If you don't want the simplifications, you have to fall back to the general model of a diode:
$$
I = I_o \left( e^{\frac{eV}{nkT}}-1 \right)
$$
This equation relates the diode current to the diode voltage (it's V-I characteristic)
- Io - is the diode reverse saturation current
- k - Boltzmann's constant = 1.38e-23 Joules per Kelvin
- T - Analysis temperature (Kelvin)
- e - Magnitude of electric charge
- n - Ideality factor (for silicon diodes, n=2 for small currents and approaches n=1 for large currents; in theory should always = 1)
You can now solve your circuit via the system of equations that it produces. Although you now have continuous V-I functions to describe your elements, a closed-form solution is not always guaranteed to exist.
It is often necessary to use an iterative solution technique such as Newton-Raphson to approximate/approach the answer. This is what SPICE solvers do in the general case... and why they ask you for initial conditions (which can dramatically speed up the solution time).
I think the easiest method to solve such problems is to assume that the diodes are off (both, and then one of the two), compute the voltages across the diodes and see if there's a contradiction with your assumption. Let's call the top left diode \$D_1\$ and the diode in the middle \$D_2\$.
Case 1: \$D_1\$ off, \$D_2\$ off: Since \$D_1\$ is off there is no current through the top 5k resistor, and since \$D_2\$ is off, there is also no current through the bottom left 10k resistor. So \$V=0\$ and the voltage at the anode of \$D_1\$ is 15 Volts. Contradiction! (\$D_1\$ should be on).
Case 2: \$D_1\$ off, \$D_2\$ on: again no current through top 5k resistor. Voltage \$V\$ is
$$V=\frac{15V\cdot(5k||10k)}{10k+(5k||10k)}=3.75V$$
Contradiction! (Because the voltage across \$D_1\$ would be \$15V-3.75V=11.25V\$ and it should be on.)
Case 3: \$D_1\$ on, \$D_2\$ off: Voltage \$V\$ is
$$V=\frac{15V\cdot 10k}{5k+10k}=10V$$
The voltage at the anode of \$D_2\$ is \$15V\cdot 5k/15k=5V\$. This agrees with our assumption, because with these voltages \$D_2\$ must be off. So your solution is
$$I=0A,\quad V=10V$$
Best Answer
Let's look at A and given that your question mentions ideal. So an ideal diode has a Vf = 0. Any voltage greater than 0 will pass through the diode, and any voltage less than 0 will be stopped.
simulate this circuit – Schematic created using CircuitLab
There are 4 possible cases. case 1: D1 = off D2 = off case 2: D1 = off D2 = on case 3: D1 = on D2 = off case 4: D1 = on D2 = on
If we assume case 1 is true. Then the math math behind it should show that we are right. If D1 and D2 are off, that means that the voltage between anode and cathode must be negative.
$$ V_{d1 anode} - V_{d1 cathode} <= 0 $$ $$ +1V - (-5V) = 6V $$
The same can be done with D2 $$ V_{d2 anode} - V_{d2 cathode} <= 0 $$ $$ +2V - (-5V) = 7V $$
But in order for the diodes to be OFF, \$ V_{anode} - V_{cathode} <= 0 \$, but this is a contradiction with our assumption. So both diodes are not off.
Our guess was wrong. So lets try another one of the cases.
Let's assume that case 2 is correct. So this means that the +2V passes straight through the diode. So the node at Vout is 2V.
$$ V_{d1 anode} - V_{d1 cathode} <= 0 $$ $$ +1V - (2) = -1V $$
This works. Case 2 is the answer.
Let's just see what happens with case 3. This means that D1 is on, so the 1V passes straight through D1 and +1V is at the Vout node.
$$ V_{d2 anode} - V_{d2 cathode} <= 0 $$ $$ +2V - (+1V) = 1 $$
This is a contradiction to our initial guess that D2 is off. So this case is incorrect.
With some intuition and some experience, you can more or less "guess" the correct diode state and then just quickly verify that the math holds.
The same can be applied to your other circuit.
When you deal with non ideal diodes, then the diode will switch on, when the voltage between anode and cathode is greater than 0.7V.