Answer
$40\%$
Work Step by Step
For ideal battery
$R_{eq} =10\Omega + 20\Omega =30\Omega $
current $I = \frac{15}{30}A = 0.5A$
potential difference across $20\Omega $
$V_i = IR =( 0.5*20)V =10V$
For real battery
$R_{eq} =1\Omega +10\Omega +20\Omega =31\Omega $
current
$I =\frac {15}{31}A =0.48A$
potential difference across $20\Omega $
$V_r = IR = 90..48 *20)V =9.6V$
percentage by which internal resistance decreases potential difference across $20\Omega $is
$\frac{V_i -V_r}{V_r}*100 \%$
$= \frac {10-9.6}{10} *100\% = 40\%$
