SORU:1 a) 𝑼𝒌 = √𝟑 ∗ 𝑬𝟎 ∗ 𝒓 ∗ 𝒎 ∗ 𝜹 ∗ (𝟏 + 𝜹 = 𝟎. 𝟑𝟖𝟔 ∗ 𝟕𝟎𝟎 = 𝟎. 𝟗𝟏𝟐𝟖 𝟐𝟕𝟑 + 𝟐𝟑 𝑼𝒌 = √𝟑 ∗ 𝟐𝟏. 𝟐 ∗ 𝟐. 𝟏 ∗ 𝟎. 𝟖𝟕 ∗ 𝟎. 𝟗𝟏𝟐𝟖 ∗ (𝟏 + 𝑼𝟎 = 𝟎. 𝟑𝟎𝟏 𝒂 ) ∗ 𝒍𝒏 𝒓 √𝒓 ∗ 𝜹 𝟎. 𝟑𝟎𝟏 𝟏𝟏𝟎𝟎 ) ∗ 𝒍𝒏 = 𝟒𝟔𝟔. 𝟕𝟔𝟓𝒌𝑽 𝟐. 𝟏 √𝟐. 𝟏 ∗ 𝟎. 𝟗𝟏𝟐𝟖 𝑼𝒌 𝟒𝟔𝟔. 𝟕𝟔𝟓 = = 𝟒𝟐𝟎. 𝟎𝟑𝒌𝑽 𝟎. 𝟑𝟎𝟏 𝟎. 𝟑𝟎𝟏 𝜹 ∗ (𝟏 + ) 𝟎. 𝟗𝟏𝟐𝟖 ∗ (𝟏 + ) √𝟐. 𝟏 ∗ 𝟎. 𝟗𝟏𝟐𝟖 √𝒓 ∗ 𝜹 b) 𝑼𝟎 > 𝑼 𝒐𝒍𝒅𝒖ğ𝒖 𝒊ç𝒊𝒏 𝒉𝒆𝒓𝒉𝒂𝒏𝒈𝒊 𝒃𝒊𝒓 𝒌𝒐𝒓𝒐𝒏𝒂 𝒌𝒂𝒚𝒃𝚤 𝒐𝒍𝒎𝒂𝒛 . 𝒌𝒐𝒓𝒐𝒏𝒂 𝒌𝒂𝒚𝚤𝒑𝒍𝒂𝒓𝚤 𝒊ş𝒍𝒆𝒕𝒎𝒆 𝒈𝒆𝒓𝒊𝒍𝒊𝒎𝒊𝒏𝒊𝒏 𝒌𝒓𝒊𝒕𝒊𝒌 𝒌𝒐𝒓𝒐𝒏𝒂 𝒈𝒆𝒓𝒊𝒍𝒊𝒎𝒊 𝒂ş𝒅𝚤ğ𝚤 𝒛𝒂𝒎𝒂𝒏 𝒈𝒆𝒓ç𝒆𝒌𝒍𝒆ş𝒎𝒆𝒚𝒆 𝒃𝒂ş𝒍𝒂𝒓 𝑹𝒌 = 𝜹 𝟏 𝒂 ∗ ∗ √ ∗ 𝟏𝟎𝟓 [𝒌Ω / 𝒌𝒎 − 𝒇𝒂𝒛] 𝟐𝟒𝟏 𝒇 + 𝟐𝟓 𝒓 c) 𝑹𝒌 = 𝟎. 𝟗𝟏𝟐𝟖 𝟏 𝟏𝟏𝟎𝟎 ∗ ∗√ ∗ 𝟏𝟎𝟓 = 𝟏𝟏𝟓. 𝟓𝟖[𝒌Ω / 𝒌𝒎 − 𝒇𝒂𝒛] 𝟐𝟒𝟏 𝟓𝟎 + 𝟐𝟓 𝟐. 𝟏 𝒑𝒌 = 𝒍 ∗ (𝑼 − 𝑼𝟎 )𝟐 (𝟒𝟒𝟎 − 𝟒𝟐𝟎)𝟐 = 𝟐𝟎𝟎 ∗ = 𝟎. 𝟔𝟗𝟐𝟏𝟔𝟏 𝑴𝑾 𝑹𝒌 𝟏𝟏𝟓. 𝟓𝟖 SORU:2 a) 𝜀𝑟 𝑡𝑎𝑛 𝑎1 𝜀𝑟1 = →→→ 𝑡𝑎𝑛𝑎2 = 1 ∗ 𝑡𝑎𝑛 𝑎1 𝑡𝑎𝑛 𝑎2 𝜀𝑟2 𝜀𝑟2 5 ∗ 𝑡𝑎𝑛37° = 𝑡𝑎𝑛𝑎2 = 0.538 →→→ 𝒂𝟐 = 𝟐𝟖. 𝟑° , 𝑎1 = 37° 7 𝑎1 + 𝛽1 = 90 →→→ 𝜷𝟏 = 𝟓𝟑° 𝑎2 + 𝛽2 = 90 →→→ 𝜷𝟐 = 𝟔𝟏. 𝟕° b) 𝐸𝑡1 = 𝐸𝑡2 |𝐸1 | ∗ 𝑠𝑖𝑛𝑎1 = |𝐸2 | ∗ 𝑠𝑖𝑛𝑎2 260 ∗ 𝑠𝑖𝑛37 = |𝐸2 | ∗ 𝑠𝑖𝑛28.3 |𝐸2 | = 330𝑘𝑉/𝑐𝑚 c) |𝐸𝑛2 | = |𝐸2 | ∗ 𝑐𝑜𝑠𝑎2 > 160𝑘𝑉 𝑖𝑠𝑒 𝑑𝑒𝑙𝑖𝑛𝑚𝑒 𝑔𝑒𝑟ç𝑒𝑘𝑙𝑒ş𝑖𝑟 |𝐸𝑛2 | = 290.6𝑘𝑉 > 160𝑘𝑉 𝑜𝑙𝑑𝑢𝑔𝑢 𝑖ç𝑖𝑛 𝑑𝑒𝑙𝑖𝑛𝑖𝑟 𝑒𝑙𝑒𝑘𝑡𝑟𝑖𝑘 𝑎𝑙𝑎𝑛𝚤𝑛 𝑖𝑘𝑖 𝑏𝑖𝑙𝑒ş𝑒𝑛𝑖 𝑣𝑎𝑟𝑑𝚤𝑟; 1: 𝑡𝑒ğ𝑒𝑡𝑠𝑒𝑙 𝑦𝑎𝑛𝑖 𝑎𝑡𝑙𝑎𝑚𝑎𝑦𝑎 𝑧𝑜𝑟𝑙𝑎𝑦𝑎𝑛 𝑏𝑖𝑙𝑒ş𝑒𝑛 , 2: 𝒏𝒐𝒓𝒎𝒂𝒍 𝒃𝒊𝒍𝒆ş𝒆𝒏 𝒚𝒂𝒏𝒊 𝒅𝒆𝒍𝒊𝒏𝒎𝒆𝒚𝒆 𝒔𝒆𝒃𝒆𝒃 𝒐𝒍𝒂𝒏 𝒃𝒊𝒍𝒆ş𝒆𝒏 , 𝑏𝑢𝑟𝑎𝑑𝑎 𝑑𝑒𝑙𝑖𝑛𝑚𝑒 𝑧𝑜𝑟𝑙𝑎𝑦𝑎𝑛 𝑏𝑖𝑙𝑒ş𝑒𝑛 𝑦𝑎𝑛𝑖 𝑐𝑜𝑠 𝑏𝑖𝑙𝑒ş𝑒𝑛𝑖 𝑒𝑙𝑒 𝑎𝑙𝚤𝑛𝚤𝑟. 𝑬𝟐 𝒏𝒐𝒓𝒎𝒂𝒍 𝒂𝒍𝒂𝒏 𝒃𝒊𝒍𝒆ş𝒆𝒏𝒊 , 𝒅𝒆𝒍𝒊𝒏𝒎𝒆 𝒌𝒓𝒊𝒕𝒊𝒌 𝒂𝒍𝒂𝒏𝚤𝒏𝒅𝒂𝒏 𝒃ü𝒚ü𝒌 𝒐𝒍𝒅𝒖ğ𝒖 𝒊ç𝒊𝒏 𝒔𝒊𝒔𝒕𝒆𝒎 𝒅𝒆𝒍𝒊𝒏𝒊𝒓 SORU 3 a) 𝑼 𝑼. 𝒓𝟐 𝑬𝒎𝒂𝒙 = 𝒓 = 𝟏 𝒓𝟏 . 𝒓𝟐 − 𝒓𝟐𝟏 𝒓𝟐 (𝒓𝟐 − 𝒓𝟏 ) 𝒅(𝑬𝐦𝐚𝐱 ) 𝒓𝟐 − 𝟐. 𝒓𝟏 = −𝑼. 𝒓𝟐 =𝟎 𝟐 𝒅𝒓𝟏 (𝒓 . 𝒓 − 𝒓𝟐 ) 𝟏 𝟐 𝟏 𝒓𝟐 − 𝟐. 𝒓𝟏 = 𝟎 𝑝𝑑 = 𝒓𝟐 = 𝟒𝟎𝒄𝒎 𝑖𝑠𝑒 𝑟2 =2 𝑟1 𝒓𝟏 = 𝟐𝟎𝒄𝒎 𝑜𝑙𝑚𝑎𝑙𝚤𝑑𝚤𝑟 b) 𝑈𝑑 = 𝐸𝑑 . 𝑼𝒅 𝒎𝒂𝒙 = 𝟑𝟎. 𝑟1 (𝑟 − 𝑟1 ) 𝑟2 2 𝟐𝟎 (𝟒𝟎 − 𝟐𝟎) = 𝟑𝟎𝟎𝒌𝑽 𝟒𝟎 c) 𝑠𝑖𝑠𝑡𝑒𝑚𝑖 𝑎ş𝑎ğ𝚤𝑑𝑎𝑘𝑖 𝑔𝑟𝑎𝑓𝑖𝑘 ü𝑧𝑒𝑟𝑖𝑛𝑑𝑒𝑛 𝑖𝑛𝑐𝑒𝑙𝑒𝑦𝑒𝑏𝑖𝑙𝑖𝑟 𝑼𝒅 𝒎𝒂𝒙 = 𝟑𝟎𝟎𝒌𝑽 , 𝑼 = 𝟐𝟕𝟎𝒌𝑽 𝑖𝑠𝑒 𝑎ş𝑎ğ𝚤𝑑𝑎𝑘𝑖 𝑠𝑜𝑛𝑢ç𝑙𝑎𝑟 𝑒𝑙𝑑𝑒 𝑒𝑑𝑖𝑙𝑖𝑟 𝑼 > 𝑼𝒅 Ö𝑵 𝑩𝑶Ş𝑨𝑳𝑴𝑨 𝑩Ö𝑳𝑮𝑬𝑺İ 𝑼 = 𝑼𝒅 𝑩𝑶Ş𝑨𝑳𝑴𝑨 𝒀𝑶𝑲𝑻𝑼𝑹 𝑼 < 𝑼𝒅 𝑻𝑨𝑴 𝑫𝑬𝑳İ𝑵𝑴𝑬 𝑩Ö𝑳𝑮𝑬𝑺İ 𝟏𝑵𝑶′𝑳𝑼 𝑩Ö𝑳𝑮𝑬 𝟐 𝑵𝑶′𝑳𝑼 𝑩Ö𝑳𝑮𝑬 𝟑 𝑵𝑶′𝑳𝑼 𝑩Ö𝑳𝑮𝑬 Soru:4 a) 𝑆 = π ∗ r12 = 150𝑚𝑚2 𝑟1 = 0.691𝑐𝑚 𝑟2 𝑝𝑑 = = 2.718 = 𝑒 𝑟1 𝒓𝟏 = 𝟎. 𝟔𝟗𝟏𝒄𝒎 𝒓𝟐 = 𝟏. 𝟖𝟕𝟖𝒄𝒎 𝐸𝑑 = 𝑼𝒅 = 𝑬𝒅 . 𝐥𝐧 ( 𝑈𝑑 𝑟 ln ( 2 ) . 𝑟1 𝑟1 𝒓𝟐 ) . 𝒓𝟏 = 𝟏𝟖𝟎 ∗ 𝟏 ∗ 𝟎. 𝟔𝟗𝟏 = 𝟏𝟐𝟒. 𝟑𝟖𝒌𝑽 𝒓𝟏 𝟐 ∗ 𝝅 ∗ 𝟐. 𝟑 ∗ 𝟖. 𝟖𝟓𝟒 ∗ 𝟏𝟎−𝟏𝟐 𝒑𝑭 𝑪= = 𝟏𝟐𝟕. 𝟗𝟓 𝟏 𝒎 𝑿𝒄 = 𝟏 = 𝟏𝟓. 𝟔𝟑𝟏𝟏𝟎𝟔 𝑴Ω 𝝎𝒄 b) 𝐸𝑑 = 𝑈𝑑 𝑟 ln ( 2 ) . 𝑟1 𝑟1 𝑼𝒅 = 𝑬𝒅 . 𝐥𝐧 ( 𝑝𝑑 = 3.3 = 𝑟2 𝑟1 𝑜𝑙𝑚𝑎𝑠𝚤 𝑑𝑢𝑟𝑢𝑚𝑢 𝑖ç𝑖𝑛 𝑈𝑑 ; 𝒓𝟐 ) . 𝒓𝟏 = 𝟏𝟖𝟎 ∗ 𝐥𝐧(𝟑. 𝟑) ∗ 𝟎. 𝟔𝟗𝟏 = 𝟏𝟒𝟖. 𝟓𝒌𝑽 𝒓𝟏 Saygılarımla….