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Four resistances 3Omega, 6Omega, 4Omega,...

Four resistances `3Omega, 6Omega, 4Omega`, and `12Omega` are connected so as to form Wheatstone's network. Shunt needed across `12Omega` resistor to balance the bridge is

A

`12Omega`

B

`24Omega`

C

`42Omega`

D

`48Omega`

Text Solution

Verified by Experts

The correct Answer is:
B
`8=(12.5)/(12+5) 96+8.5=12.5`
`5=(96)/(4)=24Omega`
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