Hi im a bit rusty my circuits. The picture of the problem is attached and i want to find the current across each diode. I think that the correct way to do it is using 4 different cases to solve for the current. Consisting of the diodes being on and off. When i have both diodes on i calculated Id1 =2mA and Id2=1ma. And when the Diode 1 is on and diode 2 is off i get Id1=2ma and Id2=0mA. However, i am not very sure if i am doing it right, if you can reteach me how to solve i would be very grateful thanks. BTW the turn on voltage is .7V but i dont think that has anything to do with it.
-Vince

 

You started with the correct thinking. In diode problems you take cases of them being open or closed, calculate the currents and then revise the state of the diode by caclulating the current through it. If it biases forward, and you supposed so, then you are correct. Otherwise you should have taken the other case. That said, the fact that it needs 0.7 volts to turn on is important.

In DC circuits where everything is stable, there is always one correct setting for the each diode. Either it will be open or not.

So, let's examine as an example, your first assumption. Remember that as the diodes have a voltage drop, then in mesh analysis we have to replace them with a voltage source of 0.7 volts against the current flow when they counduct and must use supperposition.

First we solve for the source. We have R3 and R2 in parallel, giving 3.333kΩ in series with R1. The current supplied by the source is 10V/13.333kΩ=0.75mA.
The voltage on the diode's anode is Va=10-10k*0.75m=3.5V and so is the voltage on the cathode. This gives 0.7mA for D1 and 0.35mA for D2.

Then, let's solve for the voltage source of the first diode. We nullify all other voltage sources. The two 10k resistors are now in parallel and give a 5kΩ resistor. The current through D1 is now -0.07mA and through D2 is +0.07mA.

Finally, we solve for the source of the second diode. This time, R1 and R2 are in parallel. The current through D2 is -0.0525mA and through D1 is +0.035mA.

So in total the current through D1 is 0.7-0.07+0.035= 0.665mA
and through D2 0.35+0.07-0.0525=0.42mA.
They are both positive, so we are OK and our assumption about them being conductive is correct.

One more thing. If we assumed them to be reversed biased and not conducting, then we would measure the voltage on their edges instead of the current through them, in order to find if they conduct or not.