1.

What is the lowest total resistance that can be secured by combinations of four coils of resistance 4Omega,8Omega,12Omega,24Omega

Answer»

Solution :(a) HIGHEST ( by connecting in series )
= 4+ 8 + 12 +24
series = `48 Omega`
(B) Lowest ( by connecting ) in parallel =
`(1)/(R_(p))= (1)/(4) + (1)/(8) +(1)/(12)+(1)/(24)`
`=(12)/(24) `
`(1)/(R_(p))=(1)/(2)`
`R_(p)=2 Omega`


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