Refer to the circuit diagram given in the fig | vedclass

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(A) The diode is forward-biased as the potential at $D$ is higher than at $A$. Thus,the diode acts as a short circuit (assuming an ideal diode).$1$. T

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$A, B$ and $D$ only

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(A) The diode is forward-biased as the potential at $D$ is higher than at $A$. Thus,the diode acts as a short circuit (assuming an ideal diode).$1$. The circuit simplifies to a $4 \ \Omega$ resistor in series with a parallel combination of two $4 \ \Omega$ resistors.$2$. The equivalent resistance of the two parallel $4 \ \Omega$ resistors is $R_p = \frac{4 	imes 4}{4 + 4} = 2 \ \Omega$.$3$. The total resistance of the circuit is $R_{net} = 4 \ \Omega + 2 \ \Omega = 6 \ \Omega$. (Statement $A$ is correct,$E$ is incorrect).$4$. The total current in the circuit is $I = \frac{V}{R_{net}} = \frac{6 \ V}{6 \ \Omega} = 1 \ A$. (Statement $B$ is correct).$5$. The potential across $CD$ is $V_{CD} = I 	imes R_{CD} = 1 \ A 	imes 4 \ \Omega = 4 \ V$. (Statement $D$ is correct).$6$. The current splits equally into the two parallel branches,so $0.5 \ A$ flows through the branch $AB$. The potential across $AB$ is $V_{AB} = 0.5 \ A 	imes 4 \ \Omega = 2 \ V$. (Statement $C$ is incorrect).Therefore,statements $A, B,$ and $D$ are correct.

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