For the diagram shown,the resistances between | vedclass

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(D) When the diode $D$ is forward biased,it acts as a short circuit (zero resistance). The circuit consists of two resistors of $40 \ \Omega$ and $400

Vedclass · Accepted Answer

$1: 11$

Vedclass · Answer

(D) When the diode $D$ is forward biased,it acts as a short circuit (zero resistance). The circuit consists of two resistors of $40 \ \Omega$ and $400 \ \Omega$ in parallel.Therefore,the effective resistance $R_1$ is given by:$R_1 = \frac{40 	imes 400}{40 + 400} = \frac{16000}{440} = \frac{1600}{44} = \frac{400}{11} \ \Omega$When the diode $D$ is reverse biased,it acts as an open circuit (infinite resistance). No current flows through the upper branch.Therefore,the effective resistance $R_2$ is simply the resistance of the lower branch:$R_2 = 400 \ \Omega$Now,calculating the ratio $R_1: R_2$:$\frac{R_1}{R_2} = \frac{400/11}{400} = \frac{1}{11}$Thus,the ratio $R_1: R_2$ is $1: 11$.

For the diagram shown,the resistances between points $A$ and $B$ when the ideal diode $D$ is forward biased is $R_1$ and when it is reverse biased is $R_2$. The ratio $R_1: R_2$ is

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