jaggie34 wrote:knightmoves wrote:As drawn, the 12 and 6 ohm resistors are in parallel, so can be replaced by a single 4 ohm resistor.
The right hand loop on the picture is irrelevant, as the diode is reverse biased and does not conduct (proof left as an exercise for the reader).
So you have a simple loop circuit with a diode and 7 ohms of resistance, so i = 11.25/7 = 1.607A.
Now consider the 6 and 12 ohm resistors separately again. 1/3 of the current goes through the 12 ohm resistor (1/3 of the conductance), so answer = 0.536 A.
Why would you ignore the right side, isn't the diode not in reverse bias in respect to the 4V source?
To determine whether or not the diode is forwards or reverse biased, you need to look at the voltage across the diode. Here, the anode is indeed at 4V, but we have no idea what voltage the cathode is. Thus we can't find the voltage across the diode. If the cathode is at a lower voltage, then the diode is forwards biased and conducts. Otherwise, it's reverse biased. The problem is we don't know the cathode voltage. The way I initially solved the problem is to just assume that the diode is forward biased and is dropping 0.75 V and apply KVL/KCL. But then, I ended up with current flowing backwards through the right diode. This suggests that the diode might actually be reverse biased. Using this assumption, we try again and ignore the right half (since no current flows). As it turns out, the voltage at the cathode using this assumption is 6.432 V which is consistent with the assumption that the right diode is reverse biased since the voltage across the diode is now -2.432 V.