I know this is a basic question, but I'm just not good at solving diodes in parallel circuits.
I tried applying the current divider rule to get the current flowing through the diode, such that:
Electrical Engineering Stack Exchange is a question and answer site for electronics and electrical engineering professionals, students, and enthusiasts. It only takes a minute to sign up.Sign up to join this community
but I'm just not good at solving diodes in parallel circuits
So, in general for problems like this (which you do hit in real life if you're doing circuit design, so it's not just academic), you start by replacing the diode by an open circuit:
Solve the circuit using whatever means you want. If \$V_d\$ is negative, or less r than the voltage drop you're using for your diode (0V, 0.6V, etc. -- or whatever the prof says), then you're done -- the diode isn't conducting, and life is wonderful. If \$V_d\$ is greater than the designated diode drop, then replace it with a voltage source:
And again, solve the circuit by whatever means you want. Me, I'd do it by a series of simplifications, replacing supplies with their Thevenin equivalents and parallel resistors by their parallel equivalents, to the extent that I could. Unless I have a headache and can't think, and the simplifications just aren't working -- then I'd use node analysis.
In the event that you have multiple diodes, life gets more complicated -- you probably want to do as above, replacing the opens with 0.7V sources where it's clear to do so then repeat -- you may find that as you add diodes other diodes get "turned on" (the obvious case here is a bunch of diodes in series).
As you do a lot of these, you'll either develop an intuition for it or not. I certainly have worked with accomplished analog circuit designers who cannot just glance at a schematic and get it right, so if you can't do it off the bat that doesn't mean you're in the wrong field.
You can solve such circuits by iteratively simplifying them.
Whenever you see two resistors in parallel, such as the 200\$\Omega\$ and 90\$\Omega\$ resistors, combine them. (as you have done.)
Same thing if you see to resistors in series.
Whenever you see a voltage supply, followed by a voltage divider, you may replace that with it's Thevanin equivalent, a reduced voltage with a single resistor.
If you do those operations, your circuit will now look like this (if I have done my math right)