Answer:
These resistances should be connected in parallel with the battery to obtain the maximum current. Let?s its equivalent resistance be \[R\,\Omega \]. \[\frac{1}{R}=\frac{1}{r}+\frac{1}{r}+\frac{1}{r}\] \[\frac{1}{R}=\frac{1+1+1}{r}=\frac{3}{r}\] \[R=\frac{r}{3}\] By Ohm?s law, \[V\mathrm{ }=\mathrm{ }IR\] \[\Rightarrow \] \[Current(I)=\frac{V}{R}=\frac{E}{r/3}\] \[=\frac{3E}{r}\]
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