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Two resistance 12 Omega and 4 Omega ...

Two resistance `12 Omega` and `4 Omega` are supplied to you . Find the maximum and minimum resistance that can be achieved by using them.

Text Solution

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`R_(1) = 12 Omega, R_(2) = 4 Omega`
Maximum resistance is obtained , when they are used in series
`R_(s) = R_(1) + R_(2) = (12 + 4) Omega = 16 Omega`
Minimum resistance is obtained , when they are in parallel
`R_(p) = ((1)/(R_(1)) + (1)/(R_(2)))^(-1) = ((1)/(12) + (1)/(4))^(-1) = (48)/(16) = 3 Omega`
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