Why is there a power drop in a serial circuit, but not in a parallel circuit?
You are basically on the right track.

In both circuits, a perfect battery will output 3&nbsp;V, and as much current as the load demands. Obviously a practical battery will have some internal resistance and a maximum current it can supply. The important point is that a perfect battery does not control the current - the load does.

In circuit 1, the effective resistance of the motor running at full speed is quite high. The motor consumes only the power needed to satisfy electrical and mechanical losses. Once you try to slow it down, it needs to provide mechanical output power, so it draws more current, and the effective resistance falls. So stalling the motor will make the lamp brighten.

In circuit 2, the lamp and motor each have 3&nbsp;V across them, and the current drawn will be the sum of the lamp current and the motor current. So stalling the motor will cause a high current to be drawn, but if the battery is perfect, the lamp brightness will be unaffected.