9 Admitancia y puentes AC

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ADMITANCIA
−𝑗𝑋𝐶
𝑅
ത𝑌 =
1
ത𝑅
−
1
𝑗𝑋𝐶
ത𝑌 =
1
ത𝑅
+ jω𝐶
ത𝑌 =
1
ത𝑅
+
1
𝑗𝑋𝐿
ത𝑌 =
1
ത𝑅
− 𝑗
1
𝜔𝐿
𝑅
−𝑗𝑋𝐶
+
-
ത𝑌1 ത𝑌2 ത𝑌3
𝑌= ത𝑌1+ ത𝑌2+ ത𝑌3
ത𝑌 =
1
ҧ𝑍
PUENTES EN AC
ҧ𝑍1
ҧ𝑍2
ҧ𝑍3
ҧ𝑍4
ത𝑉
ҧ𝐼
𝑬𝑿𝑷𝑹𝑬𝑺𝑰Ó𝑵 𝑬𝑵 𝑬𝑸𝑼𝑰𝑳𝑰𝑩𝑹𝑰𝑶
ҧ𝑍1. ҧ𝑍4 = ҧ𝑍2. ҧ𝑍3
𝑍1උ𝜃1𝑍1උ𝜃4=𝑍2උ𝜃2 ҧ𝑍3උ𝜃3
𝑍1. 𝑍4 = 𝑍2. 𝑍3
𝜃º1+𝜃º4=𝜃º2+𝜃º3
PUENTE MAXWELL
𝐶1
𝑅1
𝑅2
𝑅3
𝑅𝑋
𝐿𝑋
ҧ𝐼
ത𝑉
𝐸𝑛 𝑒𝑙 𝑝𝑢𝑒𝑛𝑡𝑒 𝑏á𝑠𝑖𝑐𝑜 𝐴𝐶 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑟 𝑒𝑙 𝑣𝑎𝑙𝑜𝑟 𝑑𝑒𝑅𝑋 y 𝐿𝑋
𝑹𝒆𝒔𝒐𝒍𝒖𝒄𝒊ó𝒏
ҧ𝑍1. ҧ𝑍4 = ҧ𝑍2. ҧ𝑍3
1
ത𝑌1
. ҧ𝑍4 = ҧ𝑍2. ҧ𝑍3
ҧ𝑍4 = ҧ𝑍2. ҧ𝑍3 ഥ𝑌1
𝑅𝑋 + j𝜔𝐿𝑋=(
1
𝑅1
+ jω𝐶1)𝑅2. 𝑅3
𝑅𝑋 + j𝜔𝐿𝑋= 𝑅2. 𝑅3
𝑅1
+ j𝑅2. 𝑅3ω𝐶1
𝑅𝑋=
𝑅2.𝑅3
𝑅1
𝐿𝑋= 𝑅2. 𝑅3. 𝐶1
PUENTE HAY
ҧ𝑍1. ҧ𝑍4 = ҧ𝑍2. ҧ𝑍3
𝐶1
𝑅1
𝑅2
𝑅3
𝐿𝑋
𝑅𝑋
(𝑅𝑋+ j𝜔𝐿𝑋)(𝑅1−
𝑗
𝜔𝐶1
)=𝑅2𝑅3
𝑅1𝑅𝑋+
𝐿𝑋
𝐶1
-
𝑗𝑅𝑥
𝜔𝐶1
+𝑗𝜔𝐿𝑥𝑅1=𝑅2𝑅3
𝑅1𝑅𝑋+
𝐿𝑋
𝐶1
=𝑅2𝑅3
𝑅𝑋
𝜔𝐶1
=𝜔𝐿𝑋𝑅3
𝑅𝑋 =
𝜔2𝐶21𝑅1𝑅2𝑅3
1 + 𝜔2 𝐶1
2 𝑅1
2
𝐿𝑋 =
𝐶1𝑅2𝑅3
1 + 𝜔2 𝐶1
2 𝑅1
2
PUENTE SCHERING
ҧ𝑍1. ҧ𝑍4 = ҧ𝑍2. ҧ𝑍3𝑅1
𝐶1
𝑅2
𝐶3
𝐶𝑋
𝑅𝑋
1
ത𝑌1
. ҧ𝑍4 = ҧ𝑍2. ҧ𝑍3
ҧ𝑍4 = ҧ𝑍2. ҧ𝑍3 ഥ𝑌1
(𝑅𝑋−
𝑗
𝜔𝐶𝑋
) = (
1
𝑅1
+ jω𝐶1)𝑅2(−
𝑗
𝜔𝐶3
)
(𝑅𝑋−
𝑗
𝜔𝐶𝑋
) = (−
𝑅2𝑗
𝑅1𝜔𝐶3
+
ω𝐶1𝑅2
𝜔𝐶3
))
𝑅𝑋 =
𝑅2𝐶1
𝐶3
𝐶𝑋 =
𝑅1𝐶3
𝑅2
PUENTE WIEN
𝑅1
𝐶1
𝑅2
𝐶3
𝑅3
𝑅4
ҧ𝑍1. ҧ𝑍4 = ҧ𝑍2. ҧ𝑍3
ҧ𝑍1. ҧ𝑍4 = ҧ𝑍2.
1
ത𝑌3
. 
ҧ𝑍1. ҧ𝑍4𝑌3 = ҧ𝑍2
𝑅2 = (
1
𝑅3
+ j𝜔𝑋𝐶3)𝑅4(𝑅1 −
𝑗
𝜔𝑋𝐶1
)
𝑅2 = (
1
𝑅3
+ j𝜔𝑋𝐶3) (𝑅1𝑅4 −
𝑗𝑅4
𝜔𝑋𝐶1
)
𝑅2 =
𝑅1𝑅4
𝑅3
+ j𝑅1𝑅4𝜔𝑋𝐶3 +
𝑅4𝜔𝑋𝐶3
𝜔𝑋𝐶1
−
𝑗𝑅4
𝜔𝑋𝐶1𝑅3
)
R2
R4
=
R1
R3
+
C3
C1
𝑅1𝑅4𝜔𝑋𝐶3=
𝑅4
𝜔𝑋𝐶1𝑅3
𝜔𝑋=
1
𝐶1𝐶3𝑅1𝑅3