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Q.In the circuit shown, containing two ideal diodes D1 and D2, what is the magnitude of the current flowing through the 1Ω resistor?

Tripura TbseHigher Secondary (+2 Stage) Examination 2025Subjective· 1mImportance★★★★★
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An ideal diode acts as a short circuit (0 resistance) when forward-biased and an open circuit (infinite resistance) when reverse-biased -- first find which of D1, D2 conducts for the battery's polarity, redraw the circuit keeping only that branch, then apply Ohm's law to the 1(ohm) resistor.

Method (apply once the exact printed figure is legible):

  1. Note the battery's polarity (its + and - terminals) at the bottom of the loop containing the 1(ohm) resistor.
  2. Check each diode's orientation against that polarity. A diode conducts (acts as a plain wire) only if the battery drives conventional current from its anode to its cathode; otherwise it blocks (acts as an open circuit, carrying zero current).
  3. Redraw the circuit keeping only the branch(es) whose diode conducts; remove/open the branch whose diode is reverse-biased.
  4. Reduce the remaining resistor network (the described 2(ohm) combinations) to a single equivalent resistance in series with the 1(ohm) resistor and the battery.
  5. Apply I=EMFRtotalI = \dfrac{\text{EMF}}{R_{total}} to get the current, which is the same current that flows through the 1(ohm) resistor since it is a series element of the single conducting loop. …

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