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Q.
The lowest resistance which can be obtained by connecting 10 resistors each of $\frac{1}{10}$ ohm is
Current Electricity
Solution:
Lowest resistance will be obtained when connected in parallel
$\frac{1}{R_{p}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}........ \frac{1}{R_{n}}$
$\frac{1}{R_{p}}=\frac{1}{\frac{1}{10}}+\frac{1}{\frac{1}{10}}+\frac{1}{\frac{1}{10}}+\frac{1}{\frac{1}{10}}+\frac{1}{\frac{1}{10}}+\frac{1}{\frac{1}{10}}+\frac{1}{\frac{1}{10}}+\frac{1}{\frac{1}{10}}+\frac{1}{\frac{1}{10}}+\frac{1}{\frac{1}{10}}$
$\frac{1}{R_{p}}=100$
$\Rightarrow R_{p}=\frac{1}{100}$ ohm