Compared to an ideal battery, by what percentage does the battery’s internal resistance reduce the potential difference across the $20\Omega$resistor in FIGURE EX28.20?

We have to find the percentage of battery's internal resistance reduce the potential diffrence across the $20\Omega$resistor.

The ideal battery has zero internal resistance $r=0$. So, determine the voltage across the resistor $20\Omega$for the ideal battery. For the ideal battery, the voltage across the combination of both resistors $20\Omega$and $10\Omega$is the same for the battery $△V=\epsilon =15V$.So, calculating the current through the circuit by

$I=\frac{△V}{R_{eq}}=\frac{△V}{R_1+R_2}$

Putting the values for $△V,R_1$and $R_2$into equation $\left(1\right)$to get $I$by

$I=\frac{△V}{R_1+R_2}=\frac{15V}{10\Omega +20\Omega }=0.5A$

The potential difference across the resistor $20\Omega$is

localid="1648903923181" $△V_{20\Omega }=IR=(0.5A)\left(20\Omega \right)=10V$

This the ideal case. the battery has an internal resistance $r=1\Omega$, the current in equation $\left(1\right)$will be

$I_r=\frac{\epsilon }{r+R_1+R_2}=\frac{15V}{1\Omega +10\Omega +20\Omega }=0.483A$

The potential difference across the resistor $20\Omega$is

$△V_{20\Omega ,r}=IR=(0.483A)\left(20\Omega \right)=9.67V$

The percentage of the reduction by

$$\begin{gathered} \%=\frac{△V_{20\Omega }-△V_{20\Omega },r}{△V_{20\Omega }}(100\%) \\ =\frac{10V-9.67V}{10V}(100\%) \\ =3.3\% \end{gathered}$$