The potential difference between the terminals of a battery of emf 6.0V and internal resistance $1\Omega$ drops to 5.8V when connected across an external resistor. Find the resistance of the external resistor.
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$\text{E}=6\text{V},\text{r}=1\Omega,\text{V}=5.8\text{V},\text{R}=?$
$\text{l}=\frac{\text{E}}{\text{R}+\text{r}}=\frac{6}{\text{R}+1},\text{V}=\text{E}-\text{lr}$
$\Rightarrow5.8=6-\frac{6}{\text{R}+1}\times1\Rightarrow\frac{6}{\text{R}+1}=0.2$
$\Rightarrow\text{R}+1=30\Rightarrow\text{R}=29\Omega.$
$\text{l}=\frac{\text{E}}{\text{R}+\text{r}}=\frac{6}{\text{R}+1},\text{V}=\text{E}-\text{lr}$
$\Rightarrow5.8=6-\frac{6}{\text{R}+1}\times1\Rightarrow\frac{6}{\text{R}+1}=0.2$
$\Rightarrow\text{R}+1=30\Rightarrow\text{R}=29\Omega.$
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