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A cell of emf \(\varepsilon\) and internal resistance \(r\) is connected across a variable load resistance \(R\). The graph drawn between its terminal voltage and resistance \(R\) is
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Verified Answer
The correct answer is:
According to the question, circuit diagram of a cell is given below,
As,
\(I r=E-V \Rightarrow \frac{V}{R} r=E-V\)
\(\Rightarrow V\left[1+\frac{r}{R}\right]=E \Rightarrow V=\frac{E}{R+r} \cdot R\)
General equation of a straight line passing through the origin,
\(y=m x\)
Here, \(y=V, m=\frac{E}{R+r}=\) slope of the line and \(\quad x=R\)
As this line is passing through the origin and has a slope which reduces as \(R\) increases. Hence, the graph is
As,
\(I r=E-V \Rightarrow \frac{V}{R} r=E-V\)
\(\Rightarrow V\left[1+\frac{r}{R}\right]=E \Rightarrow V=\frac{E}{R+r} \cdot R\)
General equation of a straight line passing through the origin,
\(y=m x\)
Here, \(y=V, m=\frac{E}{R+r}=\) slope of the line and \(\quad x=R\)
As this line is passing through the origin and has a slope which reduces as \(R\) increases. Hence, the graph is
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