Question

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.

Answer 1

$E=6V,r=1\Omega ,V=5.8V,R=?$

$i=\frac{E}{R+r}=\frac{6}{R+1},V=E-ir$

$\Rightarrow 5.8=6-\frac{6}{R+1}\times 1\Rightarrow \frac{6}{R+1}=0.2$

$\Rightarrow R+1=30\Rightarrow R=29\Omega$.