Consider the circuits shown in the figure. Both the circuits are taking same current from battery but current through $R$ in the second circuit is $\frac{1}{{10}}$$^{th}$ of current through $R$ in the first circuit. If $R$ is $11$ $\Omega$, the value of $ R_1$ ................ $\Omega$
Diffcult
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(a) In figure $(b)$ current through ${R_2} = i - \frac{i}{{10}} = \frac{{9i}}{{10}}$
Potential difference across ${R_2}$ = Potential difference across $R$
Potential difference across ${R_2}$ = Potential difference across $R$
$==>$ ${R_2} \times \frac{9}{{10}}i = R \times \frac{i}{{10}}$ i.e. ${R_2} = \frac{R}{9} = \frac{{11}}{9}\,\Omega $
${R_{eq}} = \frac{{{R_2} \times R}}{{({R_2} + R)}} = \frac{{\frac{{11}}{9} \times \frac{{11}}{1}}}{{\frac{{11}}{9} + \frac{{11}}{1}}} = \frac{{11}}{{10}}\,\Omega $
Total circuit resistance $ = \frac{{11}}{{10}} + {R_1} = R = 11$ $==>$ ${R_1} = 9.9\,\Omega $
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