Refer to the circuit diagram given in the figure. which of the following observations are correct?
A. Total resistance of circuit is 6 Ω
B. Current in Ammeter is 1 A
C. Potential across A B is 4 Volts.
D. Potential across CD is 4 Volts
E. Total resistance of the circuit is 8 Ω.
Choose the correct answer from the options given below:

Solution

Physics sir · 4 steps

First decide if the diode conducts. Here it does, so the two lower 4 Ω resistors share D to B in parallel.

1. Check the diode. The long plate of the cell is positive, so current leaves it, goes through the ammeter to C and down to D. The diode points from D to A, so D is at the higher potential. The diode is forward biased and acts as a plain wire, taking it as ideal.
2. Total resistance. Between D and B, the middle 4 Ω and the branch D to A to B (another 4 Ω) are in parallel, giving 2 Ω. This is in series with the 4 Ω from C to D.

   R = 4 + (4 × 4)/(4 + 4) = 6 Ω, I = (6 V)/(6 Ω) = 1 A

3. Split the voltage. Across CD: V = 1 × 4 = 4 V. Across D to B: 1 × 2 = 2 V. The AB resistor sits across D and B through the ideal diode, so it carries 0.5 A and has only 2 V across it.
4. So. A (6 Ω), B (1 A) and D (4 V across CD) are correct. C and E are wrong, so the answer is (d).

The idea to keep. With a diode in the circuit, always fix its bias first from the cell's polarity. A forward-biased ideal diode is a wire, a reverse-biased one is a break. Only then combine resistors.

Answer from the NTA final answer key. Solution by PrepWiser.