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Por lo tanto: I= V / R I= 12V / 660Ω I= 18, 18mA La corriente total que fluye por este circuito es de 18, 18mA, y por ley, en un circuito en serie la corriente es igual en cada uno de los dispositivos que lo forman. Es decir, que en ambas resistencias tendremos 18, 18mA de corriente. La ley de Ohm dice que el voltaje es igual a la corriente multiplicada por la resistencia. De este modo podremos calcular el voltaje que está presente en cada resistencia, y comenzaremos con la resistencia de 100Ω: V= I x R V= 18, 18mA x 100Ω V= 1, 818V Lo mismo hacemos con la resistencia de 560Ω: V= 18, 18mA x 560Ω V= 10, 1808V Según los cálculos teóricos, tenemos 1, 818V en la Resistencia de 100Ω y 10, 8108V en la resistencia de 560Ω. Y en base a esto se cumple otra ley de los circuitos en serie que dice que la suma de los voltajes parciales es igual al voltaje total del circuito. También es importante mencionar que el voltaje en cada resistencia será diferente si el valor de la resistencia es diferente.
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1 Chapter 9 - Cicero, Pro Archia 13. 2 Chapter 10 - Cicero, Pro Archia 14. 1 Conversational and Grammar Goals Conversational Goals Read the First Part of Caesar's Gallic Wars Grammar Goals Form 3rd Person Singular and Plural Present Indicative Actives Form 3rd Person Singular and Plural Present Indicative Passives Understand Gender and Number Agreement Use Nominative, Accusative, and Ablative Cases Read Gallic Wars 1. 2. Practice the Demonstrative Pronouns hic and is Use the Gerundive as a Gerund with ad Use the Nominative, Accusative, Ablative, Genitive, and Dative Cases Use the Relative Pronoun quī Read Caesar's Gallic Wars 1. 3 Distinguish Reflexive from Non-Reflexive Third Person Use the Nominative, Accusative, Genitive, Dative, and Ablative Cases Use the Prepositions cum, dē, and in Use aut... aut... Read Caesar's Gallic Wars 1.
