The equivalent resistance across A and B is

2 Ω

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3 Ω

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4 Ω

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5 Ω

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Solution

The correct option is C $4 Ω$
The setup is a wheat-stone bridge arrangement. Thus there would be no current in the $10 Ω$ resistor.
So, we could neglect the resistance in the equivalent circuit.
Now we have two branches, each containing two $4 Ω$ resistances in series, parallel to each other.
Thus, the equivalent resistance across A and B is $\frac{(4 + 4) \times (4 + 4)}{(4 + 4) + (4 + 4)} = 4 Ω$

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