Question

Four resistances 6Ω,6 Ω,6 Ω and 18 Ω form a Wheatstone’s meter bridge. Find the resistance which should be connected across the 18 Ω resistance will balance the network.​

Answer 1

Answer:

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Answer 2

$Given :- \\ \\ Four \: resistances \: 6Ω \: ,6 Ω \: ,6 Ω \: \\ and \: 1 8 Ω \: form \: a \\ \: \bf \: Wheatstone’s \: meter \: bridge \\ \\ \\ To Find :- \\ \\ The \:  resistance \: which \: should \: be \\ connected \: across \: the 18 Ω \: resistance \\ will \: balance   \: the \: network. \\ \\ \\ Solution :- \\ \\ Let \: the \: resistance \: be \: r \\ \sf \dfrac{6}{6} = \dfrac{6}{ 18r/(18 + r) }\\ \sf \dfrac{6}{6} = \dfrac{6}{18r} \times\dfrac{1}{18 +r}\\ \sf 1 = \dfrac{6(18 + r)}{18r}\\ \sf 1 = \dfrac{108 + 6r}{18r} \\ \sf 1= \dfrac{108 + r}{3r}\\ \sf 3r - r = 108\\ \sf 2r = 108 \\ r = 108/2 \\ r = 54 ohm$