Question

Determine a resistência equivalente, entre os terminais A e B, das associações representadas a seguir: ​

Answer 1

Explicação:

1) $\sf R~e~R~\Rightarrow~R_{eq_{1}}$

Em série

$\sf R_{eq_{1}}=R+R$

$\sf R_{eq_{1}}=2R$

2) $\sf R~e~R_{eq_{1}}~\Rightarrow~R_{eq_{2}}$

Em paralelo

$\sf R_{eq_{2}}=\dfrac{R\cdot2R}{R+2R}$

$\sf R_{eq_{2}}=\dfrac{2R^2}{3R}$

$\sf R_{eq_{2}}=\dfrac{2R}{3}$

3) $\sf R~e~R_{eq_{2}}~\Rightarrow~R_{eq_{3}}$

Em série

$\sf R_{eq_{3}}=R+\dfrac{2R}{3}$

$\sf R_{eq_{3}}=\dfrac{3R+2R}{3}$

$\sf R_{eq_{3}}=\dfrac{5R}{3}$

4) $\sf R~e~R_{eq_{3}}~\Rightarrow~R_{eq_{4}}$

Em paralelo

$\sf R_{eq_{4}}=\dfrac{R\cdot\frac{5R}{3}}{R+\frac{5R}{3}}$

$\sf R_{eq_{4}}=\dfrac{\frac{5R^2}{3}}{\frac{3R+5R}{3}}$

$\sf R_{eq_{4}}=\dfrac{\frac{5R^2}{3}}{\frac{8R}{3}}$

$\sf R_{eq_{4}}=\dfrac{5R^2}{3}\cdot\dfrac{3}{8R}$

$\sf R_{eq_{4}}=\dfrac{5R^2}{8R}$

$\sf R_{eq_{4}}=\dfrac{5R}{8}$

5) $\sf R~e~R_{eq_{4}}~\Rightarrow~R_{eq_{5}}$

Em série

$\sf R_{eq_{5}}=R+\dfrac{5R}{8}$

$\sf R_{eq_{5}}=\dfrac{8R+5R}{8}$

$\sf R_{eq_{5}}=\dfrac{13R}{8}$

6) $\sf R~e~R_{eq_{5}}~\Rightarrow~R_{eq_{6}}$

Em paralelo

$\sf R_{eq_{6}}=\dfrac{R\cdot\frac{13R}{8}}{R+\frac{13R}{8}}$

$\sf R_{eq_{6}}=\dfrac{\frac{13R^2}{8}}{\frac{8R+13R}{8}}$

$\sf R_{eq_{6}}=\dfrac{\frac{13R^2}{8}}{\frac{21R}{8}}$

$\sf R_{eq_{6}}=\dfrac{13R^2}{8}\cdot\dfrac{8}{21R}$

$\sf R_{eq_{6}}=\dfrac{13R^2}{21R}$

$\sf R_{eq_{6}}=\dfrac{13R}{21}$

7) $\sf R~e~R_{eq_{6}}~\Rightarrow~R_{eq_{7}}$

Em série

$\sf R_{eq_{7}}=R+\dfrac{13R}{21}$

$\sf R_{eq_{7}}=\dfrac{21R+13R}{21}$

$\sf R_{eq_{7}}=\dfrac{34R}{21}$

8) $\sf R~e~R_{eq_{7}}~\Rightarrow~R_{eq_{AB}}$

Em paralelo

$\sf R_{eq_{AB}}=\dfrac{R\cdot\frac{34R}{21}}{R+\frac{34R}{21}}$

$\sf R_{eq_{AB}}=\dfrac{\frac{34R^2}{21}}{\frac{21R+34R}{21}}$

$\sf R_{eq_{AB}}=\dfrac{\frac{34R^2}{21}}{\frac{55R}{21}}$

$\sf R_{eq_{AB}}=\dfrac{34R^2}{21}\cdot\dfrac{21}{55R}$

$\sf R_{eq_{AB}}=\dfrac{34R^2}{55R}$

$\sf \red{R_{eq_{AB}}=\dfrac{34R}{55}}$