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The resistors of resistances $2\Omega,4\Omega,5\Omega$ are connected in parallel.The total resistance of the combination will be :

$\begin{array}{1 1}(A)\;11\Omega&(B)\;\large\frac{19}{20}\normalsize \Omega\\(C)\;\large\frac{10}{20}\normalsize \Omega&(D)\;\large\frac{20}{19}\normalsize \Omega\end{array} $

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1 Answer

The total resistance of the combination will be $\large\frac{20}{19}\normalsize \Omega$
Hence (D) is the correct answer.
answered Apr 15, 2014 by sreemathi.v
 

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