I am having difficulty understanding how you would find the voltage across R2.
Let's say I had this circuit, with Vout as 4.4v and a voltage drop across the diode at 2v. From the mark scheme, it says do 9-2-4.4. I am not sure why this does this, can anyone explain?

 

Do you know II Kirchhoff's law (KVL)??
For your circuit KVL must hold

Vcc = Vout + VR + Vd = 9V
VR = Vcc - Vd - Vout

 

Do you know II Kirchhoff's law (KVL)??
For your circuit KVL must hold

Vcc = Vout + VR + Vd = 9V
VR = Vcc - Vd - Vout

Wait, so why does the voltage from the v out of the transistor not add to the 9v?
I have done work on it but without additional sources.

 

Well, I don't understand you question.
Can you specify your question.
Your circuit has "external 9V additional sources".

 

Wait, so why does the voltage from the v out of the transistor not add to the 9v?

Because they are of opposite polarity. Look at the arrows drawn for you.

If you put two 1.5 volt batteries in series, same polarity, they add, you see 3V across both.

If you put two 1.5 volt batteries in series, but opposite polarity, they subtract, and you see about 0V across both.

 

Wait, so why does the voltage from the v out of the transistor not add to the 9v?
I have done work on it but without additional sources.

If you are confused about the arrows, it basically indicates how one would attach a voltmeter to the circuit.

See the figure attached.

 

Because they are of opposite polarity. Look at the arrows drawn for you.

If you put two 1.5 volt batteries in series, same polarity, they add, you see 3V across both.

If you put two 1.5 volt batteries in series, but opposite polarity, they subtract, and you see about 0V across both.

Now I see it, Cheers .