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‫ﺳﻠﺴﻠﺔ ﺗﻤﺎرﯾﻦ ﺗﻨﺎﺋﻲ اﻟﻘﻄﺐ ‪RC‬‬ ‫ﺗﻤﺮﯾﻦ ‪1‬‬ ‫ﯾﺘﻜﻮن اﻟﺘﺮﻛﯿﺐ ﺟﺎﻧﺒﮫ ﻣﻦ‪ - :‬ﻣﻮﻟﺪ ﻟﻠﺘﻮﺗﺮ اﻟﻜﮭﺮﺑﺎﺋﻲ ﻗﻮﺗﮫ اﻟﻜﮭﺮﻣﺤﺮﻛﺔ ‪ E‬وﻣﻘﺎوﻣﺘﮫ اﻟﺪاﺧﻠﯿﺔ‪- r .‬‬ ‫ﻣﻜﺜﻒ ﺳﻌﺘﮫ‪ - C=4,7μA .‬ﻣﻮﺻﻞ أوﻣﻲ ﻣﻘﺎوﻣﺘﮫ‪ - R=1kΩ .‬ﻗﺎطﻊ ﻟﻠﺘﯿﺎر‪K‬‬ ‫‪-1‬ﻓﻲ ﻟﺤﻈﺔ ﺗﺎرﯾﺨﮭﺎ ‪ ، t₀=0‬ﻧﻀﻊ ﻗﺎطﻊ اﻟﺘﯿﺎر ‪ K‬ﻓﻲ اﻟﻤﻮﺿﻊ ‪ ، 1‬ﻣﺎذا ﯾﺤﺪث ﻟﻠﻤﻜﺜﻒ ؟‬ ‫‪-2‬أوﺟﺪ اﻟﻤﻌﺎدﻟﺔ اﻟﺘﻔﺎﺿﻠﯿﺔ ﻟﻠﺘﻮﺗﺮ )‪ UC(t‬ﺑﯿﻦ ﻣﺮﺑﻄﻲ اﻟﻤﻜﺜﻒ أﺛﻨﺎء ﺷﺤﻨﮫ‪.‬‬ ‫‪-3‬ﯾﻜﺘﺐ ﺣﻞ اﻟﻤﻌﺎدﻟﺔ اﻟﺘﻔﺎﺿﻠﯿﺔ ﻋﻠﻰ اﻟﺸﻜﻞ اﻟﺘﺎﻟﻲ ) ‪: UC(t)=K.(1- /‬‬ ‫أوﺟﺪ ﺗﻌﺒﯿﺮي ‪ K‬و ‪ ‬ﺑﺪﻻﻟﺔ ﻣﻌﻄﯿﺎت اﻟﺘﻤﺮﯾﻦ‪ ، .‬ﺛﻢ اﺣﺴﺐ ﻗﯿﻤﺘﯿﮭﻤﺎ‬ ‫‪-4‬ﯾﻤﺜﻞ اﻟﻤﻨﺤﻨﻰ ﺗﻐﯿﺮات ‪ UC‬ﺑﺪﻻﻟﺔ اﻟﺰﻣﻦ‬ ‫‪ -1-4‬ﺑﺜﻼت طﺮق ﺣﺪد ﻗﯿﻤﺔ ‪ ‬ﺗﺎﺑﺜﺔ اﻟﺰﻣﻦ‬ ‫‪ -2-4‬ﺣﺪد ﻗﯿﻤﺔ اﻟﻘﻮه اﻟﻜﮭﺮﻣﺤﺮﻛﺔ ‪ E‬ﻟﻠﻤﻮﻟﺪ و ﻣﻘﺎوﺗﮫ اﻟﺪاﺧﻠﯿﺔ ‪r‬‬ ‫‪ -3-4‬ﺣﺪد ﻓﻲ اي ﻟﺤﻈﺔ ﯾﺄﺧﺪ اﻟﺘﯿﺎر اﻟﻜﮭﺮﺑﺎﺋﻲ ﻓﻲ اﻟﺪارة ﻗﯿﻤﺔ ﻗﺼﻮى ﺣﺪد ﻗﯿﻤﺘﮫ‬ ‫‪ -5-4‬ﻋﻨﺪ اﻟﻠﺤﻈﺔ ‪ t=100ms‬اﺣﺴﺐ اﻟﻄﺎﻗﺔ اﻟﻤﺨﺰﻧﺔ ﻓﻲ اﻟﻤﻜﺜﻒ ﻣﺎذا ﺗﺴﺘﻨﺘﺞ‬ ‫‪-8‬ﻋﻨﺪﻣﺎ ﯾﺸﺤﻦ اﻟﻤﻜﺜﻒ ﻛﻠﯿﺎ ‪ ،‬وﻓﻲ ﻟﺤﻈﺔ ﻧﺄﺧﺬھﺎ ﻣﻦ ﺟﺪﯾﺪ أﺻﻼ ﻟﻠﺘﻮارﯾﺦ ‪ ، t₀=0‬ﻧﺆرﺟﺢ‬ ‫ﻗﺎطﻊ اﻟﺘﯿﺎر ‪ K‬اﻟﻰ اﻟﻤﻮﺿﻊ‪2‬‬ ‫‪-1-8‬أوﺟﺪ اﻟﻤﻌﺎدﻟﺔ اﻟﺘﻔﺎﺿﻠﯿﺔ ﻟﺸﺤﻨﺔ اﻟﻤﻜﺜﻒ )‪ q(t‬أﺛﻨﺎء ﺗﻔﺮﯾﻐﮫ‪.‬‬ ‫‪-2-8‬ﺣﻞ اﻟﻤﻌﺎدﻟﺔ اﻟﺘﻔﺎﺿﻠﯿﺔ‪ : q(t)=K / .‬ﺣﺪد ﺗﻌﺒﯿﺮ ‪ K‬و ‪‬‬ ‫‪-3-8‬ﺧﻂ اﻟﻤﻨﺤﻨﻰ اﻟﻤﻤﺜﻞ ﻟﺘﻐﯿﺮات )‪ q(t‬ﺷﺤﻨﺔاﻟﻤﻜﺜﻒ ﺑﺪﻻﻟﺔ اﻟﺰﻣﻦ ‪.‬‬ ‫‪-5-8‬أوﺟﺪ ﺗﻌﺒﯿﺮ ‪ Ee‬اﻟﻄﺎﻗﺔ اﻟﻜﮭﺮﺑﺎﺋﯿﺔ اﻟﻤﺨﺰوﻧﺔ ﻓﻲ اﻟﻤﻜﺜﻒ ﺑﺪﻻﻟﺔ اﻟﺰﻣﻦ‪ .‬أﺣﺴﺐ ‪ Ee‬ﻋﻨﺪ اﻟﻠﺤﻈﺔ‪t=τ .‬‬ ‫ﺗﻤﺮﯾﻦ ‪2‬‬ ‫ﻧﻨﺠﺰ اﻟﺪارة اﻟﻜﮭﺮﺑﺎﺋﯿﺔ اﻟﻤﻤﺜﻠﺔ ﻓﻲ اﻟﺸﻜﻞ واﻟﻤﻜﻮﻧﺔ ﻣﻦ ‪:‬‬ ‫)‪(G‬ﻣﻮﻟﺪ ﻛﮭﺮﺑﺎﺋﻲ ﻣﺆﻣﺜﻞ ﻟﻠﺘﻮﺗﺮ ﻗﻮﺗﮫ اﻟﻜﮭﺮﻣﺤﺮﻛﺔ ‪ (D),E‬ﻣﻮﺻﻞ اوﻣﻲ ﻣﻘﺎوﻣﺘﮫ ‪ R=100Ώ‬ﻣﻜﺜﻒ‬ ‫ﺳﻌﺘﮫ ‪ C‬وﻗﺎطﻊ اﻟﺘﯿﺎر‪ . K‬اﻟﻤﻜﺜﻒ ﻏﯿﺮ ﻣﺸﺤﻮن‪ .‬ﻧﻐﻠﻖ ﻗﺎطﻊ اﻟﺘﯿﺎر ﻋﻨﺪ ﻟﺤﻈﺔ ﻧﺨﺘﺎرھﺎ اﺻﻼ‬ ‫ﻟﻠﺘﻮارﯾﺦ )‪(t=0‬‬ ‫‪ -1‬ﺑﯿﻦ ان ‪ln(E-UC)=-t/ +lnE‬‬ ‫‪ -2‬ﯾﻌﻄﻲ اﻟﻤﻨﺤﻨﻰ ﺗﻐﯿﺮات اﻟﻤﻘﺪار )‪ ln(E-UC‬ﺑﺪﻻﻟﺔ اﻟﺰﻣﻦ ‪.t‬‬ ‫ﺑﺎﺳﺘﻐﻼل اﻟﻤﺒﯿﺎن اوﺟﺪ ﻗﯿﻤﺔ ﻛﻞ ﻣﻦ ‪ E‬و‬ ‫‪ -3‬ﻧﺮﻣﺰ ب ‪ Ee‬ﻟﻠﻄﺎﻗﺔ اﻟﻤﺨﺰوﻧﺔ ﻓﻲ اﻟﻤﻜﺜﻒ ﻋﻨﺪ اﻟﻠﺤﻈﺔ ‪ t=τ‬وﻧﺮﻣﺰ ب‬ ‫)‪ E (max‬ﻟﻠﻄﺎﻗﺔ اﻟﻘﺼﻮى اﻟﺘﻲ ﯾﺨﺘﺰﻧﮭﺎ اﻟﻤﻜﺜﻒ اﺣﺴﺐ اﻟﻨﺴﺒﺔ )‪. Ee/E(max‬‬ ‫‪ -4‬اﺣﺴﺐ ﻗﯿﻤﺔ اﻟﺴﻌﺔ ’‪ C‬ﻟﻠﻤﻜﺜﻒ اﻟﺬي ﯾﺠﺐ ﺗﺮﻛﯿﺒﮫ ﻣﻊ اﻟﻤﻜﺜﻒ اﻟﺴﺎﺑﻖ ﻓﻲ اﻟﺪارة‬ ‫ﻟﺘﺎﺧﺬ ﺛﺎﺑﺘﺔ اﻟﺰﻣﻦ اﻟﻘﯿﻤﺔ‪= /3‬‬

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‫ﻣﺒﺮزا ﻛﯿﻔﯿﺔ ﺗﺮﻛﯿﺐ ھﺬﯾﻦ اﻟﻤﻜﺜﻔﯿﻦ‬

‫ﺗﻤﺮﯾﻦ ‪3‬‬ ‫ﯾﺘﻜﻮن اﻟﺘﺮﻛﯿﺐ اﻟﺘﺠﺮﯾﺒﻲ ﻣﻦ ﻣﻮﻟﺪ ﻣﺆﻣﺜﻞ ﻟﻠﺘﻮﺗﺮ ﻗﻮﺗﮫ اﻟﻜﮭﺮﻣﺤﺮﻛﺔ ‪ ،E=12V‬ﻣﻮﺻﻠﯿﻦ اوﻣﯿﯿﻦ ‪ R=1KΩ‬و‬ ‫’‪ R‬و ﻣﻜﺜﻒ ﺳﻌﺘﮫ ‪ C‬ﻣﻔﺮغ ﺑﺪﺋﯿﺎ اﻟﺸﻜﻞ ﺟﺎﻧﺒﮫ‬ ‫‪ -1‬ﻋﻨﺪ اﻟﻠﺤﻈﺔ ‪t=0‬ﻧﻐﻠﻖ ﻗﺎطﻊ اﻟﺘﯿﺎر‬ ‫‪ -1-1‬ﺑﯿﻦ ان اﻟﻤﻌﺎدﻟﺔ اﻟﺘﻔﺎﺿﻠﯿﺔ اﻟﺘﻲ ﯾﺤﻘﻘﮭﺎ اﻟﺘﻮﺗﺮ )‪ Uc(t‬ﺗﻜﺘﺐ ﻋﻠﻰ ﺷﻜﻞ‬ ‫‪ .‬ﻣﺤﺪدا ﺗﻌﺒﯿﺮ و وﺣﺪة ﻛﻞ ﻣﻦ ‪ α‬و ‪β‬‬ ‫‪/dt + u = β‬‬ ‫‪/‬‬ ‫‪ uc(t)=A(1‬ﺣﺪد ﺗﻌﺒﯿﺮ ‪ A‬و ‪τ‬‬‫‪ -2-1‬ﺣﻞ اﻟﻤﻌﺎدﻟﺔ اﻟﺘﻔﺎﺿﻠﯿﺔ ﯾﻜﺘﺐ ﻋﻠﻰ ﺷﻜﻞ )‬ ‫‪ -3-1‬اﺳﺘﻨﺘﺞ ﺗﻌﺒﯿﺮ اﻟﺘﻮﺗﺮ ﺑﯿﻦ ﻣﺮﺑﻄﻲ اﻟﻤﻜﺜﻒ ﻓﻲ اﻟﻨﻈﺎم اﻟﺪاﺋﻢ‬ ‫‪ -4-1‬اوﺟﺪ ﺗﻌﺒﯿﺮ ﺷﺪة اﻟﺘﯿﺎر اﻟﻤﺎر ﻓﻲ اﻟﻤﻮﺻﻞ اﻻوﻣﻲ ’‪ R‬ﻓﻲ اﻟﻨﻈﺎم اﻟﺪاﺋﻢ‬ ‫‪ -2‬ﻋﻨﺪ ﻟﺤﻈﺔ ﻣﻦ ﻟﺤﻈﺎت اﻟﻨﻈﺎم اﻟﺪاﺋﻢ ﻧﻌﺘﺒﺮھﺎ اﺻﻼ ﻟﻠﺘﻮارﯾﺦ ﻧﻔﺘﺢ ﻗﺎطﻊ اﻟﺘﯿﺎر‬ ‫‪ -1-2‬اوﺟﺪ اﻟﻤﻌﺎدﻟﺔ اﻟﺘﻔﺎﺿﻠﯿﺔ اﻟﺘﻲ ﯾﺤﻘﻘﮭﺎ اﻟﺘﻮﺗﺮ) ‪ Uc(t‬ﺑﯿﻦ ﻣﺮﺑﻄﻲ اﻟﻤﻜﺜﻒ‬ ‫‪ -2-2‬اوﺟﺪ ﺗﻌﺒﺮ اﻟﺘﻮﺗﺮ )‪Uc(t‬‬ ‫‪ -3-2‬ﯾﻌﻄﻲ اﻟﻤﻨﺤﻨﻰ ﺟﺎﻧﺒﮫ ﺗﻐﯿﺮات اﻟﺘﻮﺗﺮ) ‪ Uc(t‬ﺑﺪﻻﻟﺔ اﻟﺰﻣﻦ ﺣﺪد ﻗﯿﻤﺔ ﻛﻞ ﻣﻦ ‪ τ‬و ‪ C‬و ’‪R‬‬ ‫‪ -4-2‬اﺣﺴﺐ اﻟﻄﺎﻗﺔ اﻟﻤﺨﺰوﻧﺔ ﻓﻲ اﻟﻤﻜﺜﻒ ﻋﻨﺪﻣﺎ ﯾﺼﺒﺢ اﻟﺘﻮﺗﺮ ﺑﯿﻦ ﻣﺮﺑﻄﻲ ‪ 10%‬ﻣﻦ ﻗﯿﻤﺘﮫ اﻟﻘﺼﻮى‬ ‫ﺗﻤﺮﯾﻦ ‪4‬‬ ‫ﻧﺘﻮﻓﺮ ﻋﻠﻰ ﻣﻜﺜﻒ ﻛﮭﺮﻛﯿﻤﯿﺎﺋﻲ ﺳﻌﺘﮫ ‪ C=5µF‬ﻟﺒﻮﺳﯿﮫ ﻣﻦ اﻻﻟﻮﻣﯿﻨﯿﻮم ‪ Al‬ﻣﻜﻮن ﻣﻦ ﺧﻤﺴﺔ ﻣﻜﺜﻔﺎت ﻣﺘﺸﺎﺑﮭﺔ ﯾﻤﻜﻦ اﻋﺘﺒﺎرھﺎ ﻣﺴﺘﻮﯾﺔ و ﻣﺮﻛﺒﺔ ﻋﻠﻰ‬ ‫اﻟﺘﻮازي‪ ،‬ﻋﺎزﻟﮭﻞ اﻻﺳﺘﻘﻄﺎﺑﻲ ﻣﻦ اﻻﻟﻮﻣﯿﻦ ‪ .Al2O3‬ﻧﺸﺤﻦ اﻟﻤﻜﺜﻒ ﺑﻮاﺳﻄﺔ ﺑﻮاﺳﻄﺔ ﻣﻮﻟﺪ ﻗﻮﺗﮫ اﻟﻜﮭﺮﻣﺤﺮﻛﺔ ‪E=24V‬‬ ‫‪ -1‬ﻛﯿﻒ ﯾﻤﻜﻦ اﻟﺘﺄﻛﺪ ﻣﻦ ﻧﮭﺎﯾﺔ ﺷﺤﻦ اﻟﻤﻜﺜﻒ‬ ‫‪ -2‬ﻋﻠﻤﺎ ان ﺳﻌﺔ ﻣﻜﺜﻒ ﻣﺴﺘﻮ ﯾﻌﺒﺮ ﻋﻨﮭﺎ ﺑﺎﻟﻌﻼﻗﺔ ‪ C= . /‬ﺣﯿﺚ ‪ S‬ﻣﺴﺎﺣﺔ اﻟﻠﺒﻮس ‪ d‬اﻟﻤﺴﺎﻓﺔ اﻟﻔﺎﺻﻠﺔ ﺑﯿﻦ اﻟﻠﺒﻮﺳﯿﻦ ‪ ε‬ﺛﺎﺑﺘﺔ ﺗﻤﯿﺰ اﻟﻌﺎزل‬ ‫اﻻﺳﺘﻘﻄﺎﺑﻲ‬ ‫‪ -1-2‬ﻋﻠﻤﺎ ان ﻣﺴﺎﺣﺔ اﻟﻠﺒﻮس ھﻲ ‪ S=3cm2‬اﺣﺴﺐ ﺳﻤﻚ طﺒﻘﺔ اﻻﻟﻮﻣﯿﻦ ﻓﻲ اﻟﻤﻜﺜﻔﺎت اﻟﺴﺎﺑﻘﺔ‬ ‫‪ -2-2‬اذﻛﺮ ﺑﺎﯾﺠﺎز طﺮﯾﻘﺔ ﺻﻨﻊ اﻟﻤﻜﺜﻔﺎت اﻟﺴﺎﺑﻘﺔ‪ ،‬ﻣﺎھﻲ اﻻﺣﺘﯿﺎطﺎت اﻟﻮاﺟﺐ اﻻﻧﺘﺒﺎه اﻟﯿﮭﺎ ﻋﻦ اﺳﺘﻌﻤﺎﻟﮫ ﻓﻲ اﻟﺘﺮاﻛﯿﺐ اﻟﻜﮭﺮﺑﺎﺋﯿﺔ‬ ‫‪ -3-2‬ﺗﻢ اﻋﺘﻤﺎد ﻧﻔﺲ اﻟﺘﻘﻨﯿﺔ اﻟﺴﺎﺑﻘﺔ ﻓﻲ ﺗﺼﻨﯿﻊ ﻣﻜﺜﻒ اﺧﺮ ﺳﻌﺘﮫ ‪ C=10µF‬ﻋﯿﺮ ان ﺗﻢ ﺗﻌﻮﯾﺾ اﻻﻟﻮﻣﯿﻦ ﺑﺘﯿﺘﺎﻧﺖ اﻟﺒﺎرﯾﻮم ‪ .TiBa‬اﺣﺴﺐ ﻣﺴﺎﺣﺔ‬ ‫ﻟﺒﻮس ھﺬا اﻟﻤﻜﺜﻒ ﻧﻔﺘﺮض ان ﺳﻤﻚ اﻟﻌﺎزل اﻻﺳﺘﻘﻄﺎﺑﻲ ﯾﺒﻘﻰ ﺗﺎﺑﺜﺎ‪ .‬ﻋﻠﻖ ﻋﻦ ھﺬه اﻟﻨﺘﯿﺠﺔ‬ ‫(‬ ‫‪) = 132,75 .‬‬ ‫‪) = 10620 .‬‬ ‫( ;‬ ‫ﻧﻌﻄﻲ‬

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