current-electricity Question 22
Question: Q. 12. Two cells of emfs $1.5 \mathrm{~V}$ and $2.0 \mathrm{~V}$ having internal resistances $0.2 \Omega$ and $0.3 \Omega$ respectively are connected in parallel. Calculate the emf and internal resistance of the equivalent cell.
A [Delhi I, II, III 2016]
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Solution:
Ans.
$$ \text { E.M.F., } \begin{aligned} E & =\frac{\mathrm{E}{1} r{2}+\mathrm{E}{2} r{1}}{r_{1}+r_{2}} \ & =\frac{(1.5 \times 0.3)+(2 \times 0.2)}{0.2+0.3} \end{aligned} $$
$$ \begin{aligned} & =\frac{0.45+0.4}{0.5} \ & =1.7 \mathrm{~V} \end{aligned} $$
Internal Resistance,
$$ \begin{aligned} r & =\frac{r_{1} r_{2}}{r_{1}+r_{2}} \ & =\frac{0.2 \times 0.3}{0.2+0.3}=\frac{0.06}{0.5} \end{aligned} $$
$$ \begin{equation*} =0.12 \Omega \tag{1} \end{equation*} $$
[CBSE Marking Scheme 2016]
