Sorry, I don't get you at the "it's a sinwave and you know its amplitude." Amplitude is 10, so therefore maximum voltage is 10? Then what is the point of the Vs equation?
For circuit 1, Vo = Vs - 0.6 because there is only one diode, yes? Just want to confirm.
For circuit 2, the diode current must be found for either D3/D4/D5/D6, since all four diodes share the same forward voltage drop = 0.6 V. So based on that logic, would the diode current be equal to I(R3,4,5,6) = Vo/R3, where Vo = Vs - 0.6 - 0.6 ? I assume this to be the case because there out of four diodes, two diodes are in series with each other with the other two diodes being in series with each other.
I assume this to be the case because there out of four diodes, two diodes are in series with each other with the other two diodes being in series with each other.
Just to clarify: In an usual bridge recitifier circuit:
* for the positive halfwave diodes right_top and left_bottom are active and thus these two are connected in series.
* for the negative halfwave diodes left_top and right_bottom are active and thus these two are connected in series.
... although these diodes don´t have direct connection at the bridge rectifier circuit.