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Diode circuit questions look easy until two diodes and two batteries share one network. Then students guess which diode conducts, get it wrong, and the rest of the answer follows the wrong guess. There is a reliable method that turns every such question into an ordinary resistor problem plus one check. This post explains it and works through three networks with numbers.
What does "ideal diode" mean in a problem?
The NCERT chapter on semiconductor electronics explains that a p-n junction diode has low resistance in forward bias and very high resistance in reverse bias. It also gives the threshold or cut-in voltage, the forward voltage beyond which current rises sharply: about 0.7 V for silicon and 0.2 V for germanium. Problems use one of two models:
Ideal diode: forward biased, it is a plain wire with zero voltage drop. Reverse biased, it is an open gap with zero current.
Ideal diode with an offset: forward biased, it has a fixed drop (say 0.7 V) and no resistance. Below that voltage, it is open.
Read the question to see which model it wants. "Ideal" with no voltage given means the first.
Which exams test this?
The Syllabus for JEE (Main) 2026 lists the semiconductor diode, its I-V characteristics and the diode as a rectifier in Unit 19. The physics syllabus in the JEE (Advanced) 2026 Information Brochure does not list semiconductor devices, so this is mainly JEE Main and Class 12 board material. Check the syllabus for your year.
What is the assume-and-check method?
Guess a state for each diode: ON or OFF.
Replace each ON diode by a wire (or a fixed drop), and each OFF diode by a gap.
Solve the resulting circuit with Ohm's law and Kirchhoff's rules.
Check every guess. An ON diode must carry current from anode to cathode. An OFF diode must have its anode no higher than its cathode (or less than the offset above it). If any check fails, flip that guess and solve again.
Worked example 1: two parallel branches, one diode reversed
A 12 V battery drives current through a 2 kΩ resistor into a node. From that node, two branches return to the negative terminal:
Branch A: diode D1, pointing along the current, in series with 4 kΩ.
Branch B: diode D2, pointing against it, in series with 2 kΩ.
All diodes are ideal.
Guess: D1 ON, D2 OFF.
Circuit: 12 V across 2 kΩ + 4 kΩ in series. I = 12 / 6 kΩ = 2 mA.
Check D1: 2 mA flows anode to cathode. Valid.
Check D2: the node sits at 2 mA × 4 kΩ = 8 V above the negative terminal. D2's cathode is on the node side, so it is reverse biased by 8 V. Valid.
Answer: 2 mA through D1, none through D2. The common slip is to treat branch B as a parallel path and split the current.
Worked example 2: two batteries pulling against each other
Point A connects through 1 kΩ to the +5 V terminal of a battery. Point B connects through 1 kΩ to the +2 V terminal of another battery. Both negative terminals are grounded. An ideal diode joins A to B, anode at A.
Guess ON. Then A and B are the same point at voltage V.
Current in from the 5 V side equals current out to the 2 V side: (5 − V)/1 kΩ = (V − 2)/1 kΩ, so V = 3.5 V.
Diode current = (5 − 3.5)/1 kΩ = 1.5 mA, from A to B. Positive, so ON is valid.
Now reverse the diode (anode at B). Guess ON again: the same arithmetic gives 1.5 mA flowing from A to B, which is cathode to anode. The check fails, so the diode is OFF. No current flows, A sits at 5 V, B at 2 V, and the diode holds off 3 V in reverse.
Worked example 3: silicon and germanium in parallel
A 12 V battery, a 5 kΩ resistor and a pair of diodes in parallel, both forward biased: one silicon (0.7 V offset) and one germanium (0.2 V offset), using the cut-in values from NCERT.
Guess: both ON. That needs the parallel pair to be at 0.7 V and 0.2 V at the same time. Impossible.
Guess: Ge ON, Si OFF. The pair sits at 0.2 V. Si would need 0.7 V to conduct, so it stays OFF. Valid.
Current: I = (12 − 0.2) / 5 kΩ = 2.36 mA, all through the germanium diode.
The diode with the lower cut-in voltage clamps the pair, and the other one never turns on.
Common mistakes and how to avoid them
Splitting current into a reversed branch. An OFF ideal diode is a gap, so its branch carries no current at all.
Skipping the check. A wrong guess still gives numbers. Only the check shows they are wrong.
Using 0.7 V when the question says ideal. Ideal means zero drop unless a value is given.
Forgetting the reverse voltage. Some questions ask for the voltage across the OFF diode. Find it from node voltages after solving.
For the sign rules in step 3, our post on Kirchhoff's laws and circuit mistakes covers loops and junctions. For the physics behind the curves, see p-n junction diode characteristics.
A practice routine
Take five diode-network questions from past papers. Before solving any, write a guess for every diode.
Solve each with the four steps, writing the check as a separate line.
For each one you get wrong, note which check would have caught it.
Once a week, redo one network with the diodes reversed and predict what changes before calculating.
When a check fails and you cannot see why, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Start with example 2 above: reverse both batteries and decide, before any arithmetic, which way the diode must point to conduct.
