Electronics

‫‪ (۵‬ﻗﺪﺭﺕ ﺩﺭ ﺟﺮﻳﺎﻥ ﻫﺎﯼ ﻣﺘﻨﺎﻭﺏ‬

‫)‪P(t‬‬

‫‪ (۵-۱‬ﻧﻤــﺎﻳﺶ ﻗــﺪﺭﺕ ﺩﺭ ﺣﺎﻟﺘﻬــﺎﯼ‬

‫‪+‬‬

‫‪+‬‬

‫ﻣﺨﺘﻠﻒ ﺟﺮﻳﺎﻥ ﻣﺘﻨﺎﻭﺏ‬

‫)‪V(t‬‬

‫ﺍﻟﻒ( ﻣﺪﺍﺭ ﯼ ﮐﻪ ﺩﺭ ﺁﻥ ‪ ϕ=۰‬ﺑﺎﺷﺪ‪:‬‬

‫)‪I(t‬‬

‫ﺩﺭ ﺍﻳﻦ ﺣﺎﻟﺖ ﺟﺮﻳﺎﻥ ﻭ ﻭﻟﺘﺎﮊ ﻫﻢ ﻓﺎﺯ ﺑﻮﺩﻩ ﻭ ﻟـﺬﺍ‬ ‫ﺗﻮﺍﻥ ﻣﺼﺮﻓﯽ ﻫﻤﺎﻧﻨـﺪ ﻳـﮏ ﺟﺮﻳـﺎﻥ ﻣﺴـﺘﻘﻴﻢ ﺍﺯ‬ ‫ﺣﺎﺻﻠﻀﺮﺏ ﻭﻟﺘﺎﮊ ﺩﺭ ﺟﺮﻳﺎﻥ ﺣﺎﺻﻞ ﻣﯽ ﮔﺮﺩﺩ ﻭ‬ ‫ﻫﻤﻮﺍﺭﻩ ﺗﻮﺍﻥ ﻣﻘﺪﺍﺭﯼ ﻣﺜﺒﺖ ﺧﻮﺍﻫﺪ ﺑﻮﺩ‬

‫ﺏ( ﻣﺪﺍﺭﻳﮑﻪ ﺩﺭ ‪ ϕ ≠ ۹۰‬ﻭ ‪ ϕ ≠ ۰‬ﺑﺎﺷﺪ‪.‬‬

‫)‪P(t‬‬

‫ﺩﺭ ﺍﻳﻦ ﺣﺎﻟﺖ ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﻣﻴﺰﺍﻥ ﺍﺧـﺘﻼﻑ‬ ‫ﻓــﺎﺯ ﺑــﻴﻦ ﻭﻟﺘــﺎﮊ ﻭ ﺟﺮﻳــﺎﻥ ﺩﺭ ﻣــﻮﺍﻗﻌﯽ‬ ‫ﺣﺎﺻﻠﻀﺮﺏ ﻭﻟﺘﺎﮊ ﻭ ﺟﺮﻳﺎﻥ ﻋﺪﺩﯼ ﻣﺜﺒـﺖ‬

‫‪+‬‬

‫‪+‬‬

‫)‪V(t‬‬

‫ﻭ ﻣﻮﺍﻗﻌﯽ ﻣﻘﺪﺍﺭﯼ ﻣﻨﻔﯽ ﺧﻮﺍﻫﺪ ﺑﻮﺩ‪ .‬ﻟـﺬﺍ‬

‫)‪I(t‬‬

‫ﺗﻮﺍﻥ ﻭﺍﻗﻌﯽ ﺑﺮﺍﺑﺮ ﺟﻤﻊ ﺟﺒﺮﯼ ﻣﻘﺎﺩﻳﺮ ﺗﻮﺍﻥ‬

‫‪-‬‬

‫‪-‬‬

‫ﺧﻮﺍﻫــﺪ ﺑــﻮﺩ‪ .‬ﻫﻤــﺎﻧﻄﻮﺭ ﮐــﻪ ﺩﺭ ﺷــﮑﻞ‬ ‫ﻣﺸﺨﺺ ﺍﺳﺖ ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﻣﻴﺰﺍﻥ ﺍﺧـﺘﻼﻑ‬ ‫ﻓﺎﺯ ﺑﻴﻦ ﻭﻟﺘﺎﮊ ﻭ ﺟﺮﻳـﺎﻥ ﺗـﻮﺍﻥ ﻭﺍﻗﻌـﯽ ﺍﺯ‬ ‫ﺣﺪﺍﮐﺜﺮ ﺗﻮﺍﻥ ﮐﻤﺘﺮ ﺧﻮﺍﻫﺪ ﺑﻮﺩ‪.‬‬

‫)‪P(t‬‬ ‫ﺝ( ﻣﺪﺍﺭﻳﮑﻪ ﺩﺭ ﺁﻥ‪ ϕ = ± ۹۰‬ﺑﺎﺷﺪ‪:‬‬ ‫ﺩﺭ ﺍﻳﻦ ﺣﺎﻟﺖ ﻣﻤﮑﻦ ﺍﺳﺖ ﮐﻪ ﻣﺪﺍﺭ ﺳﻠﻔﯽ‬

‫‪+‬‬

‫ﺧﺎﻟﺺ ﻭ ﻳﺎ ﺧﺎﺯﻧﯽ ﺧﺎﺹ ﺑﺎﺷﺪ‪.‬ﺩﺭ ﺍﻳـﻦ‬ ‫ﺣﺎﻟﺖ ﺣﺎﺻﻠﻀـﺮﺏ ﻭﻟﺘـﺎﮊ ﻭ ﺟﺮﻳـﺎﻥ ﺩﺭ‬

‫‪+‬‬ ‫‪V(t‬‬ ‫)‬

‫)‪I(t‬‬

‫ﻧﻴﻤﯽ ﺍﺯ ﺯﻣـﺎﻥ ﻣﺜﺒـﺖ ﻭ ﺩﺭ ﻧﻴﻤـﯽ ﺩﻳﮕـﺮ‬ ‫ﻣﻨﻔﯽ ﺧﻮﺍﻫﺪ ﺑﻮﺩ‪ .‬ﺑﻄﻮﺭﻳﮑﻪ ﺟﻤﻊ ﺟﺒـﺮﯼ‬

‫‪-‬‬

‫ﺗﻮﺍﻥ ﺑﺮﺍﺑﺮ ﺻﻔﺮ ﺧﻮﺍﻫﺪ ﺑﻮﺩ‪.‬‬

‫‪۴۵‬‬

‫‪-‬‬

‫‪ (۵-۲‬ﺍﻧﻮﺍﻉ ﻗﺪﺭﺕ ﺩﺭ ﺟﺮﻳﺎﻧﻬﺎﯼ ﻣﺘﻨﺎﻭﺏ‬ ‫ﻫﻤﺎﻧﻄﻮﺭ ﮐﻪ ﺩﺭ ﺑﺎﻻ ﺍﺷﺎﺭﻩ ﺷﺪ ﺩﺭ ﺻﻮﺭﺗﻴﮑﻪ ﺩﺭ ﻣﺪﺍﺭ ﻓﻘﻂ ﺍﺯ ﻣﺼﺮﻑ ﮐﻨﻨﺪﻩ ﺍﻫﻤﯽ ﺍﺳﺘﻔﺎﺩﻩ ﺷﻮﺩ ﺑﺎ ﺍﻧـﺪﺍﺯﻩ ﮔﻴـﺮﯼ‬ ‫ﻣﻘﺪﺍﺭ ﺟﺮﻳﺎﻥ ﻣﯽ ﺗﻮﺍﻥ ﺍﺯ ﺭﺍﺑﻄﻪ ﺯﻳﺮ ﻣﻘﺪﺍﺭ ﺗﻮﺍﻥ ﻣﺼﺮﻓﯽ ﺭﺍ ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﯼ ﻧﻤﻮﺩ‪:‬‬ ‫‪P=R I2‬‬ ‫ﻭﻟﯽ ﻫﻤﻴﺸﻪ ﻣﺼﺮﻑ ﮐﻨﻨﺪﻩ ﻫﺎ ﺍﺯ ﻧﻮﻉ ﻣﻘﺎﻭﻣﺖ ﺧﺎﻟﺺ ﻧﻴﺴﺘﻨﺪ ﮐﻪ ﺍﻧـﺮﮊﯼ ﺩﺭ ﺁﻧﻬـﺎ ﺑـﻪ ﺻـﻮﺭﺕ ﺣـﺮﺍﺭﺕ ﺗﻮﻟﻴـﺪ‬ ‫ﺷﻮﺩ‪.‬ﺑﻨﺎﺑﺮﺍﻳﻦ ﻻﺯﻡ ﺍﺳﺖ ﺭﺍﺑﻄﻪ ﺍﯼ ﺑﺮﺍﯼ ﻗﺪﺭﺕ ﺩﺭ ﺟﺮﻳﺎﻥ ﻣﺘﻨﻨﺎﻭﺏ ﺍﺭﺍﺋﻪ ﻧﻤﻮﺩ ﮐﻪ ﺑﺮﺍﯼ ﮐﻠﻴﻪ ﻣﺼـﺮﻑ ﮐﻨﻨـﺪﻩ ﻫـﺎ‬ ‫ﻧﻈﻴﺮ ﺍﻧﻮﺍﻉ ﻣﺎﺷﻴﻨﻬﺎﯼ ﺍﻟﮑﺘﺮﻳﮑﯽ ﮐﺎﺭﺑﺮﺩ ﺩﺍﺷﺘﻪ ﺑﺎﺷﺪ‪ .‬ﺑﺮﺍﯼ ﻣﺤﺎﺳﺒﻪ ﻗﺪﺭﺕ ﺩﺭ ﻳﮏ ﺟﺮﻳﺎﻥ ﻣﺘﻨﺎﻭﺏ ﺍﺑﺘﺪﺍ ﮐﺎﺭ ﺣﺎﺻﻞ‬ ‫ﺍﺯ ﻳﮏ ﺳﻴﮑﻞ ﺭﺍ ﺩﺭ ﻧﻈﺮ ﻣﯽ ﮔﻴﺮﻳﻢ‪:‬‬ ‫‪T‬‬

‫‪W = ∫ v( t ) ⋅ I (t ) dt‬‬ ‫‪0‬‬

‫‪W 1 T‬‬ ‫=‬ ‫‪v( t ) ⋅ I (t ) dt‬‬ ‫‪T T ∫0‬‬ ‫ﻋﻤﻮﻣﺎﹰ ﻭﺟﻮﺩ ﭼﻨﺪ ﻣﺼﺮﻑ ﮐﻨﻨﺪﻩ ﺩﺭ ﻣﺪﺍﺭ ﺑﺪﻟﻴﻞ ﺩﺍﺷﺘﻦ ﺳﻴﻢ ﭘﻴﭻ ﺩﺍﺭﺍﯼ ﺧﺎﺻﻴﺖ ﺳﻠﻔﯽ ﺑﻮﺩﻩ ﻟﺬﺍ ﺩﺭ ﻋﻤﻞ ﻣﺪﺍﺭﻫﺎ‬ ‫=‪⇒P‬‬

‫ﺑﻪ ﺻﻮﺭﺕ ﭘﺲ ﻓﺎﺯ ﻣﯽ ﺑﺎﺷﻨﺪ‪ .‬ﺑﻨﺎﺑﺮﺍﻳﻦ ﺑﺎ ﺍﻳﻦ ﻓﺮﺽ ﮐﻪ ‪ ϕ‬ﺯﺍﻭﻳﻪ ﺍﺧﺘﻼﻑ ﻓﺎﺯ ﺑﻴﻦ ﻭﻟﺘﺎﮊ ﻭ ﺟﺮﻳﺎﻥ ﺑﺎﺷﺪ‪ ،‬ﺩﺍﺭﻳﻢ‪:‬‬ ‫) ‪I (t ) = I e 2 sin(ωt‬‬ ‫) ‪V( t ) = Ve 2 sin(ωt + ϕ‬‬ ‫‪1 T‬‬ ‫‪I eVe 2 sin(ωt ) ⋅ sin(ωt + ϕ )dt‬‬ ‫‪T ∫0‬‬ ‫‪2I V T‬‬ ‫‪P = e e ∫ sin(ωt ) ⋅ sin(ωt + ϕ ) dt‬‬ ‫‪T 0‬‬

‫=‪⇒P‬‬

‫ﺩﺍﺭﻳﻢ‪:‬‬ ‫‪1‬‬ ‫]) ‪[cos(ϕ ) − cos(2ωt + ϕ‬‬ ‫‪2‬‬

‫= ) ‪sin(ωt ) ⋅ sin(ωt + ϕ‬‬ ‫‪2π‬‬ ‫‪T‬‬

‫=‪ω‬‬

‫ﺑﺎ ﺟﺎﻳﮕﺰﻳﻦ ﮐﺮﺩﻥ ﻭ ﺍﻧﺘﮕﺮﺍﻝ ﮔﻴﺮﯼ ﺩﺍﺭﻳﻢ‪:‬‬ ‫‪2 I eVe T‬‬ ‫) ‪× cos(ϕ‬‬ ‫‪T‬‬ ‫‪2‬‬

‫=‪P‬‬

‫) ‪P = I eVe cos(ϕ‬‬

‫)‪(۵-۱‬‬

‫ﺭﺍﺑﻄﻪ )‪ (۵-۱‬ﻗﺪﺭﺕ ﻣﺘﻮﺳﻂ ﻣﺪﺍﺭ ﻭ )‪ cos(ϕ‬ﺭﺍ ﺿﺮﻳﺐ ﻗﺪﺭﺕ‪ ١‬ﻣﯽ ﻧﺎﻣﻨﺪ‪.‬‬

‫‪Power Factor‬‬

‫‪۴۶‬‬

‫‪1‬‬

‫‪ (۵-۳‬ﻗﺪﺭﺕ ﻇﺎﻫﺮﯼ‪ ،‬ﻓﻌﺎﻝ ﻭ ﻏﻴﺮ ﻓﻌﺎﻝ )ﺭﺍﮐﺘﻴﻮ(‬ ‫ﭼﻨﺎﻧﭽﻪ ﺷﺪﺕ ﺟﺮﻳﺎﻥ ﺩﺭ ﻳﮏ ﺟﺮﻳﺎﻥ ﻣﺘﻨﺎﻭﺏ ﺭﺍ ﻣﺒﺪﺍﺀ ﻓﺎﺯ ﺩﺭ ﻣﺪﺍﺭ ﺍﻧﺘﺨﺎﺏ ﻧﻤﺎﺋﻴﻢ ﻭ ﺟﺮﻳﺎﻥ ﻣﻮﺛﺮ ﺭﺍ ﺭﻭﯼ ﺍﻥ ﺟﺪﺍ‬ ‫ﮐﻨﻴﻢ ﺑﺮﺩﺍﺭ ﻭﻟﺘﺎﮊ ﺑﻪ ﻣﻴﺰﺍﻥ ‪ ϕ‬ﻧﺴﺒﺖ ﺑﻪ ﺟﺮﻳﺎﻥ ﺗﻘﺪﻡ ﺩﺍﺷﺘﻪ ﻭ ﻟﺬﺍ ﺍﮔﺮ ﺑﺮﺩﺍﺭ ‪ Ie‬ﺭﺍ ﺩﺭ ﺭﺍﺳﺘﺎﯼ ‪ Ve‬ﻭ ﺭﺍﺳﺘﺎﯼ ﻋﻤـﻮﺩ‬ ‫ﺑﺮ ﺁﻥ ﺗﺼﻮﻳﺮ ﻧﻤﺎﺋﻴﻢ ﻣﻄﺎﺑﻖ ﺷﮑﻞ ﺩﺍﺭﻳﻢ‪:‬‬ ‫) ‪I a = I e cos(ϕ‬‬

‫‪Ve‬‬

‫) ‪I r = I e sin(ϕ‬‬ ‫ﻣﻮﻟﻔﻪ ‪ Ia‬ﮐﻪ ﺩﺭ ﺭﺍﺳﺘﺎﯼ ﻭﻟﺘﺎﮊ ‪ Ve‬ﺍﺳﺖ ﺗﻮﺍﻥ ﺁﻥ ﺑﺮﺍﺑﺮ‪:‬‬ ‫‪P = Ve I a‬‬ ‫) ‪P = Ve I e cos(ϕ‬‬ ‫ﺭﺍﺑﻄﻪ ﻓﻮﻕ ﻫﻤﺎﻥ ﺭﺍﺑﻄﻪ )‪ (۵-۱‬ﺍﺳﺖ ﮐﻪ ﺑﻪ ﻧـﺎﻡ ﻗـﺪﺭﺕ‬

‫‪Ie‬‬

‫‪I‬‬

‫ﻓﻌﺎﻝ‪ ١‬ﻧﺎﻣﻴﺪﻩ ﻣﯽ ﺷﻮﺩ‪ .‬ﺍﻳﻦ ﺗﻮﺍﻥ‪ ،‬ﻣﻘﺪﺍﺭ ﺗﻮﺍﻧﯽ ﺍﺳﺖ ﮐـﻪ‬

‫‪Ia‬‬

‫ﺑﻪ ﺻﻮﺭﺕ ﻭﺍﻗﻌﯽ ﺗﻮﺳﻂ ﻣﺼﺮﻑ ﮐﻨﻨـﺪﻩ ﻣـﻮﺭﺩ ﺍﺳـﺘﻔﺎﺩﻩ‬

‫‪ϕ‬‬

‫‪Ir‬‬

‫ﻭﺍﻗﻊ ﻣﯽ ﮔﺮﺩﺩ ﻟﺬﺍ ﺑﻪ ﻧﺎﻡ ﺗﻮﺍﻥ ﻭﺍﻗﻌﯽ ﻧﻴﺰ ﻧﺎﻣﻴﺪﻩ ﻣﯽ ﺷﻮﺩ‬ ‫ﻭ ﺑﻪ ﻭﺍﺕ ﺁﻧﺮﺍ ﺑﻴﺎﻥ ﻣﯽ ﮐﻨﻨﺪ‪.‬‬ ‫ﺷﺪﺕ ﺟﺮﻳﺎﻥ ‪ Ir‬ﺩﺭ ﺟﻬﺖ ﻋﻤﻮﺩ ﺑﺮ ﻭﻟﺘﺎﮊ ‪ Ve‬ﺑﻮﺩﻩ ﻭ ﺗﻮﺍﻥ ﻣﺮﺑﻮﻁ ﺑﻪ ﺁﻥ ﺑﺮﺍﺑﺮ‪:‬‬ ‫‪Q = Ve I r‬‬

‫)‪(۵-۲‬‬

‫) ‪Q = Ve I e sin(ϕ‬‬

‫ﺭﺍﺑﻄﻪ ﻓﻮﻕ ﺭﺍ ﻗﺪﺭﺕ ﻏﻴﺮ ﻓﻌﺎﻝ‪ ،‬ﻣﺠﺎﺯﯼ ﻳﺎ ﺭﺍﮐﺘﻴﻮ‪ ٢‬ﻣﯽ ﻧﺎﻣﻨﺪ‪ .‬ﮐﺎﺭ ﻭﺍﻗﻌﯽ ﻣﺮﺑﻮﻁ ﺑﻪ ﺍﻳﻦ ﺷﺪﺕ ﺟﺮﻳﺎﻥ ﺻﻔﺮ ﺍﺳـﺖ‬ ‫ﭼﻮﻥ ﺷﺪﺕ ﺟﺮﻳﺎﻥ ﻋﻤﻮﺩ ﺑﺮ ﻭﻟﺘﺎﮊ ﺍﺳﺖ ﺑﻪ ﻫﻤﻴﻦ ﺩﻟﻴﻞ ﺑﻪ ﺁﻥ ﺗﻮﺍﻥ ﺑﯽ ﻭﺍﺕ ﻳﺎ ﺩﻭﺍﺗﻪ ﻧﻴﺰ ﻣﯽ ﮔﻮﻳﻨـﺪ‪ .‬ﺑـﻪ ﻋﺒـﺎﺭﺕ‬ ‫ﺩﻳﮕﺮ ﺍﻳﻦ ﺗﻮﺍﻥ ﺗﻮﺳﻂ ﻣﺼﺮﻑ ﮐﻨﻨﺪﻩ ﻗﺎﺑﻞ ﻣﺼﺮﻑ ﻧﺒﻮﺩﻩ ﻭ ﻟﺬﺍ ﺑﻴﻦ ﮊﻧﺮﺍﺗﻮﺭ ﻭ ﻣﺼﺮﻑ ﮐﻨﻨﺪﻩ ﺭﺩ ﻭﺑﺪﻝ ﻣـﯽ ﺷـﻮﺩ‪.‬‬ ‫ﺍﻳﻦ ﺗﻮﺍﻥ ﺑﺎ ﻭﻟﺖ ﺁﻣﭙﺮ )‪ (VAR‬ﺑﻴﺎﻥ ﻣﯽ ﮔﺮﺩﺩ‪.‬‬

‫ﻗﺪﺭﺕ ﻇﺎﻫﺮﯼ‪:٣‬‬ ‫ﻋﺒﺎﺭﺗﺴﺖ ﺍﺯ ﺣﺎﺻﻞ ﺿﺮﺏ ﺷﺪﺕ ﺟﺮﻳﺎﻥ ﻣﻮﺛﺮ ﺩﺭ ﻭﻟﺘﺎﮊ ﻣﻮﺛﺮ ﮐﻪ ﺑﺎ ﺣﺮﻑ ‪ S‬ﻧﺸﺎﻥ ﻣﯽ ﺩﻫﻨﺪ‪ .‬ﺍﻳﻦ ﺗﻮﺍﻥ ﺑﺮ ﺣﺴﺐ‬ ‫‪ VA‬ﺑﻴﺎﻥ ﻣﯽ ﮔﺮﺩﺩ‪ .‬ﮊﻧﺮﺍﺗﻮﺭ ﻭ ﻳﺎ ﻣﻮﻟﺪ ﺑﺮﻕ ﺩﺭ ﻭﺍﻗﻊ ﻫﻤﻴﺸﻪ ‪ S‬ﺭﺍ ﺗﻮﻟﻴﺪ ﻣﯽ ﮐﻨﺪ‪.‬‬ ‫‪S = Ve I e‬‬ ‫‪VMax I Max‬‬ ‫)‪(۵-۳‬‬ ‫⋅‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪V ⋅I‬‬ ‫‪S = Max Max‬‬ ‫‪2‬‬ ‫ﻳﻌﻨﯽ ﺗﻮﺍﻥ ﻇﺎﻫﺮﯼ ﺩﺭ ﺟﺮﻳﺎﻥ ﻣﺘﻨﺎﻭﺏ ﻧﺼﻒ ﺗﻮﺍﻥ ﺣﺪﺍﮐﺜﺮ ﻣﯽ ﺑﺎﺷﺪ‪ .‬ﺍﺯ ﻃﺮﻑ ﺩﻳﮕﺮ ﻣﯽ ﺗﻮﺍﻥ ﮔﻔﺖ ﮐﻪ ﺗﻮﺍﻥ ﺍﮐﺘﻴﻮ‬ ‫=‪S‬‬

‫ﻭ ﺗﻮﺍﻥ ﺭﺍﮐﺘﻴﻮ ﻣﻮﻟﻔﻪ ﻫﺎﯼ ﺗﻮﺍﻥ ﻇﺎﻫﺮﯼ ﻫﺴﺘﻨﺪ‪ .‬ﺑﻪ ﻋﺒﺎﺭﺕ ﺩﻳﮕﺮ ﺍﮔﺮ ﻣﺜﻠﺜﯽ ﻗﺎﺋﻢ ﺍﻟﺰﺍﻭﻳﻪ ﺍﯼ ﺭﺍ ﺗﻌﺮﻳﻒ ﻧﻤـﺎﺋﻴﻢ ﮐـﻪ‬ ‫ﺗﺤﺖ ﻋﻨﻮﺍﻥ ﻣﺜﻠﺚ ﺗﻮﺍﻥ ﺑﺎﺷﺪ ﻭﺗﺮ ﺍﻳﻦ ﻣﺜﻠﺚ ‪ S‬ﻭ ﺩﻭ ﺿﻠﻊ ﺩﻳﮕﺮ ﺁﻥ ‪ P‬ﻭ‪ Q‬ﺧﻮﺍﻫﻨﺪ ﺑﻮﺩ‪.‬‬ ‫‪Active Power‬‬ ‫‪Reactive Power‬‬ ‫‪Apparent Power‬‬

‫‪۴۷‬‬

‫‪1‬‬ ‫‪2‬‬ ‫‪3‬‬

‫‪S‬‬ ‫‪Q‬‬ ‫ﺑﺮ ﺍﺳﺎﺱ ﻣﺜﻠﺚ ﺗﻮﺍﻥ ﻣﯽ ﺗﻮﺍﻥ ﻧﻮﺷﺖ‪:‬‬ ‫‪ϕ‬‬

‫‪S = P2 + Q2‬‬

‫‪Q‬‬ ‫‪P‬‬ ‫‪ P‬ﺁﻥ ﺑﺨﺶ ﺍﺯ ﺗﻮﺍﻥ ﺗﻮﻟﻴﺪﯼ ﮊﻧﺮﺍﺗﻮﺭ ﺍﺳﺖ ﮐﻪ ﺩﺭ ﻣﺼﺮﻑ ﮐﻨﻨﺪﻩ ﺍﺳﺘﻔﺎﺩﻩ ﻣﯽ ﮔﺮﺩﺩ‪ Q .‬ﺁﻥ ﺑﺨﺶ ﺍﺯ ﺗـﻮﺍﻥ ﺗﻮﻟﻴـﺪ‬ ‫‪P‬‬

‫) ( ‪ϕ = tan −1‬‬

‫ﺷﺪﻩ ﮊﻧﺮﺍﺗﻮﺭ ﺍﺳﺖ ﮐﻪ ﻧﻤﯽ ﺗﻮﺍﻧﺪ ﺑﻪ ﻣﺼﺮﻑ‪ ،‬ﻣﺼﺮﻑ ﮐﻨﻨﺪﻩ ﺑﺮﺳﺪ‪ .‬ﻫﺮ ﭼﻪ ﺿﺮﻳﺐ ﻗﺪﺭﺕ ﮐﻮﭼﮑﺘﺮ ﺑﺎﺷـﺪ ﻣﻔﻬـﻮﻡ‬ ‫ﺁﻥ ﺍﻳﻨﺴﺖ ﮐﻪ ﻣﺼﺮﻑ ﮐﻨﻨﺪﻩ ﻧﻤﯽ ﺗﻮﺍﻧﺪ ﺑﺨﺶ ﺑﻴﺸﺘﺮﯼ ﺭﺍ ﺍﺯ ﺗﻮﺍﻥ ﺗﻮﻟﻴﺪﯼ ﮊﻧﺮﺍﺗﻮﺭ ﺑﻪ ﻣﺼﺮﻑ ﺑﺮﺳﺎﻧﺪ‪ .‬ﺍﻳﻦ ﻫﻢ ﺑـﻪ‬ ‫ﺿﺮﺭ ﻣﺼﺮﻑ ﮐﻨﻨﺪﻩ ﺍﺳﺖ )ﭼﻮﻥ ﮐﻨﺘﻮﺭ ‪ S‬ﺭﺍ ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﯼ ﻣﯽ ﮐﻨﺪ( ﻭ ﻫﻢ ﺑﻪ ﺿﺮﺭ ﺷﺮﮐﺖ ﺑﺮﻕ ﭼﻮﻥ ﺑﺎﻳـﺪ ﺗـﻮﺍﻥ‬ ‫ﺑﻴﺸﺘﺮﯼ ﺭﺍ ﺗﻮﻟﻴﺪ ﻧﻤﺎﻳﺪ‪.‬ﺍﺯ ﺁﻧﺠﺎﺋﻴﮑﻪ ﻣﺼﺮﻑ ﮐﻨﻨﺪﻩ ﻫﺎ ﯼ ﺻﻨﻌﺘﯽ ﻋﻤﻮﻣﺎﹰ ﻣﺪﺍﺭ ﺭﺍ ﭘﺲ ﻓﺎﺯ ﻣﯽ ﮐﻨﻨﺪﺩﺭ ﻣﺪﺍﺭ ﻫﺎ ﺳﻌﯽ‬ ‫ﻣﯽ ﺷﻮﺩ ﻣﻘﺎﺩﻳﺮ ‪ XL‬ﻭ ‪ Xc‬ﺭﺍ ﻃﻮﺭﯼ ﺍﻧﺘﺨﺎﺏ ﻧﻤﺎﻳﻨﺪ ﺗﺎﺣﺘﯽ ﺍﻻﻣﮑﺎﻥ ﺿﺮﻳﺐ ﺗﻮﺍﻥ ﻫﺮ ﭼﻪ ﺑﻴﺸﺘﺮ ﺑـﻪ ﻳـﮏ ﻧﺰﺩﻳـﮏ‬ ‫ﺑﺎﺷﺪ‬ ‫ﻣﺜﺎﻝ‪.:‬‬ ‫ﺗﻮﺍﻥ ﻣﺼﺮﻑ ﺷﺪﻩ ﺩﺭ ﻳﮏ ﻣﻮﺗﻮﺭ ﺟﺮﻳﺎﻥ ﻣﺘﻨﺎﻭﺏ ﺗﻮﺳﻂ ﻳﮏ ﻭﺍﺗﻤﺘﺮ ﺑﺮﺍﺑﺮ ‪ ۲۵۰۰‬ﻭﺍﺕ ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﯼ ﺷﺪﻩ ﺍﺳﺖ‪ ،‬ﺩﺭ‬ ‫ﺣﺎﻟﻴﮑﻪ ﻭﻟﺘﻤﺘﺮ ﻭ ﺁﻣﭙﺮ ﻣﺘﺮ ﻧﺼﺐ ﺷﺪﻩ ﺩﺭ ﻣﺪﺍﺭ ﻣﻮﺗﻮﺭ ﺑﺘﺮﺗﻴﺐ ‪ ۲۲۰‬ﻭﻟﺖ ﻭ ‪ ۱۵‬ﺍﻣﭙﺮ ﺭﺍ ﻧﺸﺎﻥ ﻣـﯽ ﺩﻫﻨـﺪ‪ .‬ﺿـﺮﻳﺐ‬ ‫ﻗﺪﺭﺕ ﺍﻳﻦ ﻣﻮﺗﻮﺭ ﭼﻘﺪﺭ ﺍﺳﺖ؟‬ ‫) ‪P = Ve I e cos(ϕ‬‬ ‫‪2500‬‬ ‫‪P‬‬ ‫=‬ ‫‪= 0.75‬‬ ‫‪Ve I e 220 × 15‬‬

‫= ) ‪cos(ϕ‬‬

‫ﻣﺜﺎﻝ ﻳﮏ ﻣﺪﺍﺭ ‪: R-L‬‬ ‫ﻣﺪﺍﺭﯼ ﻣﻄﺎﺑﻖ ﺷﮑﻞ ﻣﻘﺎﺑﻞ ﻣﻔﺮﻭﺽ ﺍﺳﺖ ﻣﻄﻠﻮﺑﺴﺖ‪:‬‬ ‫ﺍﻣﭙﺪﺍﻧﺲ ﮐﻞ ﻣﺪﺍﺭ؟ ﺟﺮﻳﺎﻥ ﻣﻮﺛﺮ ﺩﺭ ﻣﺪﺍﺭ؟ ﺿﺮﻳﺐ ﺗﻮﺍﻥ ﻣﺪﺍﺭ؟‬ ‫ﺗﺌﺎﻥ ﻓﻌﺎﻝ‪ ،‬ﻏﻴﺮ ﻓﻌﺎﻝ ﻭ ﻇﺎﻫﺮﯼ ﻣﺪﺍﺭ؟ ﻭ ﺭﺳﻢ ﻣﺜﻠﺚ ﺗﻮﺍﻥ؟‬ ‫ﺍﮔﺮ ﻣﻘﺎﻭﻣﺖ ﺍﻫﻤﯽ ‪ R‬ﺑﺮﺍﺑﺮ ‪ ۲‬ﺍﻫﻢ ﻭ ﻣﻘﺎﻭﻣﺖ ﺳﻠﻔﯽ ‪ L‬ﺑﺮﺍﺑﺮ ‪ ۴‬ﺍﻫﻢ ﺑﺎﺷﺪ‪:‬‬

‫‪X R2 + X L2‬‬

‫=‪Z‬‬

‫‪Z = 4 + 16 = 4.47 Ω‬‬ ‫‪X‬‬ ‫‪4‬‬ ‫‪ϕ = tan −1 ( L ) = tan −1 ( ) = 63 .42 o‬‬ ‫‪XR‬‬ ‫‪2‬‬ ‫‪Ve 120‬‬ ‫=‬ ‫‪= 26 .85 A‬‬ ‫‪Z 4.47‬‬ ‫‪PF = cos(ϕ ) = cos( 63 .42 ) = 0.45‬‬

‫= ‪Ve = ZI e ⇒ I e‬‬

‫‪۴۸‬‬

‫‪P = Ve I e cos(ϕ ) = (120 )( 26.85)(.45) = 1441watt‬‬ ‫‪Q = Ve I e sin(ϕ ) = (120)( 26.85)(sin( 63.42)) = 2882VAR‬‬ ‫‪S = Ve I e = (120 )(26.85) = 3222VA‬‬

‫ﺗﻮﺟﻪ ﺷﻮﺩ ﮐﻪ ‪ P‬ﺭﺍ ﻣﯽ ﺗﻮﺍﻥ ﺍﺯ ﺭﺍﺑﻄـﻪ ‪RI2‬‬

‫‪S=3222VA‬‬

‫ﻧﻴﺰ ﻣﯽ ﺗﻮﺍﻥ ﻣﺤﺎﺳﺒﻪ ﻧﻤﻮﺩ ﻭﻟﯽ ﺑﺎﻳـﺪ ﺗﻮﺟـﻪ‬

‫‪Q=2882VAR‬‬

‫ﺩﺍﺷﺖ ﮐﻪ ﺩﺭ ﺍﻳﻨﺠﺎ ﺑﺠﺎﯼ ‪ I‬ﺑﺎﻳﺴﺘﯽ ﻣﻘﺪﺍﺭ ‪Ie‬‬

‫‪ϕ=63.42‬‬

‫ﮐﻞ ﻣﺪﺍﺭ ﺭﺍ ﻗﺮﺍﺭ ﺩﺍﺩ‪.‬‬ ‫‪P=1441 watt‬‬ ‫ﻣﺜﺎﻝ ﻳﮏ ﻣﺪﺍﺭ ‪: R-L-C‬‬

‫ﺩﺭ ﻳﮏ ﻣﺪﺍﺭ ﺗﮏ ﻓﺎﺯ ‪ AC‬ﺳﻪ ﻣﺼﺮﻑ ﮐﻨﻨﺪﻩ ﺑﻪ ﺻﻮﺭﺕ ﺳﺮﯼ ﻭ ﺑﻪ ﺷﺮﺡ ﺟﺪﻭﻝ ﺯﻳﺮ ﻗـﺮﺍﺭ ﮔﺮﻓﺘـﻪ ﺷـﺪﻩ ﺍﺳـﺖ‪.‬‬ ‫ﻣﻄﻠﻮﺑﺴﺖ ﻣﺤﺎﺳﺒﻪ ﺍﻧﻮﺍﻉ ﺗﻮﺍﻥ ﺩﺭ ﻫﺮ ﻣﺼﺮﻑ ﮐﻨﻨﺪﻩ؟ ﻭ ﺿﺮﻳﺐ ﺗﻮﺍﻥ ﮐﻞ ﻣﺪﺍﺭ ﻭ ﺭﺳﻢ ﻣﺜﻠﺚ ﺗﻮﺍﻥ‪.‬‬

‫‪PF‬‬

‫ﺗﻮﺍﻥ ﻇﺎﻫﺮﯼ ‪kVA‬‬

‫ﻧﺎﻡ ﻣﺼﺮﻑ ﮐﻨﻨﺪﻩ‬

‫‪ ۰/۹‬ﭘﺲ ﻓﺎﺯ‬

‫‪۱۵‬‬

‫‪A‬‬

‫‪ ۰/۹‬ﭘﻴﺶ ﻓﺎﺯ‬

‫‪۵‬‬

‫‪B‬‬

‫‪ ۰/۸‬ﭘﺲ ﻓﺎﺯ‬

‫‪۲۰‬‬

‫‪C‬‬

‫‪) = (15)(0.9) = 13.5kw‬‬

‫‪A‬‬

‫(‪PA = S A cos‬‬

‫‪) = (5)(0.9) = 4.5kw‬‬

‫‪B‬‬

‫(‪PB = S B cos‬‬

‫‪) = ( 20)(0.8) = 16 kw‬‬

‫‪C‬‬

‫(‪PC = S C cos‬‬

‫‪Ptotal = PA + PB + PC = 13.5 + 4.5 + 16 = 34 kw‬‬ ‫‪Q A = S A sin(φ A ) = (15)(sin(cos −1 (0.9))) = (15)(0.44) = 6.54kVAR‬‬ ‫‪QB = S B sin(φ B ) = (5)(sin(− cos −1 (0.9))) = (5)(sin( −25.84)) = −2.18kVAR‬‬ ‫‪QC = S C sin(φC ) = ( 20)(sin(cos −1 (0.8))) = ( 20)(sin(36.87)) = 12kVAR‬‬ ‫‪Qtotal = Q A + QB + QC = 6.54 − 2.18 + 12 = 16.36kVAR‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪S total = Ptotal‬‬ ‫‪+ Qtotal‬‬ ‫‪= (34) 2 + (16.36) 2 = 37.85kVA‬‬

‫‪Ptotal‬‬ ‫‪34‬‬ ‫=‬ ‫‪= 0.9‬‬ ‫‪S total 37.85‬‬

‫‪lagging‬‬

‫= ) ‪PFtotal = cos(φtotal‬‬

‫‪= cos −1 (0.9) = 25.84‬‬

‫‪۴۹‬‬

‫‪total‬‬

‫‪PF‬‬

‫‪PC‬‬

‫‪QB‬‬

‫‪PB‬‬

‫ﻣﺜﻠﺚ ﺗﻮﺍﻥ‬ ‫‪ϕtotal‬‬

‫‪Ptotal‬‬

‫‪QA‬‬ ‫‪Qtotal‬‬

‫‪Stotal‬‬

‫‪QC‬‬

‫ﺩﺭ ﻣﺜﺎﻝ ﻫﺎﯼ ﻓﻮﻕ ﻫﻤﻮﺍﺭﻩ ﻣﺼﺮﻑ ﮐﻨﻨﺪﻩ ﻫﺎ ﺑﻪ ﺻﻮﺭﺕ ﺳﺮﯼ ﺩﺭ ﻣﺪﺍﺭ ﻗﺮﺍﺭ ﺩﺍﺷﺘﻨﺪ‪ .‬ﻟﺬﺍ ﺑـﺮﺍﯼ ﺣـﻞ ﺍﻧﻬـﺎ ﭼـﻮﻥ‬ ‫ﺟﺮﻳﺎﻥ ﺩﺭ ﺗﻤﺎﻡ ﻣﺼﺮﻑ ﮐﻨﻨﺪﻩ ﻫﺎ ﻳﮑﺴﺎﻥ ﺑﻮﺩ ﻣﺒﻨﺎ ﺭﺍ ﺟﺮﻳﺎﻥ ﮔﺮﻓﺘﻪ ﻭ ﻧﺴﺒﺖ ﺑﻪ ﺣﻞ ﻣﺴﺌﻠﻪ ﺍﻗﺪﺍﻡ ﻣﯽ ﺷﺪ‪ .‬ﺣﺎﻝ ﺍﮔﺮ‬ ‫ﻣﺼﺮﻑ ﮐﻨﻨﺪﻩ ﻫﺎ ﺑﻪ ﺻﻮﺭﺕ ﻣﻮﺍﺯﯼ ﺑﺎﺷﻨﺪ‪ ،‬ﻭﻟﺘﺎﮊ ﺩﺭ ﻣﺼﺮﻑ ﮐﻨﻨﺪﻩ ﻫﺎ ﻳﮑﺴﺎﻥ ﺑﻮﺩﻩ ﻟﺬﺍ ﻭﻟﺘﺎﮊ ﺭﺍ ﻣﺒﻨﺎ ﻗـﺮﺍﺭ ﺩﺍﺩﻩ ﻭ‬ ‫ﻧﺴﺒﺖ ﺑﻪ ﺣﻞ ﻣﺴﺌﻠﻪ ﺍﻗﺪﺍﻡ ﻣﯽ ﺷﻮﺩ‪.‬‬ ‫ﻣﺜﺎﻝ‪:‬‬ ‫ﺭﻭﯼ ﻳﮏ ﺍﺧﺘﻼﻑ ﺳﻄﺢ ﻣﺘﻨﺎﻭﺏ ‪ ۲۲۰‬ﻭﻟﺘﯽ ﺑﺎ ﻓﺮﮐـﺎﻧﺲ ‪ ۵۰‬ﻫﺮﺗـﺰ ﻳـﮏ‬

‫‪I1‬‬

‫ﻣﻘﺎﻭﻣﺖ ‪ ۲‬ﺍﻫﻤﯽ‪ ،‬ﺳﻠﻒ ‪ ۰/۱۰‬ﻫـﺎﻧﺮﯼ ﻭ ﺧـﺎﺯﻥ ‪ ۳۰۰‬ﻣﻴﮑﺮﻭﻓـﺎﺭﺍﺩﯼ ﺑـﻪ‬ ‫ﺻﻮﺭﺕ ﻣﻮﺍﺯﯼ ﻧﺼﺐ ﺷﺪﻩ ﺍﻧﺪ ﻣﻄﻠﻮﺑﺴﺖ‪:‬‬

‫‪I2‬‬

‫ﺷﺪﺕ ﺟﺮﻳﺎﻥ ﺩﺭ ﻫﺮ ﺍﻧﺸﻌﺎﺏ ﻭ ﺍﺧﺘﻼﻑ ﻓﺎﺯ ﺑﻴﻦ ﺟﺮﻳﺎﻥ ﻭ ﻭﻟﺘـﺎﮊ ﺩﺭ ﻫـﺮ‬ ‫ﺍﻧﺸﻌﺎﺏ؟‬

‫‪I3‬‬

‫ﺷﺪﺕ ﺟﺮﻳﺎﻥ ﺩﺭ ﺧﻂ ﺍﺻﻠﯽ ﻭ ﺍﺧﺘﻼﻑ ﻓﺎﺯ ﺁﻥ ﺑﺎ ﻭﻟﺘﺎﮊ؟‬ ‫ﺿﺮﻳﺐ ﺗﻮﺍﻥ ﮐﻞ ﻣﺪﺍﺭ؟‬ ‫ﺍﮔﺮ ﺧﺎﺯﻥ ﺑﺎ ﻇﺮﻓﻴﺖ ﻣﺘﻐﻴﻴﺮ ﺑﺎﺷـﺪ ﺩﺭ ﭼـﻪ ﻇﺮﻓﻴﺘـﯽ ﻣـﺪﺍﺭ ﺑـﻪ ﺻـﻮﺭﺕ‬ ‫ﺭﺯﻭﻧﺎﻧﺲ ﺧﻮﺍﻫﺪ ﺑﻮﺩ‪.‬؟‬ ‫‪Ve 220‬‬ ‫=‬ ‫‪= 110 A‬‬ ‫‪R‬‬ ‫‪2‬‬ ‫‪V‬‬ ‫‪220‬‬ ‫‪= 220 × 300 × 10 −6 × 100π = 20.72 A‬‬ ‫= ‪I2 = e‬‬ ‫‪1‬‬ ‫‪xc‬‬ ‫‪Cω‬‬ ‫‪V‬‬ ‫‪220‬‬ ‫‪= 70.06 A‬‬ ‫= ‪I3 = e‬‬ ‫‪Lω 0.01 × 100π‬‬

‫= ‪I1‬‬

‫‪۵۰‬‬

‫ﭼﻮﻥ ﻭﻟﺘﺎﮊ ﺩﺭ ﺗﻤﺎﻡ ﻣﺼﺮﻑ ﮐﻨﻨﺪﻩ ﻫﺎ ﻳﮑﺴﺎﻥ ﺍﺳﺖ ﺑﻪ ﻋﻨﻮﺍﻥ ﻣﺒﻨﺎ ﺩﺭ ﻧﻈﺮ ﮔﺮﻓﺘﻪ ﻭ ﺩﻳﺎﮔﺮﺍﻡ ﺟﺮﻳﺎﻧﻬﺎ ﺭﺍ ﺭﺳـﻢ ﻣـﯽ‬ ‫ﮐﻨﻴﻢ‪:‬‬

‫‪I e2 = I12 + ( I 3 − I 2 ) 2‬‬ ‫‪I2‬‬

‫‪I e = (110) 2 + (70.06 − 20.72) 2‬‬ ‫‪I e = 120.5 A‬‬ ‫‪110‬‬ ‫‪I1‬‬ ‫=‬ ‫‪= 0.92‬‬ ‫‪I e 120.5‬‬

‫‪I1‬‬

‫‪Ve‬‬

‫‪ϕ‬‬

‫= ‪cos ϕ‬‬

‫‪I3-I2‬‬ ‫‪Ie‬‬

‫‪S = Ve I e = 220 × 120.5 = 26.4kVA‬‬

‫‪I3‬‬

‫‪P = Ve I e cos ϕ = 220 ×120.5 × 0.92 = 24.2kw‬‬ ‫‪Ve‬‬ ‫‪= VeCω‬‬ ‫‪Lω‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫= ‪⇒C‬‬ ‫=‬ ‫‪= 0.001F = 1000µF‬‬ ‫‪2‬‬ ‫‪Lω‬‬ ‫‪0.01(100π ) 2‬‬ ‫⇒ ‪I 2 = I3‬‬

‫‪ (۵-۴‬ﺟﺮﻳﺎﻧﻬﺎﯼ ﻣﺘﻨﺎﻭﺏ ﺳﻪ ﻓﺎﺯ‬ ‫ﺟﺮﻳﺎﻧﻬﺎﯼ ﻣﺘﻨﺎﻭﺏ ﺳﻪ ﻓﺎﺯ ﺍﺯ ﺳﻪ ﺟﺮﻳﺎﻥ ﺳﻴﻨﻮﺳﯽ ﺗﺮﮐﻴﺐ ﺷﺪﻩ ﺍﺳﺖ ﮐﻪ ﻫﺮ ﻓﺎﺯ ﻧﺴﺒﺖ ﺑﻪ ﻓﺎﺯ ﺩﻳﮕـﺮ ‪ ۱۲۰‬ﺩﺭﺟـﻪ‬ ‫ﺍﺧﺘﻼﻑ ﻓﺎﺯ ﺩﺍﺭﺩ‬

‫)‪V(t‬‬

‫)‪V1(t‬‬

‫‪β‬‬ ‫)‪V(t‬‬

‫‪α‬‬

‫)‪(t‬‬

‫‪β‬‬

‫‪α‬‬ ‫)‪V2(t‬‬

‫)‪V2(t‬‬

‫)‪V(t‬‬

‫)‪V1(t‬‬

‫ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﺷﮑﻞ ﺑﺎﻻ ﻣﯽ ﺗﻮﺍﻥ ﻭﻳﮋﮔﻴﻬﺎﯼ ﺯﻳﺮ ﺭﺍ ﺩﺭ ﻣﻮﺭﺩ ﺟﺮﻳﺎﻧﻬﺎﯼ ﺳﻪ ﻓﺎﺯ ﻣﺸﺨﺺ ﻧﻤﻮﺩ‬ ‫‪ -۱‬ﻭﻗﺘﯽ ﻳﮑﯽ ﺍﺯ ﻓﺎﺯﻫﺎ ﺣﺪﺍﮐﺜﺮ ﺑﺎﺷﺪ ﺩﻭ ﻓﺎﺯ ﺩﻳﮕﺮ ﺑﺮﺍﺑﺮ ﻫﻢ ﻭﻟﯽ ﻋﮑﺲ ﻓﺎﺯ ﺍﻭﻝ ﺍﺳﺖ‪.‬‬ ‫‪ -۲‬ﻫﻤﻴﺸﻪ ﺟﻤﻊ ﻭﻟﺘﺎﮊ ﻫﺎ )ﺟﺮﻳﺎﻧﻬﺎ( ﺩﺭ ﻫﺮ ﻟﺤﻈﻪ ﺻﻔﺮ ﺍﺳﺖ‬ ‫‪ -۳‬ﻭﻗﺘﯽ ﻳﮑﯽ ﺍﺯ ﻓﺎﺯﻫﺎ ﺻﻔﺮ ﺍﺳﺖ ﺩﻭ ﻓﺎﺯ ﺩﻳﮕﺮ ﺍﺯ ﻧﻈﺮ ﻣﻘﺪﺍﺭ ﺑﺮﺍﺑﺮ ﻭﻟﯽ ﻣﺨﺘﻠﻒ ﺍﻟﺠﻬﺖ ﻫﺴﺘﻨﺪ‬ ‫ﻣﺼﺮﻑ ﮐﻨﻨﺪﻩ ﻫﺎ ﻣﻤﮑﻦ ﺍﺳﺖ ﺑﻪ ﺻﻮﺭﺕ ﺳﺎﺩﻩ ﺑﻪ ﺻﻮﺭﺕ ﺗﮏ ﻓﺎﺯ ﺑﻪ ﻫﺮ ﻳﮏ ﺍﺯ ﻓﺎﺯ ﻫﺎﯼ ﻣﻮﻟﺪ ﺳـﻪ ﻓـﺎﺯ ﻭﺻـﻞ‬ ‫ﺷﺪﻩ ﻭ ﻳﺎﻣﻤﮑﻦ ﺍﺳﺖ ﻣﺼﺮﻑ ﮐﻨﻨﺪﻩ ﻫﺎﯼ ﺳﻪ ﻓﺎﺯ ﺑﻪ ﺭﻭﺷﻬﺎﯼ ﺧﺎﺻﯽ ﺑﻪ ﻣﻮﻟﺪ ﻫﺎ ﻭﺻﻞ ﮔﺮﺩﻧﺪ‪.‬‬

‫‪ (۵-۴-۱‬ﺍﺗﺼﺎﻻﺕ ﺑﺮﻕ ﺳﻪ ﻓﺎﺯ‬ ‫‪۵۱‬‬

‫™ ﺍﺗﺼﺎﻝ ﺳﺘﺎﺭﻩ )‪:(Y‬‬ ‫ﺩﺭ ﺍﻳﻦ ﻧﻮﻉ ﺍﺗﺼﺎﻝ ﺍﻧﺘﻬﺎﯼ ﻫﺮ ﺳﻪ ﺳﻴﻢ ﭘﻴﭻ ﺩﺭ ﻳﮏ ﻧﻘﻄﻪ ﺑﻪ ﻫﻤﺪﻳﮕﺮ ﻣﺘﺼﻞ ﺷﺪﻩ ﻭ ﺳﺮ ﺩﻳﮕﺮ ﺳﻴﻢ ﭘﻴﭽﻬﺎ ﺑﻪ ﻋﻨﻮﺍﻥ‬ ‫ﺧﺮﻭﺟﻲ ﻫﺎ ﺩﺭ ﻣﻮﻟﺪ ﻭ ﻭﺭﻭﺩﻳﻬﺎ ﺩﺭ ﻣﺼﺮﻑ ﮐﻨﻨﺪﻩ ﻫﺎ ﻣﻮﺭﺩ ﺍﺳﺘﻔﺎﺩﻩ ﻗﺮﺍﺭ ﻣﯽ ﮔﻴﺮﻧﺪ‪ .‬ﻫﺮ ﻳﮏ ﺍﺯ ﺍﻳﻦ ﺳﻪ ﺳﻴﻢ ﺑﻪ ﻧﺎﻡ‬ ‫ﻫﺎﯼ ‪ R ,S‬ﻭ‪ T‬ﻧﺎﻣﻴﺪﻩ ﻣﯽ ﺷﻮﻧﺪ‪ .‬ﺍﻧﺘﻬﺎﯼ ﺩﻳﮕﺮ ﺳﻴﻢ ﭘﻴﭽﻬﺎ ﮐﻪ ﺑﻪ ﻫﻢ ﻭﺻﻞ ﻣﯽ ﺷﻮﻧﺪ ﺑﻪ ﻧﺎﻡ ﻧﻘﻄﻪ ﻣﺸﺘﺮﮎ ﻳﺎ ﻧﻘﻄﻪ‬ ‫ﺻﻔﺮ‪ ،‬ﺧﻨﺜﯽ ﻭ ﻳﺎ ﻧﻮﻝ ﻧﺎﻣﻴﺪﻩ ﻣﯽ ﺷﻮﺩ ﺍﺯ ﺍﻳﻦ ﻧﻘﻄﻪ ﻧﻴﺰ ﻣﯽ ﺗﻮﺍﻥ ﻳﮏ ﺳﻴﻢ ﺧﺎﺭﺝ ﻧﻤﻮﺩ ﮐﻪ ﺑﻪ ﻧﺎﻡ ﺳﻴﻢ ﻧـﻮﻝ ﻧﺎﻣﻴـﺪﻩ‬ ‫ﻣﯽ ﺷﻮﺩ ﺯﻳﺮﺍ ﻣﺠﻤﻮﻉ ﺟﺮﻳﺎﻥ ﻫﺎ )ﻭﻟﺘﺎﮊ( ﺩﺭ ﺍﻳﻦ ﺧﻂ ﺻﻔﺮ ﻣﯽ ﺑﺎﺷﺪ‪ .‬ﺻﻔﺮ ﺷﺪﻥ ﺟﺮﻳﺎﻥ ﺩﺭ ﺍﻳﻦ ﻧﻘﻄﻪ )ﺧﻂ( ﺭﺍ ﺑﻪ‬ ‫ﺻﻮﺭﺕ ﺭﻳﺎﺿﯽ ﻧﻴﺰ ﻣﯽ ﺗﻮﺍﻥ ﻧﺸﺎﻥ ﺩﺍﺩ‪ .‬ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﺷﮑﻞ ﻓﻮﻕ ﻣﯽ ﺗﻮﺍﻥ ﺟﺮﻳﺎﻧﻬـﺎ ﺭﺍ ﺩﺭ ﺟﺮﻳﺎﻧﻬـﺎﯼ ﺳـﻪ ﻓـﺎﺯ ﺑـﻪ‬ ‫ﺻﻮﺭﺕ ﺯﻳﺮ ﻧﻮﺷﺖ‬ ‫) ‪L1 ⇒ I 1( t ) = I max sin(θ‬‬ ‫)‪L2 ⇒ I 2( t ) = I max sin(θ + 120‬‬ ‫)‪L3 ⇒ I 3( t ) = I max sin(θ + 240‬‬ ‫ﺩﺭ ﻧﻘﻄﻪ ﻧﻮﻝ ﺟﻤﻊ ﺟﺮﻳﺎﻧﻬﺎ ﺑﺮﺍﺑﺮ‪:‬‬ ‫‪I N = I1 + I 2 + I 3‬‬ ‫)) ‪I N = I Max (sin(θ ) + sin(θ + 120 ) + sin(θ + 240‬‬ ‫ﺑﺮﺍﯼ ﻫﺮ ﺯﺍﻭﻳﻪ ﺍﯼ ﻣﺜﻞ‬

‫‪ θ‬ﺩﺍﺭﻳﻢ‪:‬‬

‫‪(sin(θ ) + sin(θ + 120 ) + sin(θ + 240 )) = 0‬‬

‫ﺷﮑﻞ‪ :۵-۱‬ﺍﻧﻮﺍﻉ ﺍﺗﺼﺎﻝ ﺩﺭ ﺟﺮﻳﺎﻧﻬﺎﯼ ﺳﻪ ﻓﺎﺯ‬

‫ﻟﺬﺍ ﻧﻘﻄﻪ ﻧﻮﻝ ﺑﻪ ﻋﻨﻮﺍﻥ ﻧﻘﻄﻪ ﺧﻨﺜﯽ ﻭ ﻳـﺎ ﻣﺜـﻞ ﺧـﻂ‬ ‫ﻧﻮﻝ ﺩﺭ ﺟﺮﻳﺎﻥ ﻫﺎﯼ ﺗﮏ ﻓﺎﺯ ﻣﯽ ﺑﺎﺷﺪ‪.‬ﺑﺎ ﺍﺗﺼﺎﻝ ﺳﺘﺎﺭﻩ ﻣﯽ ﺗﻮﺍﻥ ﺍﺯ ﻫﺮﻳﮏ ﺍﺯ ﻓﺎﺯﻫﺎﯼ ﺳﻪ ﮔﺎﻧﻪ ﺟﺮﻳﺎﻥ ﺳﻪ ﻓﺎﺯ ﺑﺮﺍﯼ‬ ‫ﻣﺼﺮﻑ ﮐﻨﻨﺪﻩ ﻫﺎﯼ ﺗﮏ ﻓﺎﺯ ﺍﺳﺘﻔﺎﺩﻩ ﮐﺮﺩ ﺩﺭ ﺍﻳﻦ ﺻﻮﺭﺕ ﺧﻂ ﻓﺎﺯ ﺑﻪ ﻳﮑﯽ ﺍﺯ ﻓﺎﺯﻫﺎ ﻭ ﺧﻂ ﻧﻮﻝ ﺑﻪ ﺧﻂ ﻧﻮﻝ ﺟﺮﻳﺎﻥ‬ ‫ﺳﻪ ﻓﺎﺯ ﻭﺻﻞ ﺧﻮﺍﻫﺪ ﺷﺪ‪ .‬ﻫﻤﭽﻨﻴﻦ ﻣﯽ ﺗﻮﺍﻥ ﺍﺯ ﺍﻳﻦ ﺍﺗﺼﺎﻝ ﺑﺮﺍﯼ ﻣﺼﺮﻑ ﮐﻨﻨﺪﻩ ﻫﺎﯼ ﺳﻪ ﻓـﺎﺯ ﺑـﻪ ﻃـﻮﺭ ﻣﺴـﺘﻘﻴﻢ‬ ‫ﺍﺳﺘﻔﺎﺩﻩ ﻧﻤﻮﺩ‪ .‬ﺑﻪ ﺍﺗﺼﺎﻝ ﺳﺘﺎﺭﻩ ﺍﺗﺼﺎﻝ ﺳﻪ ﻓﺎﺯ ﭼﻬﺎﺭ ﺳﻴﻤﻪ ﻧﻴﺰ ﮔﻔﺘﻪ ﻣﯽ ﺷﻮﺩ‪ .‬ﭼﻨﺎﻧﭽﻪ ﺩﺭ ﺍﺗﺼﺎﻝ ﺳﺘﺎﺭﻩ ﺍﺯ ﻓﺎﺯﻫﺎﯼ‬ ‫ﻣﺨﺘﻠﻒ ﺑﺎﺭﻫﺎﯼ ﻣﺨﺘﻠﻔﯽ ﮔﺮﻓﺘﻪ ﺷﻮﺩ ﺩﺭ ﺍﻳﻦ ﺻﻮﺭﺕ ﺳﻴﺴﺘﻢ ﺑﻪ ﺻـﻮﺭﺕ ﻧﺎﻣﺘﻌـﺎﺩﻝ ﺩﺭ ﺧﻮﺍﻫـﺪ ﺁﻣـﺪ‪ .‬ﺧﺼﻴﺼـﻪ‬ ‫ﻋﻤﻮﻣﯽ ﺑﺎﺭﻫﺎﯼ ﻣﺘﻌﺎﺩﻝ ﺍﻳﻨﺴﺖ ﮐﻪ ﺍﺧﺘﻼﻑ ﻓﺎﺯ ﻧﺴﺒﯽ ﺑﻴﻦ ﻓﺎﺯﻫﺎ ‪ ۱۲۰‬ﺩﺭﺟﻪ ﺍﺳﺖ ﻭﺍﻧﮕﻬﯽ ﺍﻣﭙﺪﺍﻧﺲ ﺗﻤـﺎﻡ ﻓﺎﺯﻫـﺎ‬ ‫ﻳﮑﯽ ﺑﻮﺩﻩ ﻭ ﺩﺭ ﻧﺘﻴﺠﻪ ﺗﻮﺍﻧﻬﺎﯼ ﺍﮐﺘﻴﻮ ﻭ ﺭﺍﮐﺘﻴﻮ ﺩﺭ ﻫﺮ ﺳﻪ ﻓﺎﺯ ﻳﮑﺴﺎﻥ ﺧﻮﺍﻫﺪ ﺑـﻮﺩ‪ .‬ﺍﻣـﺎ ﺩﺭ ﺳﻴﺴـﺘﻤﻬﺎﯼ ﻧﺎﻣﺘﻌـﺎﺩﻝ‬

‫‪۵۲‬‬

‫ﻣﻤﮑﻦ ﺍﺳﺖ ﮐﻪ ﻳﮑﯽ ﺍﺯ ﺍﻳﻦ ﺑﺮﺍﺑﺮﻳﻬﺎ ﻣﻮﺟﻮﺩ ﻧﺒﺎﺷﺪ ﻭ ﻳﺎ ﺍﻳﻨﮑﻪ ﺍﺧﺘﻼﻑ ﻧﺴﺒﯽ ﺑـﻴﻦ ﻓﺎﺯﻫـﺎ ‪ ۱۲۰‬ﺩﺭﺟـﻪ ﻧﺒﺎﺷـﺪ‪ .‬ﺍﺯ‬ ‫ﺍﻧﺠﺎﺋﻴﮑﻪ ﺗﺠﺰﻳﻪ ﻭ ﺗﺤﻠﻴﻞ ﺳﻴﺴﺘﻤﻬﺎﯼ ﻧﺎﻣﺘﻌﺎﺩﻝ ﭘﻴﭽﻴﺪﻩ ﻣﯽ ﺑﺎﺷﺪ ﺩﺭ ﺗﻐﺬﻳﻪ ﺑﺎﺭﻫﺎﯼ ﺗﮏ ﻓﺎﺯ ﺍﺯ ﻃﺮﻳﻖ ﺧﻄﻮﻁ ﺳـﻪ‬ ‫ﻓﺎﺯ ﺳﻌﯽ ﻣﯽ ﮔﺮﺩﺩ ﮐﻪ ﺑﺎﺭ ﻫﺎ ﺍﺯ ﻧﻈﺮ ﺗﻌﺪﺍﺩ ‪ ،‬ﻣﻴﺰﺍﻥ ﺑﺎﺭ ﻭ ﮐﻴﻔﻴﺖ ﺑﺎﺭ ﺑﻪ ﻃﻮﺭ ﻣﺴﺎﻭﯼ ﺗﻘﺴﻴﻢ ﮔﺮﺩﺩ ﺗﺎ ﺳﻴﺴـﺘﻢ ﺑـﻪ‬ ‫ﺻﻮﺭﺕ ﻣﺘﻌﺎﺩﻝ ﺑﺎﻗﯽ ﺑﻤﺎﻧﺪ‪ .‬ﺑﻄﻮﺭﻳﮑﻪ ﺑﺨﺶ ﺍﻋﻈﻢ ﻣﺴﺎﺋﻞ ﻋﻤﻠﯽ ﺑﻪ ﺳﻴﺴﺘﻤﻬﺎﯼ ﻣﺘﻌﺎﺩﻝ ﻣﺮﺑﻮﻁ ﻣﯽ ﺷﻮﺩ‪ .‬ﺩﺭ ﺍﺗﺼﺎﻝ‬ ‫ﺳﺘﺎﺭﻩ ﺩﻭ ﻧﻮﻉ ﺍﺧﺘﻼﻑ ﺳﻄﺢ ﺍﻟﮑﺘﺮﻳﮑﯽ ) ﻭﻟﺘﺎﮊ( ﻗﺎﺑﻞ ﺩﺳﺘﺮﺳﯽ ﻣﯽ ﺑﺎﺷﺪ‪.‬‬ ‫ﻭﻟﺘﺎﮊ ﺧﻂ ﺑﻪ ﺧﻨﺜﯽ‪ :‬ﺍﺧﺘﻼﻑ ﭘﺘﺎﻧﺴﻴﻞ ﺑﻴﻦ ﻫﺮ ﻳﮏ ﺍﺯ ﺧﻄﻮﻁ ﺍﺻﻠﯽ ‪ R, S‬ﻭ ‪ T‬ﻭ ﺧﻂ ﺧﻨﺜﯽ ﻣﯽ ﺑﺎﺷﺪ‪ .‬ﺍﻳﻦ ﻭﻟﺘﺎﮊ‬ ‫ﺑﻪ ﺍﺧﺘﺼﺎﺭ ﻭﻟﺘﺎﮊ ﻓﺎﺯ ﻧﺎﻣﻴﺪﻩ ﻣﯽ ﺷﻮﺩ‪.‬‬ ‫ﻭﻟﺘﺎﮊ ﺧﻂ ﺑﻪ ﺧﻂ‪ :‬ﺍﺧﺘﻼﻑ ﭘﺘﺎﻧﺴﻴﻞ ﺑﻴﻦ ﺧﻄﻮﻁ ﺍﺻﻠﯽ ﻣﯽ ﺑﺎﺷﺪ ﮐﻪ ﺑﻪ ﺍﺧﺘﺼﺎﺭ ﻭﻟﺘﺎﮊ ﺧـﻂ ﻣـﯽ ﻧﺎﻣﻨـﺪ‪ .‬ﺩﺭ ﻳـﮏ‬ ‫ﺟﺮﻳﺎﻥ ﺳﻪ ﻓﺎﺯ ﻣﺘﻌﺎﺩﻝ ﺍﺯ ﻧﻈﺮ ﻣﻘﺪﺍﺭ ﻭ ﭘﺲ ﻓﺎﺯ ﻳﺎ ﭘﻴﺶ ﻓﺎﺯ ﺑﻮﺩﻥ ﺷﺒﻴﻪ ﻫﻢ ﻫﺴﺘﻨﺪ‪.‬‬ ‫™ ﺍﺗﺼﺎﻝ ﻣﺜﻠﺚ ﻳﺎ ﺩﻟﺘﺎ )∇ (‪:‬‬ ‫ﺩﺭ ﺍﻳﻦ ﺍﺗﺼﺎﻝ ﺳﺮ ﻳﮏ ﺳﻴﻢ ﭘﻴﭻ ﺑﻪ ﺗﻪ ﺳﻴﻢ ﭘﻴﭻ ﺩﻳﮕﺮ ﻣﺘﺼﻞ ﻣﯽ ﮔﺮﺩﺩ ﺗﺎ ﺍﻳﻨﮑﻪ ﺗﺸﮑﻴﻞ ﻳﮏ ﺣﻠﻘﻪ ﺑﺴﺘﻪ ﺩﺍﺩﻩ ﺷﻮﺩ‬ ‫)ﺷﮑﻞ‪ .(۵-۱‬ﺩﺭ ﺍﻳﻦ ﻧﻮﻉ ﺍﺗﺼﺎﻝ ﺳﻪ ﺧﻄﻮﻁ ﺍﺻﻠﯽ ﺳﻪ ﻓﺎﺯ ﺍﺯ ﻣﺤﻞ ﻧﻘﺎﻁ ﺍﺗﺼﺎﻝ ﺳﺮ ﺑﻪ ﺗﻪ ﺳﻴﻢ ﭘﻴﭽﻬﺎ ﺧﺎﺭﺝ ﻣـﯽ‬ ‫ﮔﺮﺩﺩ‪ .‬ﺩﺭ ﺍﻳﻦ ﺍﺗﺼﺎﻝ ﺑﻪ ﺻﻮﺭﺕ ﻣﻌﻤﻮﻝ ﺧﻂ ﺧﻨﺜﯽ ﻭﺟﻮﺩ ﻧﺪﺍﺭﺩ ﻟﺬﺍ ﺑﻪ ﺍﻳﻦ ﺍﺗﺼﺎﻝ ‪ ،‬ﺍﺗﺼﺎﻝ ﺳﻪ ﺳﻴﻤﻪ ﻧﻴﺰ ﮔﻔﺘﻪ ﻣﯽ‬ ‫ﺷﻮﺩ‪ .‬ﻟﺬﺍ ﻣﺼﺮﻑ ﮐﻨﻨﺪﻩ ﻫﺎﻳﯽ ﮐﻪ ﺑﻪ ﺻﻮﺭﺕ ﻣﺜﻠﺚ ﺑﺴﺘﻪ ﻣﯽ ﺷﻮﻧﺪ ﺑﻪ ﺻﻮﺭﺕ ﺳﻴﺴﺘﻢ ﻫﺎﯼ ﻣﺘﻌﺎﺩﻝ ﻫﺴﺘﻨﺪ‪.‬‬

‫‪ (۵-۴-۲‬ﺟﺮﻳﺎﻥ ﻭ ﻭﻟﺘﺎﮊ ﻭ ﻗﺪﺭﺕ ﺩﺭ ﺍﺗﺼﺎﻻﺕ ﺳﻪ ﻓﺎﺯ‪:‬‬ ‫ﻣﻄﺎﺑﻖ ﺷﮑﻞ ‪ ۵-۲‬ﻭﻟﺘﺎﮊ ﻫﺎﯼ ﺳﻪ ﮔﺎﻧﻪ ﻓﺎﺯ ﺑﻴﻦ ﻧﻘﻄﻪ ﺧﻨﺜﯽ ‪ O‬ﻭ ﻧﻘﺎﻁ ﺍﻧﺘﻬﺎﻳﯽ ﺳﻴﻢ ﭘﻴﭽﻬﺎ ﺑﻮﺟـﻮﺩ ﻣـﯽ ﺍﻳﻨـﺪ ﮐـﻪ‬ ‫ﻋﺒﺎﺭﺗﻨﺪ ﺍﺯ ‪ EOA, EOB EOC‬ﻭ ﻭﻟﺘﺎﮊ ﻫﺎﯼ ﺧﻂ ﻋﺒﺎﺭﺗﻨﺪ ﺍﺯ ‪ . ECA, EBC, EAB :‬ﺗﻤﺎﻡ ﻭﻟﺘﺎﮊﻫﺎﯼ ﻓﺎﺯ ﺍﺯ ﻧﻈﺮ ﻣﻘﺪﺍﺭ‬ ‫‪r‬‬ ‫‪r‬‬ ‫ﺑﺎ ﻫﻢ ﺑﺮﺍﺑﺮ ﻭ ﻟﺬﺍ ﻃﺒﻖ ﺷﮑﻞ ﻭﻟﺘﺎﮊﻫﺎﯼ ﺧﻂ ﺑﺮﺍﺑﺮ‪:‬‬ ‫‪ECA = − EOC + EOA‬‬ ‫‪r‬‬ ‫‪r‬‬ ‫‪E BC = − EOB + EOC‬‬ ‫‪r‬‬ ‫‪r‬‬ ‫‪E AB = − EOA + EOB‬‬

‫‪-EOA‬‬ ‫‪EAB‬‬

‫‪IL‬‬

‫‪EOC‬‬ ‫‪EOB‬‬

‫‪EBC‬‬

‫‪-EOB‬‬

‫‪۶۰°‬‬

‫‪Ip‬‬

‫‪O‬‬

‫‪EOA‬‬

‫‪O‬‬

‫‪-EOC‬‬

‫‪ECA‬‬

‫ﺷﮑﻞ‪ :۵-۲‬ﺍﻧﻮﺍﻉ ﻭﻟﺘﺎﮊ ﻭ ﺟﺮﻳﺎﻥ ﺩﺭ ﺍﺗﺼﺎﻝ ﺳﺘﺎﺭﻩ‬

‫‪۵۳‬‬

‫ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﺷﮑﻞ ‪ ۵-۲‬ﺑﺮﺍﯼ ﺍﺗﺼﺎﻝ ﺳﺘﺎﺭﻩ ﻣﯽ ﺗﻮﺍﻥ ﻧﻮﺷﺖ‪:‬‬ ‫‪V L = 3V p‬‬

‫)‪(۵-۴‬‬

‫‪IL = I p‬‬

‫ﻗﺪﺭﺕ ﻫﺮ ﻓﺎﺯ ﺩﺭ ﺍﺗﺼﺎﻝ ﺳﺘﺎﺭﻩ ﺑﺮﺍﺑﺮ‪:‬‬

‫) ‪p p = V p I p cos(ϕ‬‬ ‫ﻭ ﻗﺪﺭﺕ ﮐﻞ ﺳﻪ ﻓﺎﺯ ﺑﺮﺍﺑﺮ‪:‬‬

‫) ‪P = 3 p p = 3V p I p cos(ϕ‬‬ ‫ﺑﺎ ﺟﺎﻳﮕﺰﻳﻨﯽ ﺭﺍﺑﻄﻪ )‪ (۵-۴‬ﺩﺍﺭﻳﻢ‬

‫) ‪P = 3VL I L cos(ϕ‬‬

‫)‪(۵-۵‬‬

‫ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﺷﮑﻞ ‪ ۵-۳‬ﻭ ﻗﺎﻧﻮﻥ ﮐﻴﺮﺷﻒ ﺑﺮﺍﯼ ﺟﺮﻳﺎﻥ ﺩﺭ ﻫﺮ ﻳﮏ ﺍﺯ ﮔﺮﻩ ﻫﺎ ﻣﯽ ﺗﻮﺍﻥ ﻧﻮﺷﺖ‪:‬‬ ‫ﮔﺮﻩ ‪:A‬‬

‫ﮔﺮﻩ ‪: B‬‬

‫ﺩﺭ ﮔﺮﻩ ‪:C‬‬

‫‪r‬‬ ‫‪r‬‬ ‫‪r‬‬ ‫‪I ca − I ab − I L = 0‬‬ ‫‪r‬‬ ‫‪r‬‬ ‫‪r‬‬ ‫‪I L = I ca − I ab‬‬

‫‪IL‬‬

‫‪R‬‬

‫‪r‬‬ ‫‪r‬‬ ‫‪r‬‬ ‫‪− I bc + I ab − I L = 0‬‬ ‫‪r‬‬ ‫‪r‬‬ ‫‪r‬‬ ‫‪I L = I ab − I bc‬‬

‫‪Ip=Iab‬‬

‫‪VL‬‬

‫‪r‬‬ ‫‪r‬‬ ‫‪r‬‬ ‫‪I bc − I ca − I L = 0‬‬ ‫‪r‬‬ ‫‪r‬‬ ‫‪r‬‬ ‫‪I L = I bc − I ca‬‬

‫‪S‬‬ ‫‪T‬‬

‫ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﺯﻭﺍﻳﺎﯼ ﺑﻴﻦ ﺑﺮﺩﺍﺭﻫﺎ ﻭ ﺑـﺎ ﺗﻮﺟـﻪ ﺑـﻪ ﺍﻳﻨﮑـﻪ‬ ‫ﺩﺍﺭﻳﻢ‪:‬‬

‫‪-Iab‬‬

‫‪I ca = I ab = I bc = I p‬‬

‫‪Ica‬‬

‫ﻟﺬﺍ ﺩﺭ ﺍﺗﺼﺎﻝ ﻣﺜﻠﺚ ﻣﯽ ﺗﻮﺍﻥ ﻧﻮﺷﺖ‪:‬‬ ‫)‪(۵-۶‬‬

‫‪3I p‬‬

‫‪Ibc‬‬

‫= ‪IL‬‬

‫ﻭ ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﺷﮑﻞ ‪ ۵-۳‬ﺩﺍﺭﻳﻢ‪:‬‬ ‫)‪(۵-۷‬‬

‫‪°‬‬

‫‪۶۰‬‬

‫‪-Ica‬‬

‫‪Vl = V p‬‬

‫‪-Ibc‬‬ ‫‪۳۰°‬‬ ‫‪Iab‬‬

‫ﺷﮑﻞ‪ :۵-۳‬ﺍﻧﻮﺍﻉ ﻭﻟﺘﺎﮊ ﻭ ﺟﺮﻳﺎﻥ ﺩﺭ ﺍﺗﺼﺎﻝ ﻣﺜﻠﺚ‬

‫‪۵۴‬‬

‫ﻗﺪﺭﺕ ﺩﺭ ﺍﺗﺼﺎﻝ ﻣﺜﻠﺚ ﺑﺮﺍﺑﺮ‪:‬‬

‫) ‪p p = V p I p cos(ϕ‬‬

‫ﻗﺪﺭﺕ ﻳﮏ ﻓﺎﺯ‬

‫) ‪P = 3 p p = 3V p I p cos(ϕ‬‬

‫ﻗﺪﺭﺕ ﻫﺮ ﺳﻪ ﻓﺎﺯ‬ ‫ﻭ ﺑﺎ ﺟﺎﻳﮕﺰﻳﻦ ﮐﺮﺩﻥ ﻣﻘﺎﺩﻳﺮ ﺧﻂ ﺑﻪ ﺟﺎﯼ ﻣﻘﺎﺩﻳﺮ ﻓﺎﺯ ﺍﺯ ﺭﺍﺑﻄﻪ ﻫـﺎﯼ ‪۵-۶‬‬

‫) ‪P = 3VL I L cos(ϕ‬‬

‫ﻭ ‪ ۵-۷‬ﻗﺪﺭﺕ ﮐﻞ ﺑﺮﺍﺑﺮ‪:‬‬

‫ﺍﻳﻦ ﺭﺍﺑﻄﻪ ﺷﺒﻴﻪ ﻭ ﺑﺮﺍﺑﺮ ﺭﺍﺑﻄﻪ ‪ ۵-۵‬ﻣﯽ ﺑﺎﺷﺪ ﻟﺬﺍ ﻧﻮﻉ ﺍﺗﺼﺎﻝ ﺑﺮ ﺭﻭﯼ ﻗﺪﺭﺕ ﺩﺭ ﺟﺮﻳﺎﻧﻬﺎﯼ ﺳﻪ ﻓﺎﺯ ﺑﯽ ﺗﺎﺛﻴﺮ ﺍﺳﺖ‪.‬‬ ‫ﻋﻼﻭﻩ ﺑﺮ ﺍﻳﻦ ﻫﻤﺎﻥ ﺭﻭﺍﺑﻄﯽ ﮐﻪ ﺑﺮﺍﯼ ﺗﻮﺍﻥ ﺣﻘﻴﻘﯽ ﺩﺭ ﺍﺗﺼﺎﻝ ﺳﺘﺎﺭﻩ ﻭ ﻣﺜﻠﺚ ﻭﺟـﻮﺩ ﺩﺍﺭﺩ ﺑـﺮﺍﯼ ﺗـﻮﺍﻥ ﻇـﺎﻫﺮﯼ ﻭ‬ ‫ﺭﺍﮐﺘﻴﻮ ﻧﻴﺰ ﻣﯽ ﺗﻮﺍﻥ ﻧﻮﺷﺖ‪:‬‬

‫) ‪PΥ ,∆ = 3V L I L cos(ϕ‬‬ ‫) ‪QΥ ,∆ = 3VL I L sin(ϕ‬‬

‫)‪(۵-۸‬‬

‫‪S Υ , ∆ = 3VL I L‬‬ ‫‪S Υ,∆ = P 2 + Q 2‬‬

‫ﭼﻨﺎﻧﭽﻪ ﺑﺎﺭ ﺩﺭ ﻓﺎﺯ ﻫﺎ ﻧﺎﻣﺘﻌﺎﺩﻝ ﺑﺎﺷﺪ ﻻﺯﻣﺴﺖ ﮐﻪ ﻗﺪﺭﺕ ﺗﮏ ﺗﮏ ﻓﺎﺯ ﻫﺎ ﺑﺮ ﺣﺴﺐ ﻣﻘﺎﺩﻳﺮ ﻓﺎﺯ ﻣﺤﺎﺳﺒﻪ ﮔﺮﺩﺩ ﻭ ﺍﺯ‬ ‫ﻣﺠﻤﻮﻉ ﺍﻧﻬﺎ ﻗﺪﺭﺕ ﮐﻞ ﻣﺪﺍﺭ ﺭﺍ ﺑﺪﺳﺖ ﺁﻭﺭﺩ‪.‬‬ ‫) ‪p1 = V p1 I p1 cos( φ1‬‬ ‫) ‪p 2 = V p 2 I p 2 cos( φ 2‬‬ ‫) ‪p 3 = V p 3 I p 3 cos( φ3‬‬

‫ﺑﻪ ﻫﻤﻴﻦ ﺗﺮﺗﻴﺐ ﻗﺪﺭﺗﻬﺎﯼ ﺭﺍﮐﺘﻴﻮ ﻭ ﻇﺎﻫﺮﯼ ﻧﻴﺰ ﻣﺤﺎﺳﺒﻪ ﻣﯽ ﮔﺮﺩﺩ‪.‬‬

‫‪P = p1 + p 2 + p 3‬‬

‫ﻣﺜﺎﻝ‪:‬‬ ‫ﺭﻭﯼ ﭘﻼﮎ ﻳﮏ ﻣﻮﺗﻮﺭ ﺳﻪ ﻓﺎﺯ ﻋﻼﻣﺖ‪ 220/380 v‬ﺣﮏ ﺷﺪﻩ ﺍﺳﺖ‪ .‬ﺭﻭﯼ ﭼﻪ ﺧﻂ ﺍﻟﮑﺘﺮﻳﮑﯽ ﻣﯽ ﺗـﻮﺍﻥ ﻣﻮﺗـﻮﺭ‬ ‫ﻣﻮﺭﺩ ﻧﻈﺮ ﺭﺍ ﻧﺼﺐ ﻧﻤﻮﺩ‪.‬؟ ﻭ ﺑﻪ ﭼﻪ ﺻﻮﺭﺕ؟ ﺍﮔﺮ ﻗﺪﺭﺕ ﻣﻔﻴﺪ ﺍﻳﻦ ﻣﻮﺗﻮﺭ ‪ ۱۲ kw‬ﻭ ﺭﺍﻧﺪﻣﺎﻥ ﺁﻥ ‪ %۹۰‬ﻭ ﺿـﺮﻳﺐ‬ ‫ﻗﺪﺭﺕ ﺁﻥ ‪ ۰/۸۵‬ﺑﺎﺷﺪ ﺟﺮﻳﺎﻧﻬﺎﯼ ﺧﻂ ﻭ ﻓﺎﺯ ﺭﺍ ﺩﺭ ﻫﺮ ﻭﺿﻌﻴﺖ ﺍﺗﺼﺎﻝ ﺗﻌﻴﻴﻦ ﮐﻨﻴﺪ‪.‬‬ ‫ﺣﻞ‪:‬‬ ‫ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﻭﻟﺘﺎﮊ ﺣﮏ ﺷﺪﻩ ﻣﯽ ﺗﻮﺍﻥ ﻧﺘﻴﺠﻪ ﮔﺮﻓﺖ ﮐﻪ ﻣﻮﺗﻮﺭ ﻭﻟﺘﺎﮊ ﺑﻴﺸﺘﺮ ﺍﺯ ‪ ۲۲۰‬ﻭﻟﺖ ﺩﺭ ﻫﺮ ﻳﮏ ﺍﺯ ﺳﻴﻢ ﭘﻴﭽﻬـﺎ‬ ‫ﺑﻴﺸﺘﺮ ﻧﻤﯽ ﺗﻮﺍﻧﺪ ﺗﺤﻤﻞ ﻧﻤﺎﻳﺪ‪ .‬ﻟﺬﺍ ﺩﺭ ﺗﺤﺖ ﻫﻴﭻ ﺷﺮﺍﻳﻄﯽ ﻧﺒﺎﻳﺪ ﻭﻟﺘﺎﮊ ﻓﺎﺯ ﺑﻴﺸﺘﺮ ﺍﺯ ‪ ۲۲۰‬ﻭﻟﺖ ﮔﺮﺩﺩ‪ .‬ﻟـﺬﺍ ﻣﻮﺗـﻮﺭ‬ ‫ﻣﯽ ﺗﻮﺍﻧﺪ ﺑﺮ ﺭﻭﯼ ﺩﻭ ﺧﻂ ﺑﻪ ﺻﻮﺭﺕ ﺯﻳﺮ ﻣﺘﺼﻞ ﮔﺮﺩﺩ‪:‬‬ ‫ﺍﻟﻒ( ﺧﻂ ‪ 127/220v‬ﺑﺎ ﺍﺗﺼﺎﻝ ﻣﺜﻠﺚ‬ ‫ﺏ( ﺧﻂ ‪ 220/380v‬ﺑﺎ ﺍﺗﺼﺎﻝ ﺳﺘﺎﺭﻩ‬

‫‪۵۵‬‬

‫ﻧﺤﻮﻩ ﺍﺗﺼﺎﻝ ﺗﺮﻣﻴﻨﺎﻝ ﻫﺎ‪:‬‬

‫ﺍﺗﺼﺎﻝ ﻣﺜﻠﺚ‬

‫ﺍﺗﺼﺎﻝ ﺳﺘﺎﺭﻩ‬

‫‪T‬‬

‫‪R‬‬

‫‪S‬‬

‫‪12000‬‬ ‫‪= 15700vA‬‬ ‫‪0.85 × 0.9‬‬ ‫‪15700‬‬ ‫= ‪Sp‬‬ ‫‪= 5233vA‬‬ ‫‪3‬‬ ‫‪S p 5233‬‬ ‫= ‪Ip‬‬ ‫=‬ ‫‪= 23.8 A‬‬ ‫‪220‬‬ ‫‪Vp‬‬ ‫=‪S‬‬

‫‪Υ‬‬ ‫∆‬

‫‪I L = I p = 23.8 A‬‬

‫‪I L = 3I p = 3 × 23.8 = 41.2 A‬‬

‫ﭼﻨﺎﻧﭽﻪ ﻣﻮﺗﻮﺭ ﺩﺭ ﺧﻂ ‪ 220/380‬ﺑﻪ ﺻﻮﺭﺕ ﻣﺜﻠﺚ ﺑﺴﺘﻪ ﺷﻮﺩ ﻭ ﻓﺮﮐﺎﻧﺲ ﺑﺮﻕ ‪ ۵۰‬ﻫﺮﺗﺰ ﺑﺎﺷﺪ ﻭ ﺑﺨﻮﺍﻫﻴﻢ ﺿﺮﻳﺐ‬ ‫ﻗﺪﺭﺕ ﺭﺍ ﺑﻪ ﻳﮏ ﺑﺮﺳﺎﻧﻴﻢ ﻇﺮﻓﻴﺖ ﺧـﺎﺯﻥ‬ ‫ﻣﻮﺭﺩ ﻧﻴﺎﺯ ﺭﺍ ﺑﺪﺳﺖ ﺁﻭﺭﻳﺪ‪.‬‬

‫‪cos(ϕ ) = 0.85 ⇒ sin(ϕ ) = 1 − (0.85) 2 = 0.527‬‬ ‫‪Q = S ⋅ sin(ϕ ) = 15700(0.527) = 8270vA‬‬ ‫‪8270‬‬ ‫‪= 2760vA‬‬ ‫‪3‬‬ ‫‪Q p 2760‬‬ ‫=‬ ‫‪= 12.54 A‬‬ ‫= ‪I rp‬‬ ‫‪Vp‬‬ ‫‪220‬‬

‫= ‪Qp‬‬

‫‪Vc = xc I c‬‬ ‫ﻇﺮﻓﻴﺖ ﺧﺎﺯﻥ ﺩﺭ ﺣﺎﻟﺖ ﺳﺘﺎﺭﻩ‬

‫ﻇﺮﻓﻴﺖ ﺧﺎﺯﻥ ﺩﺭ ﺣﺎﻟﺖ ﻣﺜﻠﺚ‬

‫‪Ic‬‬ ‫‪12.54‬‬ ‫=‬ ‫‪= 1.81× 10 − 4 F = 181.5µF‬‬ ‫)‪Vcω 220(314‬‬

‫=‪⇒C‬‬

‫‪12.54‬‬ ‫‪I‬‬ ‫‪3‬‬ ‫= ‪⇒C= c‬‬ ‫‪= 6.05 ×10 −5 = 60.5µF‬‬ ‫)‪Vcω 380(314‬‬

‫‪۵۶‬‬

‫ﺑﺎ ﻣﻘﺎﻳﺴﻪ ﻇﺮﻓﻴﺖ ﺧﺎﺯﻥ ﺩﺭ ﺩﻭ ﺣﺎﻟﺖ ﺍﺗﺼﺎﻝ ﺑﻪ ﺻﻮﺭﺕ ﻣﺜﻠﺚ ﻭ ﺳﺘﺎﺭﻩ ﻣﻼﺣﻈﻪ ﻣﯽ ﮔﺮﺩﺩ ﮐﻪ ﻇﺮﻓﻴﺖ ﺧﺎﺯﻥ ﻣﻮﺭﺩ‬ ‫ﻧﻴﺎﺯ ﺩﺭ ﺣﺎﻟﺖ ﻣﺜﻠﺚ ﻳﮏ ﺳﻮﻡ ﺣﺎﻟﺖ ﺳﺘﺎﺭﻩ ﺍﺳﺖ ﻫﺮ ﭼﻨﺪ ﮐﻪ ﺑﺎﻳﺪ ﻭﻟﺘﺎﮊ ﺑﺎﻻﺗﺮﯼ ﺭﺍ ﺗﺤﻤﻞ ﻧﻤﺎﻳﻨﺪ‪ .‬ﺑﺎ ﺍﻳﻦ ﻭﺟـﻮﺩ‬ ‫ﺍﺗﺼﺎﻝ ﺑﻪ ﺻﻮﺭﺕ ﻣﺜﻠﺚ ﺍﺭﺯﺍﻧﺘﺮ ﺧﻮﺍﻫﺪ ﺑﻮﺩ ﺯﻳﺮﺍ ﻫﺰﻳﻨﻪ ﻭﻟﺘﺎﮊ ﺑﻪ ﺍﻧﺪﺍﺯﻩ ﻫﺰﻳﻨﻪ ﻇﺮﻓﻴﺖ ﻧﻤﯽ ﺑﺎﺷﺪ‪.‬‬ ‫ﻣﺜﺎﻝ‪:‬‬ ‫ﺗﻮﺍﻥ ﻣﺼﺮﻓﯽ ﻳﮏ ﮐﺎﺭﺧﺎﻧﻪ ‪ ۲۵۰‬ﮐﻴﻠﻮ ﻭﺍﺕ ﺍﺳﺖ ﮐﻪ ﺍﺯ ﻳﮏ ﺷﺒﮑﻪ ﺳﻪ ﻓﺎﺯ ﺳﻪ ﺳﻴﻤﻪ ﺑﺎ ﺍﺧـﺘﻼﻑ ﺳـﻄﺢ )ﻭﻟﺘـﺎﮊ(‬ ‫‪ ۲۲۰۰‬ﻭﻟﺖ ﻭ ﻓﺮﮐﺎﻧﺲ ‪ ۵۰‬ﻫﺮﺗﺰ ﺗﻐﺬﻳﻪ ﻣﯽ ﺷﻮﺩ‪ .‬ﺿﺮﻳﺐ ﻗﺪﺭﺕ ﮐﻠﯽ ﮐﺎﺭﺧﺎﻧﻪ ‪ %۷۰‬ﻫﺴﺖ ﮐﻪ ﺑﻪ ﻣﻮﺟﺐ ﻗﺮﺍﺭ ﺩﺍﺩ‬ ‫ﺑﺎ ﺷﺮﮐﺖ ﺑﺮﻕ ﺍﮔﺮ ﺿﺮﻳﺐ ﻗﺪﺭﺕ ﺍﺯ ‪ %۸۵‬ﮐﻤﺘﺮ ﺑﺎﺷﺪ ﺑﻬﺎﯼ ﺍﻧﺮﮊﯼ ﻣﺼﺮﻓﯽ ﺑﺎ ﻧﺮﺥ ﺑﺎﻻﺗﺮﯼ ﻣﺤﺎﺳﺒﻪ ﺧﻮﺍﻫـﺪ ﺷـﺪ‪.‬‬ ‫ﻇﺮﻓﻴﺖ ﺧﺎﺯﻥ ﻫﺎﯼ ﻣﻮﺭﺩ ﻧﻴﺎﺯ ﺑﺮﺍﯼ ﺍﻳﻦ ﮐﺎﺭﺧﺎﻧﻪ ﺑﻪ ﺻﻮﺭﺕ ﺳﺘﺎﺭﻩ ﻭ ﻣﺜﻠﺚ ﺭﺍ ﻣﺤﺎﺳﺒﻪ ﻧﻤﺎﺋﻴﺪ‪.‬‬ ‫ﺣﻞ‪:‬‬ ‫ﭼﻮﻥ ﺑﺎﺭ ﻣﺘﻌﺎﺩﻝ ﺑﻴﻦ ﻓﺎﺯ ﻫﺎ ﺗﻘﺴﻴﻢ ﺷﺪﻩ ﺍﺳﺖ ﻟﺬﺍ ﺍﺧﺘﻼﻑ ﻓﺎﺯ ﺑﻴﻦ ﻭﻟﺘﺎﮊ ﻭ ﺟﺮﻳﺎﻥ ﺩﺭ ﻫﺮ ﺳـﻪ ﻓـﺎﺯ ﻳﮑﺴـﺎﻥ ﻣـﯽ‬ ‫ﺑﺎﺷﺪ‪ .‬ﻟﺬﺍ ﮐﺎﻓﯽ ﺍﺳﺖ ﻣﺴﺌﻠﻪ ﺭﺍ ﺑﺮﺍﯼ ﻳﮏ ﻓﺎﺯ ﺣﻞ ﮐﺮﺩﻩ ﻭ ﻧﺘﻴﺠﻪ ﺭﺍ ﺑﺮﺍﯼ ﺩﻳﮕﺮ ﻓﺎﺯ ﻫﺎ ﺑﮑﺎﺭ ﺑﺮﺩ‪ .‬ﺿﻤﻨﺎﹰ ﭼﻮﻥ ﻧـﻮﻉ‬ ‫ﺍﺗﺼﺎﻝ ﺗﺎﺛﻴﺮﯼ ﺑﺮ ﻣﺴﺌﻠﻪ ﻧﺪﺍﺭﺩ ﻫﺮﺍﺗﺼﺎﻟﯽ ﺭﺍ ﻣﯽ ﺗﻮﺍﻥ ﺑﺮﺍﯼ ﺣﻞ ﺩﺭ ﻧﻈﺮ ﮔﺮﻓﺖ ﺑﻨﺎﺑﺮﺍﻳﻦ ﺑﺎ ﺍﻳﻦ ﻓﺮﺽ ﮐﻪ ﺍﺗﺼﺎﻝ ﺑﻪ‬ ‫ﺻﻮﺭﺕ ﺳﺘﺎﺭﻩ ﺍﺳﺖ ﺩﺍﺭﻳﻢ‪:‬‬

‫‪It‬‬ ‫‪I1‬‬

‫‪Vp‬‬ ‫‪Ic‬‬

‫‪P = 3V p I1 cos(ϕ1 ) = 250000 watt‬‬

‫ﺗﻮﺍﻥ ﻣﺼﺮﻓﯽ ﮐﻞ‬ ‫ﻭﻟﺘﺎﮊ ﻓﺎﺯ‬

‫‪2200‬‬ ‫‪= 1270 volt‬‬ ‫‪3‬‬ ‫‪250000‬‬ ‫= ‪I1‬‬ ‫‪= 93 .7 A‬‬ ‫) ‪(3)(1270 )(0.7‬‬

‫= ‪Vp‬‬

‫ﺟﺮﻳﺎﻥ ﺧﻂ ﺩﺭ ﺣﺎﻟﺖ ﺍﻭﻝ‬ ‫ﺩﺭ ﺍﻳﻦ ﺣﺎﻟﺖ ﻣﺪﺍﺭ ﭘﺲ ﻓﺎﺯ ﺍﺳﺖ ﻭ ‪ I1‬ﺑـﻪ ﻣﻴـﺰﺍﻥ ‪ ϕ1‬ﺍﺯ ﻭﻟﺘـﺎﮊ‬ ‫ﻋﻘﺐ ﺗﺮ ﺍﺳـﺖ ﺯﻳـﺮﺍ ﻋﻤﻮﻣـﺎﹰ ﻣﺼـﺮﻑ ﻭ ﺑـﺎﺭ‬ ‫ﮐﺎﺭﺧﺎﻧﻪ ﻫﺎ ﺍﻧﺪﻭﮐﺘﻴﻮ ﺍﺳـﺖ‪ .‬ﺟﺮﻳـﺎﻧﯽ ﮐـﻪ ﺍﺯ‬

‫‪Ic‬‬

‫ﺧﺎﺯﻧﻬﺎ ﻋﺒﻮﺭ ﺧﻮﺍﻫﺪ ﮐﺮﺩ ﭘﻴﺶ ﻓﺎﺯ ﺧﻮﺍﻫﺪ ﺑﻮﺩ‬ ‫ﻭ ﺑﻪ ﻣﻴﺰﺍﻥ ‪ ۹۰‬ﺩﺭﺟﻪ ﺍﺯ ﻭﻟﺘﺎﮊ ﺟﻠﻮ ﻣـﯽ ﺍﻓﺘـﺪ‪.‬‬

‫‪N‬‬ ‫‪ϕt‬‬

‫ﺑﻨﺎﺑﺮﺍﻳﻦ ﺩﻳﺎﮔﺮﺍﻡ ﺑﺮﺩﺍﺭﯼ ﺟﺮﻳﺎﻧﻬﺎ ﺑﺎﻳﺪ ﺑﻪ ﮔﻮﻧﻪ‬ ‫ﺍﯼ ﺭﺳﻢ ﺷﻮﺩ ﮐﻪ ﺑﻴﻦ ﺟﺮﻳﺎﻥ ﮐـﻞ ‪ It‬ﻭ ﻭﻟﺘـﺎﮊ‬

‫‪It‬‬

‫ﺍﺧﺘﻼﻑ ﻓﺎﺯﯼ ﺑﻮﺟﻮﺩ ﺍﻳـﺪ ﮐـﻪ ﮐﺴـﻴﻨﻮﺱ ﺍﻥ‬

‫‪Q‬‬

‫‪ ۰/۸۵‬ﺑﺎﺷﺪ‪:‬‬

‫‪P‬‬ ‫‪۵۷‬‬

‫‪I1‬‬

‫‪O‬‬ ‫‪ϕ1‬‬

‫‪ON = I 1 cos(ϕ1 ) = (93 .7 )(0.7 ) = 65.5‬‬ ‫‪PN = I 1 sin(ϕ1 ) = (93.7)(0.714 ) = 66.9‬‬ ‫‪QN = ON tan(ϕ t ) = ( 65.6) tan(31.89) = 40 .6‬‬ ‫‪I c = PQ = PN − QN = 66 .9 − 40 .6 = 26.3 A‬‬ ‫‪1‬‬ ‫‪ωC‬‬ ‫‪26 .3‬‬ ‫‪= 66 × 10 − 6 F = 66 µF‬‬ ‫‪100π × 1270‬‬ ‫‪26 .3 / 3‬‬ ‫‪= 22 × 10 − 6 F = 22 µF‬‬ ‫=‪C‬‬ ‫‪100π × 2200‬‬ ‫=‪C‬‬

‫ﺣﻞ ﺑﻪ ﺭﻭﺵ ﻣﺤﺎﺳﺒﻪ ﺗﻮﺍﻥ‪:‬‬

‫= ‪⇒ Xc‬‬ ‫‪‬‬ ‫‪ Υ :‬‬ ‫‪‬‬ ‫‪∆ :‬‬ ‫‪‬‬

‫‪Vp‬‬ ‫‪Xc‬‬

‫= ‪Ic‬‬

‫‪I‬‬ ‫‪C= c‬‬ ‫‪ωV p‬‬

‫‪P1‬‬ ‫‪250000‬‬ ‫=‬ ‫‪= 357142.8vA‬‬ ‫) ‪cos(ϕ1‬‬ ‫‪0.7‬‬

‫= ‪S1‬‬

‫‪Q1 = S1 sin(ϕ1 ) = 357142.8( 0.714) = 255050vAR‬‬ ‫‪−−−−−−−−−−−−−−−−−−−−−−−−−−‬‬ ‫‪P1‬‬ ‫‪250000‬‬ ‫=‬ ‫‪= 29411.7vA‬‬ ‫) ‪cos(ϕ 2‬‬ ‫‪0.85‬‬

‫= ‪S2‬‬

‫‪Q2 = S1 sin(ϕ 2 ) = 29411.7(0.526) = 154705.5vAR‬‬ ‫‪Qc = Q1 − Q2 = 255050 − 154705 = 100344vAR‬‬ ‫‪100344‬‬ ‫‪= 33448‬‬ ‫‪3‬‬ ‫‪Q‬‬ ‫‪33448‬‬ ‫= ) ‪= I c = p (c‬‬ ‫‪= 26.3 A‬‬ ‫)‪V p (c‬‬ ‫‪1270‬‬

‫= )‪Q p (c‬‬

‫ﺩﺭ ﺻﻮﺭﺗﻴﮑﻪ ﺍﺗﺼﺎﻝ ﺳﺘﺎﺭﻩ ﺑﺎﺷﺪ‪.‬‬

‫) ‪I p (c‬‬

‫‪ (۵-۵‬ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﯼ ﻗﺪﺭﺕ ﺩﺭ ﺟﺮﻳﺎﻥ ﻣﺘﻨﺎﻭﺏ‬ ‫ﮐﻨﺘﻮﺭ ﺗﮏ ﻓﺎﺯ‪:‬‬ ‫ﭼﻨﺎﻧﭽﻪ ﺗﻮﺍﻥ ﺑﺮ ﺣﺴﺐ ﻭﺍﺕ ﻭ ﺯﻣﺎﻥ ﺑﺮ ﺣﺴﺐ ﺳﺎﻋﺖ ﺑﺎﺷﺪ ﻣﻘﺪﺍﺭ ﮐﺎﺭ ﻳﺎ ﺍﻧﺮﮊﯼ ﻣﺼﺮﻑ ﺷـﺪﻩ ﺑـﺮ ﺣﺴـﺐ ﻭﺍﺕ‬ ‫ﺳﺎﻋﺖ ﺧﻮﺍﻫﺪ ﺑﻮﺩ‪ .‬ﺑﺮﺍﯼ ﺍﻳﻦ ﻣﻨﻈﻮﺭ ﺍﺯ ﮐﻨﺘﻮﺭ‪ ١‬ﺍﺳﺘﻔﺎﺩﻩ ﻣﯽ ﮔﺮﺩﺩ‪ .‬ﮐﻨﺘﻮﺭ ﺍﺯ ﻳﮏ ﺳﻴﻢ ﭘـﻴﭻ ﺿـﺨﻴﻢ ﺑﺎﺗﻌـﺪﺍﺩ ﺩﻭﺭ‬ ‫ﮐﻢ)‪۵۰-۴۰‬ﺩﻭﺭ( ﮐﻪ ﺑﻪ ﺻﻮﺭﺕ ﺳﺮﯼ ﺩﺭ ﻣﺪﺍﺭ ﻗﺮﺍﺭ ﻣﯽ ﮔﻴﺮﺩ ﻭ ﺑﻪ ﺳﻴﻢ ﭘﻴﭻ ﺟﺮﻳﺎﻥ‪ ٢‬ﻣﻌﺮﻭﻑ ﻣﯽ ﺑﺎﺷﺪ ﻭ ﻳﮏ ﺳﻴﻢ‬

‫‪Watt Hour Meter‬‬ ‫‪Current Coil‬‬

‫‪۵۸‬‬

‫‪1‬‬ ‫‪2‬‬

‫ﭘﻴﭻ ﮐﻪ ﻧﺎﺯﮎ ﺑﻮﺩﻩ ﻭ ﺑﺎ ﺗﻌﺪﺍﺩ ﺩﻭﺭ ﺯﻳﺎﺩ ﮐﻪ ﺑﻪ ﺻﻮﺭﺕ ﻣﻮﺍﺯﯼ ﺩﺭ ﻣﺪﺍﺭ ﻗﺮﺍﺭ ﻣﯽ ﮔﻴﺮﺩ )ﺳﻴﻢ ﭘﻴﭻ ﻭﻟﺘﺎﮊ‪ (١‬ﺗﺸـﮑﻴﻞ‬ ‫ﻣﯽ ﮔﺮﺩﺩ‪ .‬ﺩﺭ ﺑﻴﻦ ﺳﻴﻢ ﭘﻴﭻ ﺟﺮﻳﺎﻥ ﻭ ﻭﻟﺘﺎﮊ ﻳﮏ ﺻﻔﺤﻪ ﺩﻭﺍﺭ ﻗﺮﺍﺭ ﺩﺍﺭﺩ ‪ .‬ﺑﺎ ﻋﺒﻮﺭ ﺟﺮﻳـﺎﻥ ﺍﺯ ﺳـﻴﻢ ﭘـﻴﭻ ﺟﺮﻳـﺎﻧﯽ‬ ‫ﮔﺮﺩﺍﺑﯽ ﺩﺭ ﺻﻔﺤﻪ ﺍﻳﺠﺎﺩ ﻣﯽ ﮔﺮﺩﺩ ﻭ ﻣﻴﺪﺍﻥ ﻣﻐﻨﺎﻃﻴﺴﯽ ﻟﻐﺰﺍﻧﯽ ﺩﺭ ﺁﻥ ﺑﻮﺟﻮﺩ ﺁﻭﺭﺩﻩ ﮐﻪ ﺑﺎﻋـﺚ ﭼـﺮﺧﺶ ﺁﻥ ﻣـﯽ‬ ‫ﺷﻮﺩ‪ .‬ﺑﺮﺍﯼ ﺍﻳﻨﮑﻪ ﺻﻔﺤﻪ ﺩﻭﺭ ﺍﺿﺎﻓﯽ ﻧﮕﻴﺮﺩ ﻳﮏ ﻣﻐﻨﺎﻃﻴﺲ ﻃﺒﻴﻌﯽ ﺩﺭ ﺩﻭ ﻃﺮﻑ ﺁﻥ ﻗﺮﺍﺭ ﻣـﯽ ﺩﻫﻨـﺪ‪ .‬ﺍﺯ ﺍﻧﺠﺎﺋﻴﮑـﻪ‬ ‫ﺟﺮﻳﺎﻥ ﻋﺒﻮﺭﯼ ﺍﺯ ﺳﻴﻢ ﭘﻴﭻ ﺟﺮﻳﺎﻥ ﺟﺮﻳﺎﻥ ﻣﻮﺛﺮ ﮐﻞ ﮐﻪ ﺷﺎﻣﻞ ﺟﺮﻳﺎﻥ ﺍﮐﺘﻴﻮ ﻭ ﺭﺍﮐﺘﻴﻮ ﻣﯽ ﺑﺎﺷﺪ ﻟـﺬﺍ ﺗـﻮﺍﻥ ﺍﻧـﺪﺍﺯﻩ‬ ‫ﮔﻴﺮﯼ ﺷﺪﻩ ﺗﻮﺍﻥ ﻇﺎﻫﺮﯼ ﻣﺪﺍﺭ ﺧﻮﺍﻫﺪ ﺑﻮﺩ‪.‬‬

‫ﺷﮑﻞ‪ :۵-۴‬ﺍﺳﺎﺱ ﮐﺎﺭ ﮐﻨﺘﻮﺭ‬

‫‪ (۵-۵-۱‬ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﯼ ﻗﺪﺭﺕ ﺩﺭ ﺟﺮﻳﺎﻥ ﺳﻪ ﻓﺎﺯ‬ ‫ﺍﻟﻒ( ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﯼ ﺟﺮﻳﺎﻥ ﭼﻬﺎﺭ ﺳﻴﻤﻪ ) ﺳﻪ ﻭﺍﺕ ﻣﺘﺮﯼ (‪ ) :‬ﺍﺗﺼﺎﻝ ﺳـﺘﺎﺭﻩ ‪ ( Y connection‬ﺩﺭ ﺍﻳـﻦ ﺣﺎﻟـﺖ‬ ‫ﻣﺪﺍﺭ ﺩﺍﺭﺍﯼ ﺳﻪ ﺧﻂ ﻓﺎﺯ ﻭ ﻳﮏ ﺧﻂ ﻧﻮﻝ ﻣﯽ ﺑﺎﺷﺪ‪ .‬ﻟﺬﺍ ﺑﺮﺍﯼ ﻫﺮ ﻓﺎﺯ ﻳﮏ ﻭﺍﺕ ﻣﺘﺮ ﻣﻮﺭﺩ ﺍﺳﺘﻔﺎﺩﻩ ﻗﺮﺍﺭ ﻣـﯽ ﮔﻴـﺮﺩ‪.‬‬ ‫ﻫﺮ ﻭﺍﺕ ﻣﺘﺮ ﻣﺜﻞ ﮐﻨﺘﻮﺭ ﺟﺮﻳﺎﻥ ﺗﮏ ﻓﺎﺯ ﺩﺍﺭﺍﯼ ﻳﮏ ﺳﻴﻢ ﭘﻴﭻ ﺟﺮﻳﺎﻥ ﻣﯽ ﺑﺎﺷﺪ ﮐﻪ ﺩﺭ ﻣﺴﻴﺮ ﺧﻂ ﻓﺎﺯ ﺑﻄـﻮﺭ ﺳـﺮﯼ‬ ‫ﺑﺴﺘﻪ ﻣﯽ ﺷﻮﺩ ﻭ ﻳﮏ ﺳﻴﻢ ﭘﻴﭻ ﻭﻟﺘﺎﮊﮐﻪ ﺑﻴﻦ ﻫﺮ ﻳﮏ ﺍﺯ ﻓﺎﺯ ﻫﺎ ﻭ ﻧﻮﻝ ﺑﻪ ﻃﻮﺭ ﻣﻮﺍﺯﯼ ﺑﺴﺘﻪ ﻣـﯽ ﺷـﻮﺩ‪ .‬ﺍﮔـﺮ ﻣـﺪﺍﺭ‬ ‫ﮐﺎﻣﻼ" ﻣﺘﻌﺎﺩﻝ ﺑﺎﺷﺪ ﻫﺮ ﺳﻪ ﻭﺍﺕ ﻣﺘﺮ ﻳﮏ ﻣﻘﺪﺍﺭ ﺭﺍ ﻧﺸﺎﻥ ﺧﻮﺍﻫﺪ ﺩﺍﺩ‪ .‬ﻟﺬﺍ ﺑﺎ ﻳﮏ ﻭﺍﺕ ﻣﺘﺮ ﻣﯽ ﺗﻮﺍﻥ ﺩﺭ ﻳـﮏ ﺧـﻂ‬ ‫ﺍﺳﺘﻔﺎﺩﻩ ﻧﻤﻮﺩ ﻭ ﻣﻘﺪﺍﺭ ﻗﺮﺍﺋﺖ ﺷﺪﻩ ﺩﺭ ﺁﻥ ﺭﺍ ﺩﺭ ﻋﺪﺩ ‪ ۳‬ﺿﺮﺏ ﻧﻤﻮﺩ‪ .‬ﺩﺭ ﻫﺮ ﺣﺎﻝ ﻗﺪﺭﺕ ﮐﻞ ﺣﺎﺻﻞ ﺟﻤﻊ ﻗﺮﺍﺋﺖ‬ ‫ﺳﻪ ﮐﻨﺘﻮﺭ ﺧﻮﺍﻫﺪ ﺑﻮﺩ‪.‬‬

‫‪Phase Coil‬‬

‫‪۵۹‬‬

‫‪1‬‬

‫‪R‬‬ ‫‪W1‬‬ ‫‪S‬‬

‫‪W2‬‬

‫‪T‬‬

‫‪W3‬‬

‫‪N‬‬ ‫ﺍﮔﺮ ﺟﺮﻳﺎﻥ ﺭﺍ ﺩﺭ ﻳﮏ ﻣﺪﺍﺭ ﺳﻪ ﻓﺎﺯ ﺍﺗﺼﺎﻝ ﺳﺘﺎﺭﻩ ﺩﺭ ﻧﻈﺮ ﺑﮕﻴﺮﻳﻢ ﻭ ﺑﺎﺭ ﺭﻭﯼ ﻫﺮ ﺧﻂ ﺑﻪ ﺻﻮﺭﺕ ‪ L 1‬ﻭ ‪ L ۲‬ﻭ ‪L ۳‬‬ ‫ﻣﺘﻌﺎﺩﻝ ﻧﻴﺰ ﺑﺎﺷﺪ ﺟﺮﻳﺎﻧﯽ ﮐﻪ ﻫﺮ ﻟﺤﻈﻪ ﺍﺯ ﺧﻂ ﻧﻮﻝ ﻋﺒﻮﺭ ﮐﻨﺪ ﻣﺠﻤﻮﻉ ‪ L 1‬ﻭ ‪ L ۲‬ﻭ ‪ L ۳‬ﻣﯽ ﺑﺎﺷﺪ‪.‬‬ ‫ﺩﺭ ﺟﺮﻳﺎﻥ ﺳﻪ ﻓﺎﺯ ﺩﺍﺭﻳﻢ‪:‬‬

‫‪L1 → I1 (t) = Im sin( θ ) +‬‬ ‫‪L2 → I2 (t) = Im sin (θ - 120°) +‬‬ ‫‪L3 → I3 (t) = Im sin (θ - 240) +‬‬ ‫) )‪LN → Im ( sin (θ) + sin (θ - 120°) + sin (θ - 240‬‬ ‫ﻫﻤﺎﻧﻄﻮﺭ ﮐﻪ ﻣﻼﺣﻈﻪ ﻣﯽ ﺷﻮﺩ ﻣﺠﻤﻮﻉ ﺟﺮﻳﺎﻥ ﺩﺭ ﻫﺮ ﻟﺤﻈﻪ ﺑﺮﺍﺑﺮ ﺻﻔﺮ ﻣﯽ ﺑﺎﺷﺪ ﻟﺬﺍ ﭼﻮﻥ ﺧﻂ ﻧﻮﻝ ﻣﺤﻞ ﺍﺗﺼﺎﻝ‬ ‫ﺳﻪ ﻓﺎﺯ ﺍﺳﺖ ﺟﺮﻳﺎﻥ ﻫﻤﻴﺸﻪ ﺩﺭ ﺍﻳﻦ ﺧﻂ ﺍﮔﺮ ﻣﺘﻌﺎﻝ ﺑﺎﺷﺪ ﺻﻔﺮ ﺧﻮﺍﻫﺪﺑﻮﺩ ﻳﻌﻨﯽ‪:‬‬

‫‪sin θ + sin (θ - 120°) + sin (θ - 240) = 0‬‬ ‫ﺭﺍﺑﻄﻪ ﻓﻮﻕ ﺑﺮﺍﯼ ﻫﺮ ﺯﺍﻭﻳﻪ ﺍﯼ ﺻﺪﻕ ﻣﯽ ﮐﻨﺪ‪:‬‬ ‫ﻟﺬﺍ ﺩﺭ ﺟﺮﻳﺎﻥ ﺳﻪ ﻓﺎﺯﺩﺭ ﻫﺮ ﻟﺤﻈﻪ ﻳﮏ ﻳﺎ ﺩﻭ ﻓﺎﺯ ﺟﺮﻳﺎﻥ ﺭﺍ ﻭﺍﺭﺩ ﻭ ﻳﮏ ﻳﺎ ﺩﻭﻓﺎﺯ ﺩﻳﮕﺮ ﺑﻪ ﻋﻨﻮﺍﻥ ﻧﻮﻝ ﻋﻤﻞ ﻣﯽ‬ ‫ﮐﻨﺪ‪ .‬ﻟﺬﺍ ﺍﮔﺮ ﺑﺎﺭ ﻣﺘﻌﺎﻝ ﺑﺎﺷﺪ ﻣﯽ ﺗﻮﺍﻥ ﻓﻘﻂ ﺍﺯ ﻳﮏ ﮐﻨﺘﻮﺭ ﺑﺮ ﺭﻭﯼ ﻳﮏ ﻓﺎﺯ ﺍﺳﺘﻔﺎﺩﻩ ﻧﻤﻮﺩ‪.‬‬ ‫ﺍﮔﺮ ﺑﺎﺭ ﻣﺘﻌﺎﺩﻝ ﻧﺒﺎﺷﺪ ﻫﻤﺎﻧﻄﻮﺭ ﮐﻪ ﺩﺭ ﺑﺎﻻ ﺍﺷﺎﺭﻩ ﺷﺪ ﺑﺎﻳﺴﺘﯽ ﺑﺮﺍﯼ ﻫﺮ ﻓﺎﺯ ﺍﺯ ﻳﮏ ﻭﺍﺕ ﻣﺘﺮ ﻣﺠﺰﺍ ﺍﺳﺘﻔﺎﺩﻩ ﻧﻤﻮﺩ‪.‬‬

‫ﺏ( ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﯼ ﻗﺪﺭﺕ ﺩﺭ ﺍﺗﺼﺎﻝ ﻣﺜﻠﺚ ﺑﺎ‬

‫‪R‬‬

‫ﺳﻪ ﻭﺍﺕ ﻣﺘﺮ‪:‬‬

‫‪W1‬‬

‫ﺩﺭ ﺍﻳﻦ ﺣﺎﻟﺖ ﺳﻴﻢ ﻧﻮﻝ ﻭﺟﻮﺩ ﻧـﺪﺍﺭﺩ‪ .‬ﻟـﺬﺍ‬

‫‪W2‬‬

‫ﺑﺎﻳﺴﺘﯽ ﺁﻥ ﺭﺍ ﺑﻪ ﺻﻮﺭﺕ ﻣﺼـﻨﻮﻋﯽ ﺍﻳﺠـﺎﺩ‬

‫‪W3‬‬

‫ﮐﺮﺩ‪ .‬ﭼﻮﻥ ﻣﻌﻤﻮﻻﹰ ﺑـﻮﺑﻴﻦ ﻭﺍﺗﻤﺘﺮﻫـﺎ ﺩﺍﺭﺍﯼ‬ ‫ﻳﮏ ﻣﻘﺎﻭﻣﺖ ﺯﻳﺎﺩ ﺑـﻪ ﺻـﻮﺭﺕ ﺳـﺮﯼ ﻣـﯽ‬ ‫ﺑﺎﺷﺪ ﻧﻘﻄﻪ ﻧﻮﻝ ﺭﺍ ﻣﯽ ﺗﻮﺍﻥ ﺑﺪﻭﻥ ﻣﻘﺎﻭﻣﺘﻬﺎﯼ‬ ‫ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﺷﺪﻩ ﺩﺭ ﺷﮑﻞ ﻣﻘﺎﺑﻞ ﻧﻴـﺰ ﺍﻳﺠـﺎﺩ‬ ‫ﻧﻤﻮﺩ‪.‬‬

‫‪N‬‬

‫‪۶۰‬‬

‫‪S‬‬ ‫‪T‬‬

‫ﺝ( ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﯼ ﻗﺪﺭﺕ ﺩﺭ ﺟﺮﻳﺎﻥ ﻣﺘﻨﺎﻭﺏ ﺳﻪ ﻓﺎﺯ ﺑﺎ ﺩﻭ ﻭﺍﺕ ﻣﺘﺮ‪:‬‬ ‫ﭼﻨﺎﻧﭽﻪ ﻧﻘﻄﻪ ‪ N‬ﺩﺭ ﺷﮑﻞ ﻗﺒﻞ ﺑﺮ ﺭﻭﯼ ﻓﺎﺯ ﺳﻮﻡ ﺍﻧﺘﺨﺎﺏ ﮔﺮﺩﺩ ﻭﺍﺕ ﻣﺘﺮ ﺷﻤﺎﺭﻩ ‪ ۳‬ﻋﺪﺩ ﺻﻔﺮ ﺭﺍﻧﺸﺎﻥ ﺧﻮﺍﻫـﺪ ﺩﺍﺩ‪.‬‬ ‫ﻟﺬﺍ ﻣﯽ ﺗﻮﺍﻥ ﺍﻥ ﺭﺍ ﺣﺬﻑ ﻧﻤﻮﺩ‪ .‬ﺩﺭ ﺍﻳﻦ ﺻﻮﺭﺕ ﻣﺠﻤﻮﻉ ﻗﺪﺭﺕ ﻗﺮﺍﺋﺖ ﺷﺪﻩ ﺩﺭ ﺩﻭ ﻭﺍﺕ ﻣﺘﺮ ﻗﺪﺭ ﺕ ﮐـﻞ ﻣـﺪﺍﺭ‬ ‫ﺧﻮﺍﻫﺪ ﺑﻮﺩ ﻭ ﻣﯽ ﺗﻮﺍﻥ ﻧﺸﺎﻥ ﺩﺍﺩ ﮐﻪ ﻣﻘﺎﺩﻳﺮ ﻗﺮﺍﺋﺖ ﺷﺪﻩ ﺩﺭ ﻭﺍﺕ ﻣﺘﺮﻫﺎ ﺑﻪ ﭘﺘﺎﻧﺴﻴﻞ ﻧﻘﻄﻪ ﻣﺸﺘﺮﮎ ﺑﺴﺘﮕﯽ ﻧﺪﺍﺭﺩ‪:‬‬ ‫‪R‬‬

‫‪Pt = (V1 − V N ) I 1 + (V2 − V N ) I 2 + (V3 − V N ) I 3‬‬

‫‪W1‬‬

‫) ‪Pt = V1 I 1 + V2 I 2 + V3 I 3 + V N ( I 1 + I 2 + I 3‬‬

‫‪W2‬‬

‫‪S‬‬

‫) ‪I1 + I 2 + I 3 = 0 ⇒ I 3 = − ( I 1 + I 2‬‬ ‫‪Pt = V1 I 1 + V2 I 2 + V3 I 3‬‬ ‫) ‪Pt = V1 I 1 + V2 I 2 − V3 ( I 1 + I 2‬‬

‫‪T‬‬

‫‪Pt = V1 I 1 + V2 I 2 − V3 I 1 − V3 I 2‬‬ ‫) ‪Pt = I1 (V1 − V3 ) + I 2 (V2 − V3‬‬ ‫‪Pt = W1 + W2‬‬ ‫ﻗﺪﺭﺗﻬﺎﯼ ﻗﺮﺍﺋﺖ ﺷﺪﻩ ﺩﺭ ﻭﺍﺕ ﻣﺘﺮﻫﺎﯼ ‪ ۱‬ﻭ‪ ۲‬ﺣﺘﯽ ﺩﺭ ﻣﺪﺍﺭﻫﺎﯼ ﮐﺎﻣﻼﹰ ﻣﺘﻌﺎﺩﻝ ﻧﻴﺰ ﺑﺮﺍﺑـﺮ ﻧﻴﺴـﺘﻨﺪ ﻭ ﺗﻔـﺎﻭﺕ ﺁﻧﻬـﺎ‬ ‫ﺑﺴﺘﮕﯽ ﺑﻪ ﺿﺮﻳﺐ ﺗﻮﺍﻥ ﺩﺍﺭﺩ ﮐﻪ ﻣﯽ ﺗﻮﺍﻥ ﺑﻪ ﺻﻮﺭﺕ ﺯﻳﺮ ﻧﺸﺎﻥ ﺩﺍﺩ‪.‬‬

‫) ‪W1 = (V1 − V3 ) ⋅ I 1 cos(30 − ϕ‬‬ ‫) ‪W2 = (V2 − V3 ) ⋅ I 2 cos(30 + ϕ‬‬ ‫) ‪W1 = 3V p ⋅ I p cos(30 − ϕ‬‬ ‫) ‪W2 = 3V p ⋅ I p cos(30 + ϕ‬‬ ‫]) ‪W + W2 = 3V p ⋅ I p [cos(30 − ϕ ) + cos(30 + ϕ‬‬ ‫]) ‪W + W2 = 3V p ⋅ I p [cos(30) cos(ϕ ) + sin(30) sin(ϕ ) cos(30) cos(ϕ ) − sin(30) sin(ϕ‬‬

‫) ‪W + W2 = 3V p ⋅ I p 3 cos(ϕ‬‬ ‫‪V1‬‬

‫‪I1‬‬ ‫)‪(V1-V3‬‬

‫) ‪(Ι‬‬

‫‪ϕ‬‬

‫] ) ‪W1 − W = 3V p ⋅ I p [cos(30 − ϕ ) − cos(30 + ϕ‬‬

‫‪-V3‬‬

‫) ‪(ΙΙ‬‬ ‫‪ϕ‬‬

‫‪V2‬‬

‫‪ϕ‬‬

‫) ‪W + W2 = 3V p ⋅ I p cos(ϕ‬‬

‫‪I3‬‬ ‫‪V3‬‬

‫‪I2‬‬

‫) ‪W1 − W = 3V p ⋅ I p sin(ϕ‬‬ ‫‪W1 − W2‬‬ ‫‪3V p I p‬‬ ‫) ‪sin(ϕ‬‬ ‫=‬ ‫= ) ‪tan(ϕ‬‬ ‫‪cos(ϕ ) W1 + W2‬‬ ‫‪3V p I p‬‬ ‫) ‪3 (W1 − W2‬‬ ‫‪W1 + W2‬‬

‫‪۶۱‬‬

‫= ) ‪tan(ϕ‬‬

‫ﻣﺜﺎﻝ‪:‬‬ ‫ﺑﺮﺍﯼ ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﯼ ﻗﺪﺭﺕ ﻳﮏ ﻣﻮﺗﻮﺭ ﺳﻪ ﻓﺎﺯ ﺑﻪ ﺭﻭﺵ ﺩﻭ ﻭﺍﺗﻤﺘﺮﯼ ﺍﻋﺪﺍﺩﺯﻳﺮ ﻗﺮﺍﺋﺖ ﮔﺮﺩﻳﺪ‪:‬‬

‫‪W1=9.6 kw‬‬ ‫‪W2=7.5 kw‬‬ ‫ﻣﻄﻠﻮﺑﺴﺖ ﺿﺮﻳﺐ ﻗﺪﺭﺕ ﻣﻮﺗﻮﺭ؟‬ ‫‪ 3 (W1 + W2 ) ‬‬ ‫‪ 3 (9.6 − 7.5) ‬‬ ‫‪−1‬‬ ‫‪ = tan ‬‬ ‫‪ = 12.0‬‬ ‫‪9‬‬ ‫‪.‬‬ ‫‪6‬‬ ‫‪7‬‬ ‫‪.‬‬ ‫‪5‬‬ ‫‪+‬‬ ‫‪ W1 + W2 ‬‬ ‫‪‬‬ ‫‪‬‬

‫‪ϕ = tan −1 ‬‬

‫‪cos(ϕ ) = 0.978‬‬ ‫ﻣﺜﺎﻝ‪:‬‬ ‫ﺑﺎﺭ ﻳﮏ ﮊﻧﺮﺍﺗﻮﺭ ﺳﻪ ﻓﺎﺯ ‪ ۱۱۵‬ﻭﻟﺘﯽ )ﺧﻂ‪-‬ﺧﻂ( ﻣﻄﺎﺑﻖ ﺷﮑﻞ ﺯﻳﺮ ﺷﺎﻣﻞ ﺳﻪ ﺩﺳﺘﻪ ﻻﻣﭗ ﺭﻭﺷﻨﺎﻳﯽ ﻫﺮ ﺩﺳﺘﻪ ﺷﺎﻣﻞ‬ ‫‪ ۲۰‬ﻻﻣﭗ ‪ ۶۰‬ﻭﺍﺗﯽ ﻭ ﻳﮏ ﻣﻮﺗﻮﺭ ﺑﺎ ﺿﺮﻳﺐ ﻗﺪﺭﺕ ‪ ۰/۸۲‬ﻭ ﺗـﻮﺍﻥ ﺍﺳـﻤﯽ ‪ ۹‬ﺍﺳـﺐ ﺑﺨـﺎﺭ ﻭ ﺭﺍﻧـﺪﻣﺎﻥ ‪ ۰/۸۷‬ﻣـﯽ‬ ‫ﺑﺎﺷﺪ‪.‬ﺿﺮﻳﺐ ﻗﺪﺭﺕ‪ ،‬ﺟﺮﻳﺎﻥ ﺧﻂ ﺩﺭ ﮐﻞ ﻣﺪﺍﺭ ﻭ ﺗﻮﺍﻥ ﻫﺮﻳﮏ ﺍﺯ ﻭﺍﺗﻤﺘﺮﻫﺎ ﺭﺍ ﻣﺤﺎﺳﺒﻪ ﮐﻨﻴﺪ‪:‬‬

‫‪It‬‬

‫‪Im‬‬ ‫‪Iℓ‬‬ ‫‪M‬‬

‫‪115v‬‬ ‫‪G‬‬

‫ﺣﻞ‪:‬‬ ‫ﻧﺤﻮﻩ ﺍﺗﺼﺎﻝ ﺗﺎﺛﻴﺮﯼ ﺩﺭ ﺣﻞ ﻣﺴﺌﻠﻪ ﻧﺪﺍﺭﺩ ﻟﺬﺍ ﺑﺎ ﻓﺮﺽ ﺍﺗﺼﺎﻝ ﻣﺜﻠﺚ ﻣﺴﺌﻠﻪ ﺭﺍ ﺩﻧﺒﺎﻝ ﻣﯽ ﮐﻨﻴﻢ‪:‬‬

‫‪۶۲‬‬

(9)(746) = 7614watt 0.87 Pm = 3I LmV L cos(φm ) Pm =

7614

I Lm =

= 46.6 A 3 (115)(0.82) Pl = (3)(20)(100) = 6000watt Il =

Pl 3VLl

=

6000 3 (115)

= 30.1A

P = Pm + Pl = 7614 + 6000 = 13614watt ≈ 13.6kwatt I t = (ON ) 2 + ( PN ) 2 = ( I l + I m cos(

m

)) 2 + ( I m sin(

I t = (30.1 + 46.6 × 0.82) 2 + (46.6 × 0.572) 2 I t = 73.3 A ON 68.3 = ⇒ OP 73.3 W1 = V L I L cos(30 + φt ) cos(φt ) =

O t

= 21.28 o

m

)) 2

Iℓ ϕm

N V

ϕt It

W1 = (115)(73.3) cos(30 + 21.28) = 5276watt

Im

W2 = VL I L cos(30 − φt ) W2 = (115)(73.3) cos(30 − 21.28) = 8331watt P = W1 + W2 = 5276 + 8331 = 13614 ≈ 13.6kwatt

۶۳

P