1. Plot a graph showing variation of voltage vs the current drawn… - Vidyadip

Question

Plot a graph showing variation of voltage vs the current drawn from the cell. How can one get information from this plot about the emf of the cell and its internal resistance?
Two cells of emf’s $E_1$ and $E_2$ and internal resistance $r_1$ and $r_2$ are connected in parallel. Obtain the expression for the emf and internal resistance of a single equivalent cell that can replace this combination?

✓

Answer

$V=\varepsilon-Ir$
When current is zero $(I = 0), \text{V}=\epsilon$
And when $V = 0, \text{I}=\text{I}_0,\text{ }r=\frac{\epsilon}{I_0}$

$\text{V}=\text{V}(\text{B}_1)-\text{V}(\text{B}_2)=\varepsilon_1-I_1r_1$
$\text{V}=\text{V}(\text{B}_1)-\text{V}(\text{B}_2)=\varepsilon_2-I_2r_2$
$I=I_1+I_2$
$=\frac{\varepsilon_1-\text{V}}{r_1}+\frac{\varepsilon_2-\text{V}}{r_2}=\Big(\frac{\varepsilon_1}{r_1}+\frac{\varepsilon_2}{r_2}\Big)-\text{V}\Big(\frac{1}{r_1}+\frac{1}{r_2}\Big)$
$\text{V}=\frac{\varepsilon_1r_2+\varepsilon_2r_1}{r_1+r_2}-I\frac{r_1r_2}{r_1+r_2}$
On comparing with
$\text{V}=\varepsilon_{eq}-Ir_{eq}$
we get
$\varepsilon_{eq}=\frac{\varepsilon_1r_2+\varepsilon_2r_1}{r_1+r_2}$
$r_{eq}=\frac{r_1r_2}{r_1+r_2}$
Alternate Answer
A student may write the last two results in the following form.
$\frac{1}{r_{eq}}=\frac{1}{r_1}+\frac{1}{r_2}$
$\frac{\varepsilon_{eq}}{r_{eq}}=\frac{\varepsilon_1}{r_1}+\frac{\varepsilon_2}{r_2}$

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