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Given the resistances of 1 Omega, 2 Omeg...

Given the resistances of `1 Omega, 2 Omega and 3 Omega `. How will you combine them to get an equivalent resistance of
(i) `(11)/3 Omega` ?
(ii) `(11)/5 Omega` ?

Text Solution

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(i) `3 Omega` will be connected in series with parallel combination of `1 Omega and 2 Omega`.
(ii) `1 Omega` will be connected in series with parallel combination of `2 Omega and 3 Omega`
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