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n resistors each of resistance R first combine to give maximum effective resistgance and then combine to give minimum. The ratio of the maximum resistance isA. nB. `n^(2)`C. `n^(2)-1`D. `n^(3)` |
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Answer» Correct Answer - B To get maximum equivalent resistance all resistances must be connected in series `therefore (R_("eq"))_("max")=R+R+R...n"times"=nR` To get minimum equivalent resistance all resistances myst be connected in parallel. `therefore (1)/(R_("eq"))_("min")=(1)/(R)+(1)/(R)+....n"time", (1)/(R_("eq"))_("min")=(n)/(R)` `(R_("eq"))_("min")=(R)/(n)therefore (R_("eq"))_("max")/(R_("eq"))_("min")=(nR)/(R//n)=n^(2)` |
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