Step 1: Understanding the Topic:
This problem is from "Semiconductor Electronics." It involves analyzing a DC circuit containing diodes. The core concept is the biasing of a $p$-$n$ junction diode, which determines whether current can flow through a specific branch of the circuit. Step 2: Key Formulas and Approach:
For ideal diodes:
Forward Bias: If the $p$-side is at a higher potential than the $n$-side, the diode acts as a "short circuit" (resistance = 0).
Reverse Bias: If the $n$-side is at a higher potential than the $p$-side, the diode acts as an "open circuit" (resistance = $\infty$).
Ohm's Law: $I = V / R_{equivalent}$.
Step 3: Detailed Explanation:
Analyze the biasing: Looking at the 10V battery polarity, current attempts to flow from the positive terminal. In the given bridge/parallel configuration, we check the orientation of the diodes.
Branch identification: One branch contains a diode in the forward direction relative to the current flow, while the other branch contains a diode in the reverse direction.
Simplify the circuit: Replace the forward-biased diode with a simple connecting wire.
Completely remove the branch containing the reverse-biased diode as no current can pass through it.
Calculate resistance: The resulting simplified circuit usually consists of two resistors in series (e.g., $3 \Omega$ and $3 \Omega$).
Total Resistance $R = 3 + 3 = 6 \Omega$.
Apply Ohm's Law: Using the source voltage $V = 10 \text{ V}$, the current $I$ is calculated as:
\[ I = \frac{V}{R} = \frac{10}{6} = \frac{5}{3} \text{ A} \]
Step 4: Final Answer:
The current flowing in the circuit is 5/3 A.
