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Four resistors 4Omega, 8Omega, 18Omega a...

Four resistors `4Omega, 8Omega, 18Omega` and `6Omega` are connected in a cyclic order to form a Wheatstone's network. How and what value of resistance should be connected in the `18Omega` branch to balance the network?

Text Solution

Verified by Experts

For the balanced condition, `P/Q=R/S`
i.e., `4/8=6/x`
`therefore x=(8xx6)/4=12Omega 12Omegalt18Omega`
A resistor y is to connected in parallel to `18Omega` to make it `12Omega`
`=(18y)/(18+y)=12`
`216 + 12y = 18y`
`6y = 216`
`thereforey=36Omega`
A resistor of resistance `36Omega` should be connected in parallel across `18Omega` to balance the network.
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