A student measures the terminal potential difference (V) of a cell (of emf $\varepsilon$ and internal)…

A student measures the terminal potential difference (V) of a cell (of emf $\varepsilon$ and internal) resistance $r$ ) as a function of the current (I) flowing through it. The slope and intercept of the graph between $\mathrm{V}$ and $\mathrm{I}$, then respectively equal to :
  1. $-\epsilon$ and $r$
  2. $\in$ and $-r$
  3. $-\mathrm{r}$ and $\in$
  4. $r$ and $-\epsilon$

Solution

$\begin{aligned} & \quad \mathrm{E}=\mathrm{V}+\mathrm{Ir} \\ & \Rightarrow \mathrm{V}=\mathrm{E}-\mathrm{Ir} \\ & \text { Comparing with } \mathrm{y}=\mathrm{mx}+\mathrm{c} \\ & \text { Slope }=-\mathrm{r} \text {, intercept }=\mathrm{E} \end{aligned}$

Asked in: NEET 2009 (Mains)

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