msgid "" msgstr "" "Project-Id-Version: Tecno Recursos 2023Report-Msgid-Bugs-To:POT-Creation-" "Date:2023-03-06 21:10+0100PO-Revision-Date:YEAR-MO-DA HO:MI+ZONELast-" "Translator:FULL NAME Language:enLanguage-Team:en " "Plural-Forms:nplurals=2; plural=(n != 1)MIME-Version:1" ".0Content-Type:text/plain; charset=utf-8Content-Transfer-Encoding" ":8bitGenerated-By:Babel 2.9.0\n" "Report-Msgid-Bugs-To: EMAIL@ADDRESS\n" "POT-Creation-Date: 2023-05-11 19:38+0200\n" "PO-Revision-Date: YEAR-MO-DA HO:MI+ZONE\n" "Last-Translator: FULL NAME \n" "Language-Team: LANGUAGE \n" "MIME-Version: 1.0\n" "Content-Type: text/plain; charset=utf-8\n" "Content-Transfer-Encoding: 8bit\n" "Generated-By: Babel 2.9.0\n" #: ../../source/electric-resolver-circuitos.rst:9 msgid "Resolución de circuitos" msgstr "Circuit resolution" #: ../../source/electric-resolver-circuitos.rst:10 msgid "" "En esta unidad vamos a estudiar cómo resolver circuitos con resistencias " "para hallar las corrientes y tensiones que circulan por ellas." msgstr "" "In this unit we are going to study how to solve circuits with resistors " "to find the currents and voltages that flow through them." #: ../../source/electric-resolver-circuitos.rst:16 msgid "Índice de contenidos" msgstr "Index of contents" #: ../../source/electric-resolver-circuitos.rst:19 msgid "Leyes y fórmulas utilizadas" msgstr "Laws and formulas used" #: ../../source/electric-resolver-circuitos.rst:20 msgid "" "Las ecuaciones que vamos a utilizar para conseguir resolver los circuitos" " son la ley de Ohm y las leyes de los circuitos en paralelo y en serie." msgstr "" "The equations that we are going to use to solve the circuits are Ohm's " "law and the laws of parallel and series circuits." #: ../../source/electric-resolver-circuitos.rst:23 msgid "" "También podemos utilizar las fórmulas, que ya hemos estudiado, de las " "resistencias equivalentes a un circuito en serie y a un circuito en " "paralelo." msgstr "" "We can also use the formulas, which we have already studied, for the " "equivalent resistances of a series circuit and a parallel circuit." #: ../../source/electric-resolver-circuitos.rst:39 msgid "Fórmulas de la ley de Ohm" msgstr "Formulas for Ohm`s law" #: ../../source/electric-resolver-circuitos.rst:28 msgid "V = I \\cdot R" msgstr "V = I \\cdot R" #: ../../source/electric-resolver-circuitos.rst:32 msgid "I = \\cfrac{V}{R}" msgstr "I = \\cfrac{V}{R}" #: ../../source/electric-resolver-circuitos.rst:36 msgid "R = \\cfrac{V}{I}" msgstr "R = \\cfrac{V}{I}" #: ../../source/electric-resolver-circuitos.rst:46 msgid "Leyes del circuito en serie" msgstr "Series circuit laws" #: ../../source/electric-resolver-circuitos.rst:42 msgid "" "La corriente que circula por varias resistencias en serie es la misma " "para todas ellas." msgstr "" "The current flowing through several resistors in series is the same for " "all of them." #: ../../source/electric-resolver-circuitos.rst:44 msgid "" "La tensión total de un circuito en serie es igual a la suma de las " "tensiones de cada una de sus resistencias." msgstr "" "The total voltage of a series circuit is equal to the sum of the voltages" " of each of its resistances." #: ../../source/electric-resolver-circuitos.rst:52 msgid "Leyes del circuito en paralelo" msgstr "Parallel circuit laws" #: ../../source/electric-resolver-circuitos.rst:49 msgid "" "La corriente total que circula por varias resistencias en paralelo es " "igual a la suma de las corrientes que circulan por las resistencias." msgstr "" "The total current flowing through several resistors in parallel is equal " "to the sum of the currents flowing through the resistors." #: ../../source/electric-resolver-circuitos.rst:51 msgid "" "La tensión de cada una de las resistencias que están en paralelo es la " "misma para todas ellas." msgstr "" "The voltage of each of the resistors that are in parallel is the same for" " all of them." #: ../../source/electric-resolver-circuitos.rst:62 msgid "Resistencia equivalente a un circuito en serie" msgstr "Equivalent resistance of a series circuit" #: ../../source/electric-resolver-circuitos.rst:55 msgid "R12 = R1 + R2" msgstr "R12 = R1 + R2" #: ../../source/electric-resolver-circuitos.rst:59 msgid "R123 = R1 + R2 + R3" msgstr "R123 = R1 + R2 + R3" #: ../../source/electric-resolver-circuitos.rst:72 msgid "Resistencia equivalente a un circuito en paralelo" msgstr "Equivalent resistance of a parallel circuit" #: ../../source/electric-resolver-circuitos.rst:65 msgid "R12 = \\cfrac{1}{ \\cfrac{1}{R1} + \\cfrac{1}{R2} }" msgstr "R12 = \\cfrac{1}{ \\cfrac{1}{R1} + \\cfrac{1}{R2} }" #: ../../source/electric-resolver-circuitos.rst:69 msgid "R123 = \\cfrac{1}{ \\cfrac{1}{R1} + \\cfrac{1}{R2} + \\cfrac{1}{R3} }" msgstr "R123 = \\cfrac{1}{ \\cfrac{1}{R1} + \\cfrac{1}{R2} + \\cfrac{1}{R3} }" #: ../../source/electric-resolver-circuitos.rst:74 msgid "" "Cada uno de los circuitos que vamos a analizar tendrá asociada una " "cuadrícula con la tensión, la intensidad de corriente y la resistencia de" " cada componente. Esa cuadrícula habrá que rellenarla con las soluciones " "que vamos obteniendo al aplicar las fórmulas anteriores, hasta que toda " "la cuadrícula esté completa y conozcamos todos los valores del circuito." msgstr "" "Each of the circuits that we are going to analyze will have a grid " "associated with the voltage, current intensity, and resistance of each " "component. That grid will have to be filled in with the solutions that we" " are obtaining by applying the previous formulas, until the entire grid " "is complete and we know all the values ​​of the circuit." #: ../../source/electric-resolver-circuitos.rst:84 msgid "Divisor de tensión con dos resistencias" msgstr "Voltage divider with two resistors" #: ../../source/electric-resolver-circuitos.rst:85 msgid "" "Este circuito es muy usado en electrónica para conseguir tensiones más " "bajas que la tensión de alimentación. Consiste en dos resistencias " "conectadas en serie entre sí y a los dos terminales de la alimentación." msgstr "" "This circuit is widely used in electronics to achieve lower voltages than" " the supply voltage. It consists of two resistors connected in series to " "each other and to the two power supply terminals." #: ../../source/electric-resolver-circuitos.rst:89 msgid "" "Si tenemos un circuito con una tensión de alimentación de 5 voltios y " "necesitamos una tensión de 3 voltios, el divisor de tensión es la forma " "más sencilla de conseguirlo." msgstr "" "If we have a circuit with a supply voltage of 5 volts and we need a " "voltage of 3 volts, the voltage divider is the easiest way to achieve it." #: ../../source/electric-resolver-circuitos.rst:93 msgid "En la siguiente imagen podemos ver el circuito divisor de tensión." msgstr "In the following image we can see the voltage divider circuit." #: ../../source/electric-resolver-circuitos.rst:100 msgid "" "Para resolver el circuito comenzamos escribiendo en la tabla los valores " "de las resistencias que conocemos, R1 y R2." msgstr "" "To solve the circuit we begin by writing in the table the values ​​of the" " resistors that we know, R1 and R2." #: ../../source/electric-resolver-circuitos.rst:108 msgid "" "A continuación escribimos en la tabla los valores de tensión que " "conocemos, que en este caso será la tensión total de las dos resistencias" " R1 y R2 en serie, que coincide con la tensión de la pila." msgstr "" "Next we write in the table the voltage values ​​that we know, which in " "this case will be the total voltage of the two resistors R1 and R2 in " "series, which coincides with the battery voltage." #: ../../source/electric-resolver-circuitos.rst:117 msgid "" "Ahora debemos buscar si se puede solucionar alguna cuadrícula con las " "fórmulas que conocemos. La resistencia total R12 se puede calcular con la" " fórmula de las resistencias en serie, es decir, sumando las dos " "resistencias." msgstr "" "Now we must find out if any grid can be solved with the formulas we know." " The total resistance R12 can be calculated with the series resistance " "formula, that is, by adding the two resistances." #: ../../source/electric-resolver-circuitos.rst:127 msgid "" "Para continuar, en la última fila tenemos la tensión y la resistencia por" " lo que podemos hallar la intensidad con la ley de Ohm. Dividiendo la " "tensión entre la resistencia obtenemos 20 mili amperios de corriente." msgstr "" "To continue, in the last row we have the voltage and the resistance so we" " can find the current with Ohm's law. Dividing the voltage by the " "resistance we get 20 milliamps of current." #: ../../source/electric-resolver-circuitos.rst:137 msgid "" "Ahora podemos aplicar la ley de los circuitos en serie que dice que la " "corriente será la misma por todos los componentes del circuito." msgstr "" "Now we can apply the law of series circuits which says that the current " "will be the same through all the components of the circuit." #: ../../source/electric-resolver-circuitos.rst:145 msgid "" "Para terminar, con la ley de Ohm podemos hallar las tensiones en cada una" " de las resistencias multiplicando la corriente por la resistencia." msgstr "" "Finally, with Ohm's law we can find the voltages in each of the resistors" " by multiplying the current by the resistance." #: ../../source/electric-resolver-circuitos.rst:153 msgid "Y el circuito está resuelto por completo." msgstr "And the circuit is completely solved." #: ../../source/electric-resolver-circuitos.rst:155 msgid "" "La tensión de la resistencia R2 será igual a 2 voltios, una tensión menor" " que la tensión de alimentación porque este circuito ha **dividido** la " "tensión de alimentación entre 6." msgstr "" "The voltage across resistor R2 will be equal to 2 volts, a voltage less " "than the supply voltage because this circuit has **divided** the supply " "voltage by 6." #: ../../source/electric-resolver-circuitos.rst:161 msgid "Divisor de tensión con dos resistencias desconocidas" msgstr "Voltage divider with two unknown resistors" #: ../../source/electric-resolver-circuitos.rst:162 msgid "" "En este apartado vamos a resolver un circuito en serie en el que no " "conocemos el valor de las resistencias, solo conocemos la corriente que " "circula por el circuito (10mA) y la tensión que queremos obtener en R2 " "(9V)." msgstr "" "In this section we are going to solve a series circuit in which we do not" " know the value of the resistors, we only know the current that " "circulates through the circuit (10mA) and the voltage that we want to " "obtain in R2 (9V)." #: ../../source/electric-resolver-circuitos.rst:172 msgid "" "Comenzamos por rellenar la tabla con los valores que conocemos del " "circuito." msgstr "We start by filling in the table with the values ​​we know of the circuit." #: ../../source/electric-resolver-circuitos.rst:180 msgid "" "A continuación podemos calcular la resistencia total R12 aplicando la ley" " de Ohm." msgstr "Next we can calculate the total resistance R12 by applying Ohm's law." #: ../../source/electric-resolver-circuitos.rst:188 msgid "" "Para poder continuar, aplicamos la ley de los componentes en serie que " "dice que la corriente por todos los elementos del circuito es la misma." msgstr "" "In order to continue, we apply the law of components in series which says" " that the current through all the elements of the circuit is the same." #: ../../source/electric-resolver-circuitos.rst:196 msgid "" "Ahora podemos aplicar de nuevo la ley de Ohm a la segunda resistencia " "para hallar su valor." msgstr "Now we can apply Ohm's law again to the second resistor to find its value." #: ../../source/electric-resolver-circuitos.rst:204 msgid "" "En este punto podemos continuar aplicando la ley de los circuitos en " "serie que dice que la tensión total de las resistencias es igual a la " "suma de las tensiones en cada resistencia." msgstr "" "At this point we can continue applying the law of series circuits which " "says that the total voltage of the resistors is equal to the sum of the " "voltages in each resistor." #: ../../source/electric-resolver-circuitos.rst:208 msgid "Es decir: V_R1 + V_R2 = 12v -> V_R1 = 12v - 9v = 3v" msgstr "That is: V_R1 + V_R2 = 12v -> V_R1 = 12v - 9v = 3v" #: ../../source/electric-resolver-circuitos.rst:215 msgid "" "Para terminar, aplicamos la ley de Ohm a la primera resistencia y " "hallamos su valor." msgstr "To finish, we apply Ohm's law to the first resistor and find its value." #: ../../source/electric-resolver-circuitos.rst:223 msgid "" "En esta última casilla también podríamos haber aplicado la fórmula del " "equivalente de las resistencias en serie. Sabiendo que R1 + R2 = R12, se " "puede calcular fácilmente que R1 debe valer 300 Ohmios." msgstr "" "In this last box we could also have applied the formula for the " "equivalent of resistances in series. Knowing that R1 + R2 = R12, it can " "be easily calculated that R1 must be worth 300 Ohms." #: ../../source/electric-resolver-circuitos.rst:230 msgid "Circuito mixto serie-paralelo" msgstr "Mixed series-parallel circuit" #: ../../source/electric-resolver-circuitos.rst:231 msgid "" "En este apartado vamos a resolver un circuito mixto, con conexiones serie" " y paralelo, en el que conocemos el valor de todas las resistencias." msgstr "" "In this section we are going to solve a mixed circuit, with series and " "parallel connections, in which we know the value of all the resistances." #: ../../source/electric-resolver-circuitos.rst:234 msgid "" "Comenzamos por copiar en la tabla los valores de resistencia y tensión " "que ya conocemos." msgstr "" "We begin by copying into the table the values ​​of resistance and voltage" " that we already know." #: ../../source/electric-resolver-circuitos.rst:242 msgid "" "A partir de aquí no tenemos datos para resolver ninguna de las tres " "primeras filas. La primera tarea será calcular la resistencia equivalente" " de las tres resistencias del circuito." msgstr "" "From here we have no data to solve for any of the first three rows. The " "first task will be to calculate the equivalent resistance of the three " "resistors in the circuit." #: ../../source/electric-resolver-circuitos.rst:246 msgid "" "Primero hallamos el paralelo de 100 Ohmios y de 400 Ohmios que nos da un " "resultado de 80 Ohmios." msgstr "" "First we find the parallel of 100 Ohms and 400 Ohms that gives us a " "result of 80 Ohms." #: ../../source/electric-resolver-circuitos.rst:249 msgid "" "R23 = \\cfrac{1}{ \\cfrac{1}{100\\Omega} + \\cfrac{1}{400\\Omega} } = " "80\\Omega" msgstr "" "R23 = \\cfrac{1}{ \\cfrac{1}{100\\Omega} + \\cfrac{1}{400\\Omega} } = " "80\\Omega" #: ../../source/electric-resolver-circuitos.rst:253 msgid "" "A continuación calculamos el equivalente en serie de la resistencia R1, " "con 120 Ohmios, y del resultado anterior, 80 Ohmios. Sumando ambos nos da" " un resultado total de 200 Ohmios, que podemos escribir en el hueco " "correspondiente a la resistencia R123." msgstr "" "Next we calculate the series equivalent of resistor R1, with 120 Ohms, " "and from the previous result, 80 Ohms. Adding both gives us a total " "result of 200 Ohms, which we can write in the hole corresponding to " "resistor R123." #: ../../source/electric-resolver-circuitos.rst:263 msgid "" "Ahora podemos aplicar la ley de Ohm a la cuarta fila para hallar la " "intensidad total que circula por el circuito, 60 miliamperios." msgstr "" "We can now apply Ohm's law to the fourth row to find the total current " "flowing through the circuit, 60 milliamps." #: ../../source/electric-resolver-circuitos.rst:271 msgid "" "Toda la corriente que circula por el circuito circulará por R1 al estar " "en serie. Con este dato podemos rellenar la corriente de R1 copiando la " "corriente total." msgstr "" "All the current that flows through the circuit will flow through R1 to be" " in series. With this data we can fill in the current of R1 by copying " "the total current." #: ../../source/electric-resolver-circuitos.rst:280 msgid "" "En este punto podemos aplicar la ley de Ohm a la primera fila para " "calcular la tensión en la resistencia R1." msgstr "" "At this point we can apply Ohm's law to the first row to calculate the " "voltage across resistor R1." #: ../../source/electric-resolver-circuitos.rst:288 msgid "" "La tensión total de la pila, 12v, será igual a la suma de las tensiones " "de las dos ramas en serie del circuito, R1 y R23." msgstr "" "The total voltage of the battery, 12v, will be equal to the sum of the " "voltages of the two branches in series of the circuit, R1 and R23." #: ../../source/electric-resolver-circuitos.rst:291 msgid "12v = V_{R1} + V_{R23}." msgstr "12v = V_{R1} + V_{R23}." #: ../../source/electric-resolver-circuitos.rst:295 msgid "" "Despejando tenemos que la tensión en las resistencias R2 y R3 es de 12v -" " 7.2v = 4.8v, que podemos escribir en las casillas correspondientes." msgstr "" "Clearing we have that the voltage in resistors R2 and R3 is 12v - 7.2v = " "4.8v, which we can write in the corresponding boxes." #: ../../source/electric-resolver-circuitos.rst:303 msgid "" "Ahora podemos aplicar la ley de Ohm a la segunda y tercera filas para " "terminar de calcular los valores de intensidad que aún no conocemos." msgstr "" "Now we can apply Ohm's law to the second and third rows to finish " "calculating the intensity values ​​we don't know yet." #: ../../source/electric-resolver-circuitos.rst:311 msgid "" "Para terminar comprobaremos que la suma de las corrientes en R2 y en R3 " "es igual a la corriente total que circula por el circuito." msgstr "" "To finish we will verify that the sum of the currents in R2 and in R3 is " "equal to the total current that circulates through the circuit." #: ../../source/electric-resolver-circuitos.rst:314 msgid "60 mA = 48 mA + 12 mA." msgstr "60mA = 48mA + 12mA." #: ../../source/electric-resolver-circuitos.rst:320 msgid "Ejercicios" msgstr "Exercises" #: ../../source/electric-resolver-circuitos.rst:321 msgid "Ejercicios para resolver circuitos." msgstr "Exercises to solve circuits." #: ../../source/electric-resolver-circuitos.rst:326 msgid "" ":download:`Ejercicios para resolver circuitos. Formato PDF `" msgstr "" ":download:`Exercises to solve circuits. PDF format `" #: ../../source/electric-resolver-circuitos.rst:329 msgid "" ":download:`Ejercicios para resolver circuitos. Formato KiCad `" msgstr "" ":download:`Exercises to solve circuits. KiCad format `" #: ../../source/electric-resolver-circuitos.rst:335 msgid "Cuestionarios" msgstr "" #: ../../source/electric-resolver-circuitos.rst:336 msgid "Cuestionarios sobre resolución de circuitos." msgstr "Questionnaires on circuit resolution." #: ../../source/electric-resolver-circuitos.rst:338 msgid "" "`Cuestionario. Resolver circuitos I. `__" msgstr "" #: ../../source/electric-resolver-circuitos.rst:340 msgid "" "`Cuestionario. Resolver circuitos II. `__" msgstr ""