Question 298 of 380

Four resistances of $15\Omega ,12\Omega ,4\Omega$ and $10\Omega$ respectively in cyclic order to form Wheatstone’s network. The resistance that is to be connected in parallel with the resistance of $10\Omega$ to balance the network is ______________ $\Omega .$

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Answer: 10

$\dfrac{10R}{10+R}\times 12=15\times 4$
On solving
$R=10\Omega$