We have circuit like this:
There's no internal resisters in batteries.
Is \$\frac{I}{J}\$ equal to zero? If it is nonzero what is the voltage between the ends of the lamp (assuming the lamp has almost infinite resistance)
A small circuit with two diodes
diodes
Related Solutions
As shown in the original post, the lower diodes can never be turned OFF because they'll never be reverse biased.
"Ideal" diodes can be likened to switches with zero resistance between the contacts when ON, and infinite resistance between the contacts when OFF.
Redrawing the circuit from that point of view, with perfect switches and with the original post's context allowing us to turn OFF the two bottom diodes, we have:
Your options 2. and 3. are not valid, because the voltage across the OFF diode is not negative.
For your ideal diode to be OFF it must have a negative voltage and zero current, while it should have a positive current and zero voltage when it is ON. Always check for these 2 conditions.
You are stating that your ideal diode has zero current and zero voltage, which is none of its 2 available states when you are analyzing your circuit.
(Please note that this is being theoretically strict (as we should be when analyzing a circuit of ideal components). Practically your diode can have V = 0, I = 0, but it is not to much use when you are analyzing your circuit)
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Best Answer
An ideal diode analysis gives you an approximate answer:
For a real diode, there will be some reverse leakage current through the "I" diode. This means I could be anywhere from -1 pA to -1 uA, depending on the type of diode. If you choose something like a Shottky diode to minimize the drop when the diode is forward biased, you will probably increase the amount of reverse leakage when it is reverse biased.
Final answer, in a realistic scenario \$\frac{I}{J}\$ will be a small but non-zero negative number.