Question

Four resistances 4 Ω, 4 Ω, 4 Ω and 12 Ω form a Wheatstone network. Find the resistance which connected across the 12 Ω resistance will balance the network.

Answer 1

The resistance in each of the three arms of the network is 4 Ω. Hence, to balance the network, the resistance in the fourth arm must also be 4 Ω.

Hence, the resistance (R) to be connected across. i.e., in parallel to, the 12 Ω resistance should be such that their equivalent resistance is 4Ω.

∴ \(\cfrac14\) = \(\cfrac1{12}\) + \(\cfrac1R\)

∴ \(\cfrac1R\) = \(\cfrac14\) - \(\cfrac1{12}\) = \(\cfrac{3-1}{12}\) = \(\cfrac2{12}\) = \(\cfrac16\)

∴ R = 6 Ω