Question
Four $ 12~\Omega $ resistances are connected in parallel. There such combinations are connected in series. What will be the total resistance ?
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Answer
Four $ 12~\Omega $ resistances are connected (i) Parallel.
$ \Rightarrow \frac{1}{R_P} = \frac{1}{12} + \frac{1}{12} + \frac{1}{12} + \frac{1}{12} = \frac{4}{12} = \frac{1}{3} $
$ R_P = 3~\Omega $
Now, $ 3~\Omega $ resistances (3 in number) are connected in series.
$ \Rightarrow R_S = 3 + 3 + 3 = 9~\Omega $.
$ \Rightarrow \frac{1}{R_P} = \frac{1}{12} + \frac{1}{12} + \frac{1}{12} + \frac{1}{12} = \frac{4}{12} = \frac{1}{3} $
$ R_P = 3~\Omega $
Now, $ 3~\Omega $ resistances (3 in number) are connected in series.
$ \Rightarrow R_S = 3 + 3 + 3 = 9~\Omega $.
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