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- 1. ORLANDO TORRES C.I.:19572555 R= 2 Resolviendo por mallas obtenemos: Ecuacion 1 calculo de 𝑰 𝟏 −𝟏𝟐∠𝟎 − 𝟐𝑰 𝟏 + 𝑽 𝟏 = 𝟎 −𝟐𝑰 𝟏 + 𝑽 𝟏 = 𝟏𝟐∠𝟎 Ec. 1 𝑰 𝟏 = −𝟎. 𝟏𝟔𝒂𝒎𝒑 Calculo del voltaje 2 −𝑽 𝟐 + 𝟒𝑰 𝟐 + 𝟐𝟒∠𝟎 = 𝟎 −𝟒𝑰 𝟐 + 𝑽 𝟐 = 𝟐𝟒∠𝟎 Ec. 2 𝑽 𝟐 = 𝟒𝑽 𝟏 Ec. 3 𝑽 𝟐 = 𝟒( 𝟔. 𝟐𝟓) = 𝟐𝟓𝒗 Calculo de I2 𝑰 𝟏 = −𝟒𝑰 𝟐 Ec. 4 𝑰 𝟐 = −𝑰 𝟏 𝟒 = 𝟎. 𝟏𝟔𝑨 𝟒 = 𝟎. 𝟎𝟒𝒂𝒎𝒑
- 2. Sustituyendo la ec. 3 y 4 en 2 calculo de voltaje 1 𝑰 𝟏 + 𝟒𝑽 𝟏 = 𝟐𝟒∠𝟎 Ec. 5 𝑽 𝟏 = 𝟔. 𝟐𝟓𝒗 𝑰 𝟏 = −𝟎. 𝟏𝟔𝒂𝒎𝒑, 𝑰 𝟐 = 𝟎. 𝟎𝟒𝒂𝒎𝒑, 𝑽 𝟏 = 𝟔. 𝟐𝟓, 𝑽 𝟐 = 𝟐𝟓 R=2 Resolviendo por mallas −𝟑𝟐𝒗 + 𝟐𝑰 𝟏 − 𝒋𝟒𝑰 𝟏 + 𝒋𝟐𝑰 𝟏 + 𝟐𝑰 𝟏 − 𝒋𝑰 𝟐 − 𝟐𝑰 𝟐 = 𝟎 ( 𝟐 − 𝒋𝟐 + 𝟐) 𝑰 𝟏 − ( 𝟐 + 𝒋) 𝑰 𝟐 = 𝟑𝟐∠𝟎 ( 𝟒 − 𝒋𝟐) 𝑰 𝟏 − ( 𝟐 + 𝒋) 𝑰 𝟐 = 𝟑𝟐∠𝟎 Ec. 1 𝟐𝑰 𝟐 + 𝒋𝟐𝑰 𝟐 − 𝒋𝟐𝑰 𝟐 + 𝟑𝑰 𝟐 − 𝟐𝑰 𝟏 − 𝒋𝑰 𝟏 = 𝟎 −( 𝟐 + 𝒋) 𝑰 𝟏 + 𝟓𝑰 𝟐 = 𝟎 Ec.2 𝑰 𝟏 = 𝟓 𝟐 + 𝒋 𝑰 𝟐 𝑬𝒄. 𝟑 Sustituyendo Ec. 3 en 1
- 3. ( 𝟒 − 𝒋𝟐) 𝟓 𝟐 + 𝒋 𝑰 𝟐 − ( 𝟐 + 𝒋) 𝑰 𝟐 = 𝟑𝟐∠𝟎 (𝟒 − 𝒋𝟗)𝑰 𝟐 = 𝟑𝟐∠𝟎 𝑰 𝟐 = 𝟑𝟐∠𝟎 𝟒 − 𝒋𝟗 = 𝟑. 𝟐𝟓∠𝟔𝟔. 𝟎𝟑𝒂𝒎𝒑 𝑰 𝟐 = 𝑰 𝒐 = 𝟑. 𝟐𝟓∠𝟔𝟔. 𝟎𝟑𝒂𝒎𝒑 R= 2 Resolviendo por mallas ( 𝟏 − 𝒋) 𝑰 𝟏 − (−𝒋) 𝑰 𝟐 = 𝟐𝟒∠𝟎 ( 𝟏 − 𝒋) 𝑰 𝟏 + 𝒋𝑰 𝟐 = 𝟐𝟒∠𝟎 Ec. 1 −(−𝒋) 𝑰 𝟏 + (−𝒋 + 𝒋𝟐 − 𝒋𝟐 + 𝒋𝟐 + 𝟐) 𝑰 𝟐 − 𝒋𝑰 𝟐 − 𝒋𝑰 𝟐 = 𝟎 𝒋𝑰 𝟏 + ( 𝟐 + 𝒋 − 𝒋𝟐) 𝑰 𝟐 = 𝟎 𝑬𝒄. 𝟐 𝒋𝑰 𝟏 + ( 𝟐 − 𝒋) 𝑰 𝟐 = 𝟎 𝑰 𝟏 = −( 𝟐 − 𝒋) 𝑰 𝟐 𝒋 𝑰 𝟏 = ( 𝟏 + 𝒋𝟐) 𝑰 𝟐 𝑬𝒄. 𝟑 Sustituyendo Ec. 3 en Ec. 1 ( 𝟏 − 𝒋)( 𝟏 + 𝒋𝟐) 𝑰 𝟐 + 𝒋𝑰 𝟐 = 𝟐𝟒∠𝟎
- 4. ( 𝟑 + 𝒋𝟐) 𝑰 𝟐 = 𝟐𝟒∠𝟎 → 𝑰 𝟐 = 𝟐𝟒∠𝟎 ( 𝟑 + 𝒋𝟐) = 𝟔. 𝟔𝟓∠ − 𝟑𝟑. 𝟔𝟗𝒂𝒎𝒑 𝑽 𝒐 = 𝟐𝑰 𝟐 = 𝟐( 𝟔.𝟔𝟓∠ − 𝟑𝟑. 𝟔𝟗𝒂𝒎𝒑) = 𝟏𝟑. 𝟑∠ − 𝟑𝟑. 𝟔𝟗𝒗 𝑽 𝒐 = 𝟏𝟑. 𝟑∠ − 𝟑𝟑. 𝟔𝟗𝒗 R= 2 𝒂 = 𝟏 𝟐 → 𝒁 𝑹 = 𝒂 𝟐 𝒁 𝒁; 𝒁 𝒁 = 𝟒 + 𝒋𝟒 𝒁 𝑹 = 𝟏 𝟒 ( 𝟒 + 𝒋𝟒) = (𝟏 + 𝑱)Ω 𝒁 𝑷 = −𝒋(𝟏 + 𝒋) −𝒋 + 𝟏 + 𝒋 = 𝟏 − 𝒋 Calculo de I1 𝑰 𝟏 = 𝟏𝟎∠𝟑𝟎 𝟐 + 𝟏 − 𝒋 = 𝟑. 𝟏𝟔∠𝟒𝟖. 𝟒𝟑𝒂𝒎𝒑 Calculo del Voltaje V1
- 5. 𝑽 𝟏 = 𝑰 𝟏( 𝟏 − 𝒋) = (𝟑. 𝟏𝟔∠𝟒𝟖. 𝟒𝟑𝒂𝒎𝒑)( 𝟏 − 𝒋) 𝒗 𝑽 𝟏 = 𝟒. 𝟒𝟔∠𝟑. 𝟒𝟑𝒗 Calculo del Voltaje V2 𝒂 = 𝟏 𝟐 = 𝑽 𝟏 𝑽 𝟐 → 𝑽 𝟐 = 𝟐𝑽 𝟏 = 𝟐( 𝟒. 𝟒𝟔∠𝟑. 𝟒𝟑𝒗) = 𝟖. 𝟗𝟐∠𝟑. 𝟒𝟑𝒗 Calculo de la Corriente I2 𝒂 = −𝑰 𝟐 𝑰 𝟏 → 𝑰 𝟐 = −𝒂𝑰 𝟏 = − 𝟏 𝟐 ( 𝟑. 𝟏𝟔∠𝟒𝟖. 𝟒𝟑𝒂𝒎𝒑) = 𝟏. 𝟓𝟖∠ − 𝟏𝟑𝟏. 𝟓𝟕𝒂𝒎𝒑