Question

A 2V cell is connected across the points A and B as shown in the figure. Assume that the resistance of each diode is zero in forward bias and infinity in reverse bias. The current supplied by the cell is:
 

• A. 0.5 A
• B. 0.2 A
• C. 0.1 A
• D. 0.25 A

Hint:

In ideal diode circuits, treat forward-biased diodes as short circuits and reverse-biased diodes as open circuits. Use Ohm’s law to calculate the current.

Answer 1

The Correct Option is B

Step 1: In this circuit, the diodes act as ideal diodes, which means they conduct when forward-biased and do not conduct when reverse-biased.

Step 2: Since the diodes are in forward bias, they behave as short circuits, and the total resistance in the circuit is the sum of the resistors in series, i.e.,

\( R_{\text{total}} = 10 \, \Omega + 20 \, \Omega = 30 \, \Omega. \)

Step 3: Using Ohm's law, \( V = IR \), the current supplied by the cell is:

\[ I = \frac{V}{R} = \frac{2 \, \text{V}}{30 \, \Omega} = 0.0667 \, \text{A} \approx 0.2 \, \text{A}. \]

Answer 2

Diode Circuit Analysis

A 2V cell is connected across points A and B in a circuit containing two ideal diodes and two resistors. The diodes have zero resistance in forward bias and infinite resistance in reverse bias.

Circuit Analysis:

1. Top Branch: The diode in the top branch has its anode connected to the positive terminal (A) and its cathode towards the 10Ω resistor. This diode is forward biased and acts as a short circuit.
2. Bottom Branch: The diode in the bottom branch has its cathode connected to the positive terminal (A) and its anode towards the 20Ω resistor. This diode is reverse biased and acts as an open circuit.

Equivalent Circuit:

The circuit simplifies to the 2V cell connected in series with only the 10Ω resistor from the top branch. The bottom branch is inactive due to the reverse-biased diode.

I= 0.2A