You know that for any KVL loop, the sum of all the voltage drops across all resistances will be equal to each other (both equal to the unknown source voltage). Let the internal source resistance be Rs. Then the KVL for the first circuit is:
[tex]V = 0.8 R_{s} +0.8*2\Omega [/tex]
The second loop is
[tex]V = 0.4 R_{s} +0.4*3\Omega [/tex]
Set these equal since they both equal V:
[tex] 0.8 R_{s} +0.8*2\Omega = 0.4 R_{s} +0.4*3\Omega[/tex]
[tex]R_{s}(0.8-0.4)=(1.2-1.6)\Omega [/tex]
[tex]R_{s}= \frac{-0.4}{-0.4} = 1\Omega[/tex]
