$V=\varepsilon-Ir$
When current is zero (I = 0), $\text{V}=\in$
And when V = 0, $\text{I}=\text{I}_0,\text{ }r=\frac{\in}{I_0}$
$\text{V}=\text{V}(\text{B}_1)-\text{V}(\text{B}_2)=\varepsilon_1-I_1r_1$
$\text{V}=\text{V}(\text{B}_1)-\text{V}(\text{B}_2)=\varepsilon_2-I_2r_2$
$I=I_1+I_2$
$=\frac{\varepsilon_1-\text{V}}{r_1}+\frac{\varepsilon_2-\text{V}}{r_2}=\Big(\frac{\varepsilon_1}{r_1}+\frac{\varepsilon_2}{r_2}\Big)-\text{V}\Big(\frac{1}{r_1}+\frac{1}{r_2}\Big)$
$\text{V}=\frac{\varepsilon_1r_2+\varepsilon_2r_1}{r_1+r_2}-I\frac{r_1r_2}{r_1+r_2}$
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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