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 Ω Ω , the value of R 1
Text Solution
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In figure current through
$R_2 = i - \frac{i}{10} = \frac{9i}{10}$
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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