A circuit is formed by connecting a 5 V ideal DC source across terminals X and Y. Between X and Y, there is a resistor R₁ = 2 connected in series with a parallel pair of branches. Branch A contains an ideal diode D₁ and a resistor R₂ in series. Branch B contains an ideal diode D₂ and a resistor R₃ in series. The diode D₁ is oriented such that it is forward-biased when X is at a higher potential than Y, while D₂ is fo

Step-by-step solution

When terminal X is at a higher potential than Y (X is positive, Y is negative), diode D₁ is forward-biased and acts as a short circuit (ideal diode), while D₂ is reverse-biased and acts as an open circuit. The active circuit consists of R₁ in series with R₂ . Using Ohm's law: I₁ = V R₁ + R₂ 1 = 5 2 + R₂ 2 + R₂ = 5 R₂ = 3 When the battery polarity is reversed (Y is positive, X is negative), diode D₁ becomes reverse-biased (open circuit) and D₂ becomes forward-biased (short circuit). The active circuit now consists o