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Physics · 2025 · 2 marks
CBSE 2025 · Region 1 · Set 1 · Q17
A battery of emf E and internal resistance r is connected to a rheostat. When a current of 2 A is dawn from the battery, the potential difference…
A battery of emf E and internal resistance r is connected to a rheostat. When a current of $\displaystyle 2$ A is dawn from the battery, the potential difference across the rheostat is $\displaystyle 5$ V. The potential difference becomes $\displaystyle 4$ V when a current of $\displaystyle 4$ A is drawn from the battery. Calculate the value of E and r.
Marking-scheme solution
\[\text { Calculating the value of } \mathrm{r}
\]
\[\mathrm{E}=V+I \mathrm{r}
\] In first case
\[\mathrm{E}=5+2 \mathrm{r}
\] In second case
\[\mathrm{E}=4+4 \mathrm{r}
\] After solving
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