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(A) For an ideal diode,the forward bias resistance is $0 \ \Omega$ and the reverse bias resistance is $\infty \ \Omega$.$(i)$ Case $V_A > V_B$:In this

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$25, \infty$

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(A) For an ideal diode,the forward bias resistance is $0 \ \Omega$ and the reverse bias resistance is $\infty \ \Omega$.$(i)$ Case $V_A > V_B$:In this case,both diodes $D_1$ and $D_2$ are forward biased.Therefore,the resistance of each branch is $50 \ \Omega + 0 \ \Omega = 50 \ \Omega$.Since the two branches are in parallel,the equivalent resistance $R_{AB}$ is given by:$\frac{1}{R_{AB}} = \frac{1}{50} + \frac{1}{50} = \frac{2}{50} = \frac{1}{25}$$R_{AB} = 25 \ \Omega$.(ii) Case $V_B > V_A$:In this case,both diodes $D_1$ and $D_2$ are reverse biased.Therefore,the resistance of each branch is $50 \ \Omega + \infty \ \Omega = \infty \ \Omega$.Since the two branches are in parallel,the equivalent resistance $R_{AB}$ is given by:$\frac{1}{R_{AB}} = \frac{1}{\infty} + \frac{1}{\infty} = 0 + 0 = 0$$R_{AB} = \infty \ \Omega$.Thus,the equivalent resistances are $25 \ \Omega$ and $\infty \ \Omega$ respectively.

For the circuit shown in the figure,the equivalent resistance between points $A$ and $B$ for two cases $(i)$ $V_A > V_B$ and (ii) $V_B > V_A$ is $\qquad \Omega$ and $\qquad \Omega$ respectively. ($D_1$ and $D_2$ are ideal diodes)

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