Question: Q. 18. A battery of emf $E$ and internal resistance, $r$, when connected with an external resistance of $12 \Omega$ produces a current of $0.5 \mathrm{~A}$. When connected across a resistance of $25 \Omega$, it produces a current of 0.25 A. Determine (i) the emf and (ii) the internal resistance of the cell.
A [O.D. Comptt. I, II, III 2013]
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Solution:
Ans. Given : $\quad R_{1}=12 \Omega, I_{1}=0.5 \mathrm{~A}$
From I condition,
$$ \begin{align*} E & =I_{1}\left(R_{1}+r\right) \ E & =0.5(12+r) \tag{i}\ & =6+0.5 r \end{align*} $$
$$ O^{-}=0.5(12+r) $$
From II condition,
$$ \begin{align*} E & =I_{2}\left(R_{2}+r\right) \ E & =0.25(25+r) \ & =6.25+0.25 r \tag{ii} \end{align*} $$
From eqns. (i) and (ii)
$$ \begin{array}{rlrl} & & 6+0.5 r & =6.25+0.25 r \ \Rightarrow & 0.5 r-0.25 r & =6.25-6.0 \ \Rightarrow & 0.25 r & =0.25 \ \Rightarrow & r & =1 \Omega \ \text { Then } & E & =6+0.5 \times 1 \ E & =6.5 \mathrm{~V} \end{array} $$
$$ \Rightarrow \quad r=1 \Omega \quad 1 / 2 $$
