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The resistance of a Wire that must be pl...

The resistance of a Wire that must be placed parallel with a 120 resistance to obtain a combined resistance of `4Omega` is

A

`2Omega`

B

`4Omega`

C

`6Omega`

D

`8Omega`

Text Solution

Verified by Experts

The correct Answer is:
C
Let the resistance of the wire be `R_(1)`
Now, `R_("comb")=(R_(1)R_(2))/(R_(1)+R_(2))`
`4+(12xxR_(1))/(12+R_(1))impliesR_(1)=6Omega`
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