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Q.
Five equal resistances of $10 \,\Omega$ are connected between $A$ and $B$ as shown in figure. The resultant resistance is
Current Electricity
Solution:
According to the given circuit $10\,\Omega$ and $10\,\Omega$ resistances are connected in series.
$\therefore R'=10+10=20\,\Omega$
Again $10\,\Omega$ and $10\,\Omega$ resistances are connected in series
$\therefore R''=10+10=20\,\Omega$
$R'$, $R''$ and $10\,\Omega$ all connected in parallel than
$\therefore \frac{1}{R_{eq}}=\frac{1}{R'}+\frac{1}{R''}+\frac{1}{10}$
$=\frac{1}{20}+\frac{1}{20}+\frac{1}{10}=\frac{1+1+2}{20}$
$=\frac{4}{20}=\frac{1}{5}$ ;
$R_{eq}=5\,\Omega$.
