Step 1: Understanding the Question:
A diode is in forward bias if the P-side (triangle part) is at a higher potential than the N-side (line part). Step 3: Detailed Explanation:
Let's analyze the potentials at the P and N terminals for each case:
(A) P-side is connected to \(0 \text{ V}\), N-side is connected to \(-4 \text{ V}\).
Potential difference \(V_P - V_N = 0 - (-4) = +4 \text{ V}\). Since \(V_P>V_N\), it is forward biased.
(B) P-side is at \(-4 \text{ V}\), N-side is at \(-3 \text{ V}\).
Potential difference \(V_P - V_N = -4 - (-3) = -1 \text{ V}\). Since \(V_P<V_N\), it is reverse biased.
(C) P-side is at \(-2 \text{ V}\), N-side is at \(+2 \text{ V}\).
Potential difference \(V_P - V_N = -2 - 2 = -4 \text{ V}\). Since \(V_P<V_N\), it is reverse biased.
(D) P-side is at \(3 \text{ V}\), N-side is at \(5 \text{ V}\).
Potential difference \(V_P - V_N = 3 - 5 = -2 \text{ V}\). Since \(V_P<V_N\), it is reverse biased. Step 4: Final Answer:
Figure (A) represents the forward biased diode.
