Question: Q. 12. Two cells of emfs
A [Delhi I, II, III 2016]
Solution:
Ans.
$$ \text { E.M.F., } \begin{aligned} E & =\frac{\mathrm{E}{1} r{2}+\mathrm{E}{2} r{1}}{r_{1}+r_{2}} \ & =\frac{(1.5 \times 0.3)+(2 \times 0.2)}{0.2+0.3} \end{aligned} $$
Internal Resistance,
[CBSE Marking Scheme 2016]
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