Diode capacitor circuit, what is the output across the diode

Lets look at this circuit in a slightly clearer topology - below is exactly the same circuit, but rearranged to put all the elements in a way that height is signifying voltage - the higher up in the diagram, the higher the voltage. (It's a nice analogy which helps visualise what is going on).

^{simulate this circuit – Schematic created using CircuitLab}

So, at the top of the diagram is \$20 + 10 = 30 \mathrm{V}\$. At the bottom is \$0 \mathrm{V}\$. So lets see if we can work out the currents \$I_r\$ and \$I_d\$. I should note that the direction I have drawn the current arrows is the direction that conventional current would have to flow (positive to negative).

Well, for \$I_r\$ it is fairly easy. Ohm's law tells us for a resistor (or any device with 'Ohmic' behaviour), that \$V=I\times R\$, so from this we can work out:

$$I_r = \frac{V}{R} = \frac{30}{50} = 0.6 \mathrm{A}$$

Why? Well, at the top of the diagram is \$30\mathrm{V}\$, and at the bottom is \$0\mathrm{V}\$, so there must be that much voltage across the resistor.

Now lets look at the diode. For a diode to conduct the voltage at the Anode must be higher than the voltage at the cathode ^(*). Now in this diagram we can see that the Anode is at \$20\mathrm{V}\$, and the cathode is at \$30\mathrm{V}\$, so this means the voltage across the diode \$V_ak = 20 - 30 = -10\mathrm{V}\$. So from this we can see that the diode is reverse biased - the voltage at the anode is negative with respect to the cathode. So we know then that \$I_d = 0 \mathrm{A}\$.

The current can't 'split down the parallel branch', because the diode is reverse biased so is blocking the flow of current - it's acting essentially as an open circuit.

If the diode was placed the other way around (not a good idea!), then current could flow down that branch. But the current that flows would not reduce the amount of current that flows through the resistor. Instead it increases the amount of current drawn from the 10V supply ^(**).

_{(*) There is some 'reverse leakage current' for a diode, basically the amount of current that it conducts when reverse biased, but this is pretty small so for simplicity we say it is \$0\$.}

_{(**) Assuming an ideal power supply - one which has no limit on how much current you can draw.}

To solve these kinds of circuits you have to make an assumption about the state of each diode (whether it is on or off) and solve the circuit based on that assumption. If in solving the circuit you arrive at a contradiction (either the diode has a nonzero current through it but you assumed no voltage across it, or the diode has no current through it but you assumed 0.7V across it) then your assumption was wrong.

This circuit has only one diode so there are only two possible solutions: the diode is on, or it is off.

First assume that the diode is off (i.e. that the current \$I_3\$ through it is 0). By KCL that means \$I_1 = I_4\$ (you are correct that \$I_2 = 0\$ in steady state). Similarly, by KCL \$I_0 = I_1\$. \$I_0\$ is flowing through the two resistors in series so it equals

$$I_0 = \frac{U_0}{R_1 + R_4} = \frac{3.5}{280 + 350} = 5.5\text{ mA}$$

Since \$I_0 = I_4\$ the voltage across \$R_4\$ is \$U_4 = I_4 \times R_4 = 5.5\text{ mA} \times 350 = 1.94\text{ V}\$. However, \$U_4 = U_3 > 0.7\text{ V}\$ so the diode would be on. This is a contradiction so the diode must not be off as assumed.

Now assume the diode is on (the voltage \$U_3\$ across it is 0.7V). \$U_4 = U_3\$ so $$I_4 = U_4/R_4 = 0.7/350 = 2\text{ mA}$$ By KVL \$U_0 = U_1 + U_3\$, so rearranging we have $$U_1 = U_0 - U_3 = 3.5 - 0.7 = 2.8\text{ V}$$ That means $$I_1 = U_1/R_1 = 10\text{ mA}$$ By KCL \$I_1 = I_3 + I_4\$, and rearranging we have $$I_3 = I_1 - I_4 = 10\text{ mA} - 2\text{ mA} = 8\text{ mA}$$ We have a nonzero current through the diode so there is no contradiction -- the diode is on.

You should be able to figure out the other variables (like \$I_5\$) from here.