InterviewSolution
Saved Bookmarks
| 1. |
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» 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 |
|