Refer to the circuit diagram given in the figure. which of the following observations are correct?

A. Total resistance of circuit is $$6\Omega$$

B. Current in Ammeter is 1 A 

C. Potential across AB is 4 Volts.

D. Potential across CD is 4 Volts 

E. Total resistance of the circuit is $$8\Omega$$ . 

Choose the correct answer from the options given below:

Analyze the orientation of the diode in the given circuit. The positive terminal of the 6V battery is connected to the p-side of the diode through the top resistor, and the n-side is connected to the negative terminal through the resistor AB. This means the diode is forward-biased.

Assuming it is an ideal diode, it acts as a short circuit with zero resistance.

equivalent resistance of the parallel part (between node D and the negative terminal):

$$ R_p = \frac{4 \times 4}{4 + 4} $$

$$ R_p = \frac{16}{8} $$

$$ R_p = 2 \Omega $$

$$ R_{total} = R_{CD} + R_p $$

$$ R_{total} = 4 + 2 $$

$$ R_{total} = 6 \Omega $$

So A correct and E incorrect.

Calculate the total current ($$I$$) flowing through the circuit, which is the reading of the ammeter:

$$ I = \frac{E}{R_{total}} $$

$$ I = \frac{6}{6} $$

$$ I = 1 \text{ A} $$

This makes observation B correct.

Calculate the potential difference across the resistor between C and D ($$V_{CD}$$):

$$ V_{CD} = I \times R_{CD} $$

$$ V_{CD} = 1 \times 4 $$

$$ V_{CD} = 4 \text{ V} $$

This makes observation D correct.

Calculate the potential difference across AB ($$V_{AB}$$). The diode is ideal, so there is no voltage drop across it. The voltage across the branch AB is the same as the total voltage across the parallel combination.

$$ V_{AB} = I \times R_p $$

$$ V_{AB} = 1 \times 2 $$

$$ V_{AB} = 2 \text{ V} $$

This makes observation C incorrect.

Therefore, the correct observations are A, B, and D.