In this circuit:
The diode is an ideal diode
The diode is on since V(anode) > V(cathode) and then we replace it by a short circuit.
Now, how to calculate Vo when two sources are existed?
I = (10 + 2) / (2k + 4.7k) = 1.79 mA.
I tried using KVL:
-10 + 2k (1.79 mA) + Vo - (-2) = 0
=> Vo = 4.42
Vo = 4700 * (1.79 mA) = 8.413
Which answer is correct? and why the other one is not correct?