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If the potential difference across the internal resistance $r_{1}$ is equal to the $\operatorname{emf} E$ of the battery, then
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Verified Answer
The correct answer is:
$R=r_{1}-r_{2}$
Total emf of circuit, $E_{T}=E+E=2 E$
Total resistance of circuit, $R_{T}=r_{1}+r_{2}+R$
Current flowing through the circuit,
$i=\frac{E_{T}}{R_{T}}=\frac{2 E}{r_{1}+r_{2}+R}...(i)$
As per question $E=i r_{1}$
$\Rightarrow \quad i=\frac{E}{r_{1}}...(ii)$
From Eqs. (i) and (ii), we get
$\begin{array}{ll}
& \frac{E}{r_{1}}=\frac{2 E}{r_{1}+r_{2}+R} \\
\Rightarrow \quad & \quad r_{1}+r_{2}+R=2 r_{1} \\
\Rightarrow \quad & R=r_{1}-r_{2}
\end{array}$
Total resistance of circuit, $R_{T}=r_{1}+r_{2}+R$
Current flowing through the circuit,
$i=\frac{E_{T}}{R_{T}}=\frac{2 E}{r_{1}+r_{2}+R}...(i)$
As per question $E=i r_{1}$
$\Rightarrow \quad i=\frac{E}{r_{1}}...(ii)$
From Eqs. (i) and (ii), we get
$\begin{array}{ll}
& \frac{E}{r_{1}}=\frac{2 E}{r_{1}+r_{2}+R} \\
\Rightarrow \quad & \quad r_{1}+r_{2}+R=2 r_{1} \\
\Rightarrow \quad & R=r_{1}-r_{2}
\end{array}$
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