$\begin {array} {1 1} (A)\;5\Omega & \quad (B)\;10\Omega \\ (C)\;15\Omega & \quad (D)\;20\Omega \end {array}$

The ratio of upper resistances and lower resistances are same. i.e. $5 : 10 : 15 = 10 : 20 : 30.$

Hence, the resistances in the middle are not useful (balanced wheatstone bridge).So effectively the given circuit becomes eqivalent to 30 ohm resistor in parallel with 60 ohm resistor. which reduces to 20 ohms

Ans : (D)

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