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….