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assume diode forward biased, forward diode drop = 0V. Iyz = 2mA. Ixy = 1mA. our diode current direction does not contradict the sign of our current. hence, Idiode = 1mA.
assume diode forward biased, forward diode drop = 0V. Iyz = 2mA. Ixy = 1mA. our diode current direction does not contradict the sign of our current. hence, Idiode = 1mA.
You can reduce the batteries and R's to its Thevenin equivalent. After that use the diode model you like: Ideal, Constant Voltage, Constant V and R drop or any non-linear equivalent.
Hi
Actually, I think it's better to follow the " Superposition " theory to solve this ploblem.
After that, assuming diode to be whether forward-biased or so, can help make it easier to continue solving.
Anyway, the Thevenin would also be useful too.
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