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Question
How much resistance should be connected to 15 Ω resistor shown in the circuit in figure below so that the points M and N are at the same potential:
Solution
Let X be the resistance in place of 15 Ω for the points M and N to be at the same potential, then by the condition of balanced wheatstone bridge
`"P"/"Q" = "R"/"S"`
`3/30 = "X"/60`
∴ X = `( 3 xx 60 )/30` = 6 Ω
To make the effective resistance of 6 Ω, let R be a resistance connected in parallel with 15Ω resistance, then
`1/6 = 1/"R" + 1/15`
`1/"R" = 1/6 - 1/15 = ( 5 - 2 )/30 = 3/30`
R = 10 Ω
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