Answer
$44V, 43\Omega $
Work Step by Step
According to Kirchhoff's loop rule, $\epsilon-Ir-IR=0$
This can be rearranged as:
$ IR+Ir=\epsilon $
We plug in the known values to obtain:
$(0.65A)(25\Omega)+(0.65A)r=\epsilon $....eq(1)
Now for $ I=0.45A $, and $ R=55\Omega $, we have
$(0.45A)(55\Omega)+(0.45A)r=\epsilon $....eq(2)
Subtracting eq(2) from eq(1), we have
$0=8.5V-(0.20A)r $
This simplifies to:
$ r=\frac{8.5V}{0.20V}$
$ r=43\Omega $
Now we plug in $ r=43\Omega $ in eq(1) to obtain:
$(0.65A)(25\Omega)+(0.65A)(43\Omega)=\epsilon $
This simplifies to:
$\epsilon=44V $
