Q171 / 200 · AP EAMCET 2001 · Physics

When a resistor of $11 \Omega$ is connected in series with an electric cell, the current flowing in it is 0.5 A. Instead, when a resistor of $5 \Omega$ is connected to the same electric cell in series, the current increases by $0.4 \mathrm{~A}$. The internal resistance of the cell is

  • A
    $1.5 \Omega$
  • B
    $2 \Omega$
  • Correct answer
    C
    $2.5 \Omega$
  • D
    $3.5 \Omega$

Solution

Current taken from the cell, $i=\frac{E}{R+r}$ where $R=$ external resistance and $r=$ internal resistance $\begin{aligned} \text { Ist Case } i_1 & =\frac{E}{R_1+r} \\ 0.5 & =\frac{E}{11+r}\end{aligned}$ $\begin{aligned} & \text { IInd Case } 0.5+0.4=\frac{E}{5+r} \\ & \Rightarrow \quad 0.9=\frac{E}{5+r}\end{aligned}$ Dividing Eq. (ii) by (i), $\begin{aligned} \frac{0.5}{0.9} & =\frac{\frac{E}{(11+r)}}{\frac{E}{(5+r)}} \\ \frac{5}{9} & =\frac{5+r}{11+r} \\ 55+5 r & =45+9 r \\ 10 & =4 r \Rightarrow r=2.5 \Omega\end{aligned}$

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

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