For real batteries:
Assuming the battery positive terminals are
connected together, the 10v battery is charging
the 5v battery.
Let's enter the real world:Example:
R5=1 ohm internal resistance of 5v battery
R10=10 ohm internal resistance of 10v battery
Kirchhoff's voltage law:
-----electron flow in wire-->
__+10 Volt -__- 5 Volt+_
| ****************************|
| ****************************|
|___ _r10________r5___ |
******-Vr10*********-Vr5
<--electron flow in wire CW loop
10v-5v-vR5-VR10=0
10v+(-5v)+(-iR5)+(-iR10)=0
solving: i = 5/11 amps
sum of voltages: 10+ (-5) + (-5 / 11 * 1) + ((-5 / 11) * 10) = 0
terminal voltage = 9.54545455
If you add the resistor to the circuit, the net terminal voltage
will further drop. Solve for new total current.