Question

(UEL-PR) Considere o circuito representado no esquema abaixo, onde cada resistência vale 10Ω.
A resistência equivalente entre os terminais X e Y , em ohms, é igual a:
a) 10
b) 15
c) 30
d) 40
e) 90

Answer 1

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⠀⠀⠀☞ A resistência equivalente é de 15 [Ω] (opção b). ✅

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⠀⠀⠀⭐⠀Para realizar este exercício vamos encontrar a resistência equivalente das resistências em série e paralelo.⠀⭐⠀

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⠀⠀⠀➡️⠀Oi, Walisson ✌. Vamos inicialmente associar os resistores que estão em série (as imagens a seguir não são visualizáveis pelo app Brainly. ☹ Experimente compartilhar > copiar e acessar a resposta pelo navegador/ browser):

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                         $\setlength{\unitlength}{0.95cm}\begin{picture}(6,5)\thicklines\put(0,0){\line(1,0){8}}\put(1,-2){\line(1,0){6}}\put(1,-4){\line(1,0){6}}\put(1,0){\line(0,-1){4}}\put(7,0){\line(0,-1){4}}\put(2,-0.23){\Huge \sf\circ }\put(2,0.3){ \sf R }\put(3.75,-0.23){\Huge \sf\circ }\put(3.75,0.3){ \sf R }\put(5.5,-0.23){\Huge \sf\circ }\put(5.5,0.3){ \sf R }\put(3,-2.23){\Huge \sf\circ }\put(3,-1.7){ \sf R }\put(4.5,-2.23){\Huge \sf\circ }\put(4.5,-1.7){ \sf R }\put(3,-4.23){\Huge \sf\circ }\put(3,-3.7){ \sf R }\put(4.5,-4.23){\Huge \sf\circ }\put(4.5,-3.7){ \sf R }\put(0.78,-1.23){\Huge \sf\circ }\put(0.78,-0.8){ \sf R }\put(6.78,-1.23){\Huge \sf\circ }\put(6.78,-0.8){ \sf R }\end{picture}$

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                         $\setlength{\unitlength}{0.95cm}\begin{picture}(6,5)\thicklines\put(0,0){\line(1,0){8}}\put(1,-2){\line(1,0){6}}\put(1,-4){\line(1,0){6}}\put(1,0){\line(0,-1){4}}\put(7,0){\line(0,-1){4}}\put(3.75,-0.23){\Huge \sf\circ }\put(3.75,0.3){ \sf 3R }\put(3.75,-2.23){\Huge \sf\circ }\put(3.75,-1.7){ \sf 2R }\put(3.75,-4.23){\Huge \sf\circ }\put(3.75,-3.7){ \sf 2R }\put(0.78,-1.23){\Huge \sf\circ }\put(0.78,-0.8){ \sf R }\put(6.78,-1.23){\Huge \sf\circ }\put(6.78,-0.8){ \sf R }\end{picture}$

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⠀⠀⠀➡️⠀Vamos agora associar as duas resistências em paralelo da segunda e terceira linha:

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$\LARGE\blue{\text{ \sf \dfrac{1}{R_{eq}} = \dfrac{1}{2R} + \dfrac{1}{2R} }}$

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$\LARGE\blue{\text{ \sf \dfrac{1}{R_{eq}} = \dfrac{\backslash\!\!\!{2}}{\backslash\!\!\!{2}R} }}$

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$\LARGE\blue{\text{ \sf \dfrac{1}{R_{eq}} = \dfrac{1}{R} }}$

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$\LARGE\blue{\sf R_{eq} = R}$

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                         $\setlength{\unitlength}{0.95cm}\begin{picture}(6,5)\thicklines\put(0,0){\line(1,0){8}}\put(1,-2){\line(1,0){6}}\put(1,0){\line(0,-1){2}}\put(7,0){\line(0,-1){2}}\put(3.75,-0.23){\Huge \sf\circ }}\put(3.75,0.3){ \sf 3R }\put(2,-2.23){\Huge \sf\circ }\put(2,-1.7){ \sf R }\put(3.75,-2.23){\Huge \sf\circ }\put(3.75,-1.7){ \sf R }\put(5.5,-2.23){\Huge \sf\circ }\put(5.5,-1.7){ \sf R }\end{picture}$

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⠀⠀⠀➡️⠀Novamente os que estão em série:

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                         $\setlength{\unitlength}{0.95cm}\begin{picture}(6,5)\thicklines\put(0,0){\line(1,0){8}}\put(1,-2){\line(1,0){6}}\put(1,0){\line(0,-1){2}}\put(7,0){\line(0,-1){2}}\put(3.75,-0.23){\Huge \sf\circ }\put(3.75,0.3){ \sf 3R }\put(3.75,-2.23){\Huge \sf\circ }}\put(3.75,-1.7){ \sf 3R }\end{picture}$

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⠀⠀⠀➡️⠀E por fim os dois que estão em paralelo:

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$\LARGE\blue{\text{ \sf \dfrac{1}{R_{eq}} = \dfrac{1}{3R} + \dfrac{1}{3R} }}$

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$\LARGE\blue{\text{ \sf \dfrac{1}{R_{eq}} = \dfrac{2}{3R} }}$

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$\LARGE\blue{\sf R_{eq} = \dfrac{3R}{2}}$

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⠀⠀⠀⭐ Sabendo que R = 10 [Ω] então a resistência equivalente vale 3·10/2 = 30/2 = 15 [Ω], o que nos leva à opção b). ✌

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                                            $\qquad\qquad\Huge\green{\boxed{\rm~~~\red{b)}~\blue{ 15 }~~~}}$ ✅

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                             $\bf\large\red{\underline{\quad\quad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}}$☁

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⠀⠀⠀☀️ L͎̙͖͉̥̳͖̭̟͊̀̏͒͑̓͊͗̋̈́ͅeia mais sobre associação de resistores:

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                                     https://brainly.com.br/tarefa/37983721 ✈  

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                                     $\huge\blue{\text{\bf\quad Bons~estudos.}}$ ☕

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                                          $\quad\qquad(\orange{D\acute{u}vidas\ nos\ coment\acute{a}rios})$ ☄

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                             $\bf\large\red{\underline{\qquad \qquad \qquad \qquad \qquad \qquad \quad }\LaTeX}$✍

                                $\sf(\purple{+}~\red{cores}~\blue{com}~\pink{o}~\orange{App}~\green{Brainly}$ ☘☀❄☃☂☻)

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                                                          $\Huge\green{\text{ \underline{\red{\mathbb{S}}\blue{\mathfrak{oli}}~}~\underline{\red{\mathbb{D}}\blue{\mathfrak{eo}}~}~\underline{\red{\mathbb{G}}\blue{\mathfrak{loria}}~} }}$ ✞

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