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.
Answers
Answer:
If the bridge is to be balanced then:
If the bridge is to be balanced then:If a resistance is R is put in parallel across 18Ω, the effective resistance in arm AD will be
If the bridge is to be balanced then:If a resistance is R is put in parallel across 18Ω, the effective resistance in arm AD will be 18+R
If the bridge is to be balanced then:If a resistance is R is put in parallel across 18Ω, the effective resistance in arm AD will be 18+R18R
If the bridge is to be balanced then:If a resistance is R is put in parallel across 18Ω, the effective resistance in arm AD will be 18+R18R
If the bridge is to be balanced then:If a resistance is R is put in parallel across 18Ω, the effective resistance in arm AD will be 18+R18R
If the bridge is to be balanced then:If a resistance is R is put in parallel across 18Ω, the effective resistance in arm AD will be 18+R18R 6
If the bridge is to be balanced then:If a resistance is R is put in parallel across 18Ω, the effective resistance in arm AD will be 18+R18R 64
If the bridge is to be balanced then:If a resistance is R is put in parallel across 18Ω, the effective resistance in arm AD will be 18+R18R 64
If the bridge is to be balanced then:If a resistance is R is put in parallel across 18Ω, the effective resistance in arm AD will be 18+R18R 64 =
If the bridge is to be balanced then:If a resistance is R is put in parallel across 18Ω, the effective resistance in arm AD will be 18+R18R 64 = 18×R/(18+R)
If the bridge is to be balanced then:If a resistance is R is put in parallel across 18Ω, the effective resistance in arm AD will be 18+R18R 64 = 18×R/(18+R)8
If the bridge is to be balanced then:If a resistance is R is put in parallel across 18Ω, the effective resistance in arm AD will be 18+R18R 64 = 18×R/(18+R)8
If the bridge is to be balanced then:If a resistance is R is put in parallel across 18Ω, the effective resistance in arm AD will be 18+R18R 64 = 18×R/(18+R)8 ⇒R=36Ω
Explanation:
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