Q146 / 180 · COMEDK 2017 · Physics

If the potential difference across the internal resistance $r_{1}$ is equal to the $\operatorname{emf} E$ of the battery, then
Figure

  • A
    $R=r_{1}+r_{2}$
  • B
    $R=\frac{r_{1}}{r_{2}}$
  • Correct answer
    C
    $R=r_{1}-r_{2}$
  • D
    $R=\frac{r_{2}}{r_{1}^{*}}$

Solution

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 $\begin{array}{ll} & \frac{E}{r_{1}}=\frac{2 E}{r_{1}+r_{2}+R} \\ \Rightarrow \quad & \quad r_{1}+r_{2}+R=2 r_{1} \\ \Rightarrow \quad & R=r_{1}-r_{2} \end{array}$

Hence, the correct answer is option: 'C'.

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