For the diagram shown,the resistance between | vedclass

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(B) When the diode is forward biased,it acts as a short circuit (zero resistance). The two resistors $40 \ \Omega$ and $60 \ \Omega$ are in parallel.$

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$\frac{2}{5}$

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(B) When the diode is forward biased,it acts as a short circuit (zero resistance). The two resistors $40 \ \Omega$ and $60 \ \Omega$ are in parallel.$	herefore$ The effective resistance $R_1$ is given by:$R_1 = \frac{40 	imes 60}{40 + 60} = \frac{2400}{100} = 24 \ \Omega$When the diode is reverse biased,it acts as an open circuit (infinite resistance). No current flows through the upper branch.$	herefore$ The effective resistance $R_2$ is simply the resistance of the lower branch,which is $60 \ \Omega$.$	herefore \frac{R_1}{R_2} = \frac{24}{60} = \frac{2}{5}$

For the diagram shown,the resistance between points $A$ and $B$ when the ideal diode '$D$' is forward biased is '$R_1$' and that when reverse biased is '$R_2$'. The ratio $\frac{R_1}{R_2}$ is

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