﺋﻴﺔ ﺍﻟﺘﺎﻟﻴﺔ :ﺃ -ﺍﻟﺸﺪﺓ ﺍﻟﻌﻈﻤﻰ ) ( I 0ﻟﻠﺘﻴﺎﺭ ﺍﻟﻜﻬﺮﺑﺎﺋﻲ ﺍﳌﺎﺭ ﰲ ﺍﻟﺪﺍﺭﺓ . ﺏ-ﺍﳌﻘﺎﻭﻣﺔ ) ( rﻟﻠﻮﺷﻴﻌﺔ . ﺟ -ﺍﻟﺬﺍﺗﻴﺔ ) ( Lﻟﻠﻮﺷﻴﻌﺔ . ﺩ -ﺛﺎﺑﺖ ﺍﻟﺰﻣﻦ ) (tﺍﳌﻤﻴﺰ ﻟﻠﺪﺍﺭﺓ . .4ﺃ -ﻣﺎ ﻗﻴﻤﺔ ﺍﻟﻄﺎﻗﺔ ﺍﳌﺨﺰﻧﺔ ﰲ ﺍﻟﻮﺷﻴﻌﺔ ﰲ ﺣﺎﻟﺔ ﺍﻟﻨﻈﺎﻡ ﺍﻟﺪﺍﺋﻢ ؟ ﺏ -ﺃﻛﺘﺐ ﻋﺒﺎﺭﺓ ﺍﻟﺘﻮﺗﺮ ﺍﻟﻜﻬﺮﺑﺎﺋﻲ ﺍﻟﻠﺤﻈﻲ ﺑﲔ ﻃﺮﰲ ﺍﻟﻮﺷﻴﻌﺔ . ﺟ -ﺃﺣﺴﺐ ﻗﻴﻤﺔ ﺍﻟﺘﻮﺗﺮ ﺍﻟﻜﻬﺮﺑﺎﺋﻲ ﺑﲔ ﻃﺮﰲ ﺍﻟﻮﺷﻴﻌﺔ ﰲ ﺍﻟﻠﺤﻈﺔ ) . ( t = 0,3s ﺍﳊﻞ ﺍﳌﻔﺼﻞ : . .1ﺭﺳﻢ ﳐﻄﻂ ﺍﻟﺪﺍﺭﺓ .ﺍﻟﺮﺳﻢ ﺑﺎﻟﺸﻜﻞ ﺍﳌﻘﺎﺑﻞ . .2ﺃ -ﻛﺘﺎﺑﺔ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﺿﻠﻴﺔ ﺍﳌﻤﻴﺰﺓ ﻟﻠﺪﺍﺭﺓ . ﺣﺴﺐ ﻗﺎﻧﻮﻥ ﲨﻊ ﺍﻟﺘﻮﺗﺮﺍﺕ ﻟﺪﻳﻨﺎ ub ( t ) = E :ﺇﺫﻥ= E : ) di ( t ) di ( t dt . ri ( t ) + L r E ﻭﻣﻨﻪ + i ( t ) = : dt L L ﺏ -ﺍﻟﺘﺄﻛﺪ ﻣﻦ ﺣﻞ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﺻﻠﻴﺔ . . r - t ö æ di r -rt ﻟﺪﻳﻨﺎ ، i ( t ) = I 0 ç 1 - e L ÷ :ﻭﺑﺎﻟﺘﺎﱄ = I 0 . .e L : dt L è ø . r - tö di ( t ) R E r -rt r æ r ﺇﺫﻥ + i ( t ) = I 0 . .e L + I 0 ç1 - e L ÷ = I 0 : ؛ ﻟﻜﻦ : r dt L L L è ø L ﺻﻔﺤﺔ 3 = . I0 r - t ö di ( t ) R æ r E E = ) + i (t = .ﻭﻣﻨﻪ ﺍﻟﻌﺒﺎﺭﺓ i ( t ) = I 0 ç 1 - e L ÷ :ﺣﻞ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﺿﻠﻴﺔ . ﻭﺑﺎﻟﺘﺎﱄ : dt L Lr L è ø .3ﺗﻌﲔ ﻗﻴﻢ ﺍﳌﻘﺎﺩﻳﺮ ﺍﻟﻜﻬﺮﺑﺎﺋﻴﺔ ﺍﳌﻤﻴﺰﺓ ﻟﻠﺪﺍﺭﺓ . ﺃ -ﺍﻟﺸﺪﺓ ﺍﻷﻋﻈﻤﻴﺔ ﻟﻠﺘﻴﺎﺭ ﺍﳌﺎﺭ ﰲ ﺍﻟﺪﺍﺭﺓ :ﰲ ﺍﻟﻨﻈﺎﻡ ﺍﻟﺪﺍﺋﻢ ) ( t > 5tﻟﺪﻳﻨﺎ . I 0 = 0, 45 A : E 4, 5 E = = ، I 0ﻭﺑﺎﻟﺘﺎﱄ = 10W ﺏ -ﻣﻘﺎﻭﻣﺔ ﺍﻟﻮﺷﻴﻌﺔ :ﻟﺪﻳﻨﺎ I 0 0, 45 r r 10 r ﺟ -ﺫﺍﺗﻴﺔ ﺍﻟﻮﺷﻴﻌﺔ :ﻟﺪﻳﻨﺎ ، = 10ﻭﺑﺎﻟﺘﺎﱄ = = 1H 10 10 L 1 L ﺩ -ﺛﺎﺑﺖ ﺍﻟﺰﻣﻦ :ﻟﺪﻳﻨﺎ = ، tﻭﺑﺎﻟﺘﺎﱄ t = = 0,1s 10 r .4ﺃ -ﺍﻟﻄﺎﻗﺔ ﺍﳌﺨﺰﻧﺔ ﰲ ﺍﻟﻮﺷﻴﻌﺔ ﰲ ﺣﺎﻟﺔ ﺍﻟﻨﻈﺎﻡ ﺍﻟﺪﺍﺋﻢ . =. r =. L 1 1 2 ﻟﺪﻳﻨﺎ ، E ( L ) = L.I 02 :ﻭﺑﺎﻟﺘﺎﱄ . E ( L ) = ´1´ ( 0, 45) = 0,101J 2 2 ﺏ -ﻛﺘﺎﺑﺔ ﻋﺒﺎﺭﺓ ﺍﻟﺘﻮﺗﺮ ﺍﻟﻜﻬﺮﺑﺎﺋﻲ ﺍﻟﻠﺤﻈﻲ ﺑﲔ ﻃﺮﰲ ﺍﻟﻮﺷﻴﻌﺔ . ﻟﺪﻳﻨﺎ . ub = E = 4,5V : r - t di r - Lr t L ، uL ( t ) = L = L.I 0 . .e = E.eﻭﻣﻨﻪ : ﻋﺒﺎﺭﺓ ﻓﺮﻕ ﺍﻟﻜﻤﻮﻥ ﺍﻟﺬﺍﰐ : dt L -10 t . uL ( t ) = 4, 5.e ﺟ -ﺣﺴﺎﺏ ﻗﻴﻤﺔ ﺍﻟﺘﻮﺗﺮ ﺍﻟﺬﺍﰐ ﰲ ﺍﻟﻠﺤﻈﺔ ) . ( t = 0,3s ﻟﺪﻳﻨﺎ ، uL ( t ) = 4, 5.e-10t :ﻭﺑﺎﻟﺘﺎﱄ . uL ( 0,3) = 4,5.e -10´0,3 = 4,5.e -3 = 0, 224V : ﺍﻟﺘﻤﺮﻳﻦ ﺍﻟﺮﺍﺑﻊ : ﺭﻳﺎﺿﻴﺎﺕ ﻭ ﺗﻘﲏ ﺭﻳﺎﺿﻲ 2008 ﰲ ﺣﺼﺔ ﻟﻸﻋﻤﺎﻝ ﺍﳌﺨﱪﻳﺔ ﺍﻗﺘﺮﺡ ﺍﻷﺳﺘﺎﺫ ﻋﻠﻰ ﺗﻼﻣﻴﺬﻩ ﳐﻄﻂ ﺍﻟﺪﺍﺭﺓ ﺍﳌﻤﺜﻠﺔ ﰲ ﺍﻟﺸﻜﻞ -2-ﻟﺪﺭﺍﺳﺔ ﺛﻨﺎﺋﻲ ﺍﻟﻘﻄﺐ . RC ﺗﺘﻜﻮﻥ ﺍﻟﺪﺍﺭﺓ ﻣﻦ ﺍﻟﻌﻨﺎﺻﺮ ﺍﻟﻜﻬﺮﺑﺎﺋﻴﺔ ﺍﻟﺘﺎﻟﻴﺔ : § ﻣﻮﻟﺪ ﺗﻮﺗﺮﻩ ﺍﻟﻜﻬﺮﺑﺎﺋﻲ ﺛﺎﺑﺖ . E = 12V : § ﻣﻜﺜﻔﺔ )ﻏﲑ ﻣﺸﺤﻮﻧﺔ( ﺳﻌﺘﻬﺎ . C = 1,0m F § ﻧﺎﻗﻞ ﺃﻭﻣﻲ ﻣﻘﺎﻭﻣﺘﻪ . R = 5 ´103 W ﺍﻟﺸــﻜﻞ -2- § ﺑـــﺎﺩﻟﺔ . K .1ﳒﻌﻞ ﺍﻟﺒﺎﺩﻟﺔ ﰲ ﺍﻟﻠﺤﻈﺔ t = 0ﻋﻠﻰ ﺍﻟﻮﺿﻊ ). (1 ﺃ -ﻣﺎﺫﺍ ﳛﺪﺙ ﻟﻠﻤﻜﺜﻔﺔ ؟ ﺏ-ﻛﻴﻒ ﳝﻜﻦ ﻋﻤﻠﻴﺎ ﻣﺸﺎﻫﺪﺓ ﺍﻟﺘﻄﻮﺭ ﺍﻟﺰﻣﲏ ﻟﻠﺘﻮﺗﺮ ﺍﻟﻜﻬﺮﺑﺎﺋﻲ u AB؟ du AB ﺟ -ﺑﲔ ﺃﻥ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﺿﻠﻴﺔ ﺍﻟﱵ ﲢﻜﻢ ﺍﺷﺘﻐﺎﻝ ﺍﻟﺪﺍﺭﺓ ﺍﻟﻜﻬﺮﺑﺎﺋﻴﺔ ﻋﺒﺎﺭﺎ + u AB = E : dt . RC ﺩ -ﺃﻋﻂ ﻋﺒﺎﺭﺓ ) (tﺍﻟﺜﺎﺑﺖ ﺍﳌﻤﻴﺰ ﻟﻠﺪﺍﺭﺓ ،ﻭﺑﲔ ﺑﺎﺳﺘﻌﻤﺎﻝ ﺍﻟﺘﺤﻠﻴﻞ ﺍﻟﺒﻌﺪﻱ ﺃﻧﻪ ﻳﻘﺪﺭ ﺑﺎﻟﺜﺎﻧﻴﺔ ﰲ ﺍﻟﻨﻈﺎﻡ ﺍﻟﺪﻭﱄ ﻟﻠﻮﺣﺪﺍﺕ ) . ( SI t - ö æ ﻫ -ﺑﲔ ﺃﻥ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﺿﻠﻴﺔ ﺍﻟﺴﺎﺑﻘﺔ ) -1ﺟ( ﺗﻘﺒﻞ ﺍﻟﻌﺒﺎﺭﺓ u AB = E ç1 - e t ÷ :ﺣﻼ ﳍﺎ . è ø ﻭ -ﺃﺭﺳﻢ ﺷﻜﻞ ﺍﳌﻨﺤﲏ ﺍﻟﺒﻴﺎﱐ ﺍﳌﻤﺜﻞ ﻟﻠﺘﻮﺗﺮ ﺍﻟﻜﻬﺮﺑﺎﺋﻲ ) u AB = f ( tﻭﺑﲔ ﻛﻴﻔﻴﺔ ﲢﺪﻳﺪ tﻣﻦ ﺍﻟﺒﻴﺎﻥ . ﻱ -ﻗﺎﺭﻥ ﺑﲔ ﻗﻴﻤﺔ ﺍﻟﺘﻮﺗﺮ u ABﰲ ﺍﻟﻠﺤﻈﺔ t = 5tﻭ . Eﻣﺎﺫﺍ ﺗﺴﺘﻨﺘﺞ ؟ .2ﺑﻌﺪ ﺍﻻﻧﺘﻬﺎﺀ ﻣﻦ ﺍﻟﺪﺭﺍﺳﺔ ﺍﻟﺴﺎﺑﻘﺔ ،ﳒﻌﻞ ﺍﻟﺒﺎﺩﻟﺔ ﰲ ﺍﻟﻮﺿﻊ ). (2 ﺃ- ﻣﺎﺫﺍ ﳛﺪﺙ ﻟﻠﻤﻜﺜﻔﺔ ؟ ﺻﻔﺤﺔ 4 ﺏ-ﺃﺣﺴﺐ ﻗﻴﻤﺔ ﺍﻟﻄﺎﻗﺔ ﺍﻷﻋﻈﻤﻴﺔ ﺍﶈﻮﻟﺔ ﰲ ﺍﻟﺪﺍﺭﺓ ﺍﻟﻜﻬﺮﺑﺎﺋﻴﺔ . ﺍﳊﻞ ﺍﳌﻔﺼﻞ : . .1ﺃ -ﺍﻟﺘﻔﺴﲑ :ﻋﻨﺪ ﺟﻌﻞ ﺍﻟﺒﺎﺩﻟﺔ ﰲ ﺍﻟﻮﺿﻊ 1ﳝﺮ ﺗﻴﺎﺭ ﺑﺎﻟﺪﺍﺭﺓ ﳑﺎ ﻳﺆﺩﻱ ﺇﱃ ﺷﺤﻦ ﺍﳌﻜﺜﻔﺔ . ﺏ -ﺗﺒﲔ ﻛﻴﻒ ﻣﺸﺎﻫﺪﺓ ﺍﻟﺘﻄﻮﺭ ﺍﻟﺰﻣﲏ ﻟﻠﺘﻮﺗﺮ ﺍﻟﻜﻬﺮﺑﺎﺋﻲ . u AB ﳝﻜﻦ ﻣﺸﺎﻫﺪﺓ ﺍﻟﺘﻄﻮﺭ ﺍﻟﺰﻣﲏ ﻟﻠﺘﻮﺗﺮ ﺍﻟﻜﻬﺮﺑﺎﺋﻲ u ABﻭﺫﻟﻚ ﺑﺮﺑﻂ ﻃﺮﰲ ﺍﳌﻜﺜﻔﺔ ﺇﱃ ﺭﺍﺳﻢ ﺇﻫﺘﺰﺍﺯ ﻣﻬﺒﻄﻲ ﺫﻭ ﺫﺍﻛﺮﺓ ،ﺃﻭ ﺟﻬﺎﺯ ﺇﻋﻼﻡ ﺁﱄ ﻣﺰﻭﺩ ﺑﺒﻄﺎﻗﺔ ﻣﺪﺧﻞ . ﺟ -ﻛﺘﺎﺑﺔ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﺿﻠﻴﺔ ﺍﻟﱵ ﲢﻜﻢ ﺍﺷﺘﻐﺎﻝ ﺍﻟﺪﺍﺭﺓ ﺍﻟﻜﻬﺮﺑﺎﺋﻴﺔ . ﺣﺴﺐ ﻗﺎﻧﻮﻥ ﲨﻊ ﺍﻟﺘﻮﺗﺮﺍﺕ ﻟﺪﻳﻨﺎ . u R + uC = E : dq du . u R = R.i = R ﺣﻴﺚ = RC C : dt dt duC . RC ﻭﻣﻨﻪ + uC = E : dt ﺩ -ﻋﺒﺎﺭﺓ ﺛﺎﺑﺖ ﺍﻟﺰﻣﻦ . t = RC : ﺍﻟﺘﺤﻠﻴﻞ ﺍﻟﺒﻌﺪﻱ = [T ] : ] [t ] = [ R].[C ] = [U ][ I ] .[ I ][T ][U -1 -1 ﻫ -ﺇﺛﺒﺎﺕ ﺣﻞ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﺿﻠﻴﺔ . t - ö æ duC E -tt ﻟﺪﻳﻨﺎ ، uC = E ç1 - e t ÷ :ﺇﺫﻥ = e : dt t è ø . t t t - ö æ duC du E -t ﻭﺑﺎﻟﺘﺎﱄ ، RC C + uC = RC. e t + E ç1 - e t ÷ = Ee t + E - Ee t :ﺃﻱ + uC = E : dt dt t è ø t - ö æ ﻭﻣﻨﻪ ﺍﻟﻌﺒﺎﺭﺓ uC = E ç1 - e t ÷ :ﺣﻞ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﺿﻠﻴﺔ . è ø . RC ) u AB (V 12 ﻭ -ﺭﺳﻢ ﺍﳌﻨﺤﲏ ﺍﳌﻤﺜﻞ ﻟﻠﺘﻮﺗﺮ ﺍﻟﻜﻬﺮﺑﺎﺋﻲ ) . u AB = f ( t t - ö æ t : ﺣﻴﺚ ، E = 12V ﻟﺪﻳﻨﺎ uC = E ç1 - e ÷ : è ø ﻭ . t = RC = 5 ´ 103 ´1, 0 ´ 10-6 = 5 ´ 10-3 s = 5ms ﻭﺑﺎﻟﺘﺎﱄ . uC = 12 (1 - e-200.t ) : )t (s ﻛﻴﻔﻴﺔ ﲢﺪﻳﺪ tﻣﻦ ﺍﻟﺒﻴﺎﻥ . ﻫﻮ ﻓﺎﺻﻠﺔ ﻧﻘﻄﺔ ﺗﻘﺎﻃﻊ ﳑﺎﺱ ﺍﻟﺒﻴﺎﻥ ﻋﻨﺪ ﺍﳌﺒﺪﺃ ﻣﻊ ﺍﳌﻘﺎﺭﺏ ،ﻭﻓﺎﺻﻠﺔ ﺍﻟﻨﻘﻄﺔ ﺫﺍﺕ ﺍﻟﺘﺮﺗﻴﺒﺔ . u AB = 0, 63.E u AB 11,9 = ﻱ -ﺍﳌﻘـــﺎﺭﻧﺔ :ﻣﻦ ﺍﻟﺒﻴﺎﻥ ﳌﱠﺎ t = 5tﻟﺪﻳﻨﺎ ، u AB = 11,9Vﻭﺑﺎﻟﺘﺎﱄ = 0, 99 E 12 ﺍﻹﺳــﺘﻨﺘﺎﺝ :ﻋﻨﺪ ﺍﻟﻠﺤﻈﺔ t = 5tﺗﺒﻠﻎ ﺷﺤﻨﺔ ﺍﳌﻜﺜﻔﺔ ﺍﻟﻨﺴﺒﺔ 99%ﻣﻦ ﺷﺤﻨﺘﻬﺎ ﺍﻷﻋﻈﻤﻴﺔ . . .2ﺃ -ﺍﻟﺘﻔﺴﲑ :ﻋﻨﺪ ﺇﻧﺘﻬﺎﺀ ﺍﻟﺸﺤﻦ ﻭﺟﻌﻞ ﺍﻟﺒﺎﺩﻟﺔ ﰲ ﺍﻟﻮﺿﻊ 2ﺗﺘﻔﺮﻍ ﺍﳌﻜﺜﻔﺔ ﰲ ﺍﻟﻨﺎﻗﻞ ﺍﻷﻭﻣﻲ ،ﻓﺘﺘﺤﻮﻝ ﺍﻟﻄﺎﻗﺔ ﺍﳌﺨﺰﻧﺔ ﺎ ﺇﱃ ﻃﺎﻗﺔ ﺣﺮﺍﺭﻳﺔ ﺑﺎﻟﻨﺎﻗﻞ ﺍﻷﻭﻣﻲ ﺏ -ﲢﺪﻳﺪ ﻗﻴﻤﺔ ﺍﻟﻄﺎﻗﺔ ﺍﻷﻋﻈﻤﻴﺔ ﺍﶈﻮﻟﺔ ﰲ ﺍﻟﺪﺍﺭﺓ ﺍﻟﻜﻬﺮﺑﺎﺋﻴﺔ . 1 1 1 2 2 ، EC = C.U maxﻭﺑﺎﻟﺘﺎﱄ . EC = ´ 10-6 ´ (12 ) = 1, 22 ´ 10-4 J : ﻟﺪﻳﻨﺎ = C.E 2 : 2 2 2 ﺍﻟﺘﻤﺮﻳﻦ ﺍﳋﺎﻣﺲ : ﻋﻠﻮﻡ ﲡﺮﻳﺒﻴﺔ 2009 ﺗﺘﻜﻮﻥ ﺍﻟﺪﺍﺭﺓ ﺍﻟﻜﻬﺮﺑﺎﺋﻴﺔ ﺍﳌﺒﻴﻨﺔ ﰲ ﺍﻟﺸﻜﻞ -1-ﻣﻦ ﺍﻟﻌﻨﺎﺻﺮ ﺍﻟﺘﺎﻟﻴﺔ ﻣﻮﺻﻮﻟﺔ ﻋﻠﻰ ﺍﻟﺘﺴﻠﺴﻞ : § ﻣﻮﻟﺪ ﻛﻬﺮﺑﺎﺋﻲ ﺗﻮﺗﺮﻩ ﺛﺎﺑﺖ . E = 6V § ﻣﻜﺜﻔﺔ ﺳﻌﺘﻬﺎ . C = 1, 2m F ﺻﻔﺤﺔ 5 § ﻧﺎﻗﻞ ﺃﻭﻣﻲ ﻣﻘﺎﻭﻣﺘﻪ . R = 5k W § ﻗﺎﻃﻌﺔ . K ﻧﻐﻠﻖ ﺍﻟﻘﺎﻃﻌﺔ : ) duC ( t .1ﺑﺘﻄﺒﻴﻖ ﻗﺎﻧﻮﻥ ﲨﻊ ﺍﻟﺘﻮﺗﺮﺍﺕ ،ﺃﻭﺟﺪ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﺿﻠﻴﺔ ﺍﻟﱵ ﺗﺮﺑﻂ ﺑﲔ ) ، uC ( t dt 1 t ö æ RC uC ( t ) = E ç1 - eﻛﺤﻞ ﳍﺎ . .2ﲢﻘﻖ ﺇﻥ ﻛﺎﻧﺖ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﺿﻠﻴﺔ ﺍﶈﺼﻞ ﻋﻠﻴﻬﺎ ﺗﻘﺒﻞ ﺍﻟﻌﺒﺎﺭﺓ ÷ : è ø .3ﺣﺪﺩ ﻭﺣﺪﺓ ﺍﳌﻘﺪﺍﺭ ، RCﻣﺎ ﻣﺪﻟﻮﻟﻪ ﺍﻟﻌﻠﻤﻲ ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﺪﺍﺭﺓ ﺍﻟﻜﻬﺮﺑﺎﺋﻴﺔ ؟ ﺍﺫﻛﺮ ﺍﲰﻪ . R ، E ،ﻭ .C ﺍﻟﺸــﻜﻞ 1- .4ﺃﺣﺴﺐ ﻗﻴﻤﺔ ﺍﻟﺘﻮﺗﺮ ﺍﻟﻜﻬﺮﺑﺎﺋﻲ ) uC ( tﰲ ﺍﻟﻠﺤﻈﺎﺕ ﺍﳌﺪﻭﻧﺔ ﰲ ﺍﳉﺪﻭﻝ ﺍﻟﺘﺎﱄ : 18 24 6 12 0 ) t ( ms ) uC ( t )(V .5ﺃﺭﺳﻢ ﺍﳌﻨﺤﲏ ﺍﻟﺒﻴﺎﱐ . uC ( t ) = f ( t ) : .6ﺃﻭﺟﺪ ﺍﻟﻌﺒﺎﺭﺓ ﺍﳊﺮﻓﻴﺔ ﻟﻠﺸﺪﺓ ﺍﻟﻠﺤﻈﻴﺔ ﻟﻠﺘﻴﺎﺭ ﺍﻟﻜﻬﺮﺑﺎﺋﻲ ) i ( tﺑﺪﻻﻟﺔ R ، Eﻭ ، Cﰒ ﺍﺣﺴﺐ ﻗﻴﻤﺘﻬﺎ ﰲ ﺍﻟﻠﺤﻈﺘﲔ ) ( t = 0ﻭ ) . ( t ® ¥ .7ﺍﻛﺘﺐ ﻋﺒﺎﺭﺓ ﺍﻟﻄﺎﻗﺔ ﺍﻟﻜﻬﺮﺑﺎﺋﻴﺔ ﺍﳌﺨﺰﻧﺔ ﰲ ﺍﳌﻜﺜﻔﺔ ،ﺍﺣﺴﺐ ﻗﻴﻤﺘﻬﺎ ﻋﻨﺪﻣﺎ ) . ( t ® ¥ ﺍﳊﻞ ﺍﳌﻔﺼﻞ : . .1ﻛﺘﺎﺑﺔ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﺿﻠﻴﺔ . du C ﺑﺘﻄﺒﻴﻖ ﻗﺎﻧﻮﻥ ﲨﻊ ﺍﻟﺘﻮﺗﺮﺍﺕ ﳒﺪ ، uC + u R = E :ﺣﻴﺚ : dt du ﻭﺑﺎﻟﺘﺎﱄ . u C + RC C = E : dt duC 1 E + = uC ﻭﻣﻨﻪ ﺗﻜﺘﺐ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﺿﻠﻴﺔ ﺍﳌﻤﻴﺰﺓ ﻟﻠﺪﺍﺭﺓ ﻛﻤﺎ ﻳﻠﻲ : dt RC RC 1 t ö æ .2ﺇﺛﺒﺎﺕ ﺃﻥ ﺍﻟﻌﺒﺎﺭﺓ uC ( t ) = E ç1 - e RC ÷ :ﺣﻞ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﺿﻠﻴﺔ . è ø 1 1 t ö æ du C E - RC t RC = e uC = E ç1 - eﻭﺑﺎﻟﺘﺎﱄ : ﻟﺪﻳﻨﺎ ÷ : dt RC è ø . u R = R .i = RC . 1 t æ duC 1 E - RC1 t 1 + = uC e + E ç1 - e RC dt RC RC RC è ö E - RC1 t E E - RC1 t E = e + e = ﺇﺫﻥ : ÷ RC RC RC ø RC 1 t ö æ RC uC ( t ) = E ç1 - eﺣﻞ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﺿﻠﻴﺔ . ﻭﻣﻨﻪ ﺍﻟﻌﺒﺎﺭﺓ ÷ : è ø .3ﲢﺪﻳﺪ ﻭﺣﺪﺓ ﺍﳌﻘﺪﺍﺭ ، RCﻭﻣﺪﻟﻮﻟﻪ ﻭ ﺍﲰﻪ . [U ] [Q ] = [ A ][T ] = T ] [ ] [ A ] [U ] [ A = ] [ RC ] = [ R ][Cﻭﻣﻨﻪ RCﻣﺘﻨﺎﺳﺐ ﻣﻊ ﺍﻟﺰﻣﻦ . ﻣﺪﻟﻮﻟﻪ ﺍﻟﻌﻠﻤﻲ :ﻫﻮ ﺍﳌﺪﺓ ﺍﻟﻼﺯﻣﺔ ﻟﺸﺤﻦ ﺍﳌﻜﺜﻔﺔ ﺑﻨﺴﺒﺔ ، 63%ﻭﻳﺴﻤﻰ ﺑﺜﺎﺑﺖ ﺍﻟﺰﻣﻦ . .4ﻣﻸ ﺍﳉﺪﻭﻝ : 24 18 12 6 0 ) t ( ms 5,89 5,70 5,19 3,79 0 ) uC ( t )(V .5ﺭﺳﻢ ﺍﳌﻨﺤﲏ ﺍﻟﺒﻴﺎﱐ . uC ( t ) = f ( t ) : ﺃﻧﻈﺮ ﺍﻟﺸﻜﻞ ﺍﳌﻘﺎﺑﻞ . ﺻﻔﺤﺔ 6 . .6ﺇﳚﺎﺩ ﺍﻟﻌﺒﺎﺭﺓ ﺍﳊﺮﻓﻴﺔ ﻟﻠﺸﺪﺓ ﺍﻟﻠﺤﻈﻴﺔ ﻟﻠﺘﻴﺎﺭ ﺍﻟﻜﻬﺮﺑﺎﺋﻲ ) . i ( t 1 1 t ö t æ RC RC u R = E - E ç1 - e ﻭ ﺑﺎﻟﺘﺎﱄ : ÷ = Ee è ø 1 t ö æ RC uC ( t ) = E ç1 - eﻭ u R = E - u C ﻟﺪﻳﻨﺎ ÷ : è ø 1 E - t u ﻟﻜﻦ u R = R .i :ﺃﻱ i = R :ﻭﺑﺎﻟﺘﺎﱄ . i (t ) = e RC : R R 1 ´ 0 E E 6 = I 0 = e RC = . I0 ﺃﻱ = 1, 2 ´ 10-3 A : ﻭﻣﻨﻪ :ﳌﺎ ( t = 0 ) :ﻟﺪﻳﻨﺎ : 3 5 ´ 10 R R E - 1 ´¥ ﳌﺎ (t ® ¥ ) :ﻟﺪﻳﻨﺎ i ¥ = e RC = 0 : . R .7ﻛﺘﺎﺑﺔ ﻋﺒﺎﺭﺓ ﺍﻟﻄﺎﻗﺔ ﺍﻟﻜﻬﺮﺑﺎﺋﻴﺔ ﺍﳌﺨﺰﻧﺔ ﰲ ﺍﳌﻜﺜﻔﺔ . 1 1 1 ﻋﻨﺪ ) ( t ® ¥ﻳﻜﻮﻥ ﻗﺪ ﺍﻛﺘﻤﻞ ﺍﻟﺸﺤﻦ ﻭﺑﺎﻟﺘﺎﱄ E 0 = C .u C2max = C .E 2 :ﻭﻣﻨﻪ E 0 = ´1, 2 ´10-6 ´ 62 = 2,16 ´10-5 J : 2 2 2 ﺍﻟﺘﻤﺮﻳﻦ ﺍﻟﺴﺎﺩﺱ : ﻋﻠﻮﻡ ﲡﺮﻳﺒﻴﺔ 2009 ﻟﺪﻳﻨﺎ ﻣﻜﺜﻔﺔ ﺳﻌﺘﻬﺎ C = 1, 0 ´ 10-1 m Fﻣﺸﺤﻮﻧﺔ ﻣﺴﺒﻘﺎ ﺑﺸﺤﻨﺔ ﻛﻬﺮﺑﺎﺋﻴﺔ ﻣﻘﺪﺍﺭﻫﺎ ، q = 0, 6 ´ 10-6 Cﻭﻧﺎﻗﻞ ﺃﻭﻣﻲ ﻣﻘﺎﻭﻣﺘﻪ R = 15k Wﳓﻘﻖ ﺩﺍﺭﺓ ﻛﻬﺮﺑﺎﺋﻴﺔ ﻋﻠﻰ ﺍﻟﺘﺴﻠﺴﻞ ﺑﺎﺳﺘﻌﻤﺎﻝ ﺍﳌﻜﺜﻔﺔ ﻭﺍﻟﻨﺎﻗﻞ ﺍﻷﻭﻣﻲ ﻭﻗﺎﻃﻌﺔ . Kﰲ ﺍﻟﻠﺤﻈﺔ t = 0ﻧﻐﻠﻖ ﺍﻟﻘﺎﻃﻌﺔ : .1ﺃﺭﺳﻢ ﳐﻄﻂ ﺍﻟﺪﺍﺭﺓ ﺍﳌﻮﺻﻮﻓﺔ ﺳﺎﺑﻘﺎ . .2ﻣﺜﻞ ﻋﻠﻰ ﺍﳌﺨﻄﻂ : § ﺟﻬﺔ ﻣﺮﻭﺭ ﺍﻟﺘﻴﺎﺭ ﺍﻟﻜﻬﺮﺑﺎﺋﻲ ﰲ ﺍﻟﺪﺍﺭﺓ . .3ﺃﻭﺟﺪ ﻋﻼﻗﺔ ﺑﲔ u Rﻭ . uC .4ﺑﺎﻻﻋﺘﻤﺎﺩ ﻋﻠﻰ ﻗﺎﻧﻮﻥ ﲨﻊ ﺍﻟﺘﻮﺗﺮﺍﺕ ،ﺃﻭﺟﺪ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﺿﻠﻴﺔ ﺑﺪﻻﻟﺔ . uC .5ﺇﻥ ﺣﻞ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﺻﻠﻴﺔ ﺍﻟﺴﺎﺑﻘﺔ ﻫﻮ ﻣﻦ ﺍﻟﺸﻜﻞ ، uC = a ´ eb.t :ﺣﻴﺚ aﻭ bﺛﺎﺑﺘﲔ ﻳﻄﻠﺐ ﺗﻌﻴﲔ ﻗﻴﻤﺔ ﻛﻞ ﻣﻨﻬﻤﺎ . .6ﺃﻛﺘﺐ ﺍﻟﻌﺒﺎﺭﺓ ﺍﻟﺰﻣﻨﻴﺔ ﻟﻠﺘﻮﺗﺮ ) . uC = f ( t .7ﺇﻥ ﺍﻟﻌﺒﺎﺭﺓ ﺍﻟﺰﻣﻨﻴﺔ uCﺗﺴﻤﺢ ﺑﺮﺳﻢ ﺍﻟﺒﻴﺎﻥ ﺍﻟﺸﻜﻞ: 1- § ﺍﺷﺮﺡ ﻋﻠﻰ ﺍﻟﺒﻴﺎﻥ ﺍﻟﻄﺮﻳﻔﺔ ﺍﳌﺘﺒﻌﺔ ﻟﻠﺘﺄﻛﺪ ﻣﻦ ﺍﻟﻘﻴﻢ ﺍﶈﺴﻮﺑﺔ ﺳﺎﺑﻘﺎ ) ﺍﻟﺴﺆﺍﻝ . ( 5 ﺍﻟــﺸﻜﻞ 1- ﺍﳊﻞ ﺍﳌﻔﺼﻞ : . .1ﺭﺳﻢ ﳐﻄﻂ ﻟﻠﺪﺍﺭﺓ ﺍﳌﻮﺻﻮﻓﺔ .ﺃﻧﻈﺮ ﺍﳌﺨﻄﻂ ﺍﳌﻘﺎﺑﻞ . .2ﺍﳉﻬﺔ ﺍﳊﻘﻴﻘﻴﺔ ﳌﺮﻭﺭ ﺍﻟﺘﻴﺎﺭ ﺍﻟﻜﻬﺮﺑﺎﺋﻲ ﰲ ﺍﻟﺪﺍﺭﺓ :ﺃﻧﻈﺮ ﺍﳌﺨﻄﻂ ﺍﳌﻘﺎﺑﻞ . .3ﺇﳚﺎﺩ ﻋﻼﻗﺔ ﺑﲔ u Rﻭ . uC dq ، u R = R.i = Rﻭ ﻟﺪﻳﻨﺎ q = C .uC :ﻭﺑﺎﻟﺘﺎﱄ : ﻟﺪﻳﻨﺎ : dt .4ﻛﺘﺎﺑﺔ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﺿﻠﻴﺔ ﺑﺪﻻﻟﺔ . uC duC dt duC ﺣﺴﺐ ﻗﺎﻧﻮﻥ ﲨﻊ ﺍﻟﺘﻮﺗﺮﺍﺕ ﻟﺪﻳﻨﺎ uC + uR = 0 :ﻭﺑﺎﻟﺘﺎﱄ = 0 : dt du 1 . C+ uC = 0 dt RC .5ﺗﻌﻴﲔ ﻗﻴﻤﺔ ﺍﻟﺜﺎﺑﺘﲔ aﻭ bﺛﺎﺑﺘﲔ . 0, 6 ´10-6 ﻣﻦ ﺍﻟﺸﺮﻭﻁ ﺍﻹﺑﺘﺪﺍﺋﻴﺔ ﻟﺪﻳﻨﺎ = 6V : 0,1´10-6 = )q ( 0 C . u R = RC uC + RCﻭﻣﻨﻪ : = uC ( 0 ) = E = a duC ﻟﺪﻳﻨﺎ uC = a ´ eb.t :ﻭﺑﺎﻟﺘﺎﱄ = ab.eb.t : dt 1 duC 1 ab.eb .t +ﻭﺑﺎﻟﺘﺎﱄ = 0 : ﺇﺫﻥ a.eb .t = 0 : + ﻭﻟﺪﻳﻨﺎ uC = 0 : RC dt RC ﺻﻔﺤﺔ 7 1 ö b .t æ açb + ÷e RC ø è 1 1 1 2000 -1 1 =-= b +ﻭﺑﺎﻟﺘﺎﱄ s : ﺇﺫﻥ = 0 : 3 -6 RC t 15 ´10 ´ 0,1´ 10 3 RC ﻭﻣﻨﻪ : 1 t RC - uC = E.eﺃﻱ : 2000 .t 3 .6ﲢﺪﻳﺪ ﺍﻟﺜﺎﺑﺘﲔ aﻭ bﺑﻴﺎﻧﻴﺎ . - . b=- . uC = 6.e ﻣﻦ ﺍﻟﺒﻴﺎﻥ :ﳌﺎ t = 0ﳒﺪ . uC ( 0 ) = a = 6V : ﳌﺎ uC = 0, 37 ´ 6 = 2, 22Vﳒﺪ . t = 0, 75 ´ 0, 002 = 1, 5 ´ 10 s : 1 1 2000 -1 =ﻭﺑﺎﻟﺘﺎﱄ s : = . b=--3 t 1, 5 ´ 10 3 -3 ﺍﻟﺘﻤﺮﻳﻦ ﺍﻟﺴﺎﺑﻊ : ﺭﻳﺎﺿﻴﺎﺕ ﻭ ﺗﻘﲏ ﺭﻳﺎﺿﻲ 2009 ﻧﺮﺑﻂ ﻋﻠﻰ ﺍﻟﺘﺴﻠﺴﻞ ﺍﻟﻌﻨﺎﺻﺮ ﺍﻟﻜﻬﺮﺑﺎﺋﻴﺔ ﺍﻟﺘﺎﻟﻴﺔ : § ﻣﻮﻟﺪ ﺫﻱ ﺗﻮﺗﺮ ﺛﺎﺑﺖ ) . ( E = 12V § ﻭﺷﻴﻌﺔ ﺫﺍﺗﻴﺘﻬﺎ ) ( L = 300mHﻭﻣﻘﺎﻭﻣﺘﻬﺎ ) . ( r = 10W § ﻧﺎﻗﻞ ﺃﻭﻣﻲ ﻣﻘﺎﻭﻣﺘﻪ ) . ( R = 110W § ﻗﺎﻃﻌﺔ ) ) ( kﺍﻟﺸﻜﻞ (-1- .1ﰲ ﺍﻟﻠﺤﻈﺔ ) ( t = 0 sﻧﻐﻠﻖ ﺍﻟﻘﺎﻃﻌﺔ ) : ( k ﺃﻭﺟﺪ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﺿﻠﻴﺔ ﺍﻟﱵ ﺗﻌﻄﻲ ﺷﺪﺓ ﺍﻟﺘﻴﺎﺭ ﺍﻟﻜﻬﺮﺑﺎﺋﻲ ﰲ ﺍﻟﺪﺍﺭﺓ . .2ﻛﻴﻒ ﻳﻜﻮﻥ ﺳﻠﻮﻙ ﺍﻟﻮﺷﻴﻌﺔ ﰲ ﺍﻟﻨﻈﺎﻡ ﺍﻟﺪﺍﺋﻢ ؟ ﻭﻣﺎ ﻫﻲ ﻋﻨﺪﺋﺬ ﻋﺒﺎﺭﺓ ﺷﺪﺓ ﺍﻟﺘﻴﺎﺭ ﺍﻟﻜﻬﺮﺑﺎﺋﻲ I 0ﺍﻟﺬﻱ ﳚﺘﺎﺯ ﺍﻟﺪﺍﺭﺓ ؟ t - ö æ .3ﺑﺎﻋﺘﺒﺎﺭ ﺍﻟﻌﻼﻗﺔ ÷ i = A ç1 - e tﺣﻼ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﺿﻠﻴﺔ ﺍﳌﻄﻠﻮﺑﺔ ﰲ ﺍﻟﺴﺆﺍﻝ -1- è ø ﺃ -ﺃﻭﺟﺪ ﺍﻟﻌﺒﺎﺭﺓ ﺍﳊﺮﻓﻴﺔ ﻟﻜﻞ ﻣﻦ Aﻭ . t ﺏ-ﺍﺳﺘﻨﺘﺞ ﻋﺒﺎﺭ ﺍﻟﺘﻮﺗﺮ ﺍﻟﻜﻬﺮﺑﺎﺋﻲ uBCﺑﲔ ﻃﺮﰲ ﺍﻟﻮﺷﻴﻌﺔ . .4ﺃ -ﺃﺣﺴﺐ ﻗﻴﻤﺔ ﺍﻟﺘﻮﺗﺮ ﺍﻟﻜﻬﺮﺑﺎﺋﻲ uBCﰲ ﺍﻟﻨﻈﺎﻡ ﺍﻟﺪﺍﺋﻢ . ﺏ -ﺃﺭﺳﻢ ﻛﻴﻔﻴﺎ ﺷﻜﻞ ﺍﻟﺒﻴﺎﻥ ) . uBC = f ( t ﺍﳊﻞ ﺍﳌﻔﺼﻞ : . .1ﻛﺘﺎﺑﺔ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﺿﻠﻴﺔ ﺍﳌﻤﻴﺰﺓ ﻟﻠﺪﺍﺭﺓ . ﺣﺴﺐ ﻗﺎﻧﻮﻥ ﲨﻊ ﺍﻟﺘﻮﺗﺮﺍﺕ ﻟﺪﻳﻨﺎ u R ( t ) + ub ( t ) = E : ﺇﺫﻥ= E : ) di ( t dt Ri ( t ) + ri ( t ) + Lﺃﻱ = E : ) di ( t R+r E ﻭﻣﻨﻪ i ( t ) = : dt L L .2ﲢﺪﻳﺪ ﺳﻠﻮﻙ ﺍﻟﻮﺷﻌﺔ ﰲ ﺍﻟﻨﻈﺎﻡ ﺍﻟﺪﺍﺋﻢ . + ) di ( t dt ( R + r ) i (t ) + L . ) di ( t ﰲ ﺍﻟﻨﻈﺎﻡ ﺍﻟﺪﺍﺋﻢ ﻳﻜﻮﻥ ، i ( t ) = I 0 = Cte :ﻭﺑﺎﻟﺘﺎﱄ = 0 : dt E 12 R+r E = ﺃﻱ= 0,1A : ﻭﺑﺎﻟﺘﺎﱄ I 0 = : = . I0 R + r 120 L L .3ﺃ -ﺇﳚﺎﺩ ﺍﻟﻌﺒﺎﺭﺓ ﺍﳊﺮﻓﻴﺔ ﻟﻜﻞ ﻣﻦ Aﻭ . t ö ﻟﺪﻳﻨﺎ ÷ : ø t æ di A -tt ، i = A ç1 - e tﻭﺑﺎﻟﺘﺎﱄ = e : dt t è ،ﺇﺫﻥ . ub = r.i :ﻭﻣﻨﻪ ﺍﻟﻮﺷﻴﻌﺔ ﺗﺴﻠﻚ ﺳﻠﻮﻙ ﻧﺎﻗﻞ ﺃﻭﻣﻲ ﻣﻘﺎﻭﻣﺘﻪ . r . ﺻﻔﺤﺔ 8 t di ( t ) R + r E A -tt R + r R + r -tt E E æ 1 R + r ö -t R + r e + A = e + A ﻭﺑﺎﻟﺘﺎﱄ Ae = : + = ) i (t . Aç: ﺃﻱ ، ﻭﻟﺪﻳﻨﺎ ÷ L ø L L t L L L dt L L èt L E R+r E 1 R+r ﺇﺫﻥ = 0 : =. A = tﻭ = I0 ،ﻭﻣﻨﻪ : =A ﻭR+r R+r L L t L ﺏ -ﺍﺳﺘﻨﺘﺞ ﻋﺒﺎﺭ ﺍﻟﺘﻮﺗﺮ ﺍﻟﻜﻬﺮﺑﺎﺋﻲ uBCﺑﲔ ﻃﺮﰲ ﺍﻟﻮﺷﻴﻌﺔ . R+ r t ö di E - RL+ r t E æ . = e = ) ، i ( tﻭﺑﺎﻟﺘﺎﱄ: ﻟﺪﻳﻨﺎ ç1 - e L ÷ : dt L R+r è ø R+ r R+r t t ö r r di E - R +r t E æ E . uBC = L + r.i = L. e L + r.ﺃﻱ Ee L : ﺇﺫﻥ ç1 - e L ÷ : R+r R+r dt L R+r è ø R+r t r R = E+ ﻭﻣﻨﻪ Ee L : R+r R+r + R+ r t L . uBC - . uBC = Ee ) uBC (V 120 t 10 110 = .12 + ﺃﻱ .12e 0,3 : 120 120 ، uBCﻭﻣﻨﻪ : -400.t . uBC = 1 + 11.e .4ﺃ -ﺣﺴﺎﺏ ﻗﻴﻤﺔ ﺍﻟﺘﻮﺗﺮ ﺍﻟﻜﻬﺮﺑﺎﺋﻲ uBCﰲ ﺍﻟﻨﻈﺎﻡ ﺍﻟﺪﺍﺋﻢ . ﰲ ﺍﻟﻨﻈﺎﻡ ﺍﻟﺪﺍﺋﻢ ﻟﺪﻳﻨﺎ I 0 = 0,1A :ﻭ . uBC = r.I 0 ،ﻭﻣﻨﻪ . uBC = 10 ´ 0,1 = 1V ﺏ -ﺭﺳﻢ ﻛﻴﻔﻲ ﻟﺸﻜﻞ ﺍﻟﺒﻴﺎﻥ ) ) . uBC = f ( tﺍﻟﺒﻴﺎﻥ ﺑﺎﻟﺸﻜﻞ ﺍﳌﻘﺎﺑﻞ( 2 )t ( ms ﺍﻟﺘﻤﺮﻳﻦ ﺍﻟﺜﺎﻣﻦ : 2,5 ﺭﻳﺎﺿﻴﺎﺕ ﻭ ﺗﻘﲏ ﺭﻳﺎﺿﻲ 2009 ﳓﻘﻖ ﺍﻟﺘﺮﻛﻴﺐ ﺍﻟﻜﻬﺮﺑﺎﺋﻲ ﺍﻟﺘﺠﺮﻳﱯ ﺍﳌﺒﲔ ﰲ ﺍﻟﺸﻜﻞ ﺍﳌﻘﺎﺑﻞ ﺑﺎﺳﺘﻌﻤﺎﻝ ﺍﻟﺘﺠﻬﻴﺰ : · ﻣﻜﺜﻔﺔ ﺳﻌﺘﻬﺎ ) ( Cﻏﲑ ﻣﺸﺤﻮﻧﺔ . · ﻧﺎﻗﻠﲔ ﺃﻭﻣﻴﲔ ﻣﻘﺎﻭﻣﺘﻴﻬﻤﺎ ) . ( R = R ' = 470W · ﻣﻮﻟﺪ ﺫﻱ ﺗﻮﺗﺮ ﺛﺎﺑﺖ · ﺑﺎﺩﻟﺔ ) (k ) (E . ،ﺃﺳﻼﻙ ﺗﻮﺻﻴﻞ . .1ﻧﻀﻊ ﺍﻟﺒﺎﺩﻟﺔ ﻋﻨﺪ ﺍﻟﻮﺿﻊ ) (1ﰲ ﺍﻟﻠﺤﻈﺔ ) : ( t = 0 ﺃ -ﺑﲔ ﻋﻠﻰ ﺍﻟﺸﻜﻞ ﺟﻬﺔ ﺍﻟﺘﻴﺎﺭ ﺍﻟﻜﻬﺮﺑﺎﺋﻲ ﺍﳌﺎﺭ ﰲ ﺍﻟﺪﺍﺭﺓ ﰒ ﻣﺜﻞ ﺑﺎﻷﺳﻬﻢ ﺍﻟﺘﻮﺗﺮﻳﻦ . uR ، uC ﺏ-ﻋﱪ ﻋﻦ uCﻭ uRﺑﺪﻻﻟﺔ ﺷﺤﻨﺔ ﺍﳌﻜﺜﻔﺔ ، q = q Aﰒ ﺃﻭﺟﺪ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﺿﻠﻴﺔ ﺍﻟﱵ ﲢﻘﻘﻬﺎ ﺍﻟﺸﺤﻨﺔ . q ﺟ -ﺗﻘﺒﻞ ﻫﺬﻩ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﺿﻠﻴﺔ ﺣﻼ ﻣﻦ ﺍﻟﺸﻜﻞ . q ( t ) = A (1 - e-a .t ) :ﻋﱪ ﻋﻦ Aﻭ aﺑﺪﻻﻟﺔ . E ، R ، C ﺩ -ﺇﺫﺍ ﻛﺎﻧﺖ ﻗﻴﻤﺔ ﺍﻟﺘﻮﺗﺮ ﺍﻟﻜﻬﺮﺑﺎﺋﻲ ﻋﻨﺪ ﺎﻳﺔ ﺍﻟﺸﺤﻦ ﺑﲔ ﻃﺮﰲ ﺍﳌﻜﺜﻔﺔ ) ، ( 5Vﺍﺳﺘﻨﺘﺞ ﻗﻴﻤﺔ ) . ( E ﻫ -ﻋﻨﺪﻣﺎ ﺗﺸﺤﻦ ﺍﳌﻜﺜﻔﺔ ﻛﻠﻲ ﲣﺰﻥ ﻃﺎﻗﺔ ) ، ( EC = 5mJﺍﺳﺘﻨﺘﺞ ﺳﻌﺔ ﺍﳌﻜﺜﻔﺔ ) . ( C .2ﳒﻌﻞ ﺍﻵﻥ ﺍﻟﺒﺎﺩﻟﺔ ﻋﻨﺪ ﺍﻟﻮﺿﻊ ): (2 ﺃ -ﻣﺎﺫﺍ ﳛﺪﺙ ﻟﻠﻤﻜﺜﻔﺔ ؟ ﺏ-ﻗﺎﺭﻥ ﺑﲔ ﻗﻴﻤﱵ ﺛﺎﺑﺖ ﺍﻟﺰﻣﻦ ﺍﳌﻮﺍﻓﻖ ﻟﻠﻮﺿﻌﲔ ) (1ﰒ ) (2ﻟﻠﺒﺎﺩﻟﺔ . k ﺍﳊﻞ ﺍﳌﻔﺼﻞ : . .1ﺃ -ﲤﺜﻴﻞ ﺍﻷﺳﻬﻢ ﺍﳌﻤﺜﻠﺔ ﻟﻔﺮﻭﻕ ﺍﻟﻜﻤﻮﻥ .ﺍﻟﺘﻤﺜﻴﻞ ﺑﺎﻟﺼﻔﺤﺔ ﺍﳌﻮﺍﻟﻴﺔ . ﺏ -ﻛﺘﺎﺑﺔ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﺿﻠﻴﺔ ﺍﳌﻤﻴﺰﺓ ﻟﻠﺪﺍﺭﺓ . q dq u R = R .i = Rﻭ ﻟﺪﻳﻨﺎ : C dt ﺣﺴﺐ ﻗﺎﻧﻮﻥ ﲨﻊ ﺍﻟﺘﻮﺗﺮﺍﺕ ﻟﺪﻳﻨﺎ . u R + uC = E : = . uC ﺻﻔﺤﺔ 9 dq 1 E dq q + =q ، Rﻭﻣﻨﻪ : ﻭﺑﺎﻟﺘﺎﱄ + = E : dt RC R dt C ﺟ -ﺇﳚﺎﺩ ﻋﺒﺎﺭﺓ Aﻭ aﺑﺪﻻﻟﺔ . E ، R ، C . ) dq ( t ﻟﺪﻳﻨﺎ q ( t ) = A (1 - e-a .t ) :ﻭﺑﺎﻟﺘﺎﱄ = Aa e-a .t : dt 1 E dq 1 E ، Aa e -a .t + ،ﺇﺫﻥ A (1 - e -a .t ) = : + =q ﻟﺪﻳﻨﺎ : RC R dt RC R 1 ö -a .t A E æ . A ça= ﺃﻱ : ÷e + RC ø RC R è 1 A E 1 = aﻭ . A = CE ﻭﻣﻨﻪ : aﻭ =ﺇﺫﻥ = 0 : RC RC R RC t æ ö ﻭﺑﺎﻟﺘﺎﱄ . q ( t ) = CE ç1 - e RC ÷ : è ø ﺩ -ﺇﳚﺎﺩ ﻗﻴﻤﺔ . Eﻋﻦ ﺎﻳﺔ ﺍﻟﺸﺤﻦ ﺗﻜﻮﻥ ﺍﻟﺪﺍﺭﺓ ﰲ ﺍﻟﻨﻈﺎﻡ ﺍﻟﺪﺍﺋﻢ ،ﺇﺫﻥ i = 0 :ﻭﺑﺎﻟﺘﺎﱄ u R = 0 :ﺇﺫﻥ ، uC = E :ﻭﻣﻨﻪ . E = 5V : . ﻫ -ﺇﳚﺎﺩ ﺳﻌﺔ ﺍﳌﻜﺜﻔﺔ . C 2 EC 2 ´ 5 ´10-3 1 = ﻟﺪﻳﻨﺎ EC = CE 2 :ﻭﺑﺎﻟﺘﺎﱄ = 4 ´10-4 F : 2 2 E 5 2 .3ﺃ -ﺍﻟﺘﻔﺴﲑ :ﻋﻨﺪ ﺟﻌﻞ ﺍﻟﺒﺎﺩﻟﺔ ﰲ ﺍﻟﻮﺿﻌﻴﺔ 2ﺗﺘﻔﺮﻍ ﺍﻟﻄﺎﻗﺔ ﺍﳌﺨﺰﻧﺔ ﺑﺎﳌﻜﺜﻔﺔ ﰲ ﺍﻟﻨﺎﻗﻞ ﺍﻷﻭﻣﻲ ﻭﺗﺘﺤﻮﻝ ﺇﱃ ﻃﺎﻗﺔ ﺣﺮﺍﺭﻳﺔ . = . Cﻭﻣﻨﻪ . C = 400m F : ﺏ -ﻣﻘﺎﺭﻧﺔ ﺛﺎﺑﱵ ﺍﻟﺰﻣﻦ ﻟﻠﻮﺿﻌﲔ 1ﻭ . 2 ﺩﺍﺭﺓ ﺍﻟﺸﺤﻦ . t 1 = RC = 470 ´ 4 ´ 10-4 = 0,188s :ﺩﺍﺭﺓ ﺍﻟﺘﻔﺮﻳﻎ . t 2 = ( R + R ') C = 2 RC = 2 ´ 470 ´ 4 ´10-4 = 0, 276s : ﺛﺎﺑﺖ ﺍﻟﺰﻣﻦ ﻟﺪﺍﺭﺓ ﺍﻟﺘﻔﺮﻳﻎ ﺿﻌﻒ ﺛﺎﺑﺖ ﺍﻟﺰﻣﻦ ﰲ ﺩﺍﺭﺓ ﺍﻟﺸﺤﻦ . ﺍﻟﺘﻤﺮﻳﻦ ﺍﻟﺘﺎﺳﻊ : ﻋﻠﻮﻡ ﲡﺮﻳﺒﻴﺔ 2010 ﻧﺮﻳﺪ ﺗﻌﲔ ) ( L, rﳑﻴﺰﰐ ﻭﺷﻴﻌﺔ ،ﻧﺮﺑﻄﻬﺎ ﰲ ﺩﺍﺭﺓ ﻛﻬﺮﺑﺎﺋﻴﺔ ﻋﻠﻰ ﺍﻟﺘﺴﻠﺴﻞ ﻣﻊ : ﻣﻮﻟﺪ ﻛﻬﺮﺑﺎﺋﻲ ﺫﻱ ﺗﻮﺗﺮ ﻛﻬﺮﺑﺎﺋﻲ ﺛﺎﺑﺖ . E = 6V ﻧﺎﻗﻞ ﺃﻭﻣﻲ ﻣﻘﺎﻭﻣﺘﻪ . R = 10W ﻗﺎﻃﻌﺔ ) kﺍﻟﺸﻜﻞ. ( 1- .1ﻧﻐﻠﻖ ﺍﻟﻘﺎﻃﻌﺔ ، kﺃﻛﺘﺐ ﻋﺒﺎﺭﺓ ﻛﻞ ﻣﻦ : ﺍﻟﺸــﻜﻞ 1- : u Rﺍﻟﺘﻮﺗﺮ ﺍﻟﻜﻬﺮﺑﺎﺋﻲ ﺑﲔ ﻃﺮﰲ ﺍﻟﻨﺎﻗﻞ ﺍﻷﻭﻣﻲ . R : ubﺍﻟﺘﻮﺗﺮ ﺍﻟﻜﻬﺮﺑﺎﺋﻲ ﺑﲔ ﻃﺮﰲ ﺍﻟﻮﺷﻴﻌﺔ . .2ﺑﺘﻄﺒﻴﻖ ﻗﺎﻧﻮﻥ ﲨﻊ ﺍﻟﺘﻮﺗﺮﺍﺕ ،ﺃﻭﺟﺪ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﺿﻠﻴﺔ ﻟﻠﺘﻴﺎﺭ ﺍﻟﻜﻬﺮﺑﺎﺋﻲ ) i ( tﺍﳌﺎﺭ ﺑﺎﻟﺪﺍﺭﺓ . .3ﺑﲔ ﺃﻥ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﺿﻠﻴﺔ ﺍﻟﺴﺎﺑﻘﺔ ﺗﻘﺒﻞ ﺣﻼ ﻣﻦ ﺍﻟﺸﻜﻞ : ( R+r) ö t E æ ÷÷ çç1 - e L R+r è ø = ) . i (t .4ﻣﻜﻨﺖ ﺍﻟﺪﺭﺍﺳﺔ ﺍﻟﺘﺠﺮﻳﺒﻴﺔ ﲟﺘﺎﺑﻌﺔ ﺗﻄﻮﺭ ﺷﺪﺓ ﺍﻟﺘﻴﺎﺭ ﺍﻟﻜﻬﺮﺑﺎﺋﻲ ﺍﳌﺎﺭ ﰲ ﺍﻟﺪﺍﺭﺓ ﻭﺭﺳﻢ ﺍﻟﺒﻴﺎﻥ ﺍﳌﻤﺜﻞ ﻟﻪ ﰲ ) ﺍﻟﺸﻜﻞ . ( 2-ﺑﺎﻻﺳﺘﻌﺎﻧﺔ ﺑﺎﻟﺒﻴﺎﻥ ﺃﺣﺴﺐ : ﺃ -ﺍﳌﻘﺎﻭﻣﺔ rﻟﻠﻮﺷﻴﻌﺔ . ﺏ-ﻗﻴﻤﺔ tﺛﺎﺑﺖ ﺍﻟﺰﻣﻦ ،ﰒ ﺍﺳﺘﻨﺘﺞ ﻗﻴﻤﺔ Lﺫﺍﺗﻴﺔ ﺍﻟﻮﺷﻴﻌﺔ . ﺍﻟﺸــﻜﻞ 2- .5ﺃﺣﺴﺐ ﻗﻴﻤﺔ ﺍﻟﻄﺎﻗﺔ ﺍﻟﻜﻬﺮﺑﺎﺋﻴﺔ ﺍﳌﺨﺰﻧﺔ ﰲ ﺍﻟﻮﺷﻴﻌﺔ ﰲ ﺣﺎﻟﺔ ﺍﻟﻨﻈﺎﻡ ﺍﻟﺪﺍﺋﻢ . ﺍﳊﻞ ﺍﳌﻔﺼﻞ : . .1ﻛﺘﺎﺑﺔ ﻋﺒﺎﺭﺓ ﻛﻞ ﻣﻦ u Rﻭ . ub ﺻﻔﺤﺔ 10 ﻋﺒﺎﺭﺓ ﺍﻟﺘﻮﺗﺮ ﺍﻟﻜﻬﺮﺑﺎﺋﻲ ﺑﲔ ﻃﺮﰲ ﺍﻟﻨﺎﻗﻞ ﺍﻷﻭﻣﻲ Rﻫﻮ . u R = R.i : di ﻋﺒﺎﺭﺓ ﺍﻟﺘﻮﺗﺮ ﺍﻟﻜﻬﺮﺑﺎﺋﻲ ﺑﲔ ﻃﺮﰲ ﺍﻟﻮﺷﻴﻌﺔ ﻫﻮ : dt . ub = r.i + L .2ﺇﳚﺎﺩ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﺿﻠﻴﺔ ﺍﳌﻤﻴﺰﺓ ﻟﻠﺪﺍﺭﺓ . ﺣﺴﺐ ﻗﺎﻧﻮﻥ ﲨﻊ ﺍﻟﺘﻮﺗﺮﺍﺕ ﻟﺪﻳﻨﺎ uR + ub = E : di di ﻭﺑﺎﻟﺘﺎﱄ R.i + r.i + L = E :؛ ﺇﺫﻥ = E : dt dt ) di ( R + r E . + =i ﻭﻣﻨﻪ : dt L L ، (R + r )i + L .3ﺇﺛﺒﺎﺕ ﺣﻞ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﺿﻠﻴﺔ . ) (R+r ( R+r) ö t di E æ (R + r) - L t ö E æ ؛ ﺃﻱ : = .e çç = ) ، i ( tﻭﺑﺎﻟﺘﺎﱄ ÷÷ : ﻟﺪﻳﻨﺎ çç1 - e L ÷÷ : dt R + r è L R+r è ø ø (R+r ) ö t E æ L 1 e çç ﺇﺫﻥ ÷÷ : R+rè ø ﻭﺑﺎﻟﺘﺎﱄ : t ) ( R+ r L E E+ - e L L (R + r) . L )R+r t L + t )( R+r L ) di ( R + r E+ i = .e dt L L t )(R+r L di E. = .e dt L . ( ) di ( R + r E ) di ( R + r E. + =i ؛ ﻭﻣﻨﻪ : + i = .e dt L L dt L L ( R+r) ö t E æ L 1 e çç ﺇﺫﻥ ﻧﺴﺘﻨﺘﺞ ﺃﻥ ﺣﻞ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﺿﻠﻴﺔ ﻣﻦ ﺍﻟﺸﻜﻞ ÷÷ : R+r è ø = ) . i (t .4ﺃ -ﲢﺪﻳﺪ ﺍﳌﻘﺎﻭﻣﺔ rﻟﻠﻮ ﺷﻴﻌﺔ . 6 E E = ، I 0ﺇﺫﻥ ، r = - R :ﻭﻟﺪﻳﻨﺎ ﻣﻦ ﺍﻟﺒﻴﺎﻥ I 0 = 0,5 A :ﻭﺑﺎﻟﺘﺎﱄ - 10 = 2W : ﻟﺪﻳﻨﺎ ﰲ ﺍﻟﻨﻈﺎﻡ ﺍﻟﺪﺍﺋﻢ : 0,5 I0 )(R + r =r . ﺏ -ﲢﺪﻳﺪ ، tﻭ ﺍﺳﺘﻨﺘﺎﺝ ﻗﻴﻤﺔ . L ﻣﻦ ﺍﻟﺒﻴﺎﻥ ﻟﺪﻳﻨﺎ ﻓﺎﺻﻠﺔ ﻧﻘﻄﺔ ﺗﻘﺎﻃﻊ ﳑﺎﺱ ﺍﻟﺒﻴﺎﻥ ﻋﻨﺪ ﺍﳌﺒﺪﺃ ﻭﺍﳌﻘﺎﺭﺏ ﻫﻲ ) t = 10ms :ﺃﻭ ﻫﻲ ﻓﺎﺻﻠﺔ ﺍﻟﻨﻘﻄﺔ ﺫﺍﺕ ﺍﻟﺘﺮﺗﻴﺐ . ( i = 0, 63.I 0 = 0,315 A L ﻭﻟﺪﻳﻨﺎ : R+r ﺟ -ﺣﺴﺎﺏ ﻗﻴﻤﺔ ﺍﻟﻄﺎﻗﺔ ﺍﻟﻜﻬﺮﺑﺎﺋﻴﺔ ﺍﳌﺨﺰﻧﺔ ﰲ ﺍﻟﻮ ﺷﻴﻌﺔ ﰲ ﺣﺎﻟﺔ ﺍﻟﻨﻈﺎﻡ ﺍﻟﺪﺍﺋﻢ . 1 2 ﻟﺪﻳﻨﺎ . Emax ( L ) = 1,5 ´10-2 J ، Emax ( L ) = L.I 02 = 0,12 ´10-2 ´ ( 0,5) : 2 = ، tﻭﺑﺎﻟﺘﺎﱄ ، L = t ( R + r ) :ﺇﺫﻥ ، L = 10 ´10-3 ´12 :ﻭﻣﻨﻪ . L = 0,12 H : ﺍﻟﺘﻤﺮﻳﻦ ﺍﻟﻌﺎﺷﺮ : ﻋﻠﻮﻡ ﲡﺮﻳﺒﻴﺔ 2010 ﳓﻘﻖ ﺩﺍﺭﺓ ﻛﻬﺮﺑﺎﺋﻴﺔ ﻋﻠﻰ ﺍﻟﺘﺴﻠﺴﻞ ﺗﺘﻜﻮﻥ ﻣﻦ : § ﻣﻮﻟﺪ ﺫﻭ ﺗﻮﺗﺮ ﻛﻬﺮﺑﺎﺋﻲ ﺛﺎﺑﺖ . E = 5V § ﻧﺎﻗﻞ ﺃﻭﻣﻲ ﻣﻘﺎﻭﻣﺘﻪ . R = 100W § ﻣﻜﺜﻔﺔ ﺳﻌﺘﻬﺎ . C § ﻗﺎﻃﻌﺔ . k ﻧﻮﺻﻞ ﻃﺮﰲ ﺍﳌﻜﺜﻔﺔ B ، Aﺇﱃ ﻭﺍﺟﻬﺔ ﺩﺧﻮﻝ ﳉﻬﺎﺯ ﺇﻋﻼﻡ ﺁﱄ ﻭﻋﻮﳉﺖ ﺍﳌﻌﻄﻴﺎﺕ ﺑﱪﳎﻴﺔ " "Microsoft Excelﻭﲢﺼﻠﻨﺎ ﻋﻠﻰ ﺍﳌﻨﺤﲏ ﺍﻟﺒﻴﺎﱐ ) ) uC = u AB = f ( tﺍﻟﺸﻜﻞ . (2- .1ﺇﻗﺘﺮﺡ ﳐﻄﻄﺎ ﻟﻠﺪﺍﺭﺓ ﻣﻮﺿﺤﺎ ﺇﲡﺎﻩ ﺍﻟﺘﻴﺎﺭ ﰒ ﻣﺜﻞ ﺑﺴﻬﻢ ﻛﻼ ﻣﻦ ﺍﻟﺘﻮﺗﺮﻳﻦ u Rﻭ . uC .2ﻋﲔ ﻗﻴﻤﺔ ﺛﺎﺑﺖ ﺍﻟﺰﻣﻦ tﻟﻠﺪﺍﺭﺓ ﻭﻣﺎ ﻣﺪﻟﻮﻟﻪ ﺍﻟﻔﻴﺰﻳﺎﺋﻲ ؟ ﺍﺳﺘﻨﺘﺞ ﻗﻴﻤﺔ ﺳﻌﺔ ﺍﳌﻜﺜﻔﺔ . C ﺻﻔﺤﺔ 11 ﺍﻟﺸــﻜﻞ 2- .3ﺃﺣﺴﺐ ﺷﺤﻨﺔ ﺍﳌﻜﺜﻔﺔ ﻋﻨﺪ ﺑﻠﻮﻍ ﺍﻟﺪﺍﺭﺓ ﻟﻠﻨﻈﺎﻡ ﺍﻟﺪﺍﺋﻢ . .4ﻟﻮ ﺍﺳﺘﺒﺪﻟﻨﺎ ﺍﳌﻜﺜﻔﺔ ﺍﻟﺴﺎﺑﻘﺔ ﲟﻜﺜﻔﺔ ﺃﺧﺮﻯ ﺳﻌﺘﻬﺎ ، C ' = 2Cﺃﺭﺳﻢ ﻛﻴﻔﻴﺎ ﰲ ﻧﻔﺲ ﺍﳌﻌﻠﻢ ﺍﻟﺴﺎﺑﻖ ﺷﻜﻞ ﺍﳌﻨﺤﲎ ) uC ' = g ( tﺍﻟﺬﻱ ﳝﻜﻦ ﻣﺸﺎﻫﺪﺗﻪ ﻋﻠﻰ ﺷﺎﺷﺔ ﺍﳉﻬﺎﺯ ﻣﻊ ﺍﻟﺘﻌﻠﻴﻞ . ﺍﳊﻞ ﺍﳌﻔﺼﻞ : . .1ﳐﻄﻂ ﺍﻟــﺪﺍﺭﺓ ﺍﻟﻜﻬﺮﺑﺎﺋﻴﺔ . ﺍﻟﺘﻤﺜﻴﻞ ﺍﳌﻮﺍﻓﻖ ﻟﻠﺪﺍﺭﺓ ﺑﺎﻟﺸﻜﻞ ﺍﳌﻘﺎﺑﻞ . .2ﺗﻌﻴﲔ ﻗﻴﻤﺔ ﺛﺎﺑﺖ ﺍﻟﺰﻣﻦ . t ﻣﻦ ﺍﻟﺒﻴﺎﻥ ﻟﺪﻳﻨﺎ ، t = 1ms :ﻭﻫﻮ ﺍﻟﺰﻣﻦ ﺍﻟﻼﺯﻡ ﻟﺒﻠﻮﻍ ﺷﺤﻨﺔ ﺍﳌﻜﺜﻔﺔ ﺍﻟﻘﻴﻤﺔ 63%ﻣﻦ ﻗﻴﻤﺔ ﺷﺤﻨﺘﻬﺎ ﺍﻟﻌﻈﻤﻰ . 10-3 t ﻟﺪﻳﻨﺎ t = R.C :ﻭﺑﺎﻟﺘﺎﱄ C = :ﻭﻣﻨﻪ = 10-5 F = 10 m F : 100 R .3ﺣﺴﺎﺏ ﺷﺤﻨﺔ ﺍﳌﻜﺜﻔﺔ ﻋﻨﺪ ﺑﻠﻮﻍ ﺍﻟﺪﺍﺭﺓ ﻟﻠﻨﻈﺎﻡ ﺍﻟﺪﺍﺋﻢ . =.C ﻟﺪﻳﻨﺎ ، Qmax = C.E :ﻭﺑﺎﻟﺘﺎﱄ Qmax = 5 ´ 10-5 C : .4ﺭﺳﻢ ﻣﻨﺤﲏ . uC ﻟﺪﻳﻨﺎ t = RC :ﻭ t ' = RC ' = 2 RCﻭﺑﺎﻟﺘﺎﱄ t ' = 2t : ﺍﻟﺘﻤﺮﻳﻦ ﺍﳊﺎﺩﻱ ﻋﺸﺮ : ﺭﻳﺎﺿﻴﺎﺕ ﻭ ﺗﻘﲏ ﺭﻳﺎﺿﻲ 2010 ﺑﻐﺮﺽ ﺷﺤﻦ ﻣﻜﺜﻔﺔ ﻓﺎﺭﻏﺔ ،ﺳﻌﺘﻬﺎ ، Cﻧﺼﻠﻬﺎ ﻋﻠﻰ ﺍﻟﺘﺴﻠﺴﻞ ﻣﻊ ﺍﻟﻌﻨﺎﺻﺮ ﺍﻟﻜﻬﺮﺑﺎﺋﻴﺔ ﺍﻟﺘﺎﻟﻴﺔ : ﻣﻮﻟﺪ ﺫﻭ ﺗﻮﺗﺮ ﻛﻬﺮﺑﺎﺋﻲ ﺛﺎﺑﺖ E = 5Vﻭﻣﻘﺎﻭﻣﺘﻪ ﺍﻟﺪﺍﺧﻠﻴﺔ ﻣﻬﻤﻠﺔ . ﻧﺎﻗﻞ ﺃﻭﻣﻲ ﻣﻘﺎﻭﻣﺘﻪ . R = 120W ﺑﺎﺩﻟﺔ ) Kﺍﻟﺸﻜﻞ . (2- .1ﳌﺘﺎﺑﻌﺔ ﺗﻄﻮﺭ ﺍﻟﺘﻮﺗﺮ ﺍﻟﻜﻬﺮﺑﺎﺋﻲ uCﺑﲔ ﻃﺮﰲ ﺍﳌﻜﺜﻔﺔ ﺑﺪﻻﻟﺔ ﺍﻟﺰﻣﻦ ،ﻧﻮﺻﻞ ﻣﻘﻴﺎﺱ ﻓﻮﻟﻄﻤﺘﺮ ﺭﻗﻤﻲ ﺑﲔ ﻃﺮﰲ ﺍﳌﻜﺜﻔﺔ ﻭﰲ ﺍﻟﻠﺤﻈﺔ ، t = 0ﻧﻀﻊ ﺍﻟﺒﺎﺩﻟﺔ ﰲ ﺍﻟﻮﺿﻊ ) . (1ﻭﺑﺎﻟﺘﺼﻮﻳﺮ ﺍﳌﺘﻌﺎﻗﺐ ﰎ ﺗﺼﻮﻳﺮ ﺷﺎﺷﺔ ﺟﻬﺎﺯ ﺍﻟﻔﻮﻟﻄﻤﺘﺮ ﺍﻟﺮﻗﻤﻲ ﳌﺪﺓ ﻣﻌﻴﻨﺔ ﻭﲟﺸﺎﻫﺪﺓ ﺷﺮﻳﻂ ﺍﻟﻔﻴﺪﻳﻮ ﺑﺒﻂﺀ ﺳﺠﻠﻨﺎ ﺍﻟﻨﺘﺎﺋﺞ ﺍﻟﺘﺎﻟﻴﺔ : 80 68 60 48 40 32 24 20 16 8 4 0 ) t ( ms 5,0 5,0 5,0 4,9 4,8 4,5 4,1 3,8 3,3 2,0 1,0 0 ) uC (V ﺃ -ﺍﺭﺳﻢ ﺍﻟﺒﻴﺎﻥ ) . uC = f ( t ﺏ-ﻋﲔ ﺑﻴﺎﻧﻴﺎ ﻗﻴﻤﺔ ﺛﺎﺑﺖ ﺍﻟﺰﻣﻦ tﻟﺜﻨﺎﺋﻲ ﺍﻟﻘﻄﺐ RCﻭﺍﺳﺘﻨﺘﺞ ﻗﻴﻤﺔ ﺍﻟﺴﻌﺔ Cﻟﻠﻤﻜﺜﻔﺔ . .2ﻛﻴﻒ ﺗﺘﻐﲑ ﻗﻴﻤﺔ ﺛﺎﺑﺖ ﺍﻟﺰﻣﻦ tﰲ ﺍﳊﺎﻟﺘﲔ ؟ ﺍﳊـﺎﻟﺔ )ﺃ( :ﻣﻦ ﺃﺟﻞ ﻣﻜﺜﻔﺔ ﺳﻌﺘﻬﺎ ' Cﺣﻴﺚ C ' > Cﻭ . R = 120W -ﺍﳊﺎﻟﺔ )ﺏ( :ﻣﻦ ﺃﺟﻞ ﻣﻜﺜﻔﺔ ﺳﻌﺘﻬﺎ '' Cﺣﻴﺚ C '' = Cﻭ . R < 120W ﺍﺭﺳﻢ ﻛﻴﻔﻴﺎ ،ﰲ ﻧﻔﺲ ﺍﳌﻌﻠﻢ ﺍﳌﻨﺤﻨﻴﲔ ) (1ﻭ ) (2ﺍﳌﻌﱪﻳﻦ ﻋﻦ ) uC ( tﰲ ﺍﳊﺎﻟﺘﲔ )ﺃ( ﻭ )ﺏ( ﺍﻟﺴﺎﺑﻘﺘﲔ . ) dq ( t 1 E .3ﺃ -ﺑﲔ ﺃﻥ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﺿﻠﻴﺔ ﺍﳌﻌﱪﺓ ﻋﻦ ) q ( tﺗﻌﻄﻰ ﺑﺎﻟﻌﺒﺎﺭﺓ q ( t ) = : dt RC R ﺏ -ﻳﻌﻄﻰ ﺣﻞ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﺿﻠﻴﺔ ﺑﺎﻟﻌﺒﺎﺭﺓ q ( t ) = Aea t + Bﺣﻴﺚ Aﻭ aﻭ Bﺛﻮﺍﺑﺖ ﻳﻄﻠﺐ ﺗﻌﻴﻨﻬﺎ ،ﻋﻠﻤﺎ ﺃﻧﻪ ﰲ ﺍﻟﻠﺤﻈﺔ t = 0ﺗﻜﻮﻥ . q ( 0 ) = 0 + . .4ﺍﳌﻜﺜﻔﺔ ﻣﺸﺤﻮﻧﺔ ﻧﻀﻊ ﺍﻟﺒﺎﺩﻟﺔ ﰲ ﺍﻟﻮﺿﻊ ) (2ﰲ ﳊﻈﺔ ﻧﻌﺘﱪﻫﺎ ﻛﻤﺒﺪﺃ ﻟﻸﺯﻣﻨﺔ . ﺃ -ﺍﺣﺴﺐ ﰲ ﺍﻟﻠﺤﻈﺔ t = 0ﺍﻟﻄﺎﻗﺔ ﺍﻟﻜﻬﺮﺑﺎﺋﻴﺔ ﺍﳌﺨﺰﻧﺔ E0ﰲ ﺍﳌﻜﺜﻔﺔ . E ﺏ-ﻣﺎ ﻫﻮ ﺍﻟﺰﻣﻦ ﺍﻟﻼﺯﻡ ﺍﻟﺬﻱ ﻣﻦ ﺃﺟﻠﻪ ﺗﺼﺒﺢ ﺍﻟﻄﺎﻗﺔ ﺍﳌﺨﺰﻧﺔ ﰲ ﺍﳌﻜﺜﻔﺔ E = 0؟ 2 ﺍﳊﻞ ﺍﳌﻔﺼﻞ : . .1ﺃ -ﺭﺳﻢ ﺍﻟﺒﻴﺎﻥ . uC = f (t ) :ﺍﻟﺸﻜﻞ ﺑﺎﻟﺼﻔﺤﺔ ﺍﳌﻮﺍﻟﻴﺔ . ﺻﻔﺤﺔ 12 ﺏ -ﲢﺪﻳﺪ ﺛﺎﺑﺖ ﺍﻟﺰﻣﻦ . ﻣﻦ ﺍﻟﺒﻴﺎﻥ ﳌﺎ ، uC = 0, 63 ´ 5 = 3,15V :ﳒﺪ . t = 16ms : t 16 ´10-3 ﻭﻟﺪﻳﻨﺎ t = R .C :ﻭﺑﺎﻟﺘﺎﱄ = 13,3 ´10-5 F : R 120 .2ﺍﳊــﺎﻟﺔ ﺃ : ﺛﺎﺑﺖ ﺍﻟﺰﻣﻦ tﻭﺳﻌﺔ ﺍﳌﻜﺜﻔﺔ ﻣﺘﻨﺎﺳﺒﺎﻥ ﻃﺮﺩﺍ ﻭﺑﺎﻟﺘﺎﱄ :ﳌﺎ ، C ' > Cﻳﻜﻮﻥ . t ' > t : = = .C ﺍﳊــﺎﻟﺔ ﺏ : ﺛﺎﺑﺖ ﺍﻟﺰﻣﻦ tﻣﻘﺎﻭﻣﺔ ﺍﻟﻨﺎﻗﻞ ﺍﻷﻭﻣﻲ ﻣﺘﻨﺎﺳﺒﺎﻥ ﻃﺮﺩﺍ ﻭﺑﺎﻟﺘﺎﱄ :ﳌﺎ R ' < Rﻳﻜﻮﻥ . t ' < t : ﺍﻟﺘﻤﺜﻴﻞ ﺍﻟﻜﻴﻔﻲ :ﺍﻟﺒﻴﺎﻥ ﺑﺎﻟﺸﻜﻞ ﺍﳌﻘﺎﺑﻞ . .3ﺃ -ﻛﺘﺎﺑﺔ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﺿﻠﻴﺔ ﺍﳌﻤﻴﺰﺓ ﻟﻠﺪﺍﺭﺓ . ﺣﺴﺐ ﻗﺎﻧﻮﻥ ﲨﻊ ﺍﻟﺘﻮﺗﺮﺍﺕ ﻟﺪﻳﻨﺎ . u R + uC = E : q dq u R = R .i = Rﻭ ﺣﻴﺚ : C dt dq 1 E dq q . + =q Rﻭﻣﻨﻪ : ﻭﺑﺎﻟﺘﺎﱄ + = E : dt RC R dt C = . uC ﺏ -ﲢﺪﻳﺪ ﻋﺒﺎﺭﺓ B ، Aﻭ . α dq ﻟﺪﻳﻨﺎ ، q (t ) = Ae at + B :ﻭﺑﺎﻟﺘﺎﱄ = A ae at : dt 1 E dq 1 E + = ) Ae at + B ﺇﺫﻥ : + =q ﻭﻟﺪﻳﻨﺎ : ( RC R dt RC R 1 B E 1 ،ﻭﻣﻨﻪ : = a+ﻭ ﺇﺫﻥ = 0 : RC RC R RC 1 ö at B E æ . A ça + = . A a e atﻭﺑﺎﻟﺘﺎﱄ : ÷e + RC ø RC R è a = -ﻭ . B = CE t æ ö RC q (t ) = CE ç1 - e ﻭﻟﺪﻳﻨﺎ ﻣﻦ ﺍﻟﺸﺮﻭﻁ ﺍﻹﻳﺘﺪﺍﺋﻴﺔ ) q ( 0 ) = A + B = 0 :ﺍﳌﻜﺜﻔﺔ ﻓﺎﺭﻏﺔ ( ،ﻭﺑﺎﻟﺘﺎﱄ . A = -B = -CE :ﻭﻣﻨﻪ ÷ : è ø .4ﺃ -ﺣﺴﺎﺏ ﺍﻟﻄﺎﻗﺔ ﺍﻟﻜﻬﺮﺑﺎﺋﻴﺔ ﺍﻟﻌﻈﻤﻰ . 1 1 ﻟﺪﻳﻨﺎ ، E 0 = CE 2 :ﻭﺑﺎﻟﺘﺎﱄ . E0 = ´ 13,3 ´ 10-5 ´ 52 = 1, 66 ´ 10-3 J : 2 2 ﺏ -ﲢﺪﻳﺪ ﺯﻣﻦ ﺍﻟﻨﺼﻒ . ﻟﺪﻳﻨﺎ ln 2 : t 2 = t1/ 2 16 ﻭﻣﻨﻪ ´ 0, 693 = 5,54ms : 2 = . t1/ 2 ﺍﻟﺘﻤﺮﻳﻦ ﺍﻟﺜﺎﱐ ﻋﺸﺮ : ﺭﻳﺎﺿﻴﺎﺕ ﻭﺗﻘﲏ ﺭﻳﺎﺿﻲ 2010 ﺗﺘﻜﻮﻥ ﺩﺍﺭﺓ ﻛﻬﺮﺑﺎﺋﻴﺔ ﻣﻦ ﺍﻟﻌﻨﺎﺻﺮ ﺍﻟﺘﺎﻟﻴﺔ ﻣﺮﺑﻮﻃﺔ ﻋﻠﻰ ﺍﻟﺘﺴﻠﺴﻞ : ﻭﺷﻴﻌﺔ ﺫﺍﺗﻴﺘﻬﺎ Lﻭﻣﻘﺎﻭﻣﺘﻬﺎ ، rﻧﺎﻗﻞ ﺃﻭﻣﻲ ﻣﻘﺎﻭﻣﺘﻪ ، R = 17,5Wﻣﻮﻟﺪ ﺫﻱ ﺗﻮﺗﺮ ﺛﺎﺑﺖ ، E = 6,00V ﻗﺎﻃﻌﺔ ﻛﻬﺮﺑﺎﺋﻴﺔ ) Kﺍﻟﺸﻜﻞ (3-ﻧﻐﻠﻖ ﺍﻟﻘﺎﻃﻌﺔ ﰲ ﺍﻟﻠﺤﻈﺔ . t = 0 ﲰﺤﺖ ﺑﺮﳎﻴﺔ ﻟﻺﻋﻼﻡ ﺍﻵﱄ ﲟﺘﺎﺑﻌﺔ ﺗﻄﻮﺭ ﺷﺪﺓ ﺍﻟﺘﻴﺎﺭ ﺍﻟﻜﻬﺮﺑﺎﺋﻲ ﺍﳌﺎﺭ ﰲ ﺍﻟﺪﺍﺭﺓ ﻣﻊ ﻣﺮﻭﺭ ﺍﻟﺰﻣﻦ ﻭﻣﺸﺎﻫﺪﺓ ﺍﻟﺒﻴﺎﻥ : ) ) i = f ( tﺍﻟﺸﻜﻞ . (4- .1ﺑﺎﻻﻋﺘﻤﺎﺩ ﻋﻠﻰ ﺍﻟﺒﻴﺎﻥ : ﺃ -ﺍﺳﺘﻨﺘﺞ ﻗﻴﻢ ﻛﻞ ﻣﻦ ﺷﺪﺓ ﺍﻟﺘﻴﺎﺭ ﺍﻟﻜﻬﺮﺑﺎﺋﻲ ﰲ ﺍﻟﻨﻈﺎﻡ ﺍﻟﺪﺍﺋﻢ ،ﻗﻴﻤﺔ ﺛﺎﺑﺖ ﺍﻟﺰﻣﻦ tﻟﻠﺪﺍﺭﺓ . ﺏ-ﺍﺣﺴﺐ ﻛﻞ ﻣﻦ ﺍﳌﻘﺎﻭﻣﺔ rﻭ ﺍﻟﺬﺍﺗﻴﺔ Lﻟﻠﻮﺷﻴﻌﺔ . .2ﰲ ﺍﻟﻨﻈﺎﻡ ﺍﻹﻧﺘﻘﺎﱄ : ﺻﻔﺤﺔ 13 ﺍﻟﺸــﻜﻞ 3- )i ( A di i I 0 = + ﺃ -ﺑﺘﻄﺒﻴﻖ ﻗﺎﻧﻮﻥ ﺍﻟﺘﻮﺗﺮﺍﺕ ﺃﺛﺒﺖ ﺃﻥ: dt t t t - ö æ ﺏ-ﺑﲔ ﺃﻥ ﺣﻞ ﺍﳌﻌﺎﺩﻟﺔ ﻫﻮ ﻣﻦ ﺍﻟﺸﻜﻞ . i = I 0 ç 1 - e t ÷ : è ø .3ﻧﻌﺘﱪ ﺍﻵﻥ ﻗﻴﻤﺔ ﺍﻟﺬﺍﺗﻴﺔ Lﻟﻠﻮﺷﻴﻌﺔ ﻭﲟﻌﺎﳉﺔ ﺍﳌﻌﻄﻴﺎﺕ ﺑﱪﳎﻴﺔ ﺇﻋﻼﻡ ﺁﱄ ﻧﺴﺠﻞ ﻗﻴﻢ tﺛﺎﺑﺖ ﺍﻟﺰﻣﻦ ﻟﻠﺪﺍﺭﺓ ﻟﻨﺤﺼﻞ ﻋﻠﻰ ﺟﺪﻭﻝ ﺍﻟﻘﻴﺎﺳﺎﺕ ﺍﻟﺘﺎﱄ : ﺣﻴﺚ I 0ﺷﺪﺓ ﺍﻟﺘﻴﺎﺭ ﰲ ﺍﻟﻨﻈﺎﻡ ﺍﻟﺪﺍﺋﻢ 20 12 8 4 ) t ( ms 0,5 0,3 0,2 0,1 ) L(H ﺃ -ﺍﺭﺳﻢ ﺍﻟﺒﻴﺎﻥ . L = h (t ) : 0,05 ) t ( ms ﺏ-ﺍﻛﺘﺐ ﻣﻌﺎﺩﻟﺔ ﺍﻟﺒﻴﺎﻥ . ﺟ -ﺍﺳﺘﻨﺘﺞ ﻗﻴﻤﺔ ﻣﻘﺎﻭﻣﺔ ﺍﻟﻮﺷﻴﻌﺔ ، rﻫﻞ ﺗﺘﻮﺍﻓﻖ ﻫﺬﻩ ﺍﻟﻘﻴﻤﺔ ﻣﻊ ﺍﻟﻘﻴﻤﺔ ﺍﶈﺴﻮﺑﺔ ﰲ ﺍﻟﺴﺆﺍﻝ -1ﺏ؟ ﺍﻟﺸــﻜﻞ 4- ﺍﳊﻞ ﺍﳌﻔﺼﻞ : 10 . .1ﺃ -ﺇﳚﺎﺩ ﺷﺪﺓ ﺍﻟﺘﻴﺎﺭ ﺍﻟﻜﻬﺮﺑﺎﺋﻲ ﰲ ﺍﻟﻨﻈﺎﻡ ﺍﻟﺪﺍﺋﻢ ﻭﲢﺪﻳﺪ ﺛﺎﺑﺖ ﺍﻟﺰﻣﻦ . ﻣﻦ ﺍﻟﺒﻴﺎﻥ ﰲ ﺍﻟﻨﻈﺎﻡ ﺍﻟﺪﺍﺋﻢ ﻟﺪﻳﻨﺎ . I 0 = 0, 24A : ﻣﻦ ﺍﻟﺒﻴﺎﻥ ﳌﺎ i = 0, 24 ´ 0, 63 » 0,15 A :ﻟﺪﻳﻨﺎ . t » 10ms : ﺏ -ﺣﺴﺎﺏ ﻣﻘﺎﻭﻣﺔ ﺍﻟﻮﺷﻴﻌﺔ ﻭﺫﺍﺗﻴﺘﻬﺎ . 6 E E = I 0ﻭﺑﺎﻟﺘﺎﱄ r = - R :ﺇﺫﻥ - 17,5 = 7,5W : =. r ﻟﺪﻳﻨﺎ ﰲ ﺍﻟﻨﻈﺎﻡ ﺍﻟﺪﺍﺋﻢ : 0, 24 R +r I0 L = tﻭﺑﺎﻟﺘﺎﱄ L = t ( R + r ) :ﺇﺫﻥ . L = 10 ´10 -3 (17,5 + 7,5 ) = 0, 25H : ﻭﻟﺪﻳﻨﺎ : R +r .2ﻛﺘﺎﺑﺔ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﺿﻠﻴﺔ ﺍﳌﻤﻴﺰﺓ ﻟﻠﺪﺍﺭﺓ . di . u b = r .i + L ﺣﺴﺐ ﻗﺎﻧﻮﻥ ﲨﻊ ﺍﻟﺘﻮﺗﺮﺍﺕ ﻟﺪﻳﻨﺎ ، E = u R + u b :ﺣﻴﺚ u R = R .iﻭ dt ) di ( R + r E di . + = i ( R + r ) i + Lﺇﺫﻥ : ﻭﺑﺎﻟﺘﺎﱄ = E : dt L L dt di i I 0 L .I 0 L . = + = . E = ( R + r ) I 0ﻭﻣﻨﻪ ﺗﻜﺘﺐ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﺿﻠﻴﺔ ﻋﻠﻰ ﺍﻟﺸﻜﻞ : = tﻭ ﻟﻜﻦ : dt t t t R +r ﺇﺛﺒﺎﺕ ﺣﻞ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﺿﻠﻴﺔ . ﺏ- t - ö æ di I 0 -tt t . ﻟﺪﻳﻨﺎ ، i (t ) = I 0 ç1 - e ÷ .ﻭﺑﺎﻟﺘﺎﱄ = .e : dt t è ø t t - ö di i I 0 -t I 0 æ ﺇﺫﻥ + = .e + ç 1 - e t ÷ : dt t t t è ø di i I 0 -tt I 0 I 0 -tt I 0 = + = .e + - e ﺃﻱ : dt t t t t t t - ö æ ﻭﻣﻨﻪ :ﺍﻟﻌﺒﺎﺭﺓ ÷ i (t ) = I 0 ç1 - e tﺣﻞ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﺿﻠﻴﺔ . è ø .3ﺃ -ﺭﺳﻢ ﺍﻟﺒﻴﺎﻥ ) : L = h (tﺃﻧﻈﺮ ﺍﻟﺸﻜﻞ ﺍﳌﻘﺎﺑﻞ . . ﺏ-ﻛﺘﺎﺑﺔ ﻣﻌﺎﺩﻟﺔ ﺍﻟﺒﻴﺎﻥ : ﺍﻟﺒﻴﺎﻥ ﻋﺒﺎﺭﺓ ﻋﻦ ﺧﻂ ﻣﺴﺘﻘﻴﻢ ﲤﺪﻳﺪﻩ ﳝﺮ ﺑﺎﳌﺒﺪﺃ ﻣﻌﺎﺩﻟﺘﻪ ﻣﻦ ﺍﻟﺸﻜﻞ . L = a.t : DL 0,5 - 0,1 = ، aﻭﺑﺎﻟﺘﺎﱄ ﻣﻌﺎﺩﻟﺔ ﺍﻟﺒﻴﺎﻥ ﻫﻲ . L = 25.t : = ﺣﻴﺚ a :ﻣﻴﻠﻪ ﻭ = 25H .s -1 = 25W Dt ( 20 - 4 ) ´10-3 ﻟﺪﻳﻨﺎ L = ( R + r ) .t :ﺃﻱ ، a = R + r :ﻭﺑﺎﻟﺘﺎﱄ ، r = a - R = 25 - 17,5 = 7,5W :ﻭﻫﻲ ﻧﻔﺴﻬﺎ ﺍﻟﻘﻴﻤﺔ ﺍﶈﺴﻮﺑﺔ ﺳﺎﺑﻘﺎ . ﺻﻔﺤﺔ 14 ﺍﻟﺘﻤﺮﻳﻦ ﺍﻟﺜﺎﻟﺚ ﻋﺸﺮ : ﺭﻳﺎﺿﻴﺎﺕ ﻭﺗﻘﲏ ﺭﻳﺎﺿﻲ 2010 ﻧﺮﺑﻂ ﻋﻠﻰ ﺍﻟﺘﺴﻠﺴﻞ ﺍﻟﻌﻨﺎﺻﺮ ﺍﻟﻜﻬﺮﺑﺎﺋﻴﺔ ﺍﻟﺘﺎﻟﻴﺔ : ﻧﺎﻗﻞ ﺃﻭﻣﻲ ﻣﻘﺎﻭﻣﺘﻪ . R = 500W ﻣﻜﺜﻔﺔ ﺳﻌﺘﻬﺎ Cﻏﲑ ﻣﺸﺤﻮﻧﺔ . ﻣﻮﻟﺪ ﺫﻱ ﺗﻮﺗﺮ ﻛﻬﺮﺑﺎﺋﻲ ﺛﺎﺑﺖ . E -ﻗﺎﻃﻌﺔ ) Kﺍﻟﺸﻜﻞ . (2- ﻣﻜﻨﺖ ﻣﺘﺎﺑﻌﺔ ﺗﻄﻮﺭ ﺍﻟﺘﻮﺗﺮ ﺍﻟﻜﻬﺮﺑﺎﺋﻲ ) uC ( tﺑﲔ ﻟﺒﻮﺳﻲ ﺍﳌﻜﺜﻔﺔ ﺑﺮﺳﻢ ﺍﻟﺒﻴﺎﻥ )ﺍﻟﺸﻜﻞ .(3- ﺍﻟﺸـﻜﻞ 2- .1ﻋﻤﻠﻴﺎ ﻳﻜﺘﻤﻞ ﺷﺤﻦ ﺍﳌﻜﺜﻔﺔ ﻋﻨﺪﻣﺎ ﻳﺒﻠﻎ ﺍﻟﺘﻮﺗﺮ ﺍﻟﻜﻬﺮﺑﺎﺋﻲ ﺑﲔ ﻃﺮﻓﻴﻬﺎ 99%ﻣﻦ ﻗﻴﻤﺔ ﺍﻟﺘﻮﺗﺮ ﺍﻟﻜﻬﺮﺑﺎﺋﻲ ﺑﲔ ﻃﺮﰲ ﺍﳌﻮﻟﺪ . ﺍﻋﺘﻤﺎﺩﺍ ﻋﻠﻰ ﺍﻟﺒﻴﺎﻥ : ﺃ -ﻋﲔ ﻗﻴﻤﺔ ﺛﺎﺑﺖ ﺍﻟﺰﻣﻦ tﻭﻗﻴﻤﺔ ﺍﻟﺘﻮﺗﺮ ﺍﻟﻜﻬﺮﺑﺎﺋﻲ ﺑﲔ ﻃﺮﰲ ﺍﳌﻮﻟﺪ ﰒ ﺍﺣﺴﺐ ﺳﻌﺔ ﺍﳌﻜﺜﻔﺔ C . ﺏ -ﺣﺪﺩ ﺍﳌﺪﺓ ﺍﻟﺰﻣﻨﻴﺔ ' tﻻﻛﺘﻤﺎﻝ ﻋﻤﻠﻴﺔ ﺷﺤﻦ ﺍﳌﻜﺜﻔﺔ . ﺟ -ﻣﺎ ﻫﻲ ﺍﻟﻌﻼﻗﺔ ﺑﲔ ' tﻭ . t .2ﺑﺘﻄﺒﻴﻖ ﻗﺎﻧﻮﻥ ﲨﻊ ﺍﻟﺘﻮﺗﺮﺍﺕ ﺃﻭﺟﺪ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﺿﻠﻴﺔ ﺑﺪﻻﻟﺔ ﺍﻟﺘﻮﺗﺮ ﺍﻟﻜﻬﺮﺑﺎﺋﻲ ﺑﲔ ﻃﺮﰲ ﺍﳌﻜﺜﻔﺔ ) = uC ( t AB ، uﰒ ﺑﲔ ﺃﺎ ﺗﻘﺒﻞ ﺣﻼ ﻣﻦ ﺍﻟﺸﻜﻞ ) : t -t ( uC ( t ) = E 1 - e . .3ﺃﻭﺟﺪ ﻗﻴﻤﺔ ﺍﻟﻄﺎﻗﺔ ﺍﻟﻜﻬﺮﺑﺎﺋﻴﺔ ﺍﳌﺨﺰﻧﺔ ECﰲ ﺍﳌﻜﺜﻔﺔ ﻋﻨﺪ ﺍﻟﻠﺤﻈﺎﺕ t2 = 5t ، t1 = t ، t0 = 0 . ) u C (V 2 ) t ( ms ﺍﻟﺸـﻜﻞ 3- 10 .4ﺗﻮﻗﻊ )ﺭﺳﻢ ﻛﻴﻔﻲ( ﺷﻜﻞ ﺍﳌﻨﺤﲎ ) . EC = f ( t ﺍﳊﻞ ﺍﳌﻔﺼﻞ : . .1ﺃ -ﺗﻌﲔ ﺛﺎﺑﺖ ﺍﻟﺰﻣﻦ tﻭ ﺍﻟﺘﻮﺗﺮ ﺑﲔ ﻃﺮﰲ ﺍﳌﻮﻟﺪ ، Eﰒ ﺣﺴﺎﺏ ﺳﻌﺔ ﺍﳌﻜﺜﻔﺔ . C ﳑﺎﺱ ﺍﻟﺒﻴﺎﻥ ﻋﻨﺪ ﺍﳌﺒﺪﺃ ،ﻳﻘﻄﻊ ﺍﳌﻘﺎﺭﺏ ﰲ ﺍﻟﻨﻘﻄﺔ ﺫﺍﺕ ﺍﻟﻔﺎﺻﻠﺔ . t = 14ms : ﻳﺒﻠﻎ ﺍﻟﺘﻮﺗﺮ ﺑﲔ ﻃﺮﰲ ﺍﳌﻜﺜﻔﺔ ﻗﻴﻤﺔ ﻋﻈﻤﻰ . E = 7, 4 ´ 2 = 14,8V : t 14 ´10-3 ﻟﺪﻳﻨﺎ t = RC :ﻭﺑﺎﻟﺘﺎﱄ = 2,8 ´10-5 F = 28m F : = =. C R 500 ﺏ -ﲢﺪﻳﺪ ﺍﳌﺪﺓ ﺍﻟﺰﻣﻨﻴﺔ ' tﻻﻛﺘﻤﺎﻝ ﻋﻤﻠﻴﺔ ﺷﺤﻦ ﺍﳌﻜﺜﻔﺔ . ﻳﺒﻠﻎ ﺍﻟﺘﻮﺗﺮ ﺑﲔ ﻃﺮﰲ ﺍﳌﻜﺜﻔﺔ ﻗﻴﻤﺘﻪ ﺍﻟﻌﻈﻤﻰ ) ( UC = 0,99.E = 14,65Vﺇﺑﺘﺪﺍﺀ ﻣﻦ ﺍﻟﻠﺤﻈﺔ . t ' = 70 ms : ﺟ -ﲢﺪﻳﺪ ﺍﻟﻌﻼﻗﺔ ﺑﲔ ' tﻭ : tﻧﻼﺣﻆ ﺃﻥ . t ' = 5t .2ﻛﺘﺎﺑﺔ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﺿﻠﻴﺔ ﺍﳌﻤﻴﺰﺓ ﻟﻠﺪﺍﺭﺓ . ﺣﺴﺐ ﻗﺎﻧﻮﻥ ﲨﻊ ﺍﻟﺘﻮﺗﺮﺍﺕ ﻟﺪﻳﻨﺎ . u R + uC = E : dq du ﺣﻴﺚ = RC C : dt dt du 1 E du . C+ = uC ﻭﺑﺎﻟﺘﺎﱄ RC C + uC = E :ﻭﻣﻨﻪ : dt RC RC dt u R = R.i = R ) -ﺇﺛﺒﺎﺕ ﺣﻞ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﺿﻠﻴﺔ . ( duC E - t t -t = e ﻟﺪﻳﻨﺎ uC ( t ) = E 1 - e tﻭﺑﺎﻟﺘﺎﱄ : dt t t 1 E -t E E - tt E + .E 1 - e t = e t + = e ﺇﺫﻥ : RC t RC RC RC .3ﺇﳚﺎﺩ ﺃﻭﺟﺪ ﻗﻴﻤﺔ ﺍﻟﻄﺎﻗﺔ ﺍﻟﻜﻬﺮﺑﺎﺋﻴﺔ ﺍﳌﺨﺰﻧﺔ ECﰲ ﺍﳌﻜﺜﻔﺔ . . ) ( duC 1 E -t + uC = e t dt RC t ﺻﻔﺤﺔ 15 ) .ﺣﻴﺚ ( t = RC : 1 1 ﻟﺪﻳﻨﺎ ، EC = C .uC2 = ´ 2,8 ´ 10 -5.uC2 : 2 2 ﻭﺑﺎﻟﺘﺎﱄ . EC = 1, 4 ´ 10-5.uC2 : ) t ( ms ) uC (V t0 = 0 t1 = t = 14 0 9,324 t2 = 5t = 70 14,652 ) EC (´10-3 J ) EC ( Joule 0 1, 22 ´ 10-3 -3 3, 01´ 10 .4ﺭﺳﻢ ﻛﻴﻔﻲ ﻟﺸﻜﻞ ﺍﳌﻨﺤﲎ ) . EC = f ( t 0,5 ﺃﻧﻈﺮ ﺍﻟﺒﻴﺎﻥ ﺍﳌﻘﺎﺑﻞ ) t ( ms 10 ﺍﻟﺘﻤﺮﻳﻦ ﺍﻟﺮﺍﺑﻊ ﻋﺸﺮ : ﻋﻠﻮﻡ ﲡﺮﻳﺒﻴﺔ 2011 ﻣﻜﺜﻔﺔ ﺳﻌﺘﻬﺎ Cﺷﺤﻨﺖ ﻛﻠﻴﺎ ﲢﺖ ﺗﻮﺗﺮ ﺛﺎﺑﺖ . E = 6Vﻣﻦ ﺃﺟﻞ ﻣﻌﺮﻓﺔ ﺳﻌﺘﻬﺎ Cﻧﻘﻮﻡ ﺑﺘﻔﺮﻳﻐﻬﺎ ﰲ ﻧﺎﻗﻞ ﺃﻭﻣﻲ ﻣﻘﺎﻭﻣﺘﻪ . R = 4k W .1ﺍﺭﺳﻢ ﳐﻄﻂ ﺩﺍﺭﺓ ﺍﻟﺘﻔﺮﻳﻎ . .2ﳌﺘﺎﺑﻌﺔ ﺗﻄﻮﺭ ﺍﻟﺘﻮﺗﺮ ) uC ( tﺑﲔ ﻃﺮﰲ ﺍﳌﻜﺜﻔﺔ ﺧﻼﻝ ﺍﻟﺰﻣﻦ ﻧﺴﺘﻌﻤﻞ ﺟﻬﺎﺯ ﻓﻮﻟﻂ ﻣﺘﺮ ﺭﻗﻤﻲ ﻭﻣﻴﻘﺎﺗﻴﺔ ﺇﻟﻜﺘﺮﻭﻧﻴﺔ . ﺃ -ﻛﻴﻒ ﻳﺘﻢ ﺭﺑﻂ ﺟﻬﺎﺯ ﺍﻟﻔﻮﻟﻂ ﻣﺘﺮ ﰲ ﺍﻟﺪﺍﺭﺓ ؟ ﻧﻐﻠﻖ ﺍﻟﻘﺎﻃﻌﺔ ﰲ ﺍﻟﻠﺤﻈﺔ t = 0msﻭﻧﺴﺤﻞ ﻧﺘﺎﺋﺞ ﺍﳌﺘﺎﺑﻌﺔ ﰲ ﺍﳉﺪﻭﻝ ﺍﻟﺘﺎﱄ : 120 100 80 60 40 30 20 10 0 ) t ( ms 0,54 0,81 1,21 1,81 2,69 3,21 4,02 4,91 6,00 ) uC (V ﺏ-ﺍﺭﺳﻢ ﺍﳌﻨﺤﲎ ﺍﻟﺒﻴﺎﱐ ﺍﳌﻤﺜﻞ ﻟﻠﺪﺍﻟﺔ ) uC = f ( tﻋﻠﻰ ﻭﺭﻗﺔ ﻣﻴﻠﻴﻤﺘﺮﻳﺔ ،ﺃﺭﻓﻘﻬﺎ ﻣﻊ ﻭﺭﻗﺔ ﺇﺟﺎﺑﺘﻚ . ﺟ -ﻋﲔ ﺑﻴﺎﻧﻴﺎ ﻗﻴﻤﺔ ﺛﺎﺑﺖ ﺍﻟﺰﻣﻦ . t ﺩ -ﺍﺣﺴﺐ ﺳﻌﺔ ﺍﳌﻜﺜﻔﺔ . C .3ﺃ -ﺑﺘﻄﺒﻴﻖ ﻗﺎﻧﻮﻥ ﲨﻊ ﺍﻟﺘﻮﺗﺮﺍﺕ ،ﺍﻛﺘﺐ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﺿﻠﻴﺔ ﻟﻠﺘﻮﺗﺮ ﺍﻟﻜﻬﺮﺑﺎﺋﻲ ) . uC ( t ﺏ -ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﺿﻠﻴﺔ ﺍﻟﺴﺎﺑﻘﺔ ﺗﻘﺒﻞ ﺍﻟﻌﺒﺎﺭﺓ uC ( t ) = Ae-a tﺣﻼ ﳍﺎ ،ﺣﻴﺚ A ، aﺛﺎﺑﺘﺎﻥ ﻳﻄﻠﺐ ﺗﻌﻴﻨﻬﻤﺎ . ﺍﳊﻞ ﺍﳌﻔﺼﻞ : . .1ﳐﻄﻂ ﺩﺍﺭﺓ ﺍﻟﺘﻔﺮﻳﻎ :ﺍﻟﺸﻜﻞ ﺍﳌﻘﺎﺑﻞ .2ﺃ -ﻳﺘﻢ ﺭﺑﻂ ﺟﻬﺎﺯ ﺍﻟﻔﻮﻟﻂ ﻣﺘﺮ ﰲ ﺍﻟﺪﺍﺭﺓ :ﻋﻠﻰ ﺍﻟﺘﻔﺮﻉ ﺍﻟﺸﻜﻞ ﺍﳌﻘﺎﺑﻞ . ) uC (V ﺏ -ﺭﺳﻢ ﺍﻟﺒﻴﺎﻥ ) . uC = f ( t ﺍﻟﺒﻴﺎﻥ ﻣﻮﺿﺢ ﺑﺎﻟﺸﻜﻞ ﺍﳌﻘﺎﺑﻞ . ﺟ -ﺗﻌﲔ ﻗﻴﻤﺔ ﺛﺎﺑﺖ ﺍﻟﺰﻣﻦ . t ﻣﻦ ﺍﻟﺒﻴﺎﻥ ﳌﺎ . uC = 0,37 ´ E = 0,37 ´ 6 = 2, 22Vﳒﺪ . t = 50ms : ﺩ -ﺍﺣﺴﺐ ﺳﻌﺔ ﺍﳌﻜﺜﻔﺔ C 50 ´10-3 ﻟﺪﻳﻨﺎ t = RC :ﻭﺑﺎﻟﺘﺎﱄ = 12,5 ´10-6 F : 4 ´103 .3ﺃ -ﻛﺘﺎﺑﺔ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﺿﻠﻴﺔ ﻟﻠﺘﻮﺗﺮ ﺍﻟﻜﻬﺮﺑﺎﺋﻲ ) . uC ( t = t R =C ) t ( ms duC 1 du + ﺣﺴﺐ ﻗﺎﻧﻮﻥ ﲨﻊ ﺍﻟﺘﻮﺗﺮﺍﺕ ﻟﺪﻳﻨﺎ . uC + uR = 0 :ﻭﺑﺎﻟﺘﺎﱄ uC + RC C = 0 :ﻭﻣﻨﻪ uC = 0 : dt RC dt ﺏ -ﺗﻌﲔ ﺍﻟﺜﺎﺑﺘﺎﻥ . A ، a ﺻﻔﺤﺔ 16 . ﻣﻦ ﺍﻟﺸﺮﻭﻁ ﺍﻹﺑﺘﺪﺍﺋﻴﺔ ﻟﺪﻳﻨﺎ uC ( 0 ) = E = A : duC ﻟﺪﻳﻨﺎ uC ( t ) = Ae-a t :ﻭﺑﺎﻟﺘﺎﱄ = -a Ae-a t : dt 1 duC 1 + ﻭﻟﺪﻳﻨﺎ uC = 0 : + ﺇﺫﻥ Ae -a t = 0 : dt RC RC ﺇﺫﻥ : 1 t RC - 1 ö -a . t 1 æ ، A ç -a +ﻭﻣﻨﻪ : -a Ae -a .tﻭﺑﺎﻟﺘﺎﱄ = 0 : ÷e RC ø RC è =a ، uC = E.eﺃﻱ . uC ( t ) = 6.e -20.t : ﺍﻟﺘﻤﺮﻳﻦ ﺍﳋﺎﻣﺲ ﻋﺸﺮ : ﻋﻠﻮﻡ ﲡﺮﻳﺒﻴﺔ 2011 ﲢﺘﻮﻱ ﺩﺍﺭﺓ ﻋﻠﻰ ﺍﻟﻌﻨﺎﺻﺮ ﺍﻟﻜﻬﺮﺑﺎﺋﻴﺔ ﺍﻟﺘﺎﻟﻴﺔ ﻣﺮﺑﻮﻃﺔ ﻋﻠﻰ ﺍﻟﺘﺴﻠﺴﻞ ) ﺍﻟﺸﻜﻞ( 2- ﻣﻮﻟﺪ ﺫﻱ ﺗﻮﺗﺮ ﺛﺎﺑﺖ . E -ﻭﺷﻴﻌﺔ ﺫﺍﺗﻴﺘﻬﺎ Lﻭﻣﻘﺎﻭﻣﺘﻬﺎ . r ﺍﻟﺸــﻜﻞ2- ﻧﺎﻗﻞ ﺃﻭﻣﻲ ﻣﻘﺎﻭﻣﺘﻪ . R = 100W -ﻗﺎﻃﻌﺔ . K ﻟﻠﻤﺘﺎﺑﻌﺔ ﺍﻟﺰﻣﻨﻴﺔ ﻟﺘﻄﻮﺭ ﺍﻟﺘﻮﺗﺮ ﺑﲔ ﻃﺮﰲ ﻛﻞ ﻣﻦ ﺍﻟﻮﺷﻴﻌﺔ ) ub ( tﻭﺍﻟﻨﺎﻗﻞ ﺍﻷﻭﻣﻲ ) uR ( tﻧﺴﺘﻌﻤﻞ ﺭﺍﺳﻢ ﺍﻫﺘﺰﺍﺯ ﻣﻬﺒﻄﻲ ﺫﻱ ﺫﺍﻛﺮﺓ . .1ﺃ -ﺑﻴﻦ ﻛﻴﻒ ﳝﻜﻦ ﺭﺑﻂ ﺭﺍﺳﻢ ﺍﻻﻫﺘﺰﺍﺯ ﺍﳌﻬﺒﻄﻲ ﺑﺎﻟﺪﺍﺭﺓ ﳌﺸﺎﻫﺪﺓ ﻛﻞ ﻣﻦ ) ub ( tﻭ ) uR ( t؟ ﺏ -ﻧﻐﻠﻖ ﺍﻟﻘﺎﻃﻌﺔ ﰲ ﺍﻟﻠﺤﻈﺔ t = 0sﻓﻨﺸﺎﻫﺪ ﻋﻠﻰ ﺍﻟﺸﺎﺷﺔ ﺍﻟﺒﻴﺎﻧﻴﲔ ﺍﳌﻤﺜﻠﲔ ﻟﻠﺘﻮﺗﺮﻳﻦ ) ub ( tﻭ ) ) uR ( tﺍﻟﺸﻜﻞ. ( 3- -ﺍﻧﺴﺐ ﻛﻞ ﻣﻨﺤﲎ ﻟﻠﺘﻮﺗﺮ ﺍﳌﻮﺍﻓﻖ ﻟﻪ .ﻣﻊ ﺍﻟﺘﻌﻠﻴﻞ . ﺍﻟﺸــﻜﻞ3- ) di ( t .2ﺃ -ﺍﺛﺒﺖ ﺃﻥ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﺿﻠﻴﺔ ﻟﺸﺪﺓ ﺍﻟﺘﻴﺎﺭ ﺍﳌﺎﺭ ﰲ ﺍﻟﺪﺍﺭﺓ ﺗﻜﻮﻥ ﻣﻦ ﺍﻟﺸﻜﻞ + Ai ( t ) = B : dt ﺏ -ﺃﻋﻂ ﻋﺒﺎﺭﺓ ﻛﻞ ﻣﻦ Aﻭ Bﺑﺪﻻﻟﺔ Eﻭ Lﻭ rﻭ . R B ﺟ -ﲢﻘﻖ ﻣﻦ ﺃﻥ ﺍﻟﻌﺒﺎﺭﺓ ) 1 - e - At ( A . = ) i ( tﻫﻲ ﺣﻼ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﺿﻠﻴﺔ ﺍﻟﺴﺎﺑﻘﺔ . ﺩ -ﺍﺣﺴﺐ ﺷﺪﺓ ﺍﻟﺘﻴﺎﺭ ﰲ ﺍﻟﻨﻈﺎﻡ ﺍﻟﺪﺍﺋﻢ . I 0 ﻫ -ﺍﺣﺴﺐ ﻛﻞ ﻣﻦ Eﻭ rﻭ tﻭ . L ﻭ -ﺃﺣﺴﺐ ﺍﻟﻄﺎﻗﺔ ﺍﻷﻋﻈﻤﻴﺔ ﺍﳌﺨﺰﻧﺔ ﺑﺎﻟﻮﺷﻴﻌﺔ . ﺍﳊﻞ ﺍﳌﻔﺼﻞ : . .1ﺃ -ﺗﻮﺿﻴﺢ ﻛﻴﻔﺔ ﺭﺑﻂ ﺭﺍﺳﻢ ﺍﻻﻫﺘﺰﺍﺯ ﺍﳌﻬﺒﻄﻲ ﺑﺎﻟﺪﺍﺭﺓ . ﺍﻟﺸﻜﻞ ﺍﳌﻘﺎﺑﻞ :ﻋﻠﻰ ﺍﳌﺪﺧﻞ y1ﻧﺸﺎﻫﺪ ) ub ( t؛ ﻭ ﻋﻠﻰ ﺍﳌﺪﺧﻞ y1ﻧﺸﺎﻫﺪ ) . uR ( t ﺏ -ﺍﻧﺴﺎﺏ ﻛﻞ ﻣﻨﺤﲎ ﻟﻠﺘﻮﺗﺮ ﺍﳌﻮﺍﻓﻖ . ﺑﻌﺪ ﻏﻠﻖ ﺍﻟﻘﺎﻃﻌﺔ ﺗﺘﺰﺍﻳﺪ ﺷﺪﺓ ﺍﻟﺘﻴﺎﺭ ﺍﻟﻜﻬﺮﺑﺎﺋﻲ ﺑﺎﻟﺪﺍﺭﺓ ﻭﺑﺎﻟﺘﺎﱄ ﻳﺘﺰﺍﻳﺪ ﻓﺮﻕ ﺍﻟﻜﻤﻮﻥ ) ) uR ( tﻷﻥ ، ( u R = R.i : ﻭﻳﺘﻨﺎﻗﺺ ﻓﺮﻕ ﺍﻟﻜﻤﻮﻥ ) ) ub ( tﻷﻥ . ( ub = E - u R : ﺻﻔﺤﺔ 17 ﻭﻣﻨﻪ :ﺍﳌﻨﺤﲎ 1-ﻳﻮﺍﻓﻖ ﻓﺮﻕ ﺍﻟﻜﻤﻮﻥ ) uR ( t؛ ﻭ ﺍﳌﻨﺤﲎ 2-ﻳﻮﺍﻓﻖ ﻓﺮﻕ ﺍﻟﻜﻤﻮﻥ ) . uR ( t .2ﺃ -ﻛﺘﺎﺑﺔ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﺿﻠﻴﺔ ﺍﳌﻤﻴﺰﺓ ﻟﻠﺪﺍﺭﺓ . ﺣﺴﺐ ﻗﺎﻧﻮﻥ ﲨﻊ ﺍﻟﺘﻮﺗﺮﺍﺕ ﻟﺪﻳﻨﺎ u R ( t ) + ub ( t ) = E :ﺇﺫﻥ= E : ) di ( t dt Ri ( t ) + ri ( t ) + Lﺃﻱ = E : di ( t ) R + r ) di ( t E + ﻭﻣﻨﻪ i ( t ) = : ؛ ﻭﻫﻲ ﻣﻦ ﺍﻟﺸﻜﻞ + Ai ( t ) = B : dt dt L L E R+r ﺏ -ﺇﻋﻂ ﻋﺒﺎﺭﺓ ﻛﻞ ﻣﻦ Aﻭ . Bﺑﺎﳌﻄﺎﺑﻘﺔ ﳒﺪ : ﻭ =B =A L L ﺟ -ﺍﻟﺘﺤﻘﻖ ﻣﻦ ﺣﻞ ﺍﳌﻌﺎﺩﻟﺔ . ) di ( t B ﻟﺪﻳﻨﺎ i ( t ) = (1 - e - At ) :ﻭﺑﺎﻟﺘﺎﱄ = B.e - At : A dt ) di ( t dt ( R + r ) i (t ) + L . . ) di ( t ) di ( t B ؛ ﺃﻱ + Ai ( t ) = B : ﺇﺫﻥ + Ai ( t ) = B.e - At + A. (1 - e - At ) = B.e - At + B - B.e - At : dt dt A . R+r t ö B E æ - At L = ) . i (t ç1 - e ﻭﻣﻨﻪ i ( t ) = (1 - e ) :ﺣﻞ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﺿﻠﻴﺔ ،ﺃﻱ ÷ : A R+rè ø ﺩ -ﺍﺣﺴﺐ ﺷﺪﺓ ﺍﻟﺘﻴﺎﺭ ﰲ ﺍﻟﻨﻈﺎﻡ ﺍﻟﺪﺍﺋﻢ . I 0 U R 10 = ﻣﻦ ﺍﳌﻨﺤﲏ : 1-ﰲ ﺍﻟﻨﻈﺎﻡ ﺍﻟﺪﺍﺋﻢ ﳒﺪ U R = 10V :؛ ﻭﻟﺪﻳﻨﺎ U R = R.I 0 :؛ ﻭﺑﺎﻟﺘﺎﱄ = 0,1A : R 100 = . I0 ﻫ -ﺣﺴﺎﺏ ﻛﻞ ﻣﻦ Eﻭ rﻭ tﻭ . L ﻣﻦ ﺍﳌﻨﺤﲏ 1-ﻭ 2ﰲ ﺍﻟﻨﻈﺎﻡ ﺍﻟﺪﺍﺋﻢ ﻟﺪﻳﻨﺎ U R = 10V :ﻭ ، U b = 2Vﻭﺑﺎﻟﺘﺎﱄ . E = U b + U R = 10 + 2 = 12V : Ub 2 = ﻭﻟﺪﻳﻨﺎ ، U b = r.I 0 :ﻭﺑﺎﻟﺘﺎﱄ = 20W : I 0 0,1 ﻣﻦ ﺍﳌﻨﺤﲏ : 1-ﳌﺎ ، uR = 0, 63 ´ U R = 6,3Vﳒﺪ t = 10ms : L ﻭﻟﺪﻳﻨﺎ : R+r =r . . = tﻭﺑﺎﻟﺘﺎﱄ . L = t ( R + r ) = 10 ´10 -3 ´120 = 0,12 H : ﻭ -ﺣﺴﺐ ﺍﻟﻄﺎﻗﺔ ﺍﻷﻋﻈﻤﻴﺔ ﺍﳌﺨﺰﻧﺔ ﺑﺎﻟﻮﺷﻴﻌﺔ . 1 2 1 2 2 ﻟﺪﻳﻨﺎ ، E ( L ) = L.I 02 :ﻭﺑﺎﻟﺘﺎﱄ . E ( L ) = ´ 0,12 ´ ( 0,1) = 6 ´ 10-2 J : ﺍﻟﺘﻤﺮﻳﻦ ﺍﻟﺴﺎﺩﺱ ﻋﺸﺮ : ﺭﻳﺎﺿﻴﺎﺕ ﻭﺗﻘﲏ ﺭﻳﺎﺿﻲ 2010 ﳓﻘﻖ ﺍﻟﺪﺍﺭﺓ ) ﺍﻟﺸﻜﻞ (5-ﻭﺍﻟﱵ ﺗﺘﻜﻮﻥ ﻣﻦ ﻣﻮﻟﺪ ﻟﺘﻮﺗﺮ ﺛﺎﺑﺖ ، E = 6,0Vﻭﻣﻜﺜﻔﺔ ﺳﻌﺘﻬﺎ C = 250m Fﻭﻧﺎﻗﻠﲔ ﺃﻭﻣﻴﲔ ﻣﺘﻤﺎﺛﻠﲔ ﻣﻘﺎﻭﻣﺔ ﻛﻞ ﻣﻨﻬﻤﺎ R = 200W ﻭﺑﺎﺩﻟﺔ . K ﺃﻭﻻ :ﻧﻀﻊ ﺍﻟﺒﺎﺩﻟﺔ ﻋﻠﻰ ﺍﻟﻮﺿﻊ : 1 .1ﺃ -ﺃﻋﻂ ﺭﺳﻢ ﺍﻟﺪﺍﺭﺓ ) ﺍﻟﺸﻜﻞ (5-ﻣﺒﻴﻨﺎ ﻋﻠﻴﻬﺎ ﺟﻬﺔ ﺍﻧﺘﻘﺎﻝ ﺣﺎﻣﻼﺕ ﺍﻟﺸﺤﻨﺔ ﻭﻣﺎ ﻃﺒﻴﻌﺘﻬﺎ ؟ ﺣﺪﺩ ﺷﺤﻨﺔ ﻛﻞ ﻟﺒﻮﺱ ﻭﺟﻬﺔ ﺍﻟﺘﻴﺎﺭ . ﺏ -ﺫﻛﺮ ﺑﺎﻟﻌﻼﻗﺔ ﺑﲔ ) i ( tﻭ ) q ( tﻭﺍﻟﻌﻼﻗﺔ ﺑﲔ ) uC ( tﻭ ) q ( tﰒ ﺍﺳﺘﻨﺘﺞ ﺍﻟﻌﻼﻗﺔ ﺑﲔ ) i ( tﻭ ) . uC ( t .2ﺃ -ﺃﻭﺟﺪ ﺍﻟﻌﻼﻗﺔ ﺑﲔ ) uR ( tﻭ ) uC ( tﻭﺑﲔ ﺃﻥ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﺿﻠﻴﺔ ﺍﻟﱵ ﳛﻘﻘﻬﺎ ) uC ( tﻫﻲ ﻣﻦ ) duC ( t ﺍﻟﺸﻜﻞ + uC ( t ) = A : dt ﺏ -ﺃﻭﺟﺪ ﺍﻟﻘﻴﻤﺔ ﺍﻟﻌﺪﺩﻳﺔ ﻟﻜﻞ ﻣﻦ . A ، t 1 ﺟ -ﺃﻭﺟﺪ ﻣﻦ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﺿﻠﻴﺔ ﻭﺣﺪﺓ . t 1ﻋﺮﻓﻪ . . t1 ﺻﻔﺤﺔ 18 ﺍﻟﺸـــﻜﻞ -5- .3ﺃ -ﺇﻗﺮﺃ ﻋﻠﻰ ﺍﳌﻨﺤﲎ ﺍﻟﺒﻴﺎﱐ ) ﺍﻟﺸﻜﻞ ( 6-ﻗﻴﻤﺔ ﺛﺎﺑﺖ ﺍﻟﺰﻣﻦ ، t 1ﻭﻗﺎﺭﺎ ﺑﺎﻟﻘﻴﻤﺔ ﺍﶈﺴﻮﺑﺔ ﺳﺎﺑﻘﺎ . ﺏ -ﺣﺪﺩ ﺑﻴﺎﻧﻴﺎ ﺍﳌﺪﺓ ﺍﻟﺰﻣﻨﻴﺔ Dtﺍﻟﺼﻐﺮﻯ ﺍﻟﻼﺯﻣﺔ ﻻﻋﺘﺒﺎﺭ ﺍﳌﻜﺜﻔﺔ ﻋﻤﻠﻴﺎ ﻣﺸﺤﻮﻧﺔ .ﻗﺎﺭﺎ ﻣﻊ t 1 ﺛﺎﻧﻴﺎ :ﻧﻀﻊ ﺍﻟﺒﺎﺩﻟﺔ ﻋﻠﻰ ﺍﻟﻮﺿﻊ . 2 ﺃ -ﻣﺎ ﺍﻟﻈﺎﻫﺮﺓ ﺍﻟﻔﻴﺰﻳﺎﺋﻴﺔ ﺍﻟﱵ ﲢﺪﺙ ؟ ﺃﻛﺘﺐ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﺻﻠﻴﺔ ﻟـ ) uC ( tﺍﳌﻮﺍﻓﻘﺔ . ﺏ -ﺃﺣﺴﺐ ، t 2ﻗﺎﺭﺎ ﺑـ . t 1ﻣﺎﺫﺍ ﺗﺴﺘﻨﺘﺞ ؟ ﺟ -ﻣﺜﻞ ﺑﺸﻜﻞ ﺗﻘﺮﻳﱯ ﺍﳌﻨﺤﲏ ﺍﻟﺒﻴﺎﱐ ﻟﺘﻐﲑ ) uC ( tﻣﺴﺘﻌﻴﻨﺎ ﺑﺎﻟﻘﻴﻢ ﺍﳌﻤﻴﺰﺓ . ﺍﳊﻞ ﺍﳌﻔﺼﻞ : . ﺃﻭﻻ :ﺍﻟﺒﺎﺩﻟﺔ ﻋﻠﻰ ﺍﻟﻮﺿﻊ : 1 .1ﺃ -ﳐـﻄﻂ ﺍﻟـﺪﺍﺭﺓ .ﺍﻟـﺸﻜﻞ ﺍﳌﻘﺎﺑﻞ . ﺣﺎﻣﻼﺕ ﺍﻟﺸﺤﻨﺔ ﻫﻲ ﺍﻹﻟﻜﺘﺮﻭﻧﺎﺕ . ﺏ -ﻛﺘﺎﺑﺔ ﺍﻟﻌﻼﻗﺎﺕ ﺍﳌﻄﻠﻮﺑﺔ . ) dq ( t = ) i ( tﻭ ) ، q ( t ) = C.uC ( tﻭﺑﺎﻟﺘﺎﱄ : ﻟﺪﻳﻨﺎ : dt .2ﺃ -ﻛﺘﺎﺑﺔ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﺿﻠﻴﺔ . ﻟﺪﻳﻨﺎ : ) duC ( t dt uR = R.i ( t ) = RC. ) duC ( t dt . i ( t ) = C. .ﺣﺴﺐ ﻗﺎﻧﻮﻥ ﲨﻊ ﺍﻟﺘﻮﺗﺮﺍﺕ ﻟﺪﻳﻨﺎ . u R ( t ) + uC ( t ) = E : ) duC ( t ﻭﺑﺎﻟﺘﺎﱄ + uC ( t ) = E : dt ﺏ -ﺇﳚﺎﺩ ﺍﻟﻘﻴﻤﺔ ﺍﻟﻌﺪﺩﻳﺔ ﻟﻜﻞ ﻣﻦ . A ، t 1 . RC.ﻭﻫﻲ ﻣﻦ ﺍﻟﺸﻜﻞ + uC ( t ) = A : ) duC ( t dt . t1 ﺑﺎﳌﻄﺎﺑﻘﺔ t 1 = RC = 200 ´ 250 ´ 10-6 = 5 ´ 10-2 s :ﻭ . A = E = 6V ﺟ -ﺇﳚﺎﺩ ﻭﺣﺪﺓ t 1؛ ﻭﺗﻌﺮﻳﻔﻪ . ﻣﻦ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﺿﻠﻴﺔ ﻧﺴﺘﻨﺘﺞ ﺃﻥ ، t é duC ( t ) ù = U :ﺃﻱ [U ] = U : [ 1]ê ] [ ] [ ú ] [T ë dt û ] [t 1 ﻭﻣﻨﻪ . [t 1 ] = [T ] : ﺗﻌﺮﻳﻔﻪ :ﻫﻮ ﺍﳌﺪﺓ ﺍﻟﻀﺮﻭﺭﻳﺔ ﻟﺸﺤﻦ ﺍﳌﻜﺜﻔﺔ ﺑﻨﺴﺒﺔ . 63% .3ﺃ -ﲢﺪﻳﺪ ﻗﻴﻤﺔ . t 1 ﳑﺎﺱ ﺍﻟﺒﻴﺎﻥ ﻳﻘﻄﻊ ﺍﳌﻘﺎﺭﺏ ﰲ ﺍﻟﻨﻘﻄﺔ ﺫﺍﺕ ﺍﻟﻔﺎﺻﻠﺔ t 1 = 0, 05s :ﻭﻫﻲ ﻣﻄﺎﺑﻘﺔ ﻟﻠﻘﻴﻤﺔ ﺍﶈﺴﻮﺑﺔ ﺳﺎﺑﻘﺎ . ﺏ -ﲢــﺪﻳﺪ . Dt ﻳﺴﺘﻘﺮ ﺍﻟﺒﻴﺎﻥ ﺇﺑﺘﺪﺍﺀ ﻣﻦ ﺍﻟﻠﺤﻈﺔ Dt » 0, 25s :ﻭﻫﻲ ﺗﻮﺍﻓﻖ ﺍﻟﻘﻴﻤﺔ . Dt » 5.t1 : ﺛﺎﻧﻴﺎ :ﺍﻟﺒﺎﺩﻟﺔ ﻋﻠﻰ ﺍﻟﻮﺿﻊ . 2 ﺕ -ﺍﻟﻈﺎﻫﺮﺓ ﺍﻟﻔﻴﺰﻳﺎﺋﻴﺔ ﺍﳊﺎﺩﺛﺔ :ﺗﻔﺮﻳﻎ ﺍﳌﻜﺜﻔﺔ ﰲ ﺍﻟﻨﺎﻗﻠﲔ ﺍﻷﻭﻣﻴﲔ . ) ﲢﻮﻝ ﺍﻟﻄﺎﻗﺔ ﺍﻟﻜﻬﺮﺑﺎﺋﻴﺔ ﺍﳌﺨﺰﻧﺔ ﰲ ﺍﳌﻜﺜﻔﺔ ﺇﱃ ﻃﺎﻗﺔ ﺣﺮﺍﺭﻳﺔ ﰲ ﺍﻟﻨﺎﻗﻠﲔ ﺍﻷﻭﻣﻴﲔ ﺑﻔﻌﻞ ﺟﻮﻝ ( ) uC (V ﻛﺘﺎﺑﺔ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﺻﻠﻴﺔ ﺍﳌﻮﺍﻓﻘﺔ . ﺣﺴﺐ ﻗﺎﻧﻮﻥ ﲨﻊ ﺍﻟﺘﻮﺗﺮﺍﺕ ﻟﺪﻳﻨﺎ . uC ( t ) + 2uR ( t ) = 0 : ) du ( t 1 . uC ( t ) + 2RC. Cﻭﻣﻨﻪ uC ( t ) = 0 : ﻭﺑﺎﻟﺘﺎﱄ = 0 : 2RC dt + ) duC ( t dt ﺙ -ﺣﺴﺎﺏ t 2ﻭﻣﻘﺎﺭﻧﺘﻪ ﻣﻊ . t1 ﻟﺪﻳﻨﺎ ، t 2 = 2RC = 2 ´ 200 ´ 250 ´10-6 = 0,1s :ﻧﻼﺣﻆ ﺃﻥ . t 2 = 2t 1 : ﺍﻹﺳﺘﻨﺘﺎﺝ :ﻣﺪﺓ ﺗﻔﺮﻳﻎ ﺍﳌﻜﺜﻔﺔ ﺗﺴﺎﻭﻱ ﺿﻌﻒ ﻣﺪﺓ ﺷﺤﻨﻬﺎ . 1 ) t ( ms ﺟ -ﲤﺜﻴﻞ ﺷﻜﻞ ﺗﻘﺮﻳﱯ ﻟﻠﻤﻨﺤﲏ ﺍﻟﺒﻴﺎﱐ ﻟﺘﻐﲑ ) . uC ( t ﺍﻟﺒﻴﺎﻥ ﺑﺎﻟﺸﻜﻞ ﺍﳌﻘﺎﺑﻞ . ﺣﻴﺚ . uC ( 5t 2 ) = 0 ، uC (t 2 ) = 0,37 E = 2, 22V ، uC ( 0 ) = E = 6V : ﺻﻔﺤﺔ 19 0,05 ﺍﻟﺘﻤﺮﻳﻦ ﺍﻟﺴﺎﺑﻊ ﻋﺸﺮ : ﺪﻑ ﺗﻌﲔ ﺍﻟﺜﺎﺑﺘﲔ ) ( L, r ﺭﻳﺎﺿﻴﺎﺕ ﻭﺗﻘﲏ ﺭﻳﺎﺿﻲ 2010 ﺍﳌﻤﻴﺰﻳﻦ ﻟﻮﺷﻴﻌﺔ ،ﳓﻘﻖ ﺍﻟﺪﺍﺭﺓ ﺍﻟﻜﻬﺮﺑﺎﺋﻴﺔ ) ﺍﻟﺸﻜﻞ. ( 1- ﺣﻴﺚ E = 9V :ﻭ . R = 45Wﰲ ﺍﻟﻠﺤﻈﺔ t1 = 0 sﻧﻐﻠﻖ ﺍﻟﻘﺎﻃﻌﺔ . K .1ﺑﺎﺳﺘﺨﺪﺍﻡ ﻗﺎﻧﻮﻥ ﲨﻊ ﺍﻟﺘﻮﺗﺮﺍﺕ ،ﺑﲔ ﺃﻥ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﺿﻠﻴﺔ ﻟﺸﺪﺓ ﺍﻟﺘﻴﺎﺭ ﺍﻟﻜﻬﺮﺑﺎﺋﻲ ﻫﻲ : di ( t ) 1 E . = ) + i (t dt L t ö .2ﺍﻟﻌﺒﺎﺭﺓ ÷ ø ﻣﺎﺫﺍ ﳝﺜﻞ ؟ ﺍﻟﺸــﻜﻞ 1- t æ i ( t ) = A ç 1 - e tﻫﻲ ﺣﻞ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﺿﻠﻴﺔ ﺍﻟﺴﺎﺑﻘﺔ ،ﺃﻭﺟﺪ ﺍﻟﺜﺎﺑﺖ ، A è .3ﻋﱪ ﻋﻦ ﺛﺎﺑﺖ ﺍﻟﺰﻣﻦ tﺑﺪﻻﻟﺔ r ، Lﻭ Rﻭﺑﲔ ﺑﺎﻟﺘﺤﻠﻴﻞ ﺍﻟﺒﻌﺪﻱ ﺃﻧﻪ ﻣﺘﺠﺎﻧﺲ ﻣﻊ ﺍﻟﺰﻣﻦ . .4ﺑﻮﺍﺳﻄﺔ ﻻﻗﻂ ﺁﻣﺒﲑ ﻣﺘﺮ ﻣﻮﺻﻮﻝ ﺑﺎﻟﺪﺍﺭﺓ ﻭﻣﺮﺗﺒﻂ ﺑﻮﺍﺟﻬﺔ ﺩﺧﻮﻝ ﳉﻬﺎﺯ ﺇﻋﻼﻡ ﺁﱄ ﻣﺰﻭﺩ ﺑﱪﳎﻴﺔ ﻣﻨﺎﺳﺒﺔ ﳓﺼﻞ ﻋﻠﻰ ﺍﻟﺘﻄﻮﺭ ﺍﻟﺰﻣﲏ ﻟﻠﺘﻴﺎﺭ ﺍﻟﻜﻬﺮﺑﺎﺋﻲ ) ) i ( tﺍﻟﺸﻜﻞ. (2- ﺃ -ﺃﻭﺟﺪ ﺑﻴﺎﻧﻴﺎ ﻗﻴﻤﺔ ﺛﺎﺑﺖ ﺍﻟﺰﻣﻦ ، tﻣﻊ ﺷﺮﺡ ﺍﻟﻄﺮﻳﻘﺔ ﺍﳌﺘﺒﻌﺔ . ﺏ-ﺃﻭﺟﺪ ﻗﻴﻤﺔ ﺍﳌﻘﺎﻭﻣﺔ ، rﰒ ﺃﺣﺴﺐ ﻗﻴﻤﺔ ﺫﺍﺗﻴﺔ ﺍﻟﻮﺷﻴﻌﺔ L .5ﺃﺣﺴﺐ ﺍﻟﻄﺎﻗﺔ ﺍﻷﻋﻈﻤﻴﺔ ﺍﳌﺨﺰﻧﺔ ﰲ ﺍﻟﻮﺷﻌﺔ . ﺍﻟﺸــﻜﻞ 2- ﺍﳊﻞ ﺍﳌﻔﺼﻞ : . .1ﻛﺘﺎﺑﺔ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﺿﻠﻴﺔ ﺍﳌﻤﻴﺰﺓ ﻟﻠﺪﺍﺭﺓ . ﺣﺴﺐ ﻗﺎﻧﻮﻥ ﲨﻊ ﺍﻟﺘﻮﺗﺮﺍﺕ ﻟﺪﻳﻨﺎ u R ( t ) + ub ( t ) = E :ﺇﺫﻥ= E : ﺃﻱ = E : ) di ( t dt ) di ( t dt Ri ( t ) + ri ( t ) + L ( R + r ) i (t ) + L di ( t ) R + r E di ( t ) 1 L E ﻭﻣﻨﻪ i ( t ) = : + ﺣﻴﺚ : ؛ ﻭﻫﻲ ﻣﻦ ﺍﻟﺸﻜﻞ + i ( t ) = : dt L L R+r t dt L .2ﺇﳚﺎﺩ ﺍﻟﺜﺎﺑﺖ . A =t t - ö æ di ( t ) 1 di ( t ) A - tt E t .ﻭﻟﺪﻳﻨﺎ : = ) + i (t ﻟﺪﻳﻨﺎ ، i ( t ) = A ç 1 - e ÷ :ﻭﺑﺎﻟﺘﺎﱄ = e : t t dt L dt è ø t t t t - ö A -t 1 æ E A A AE A E . = . e t + - e tﺇﺫﻥ : ،ﺃﻱ : ﺇﺫﻥ : = ÷ e + .A ç1 - e t = t t t t L L t t è ø L E L E E ﻭﻣﻨﻪ : = = L R+r L R+r .3ﻋﺒﺎﺭﺓ ﺛﺎﺑﺖ ﺍﻟﺰﻣﻦ . t . ﺍﻟﺸــﻜﻞ 1- . A = I 0 = 4, 5 ´ 0, 04 = 0,18 A ، A = tﻭﻫﻲ ﲤﺜﻞ ﺍﻟﺸﺪﺓ ﺍﻷﻋﻈﻤﻴﺔ ﻟﻠﺘﻴﺎﺭ ﺍﳌﺎﺭ ﰲ ﺍﻟﺪﺍﺭﺓ ) ﰲ ﺍﻟﻨﻈﺎﻡ ﺍﻟﺪﺋﻢ ( L L ﺗﻌﻄﻰ ﻋﺒﺎﺭﺓ ﺛﺎﺑﺖ ﺍﻟﺰﻣﻦ ﻛﻤﺎﻳﻠﻲ : = R + r Réq = . tﻭﺑﺎﻟﺘﺎﱄ = [T ] : ] [U ][T ][ I = = ] [t -1 éë Réq ùû ] [U ][ I -1 .4ﺃ -ﲢﺪﻳﺪ ﻗﻴﻤﺔ ﺛﺎﺑﺖ ﺍﻟﺰﻣﻦ . t ﻣﻦ ﺍﻟﺒﻴﺎﻥ ﳌﱠﺎ i = 0, 63.I 0 = 0, 63 ´ 0,18 = 0,11A :ﳒﺪ . t = 0, 2ms : ][ L ﺏ -ﲢﺪﻳﺪ ﻗﻴﻤﺔ ﺍﳌﻘﺎﻭﻣﺔ ، rﰒ ﺣﺴﺎﺏ ﻗﻴﻤﺔ ﺫﺍﺗﻴﺔ ﺍﻟﻮﺷﻴﻌﺔ . L L E E 9 = tﺇﺫﻥ . L = t ( R + r ) = 0, 2 ´ 10 -3 ´ 50 = 10 -2 H : = . r = - Rﻭﻟﺪﻳﻨﺎ : ﻟﺪﻳﻨﺎ : = I 0ﺇﺫﻥ - 45 = 5W : R+r R+r I0 0,18 1 1 2 .5ﺣﺴﺎﺏ ﺍﻟﻄﺎﻗﺔ ﺍﻷﻋﻈﻤﻴﺔ ﺍﳌﺨﺰﻧﺔ ﰲ ﺍﻟﻮﺷﻌﺔ .ﻟﺪﻳﻨﺎ E ( L ) = L.I 02 = ´ 10 -2 ´ ( 0,18 ) = 1, 62 ´ 10 -4 : 2 2 ﺻﻔﺤﺔ 20 ﻫﺎﻡ ﺟــﺪﺍ :ﻭﺭﺩ ﺧﻂ ﰲ ﺍﻟﺴﻠﺴﻠﺔ ﺭﻗﻢ 01ﻟﻠﻮﺣﺪﺓ ﺍﻷﻭﱃ ﰲ ﺍﻟﺘﻤﺮﻳﻦ 10ﺷﻌﺒﺔ ﺍﻟﻌﻠﻮﻡ ﺍﻟﺘﺠﺮﻳﺒﻴﺔ . 2010ﻓﺄﺭﺟﻮ ﺍﳌﻌﺬﺭﺓ . ﺍﳋﻄﺄ ﺑﺎﻟﺘﺤﺪﻳﺪ ﰲ ﺍﻟﺘﺠﺮﺑﺔ ﺍﻟﺜﺎﻧﻴﺔ ﻭﻫﻮ ﻧﻘﻄﺔ ﺑﺪﺍﻳﺔ ﺍﻟﺒﻴﺎﻥ . 02 .3ﺍﻟﺘﺠﺮﺑﺔ ﺍﻟﺜﺎﻧﻴﺔ :ﻳﺘﻨﺎﻗﺺ ﺍﻟﺒﻴﺎﻥ ﺑﺴﺮﻋﺔ ﺃﻗﻞ ﻣﻦ ﺍﳊﺎﻟﺔ ﺍﻷﻭﱃ ،ﻭﺫﻟﻚ ﻷﻥ ﺗﺮﻛﻴﺰ ﺛﻨﺎﺋﻲ ﺍﻟﻴﻮﺩ ﺍﳌﺴﺘﻌﻤﻞ ﰲ ﻫﺬﻩ ﺍﳊﺎﻟﺔ ﺃﻗﻞ ﻣﻦ ﺗﺮﻛﻴﺰ ﺛﻨﺎﺋﻲ ﺍﻟﻴﻮﺩ ﰲ ﺍﳊﺎﻟﺔ ﺍﻷﻭﱃ)ﳏﻠﻮﻝ ﳐﻔﻒ( . C 20 V 100 = = F = f؛ ﻭﺑﺎﻟﺘﺎﱄ = 10mmol .L -1 : ﻟﺪﻳﻨﺎ = 2 : F 2 Vi 50 = . [ I 2 ]i .4ﺍﻟﺘﺠﺮﺑﺔ ﺍﻟﺜﺎﻟﺜﺔ :ﻳﺘﻨﺎﻗﺺ ﺍﻟﺒﻴﺎﻥ ﺑﺴﺮﻋﺔ ﺃﻛﱪ ﻣﻦ ﺍﳊﺎﻟﺘﲔ ،ﻭﺫﻟﻚ ﻻﺭﺗﻔﺎﻉ ﺩﺭﺟﺔ ﺍﳊﺮﺍﺭﺓ .5ﺍﻟﻌﻮﺍﻣﻞ ﺍﳊﺮﻛﻴﺔ ﺍﻟﱵ ﺗﱪﺯﻫﺎ ﻫﺬﻩ ﺍﻟﺘﺠﺎﺭﺏ ﻫﻲ ﺗﺄﺛﲑ ﺗﺮﺍﻛﻴﺰ ﺍﳌﺘﻔﺎﻋﻼﺕ ﻭﺩﺭﺟﺔ ﺍﳊﺮﺍﺭﺓ ﻋﻠﻰ ﺳﺮﻋﺔ ﺍﻟﺘﻔﺎﻋﻞ . ﺍﻹﺳﺘﻨﺘﺎﺝ : · ﻛﻠﻤﺎ ﻛﺎﻧﺖ ﺗﺮﺍﻛﻴﺰ ﺍﳌﺘﻔﺎﻋﻼﺕ ﺃﻛﱪ ﻛﻠﻤﺎ ﻛﺎﻧﺖ ﺳﺮﻋﺔ ﺍﻟﺘﻔﺎﻋﻞ ﺃﻛﱪ . · ﻛﻠﻤﺎ ﻛﺎﻧﺖ ﺩﺭﺟﺔ ﺣﺮﺍﺭﺓ ﺍﻟﻮﺳﻂ ﺃﻋﻠﻰ ﻛﻠﻤﺎ ﻛﺎﻧﺖ ﺳﺮﻋﺔ ﺍﻟﺘﻔﺎﻋﻞ ﺃﻛﱪ . ﻣﻼﺣﻈﺔ :ﻳﻜﻮﻥ ﺯﻣﻦ ﻧﺼﻒ ﺍﻟﺘﻔﺎﻋﻞ ﺃﺻﻐﺮ ﰲ ﺍﻟﺘﺤﻮﻝ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ ﺍﻷﺳﺮﻉ . ﺻﻔﺤﺔ 21
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