Consider the circuits shown in the figure. Bo | vedclass

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(A) In circuit $(a)$,the total resistance is $R = 11 \ \Omega$. Since both circuits draw the same current $i$ from the battery $E$,the equivalent resi

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$9.9$

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(A) In circuit $(a)$,the total resistance is $R = 11 \ \Omega$. Since both circuits draw the same current $i$ from the battery $E$,the equivalent resistance of circuit $(b)$ must also be $R = 11 \ \Omega$.In circuit $(b)$,the current through $R$ is $i' = \frac{i}{10}$.Therefore,the current through $R_2$ is $i_2 = i - \frac{i}{10} = \frac{9i}{10}$.Since $R_2$ and $R$ are in parallel,the potential difference across them is the same:$R_2 	imes \frac{9i}{10} = R 	imes \frac{i}{10}$$R_2 	imes 9 = R$$R_2 = \frac{R}{9} = \frac{11}{9} \ \Omega$.The equivalent resistance of the parallel combination of $R_2$ and $R$ is:$R_p = \frac{R_2 	imes R}{R_2 + R} = \frac{(\frac{11}{9}) 	imes 11}{\frac{11}{9} + 11} = \frac{\frac{121}{9}}{\frac{110}{9}} = \frac{121}{110} = 1.1 \ \Omega$.The total resistance of circuit $(b)$ is $R_1 + R_p = 11 \ \Omega$.$R_1 + 1.1 = 11$$R_1 = 11 - 1.1 = 9.9 \ \Omega$.

Consider the circuits shown in the figure. Both the circuits draw the same current from the battery,but the current through $R$ in the second circuit is $\frac{1}{10}$th of the current through $R$ in the first circuit. If $R = 11 \ \Omega$,the value of $R_1$ is ................ $\Omega$.

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