The circuit has two oppositely connected ideal diodes in parallel as shown in the figure. What is the current flowing in the circuit?
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To analyze diode circuits quickly, replace each ideal diode with a switch: open if the diode points against the conventional current flow (from positive to negative terminal), and closed if it points along the current flow. This simplifies the schematic into a standard resistor network instantly.
Step 1: Understanding the Question:
The circuit consists of a \( 12\text{ V} \) battery, a series resistor of \( 4\ \Omega \), and two parallel branches, each containing an ideal diode and a resistor. We need to determine the total current flowing in the circuit. Step 2: Key Formula or Approach:
1. Ideal Diode Behavior:
- In forward bias, an ideal diode behaves as a short circuit (closed switch, \( R_D = 0 \)).
- In reverse bias, an ideal diode behaves as an open circuit (open switch, \( R_D = \infty \)).
2. Ohm's Law:
\[ I = \frac{V}{R_{\text{eq}}} \] Step 3: Detailed Explanation:
- The positive terminal of the \( 12\text{ V} \) battery is at the top, so current tends to flow clockwise through the circuit, entering the parallel branch from the top.
- Analyzing Diode \( D_1 \): The diode \( D_1 \) is oriented with its cathode pointing upwards (anode on the bottom). Since current tries to flow from top to bottom, \( D_1 \) is reverse-biased. Thus, it acts as an open circuit, and no current flows through the \( 3\ \Omega \) branch.
- Analyzing Diode \( D_2 \): The diode \( D_2 \) is oriented with its anode pointing upwards (cathode on the bottom). Since current flows from top to bottom, \( D_2 \) is forward-biased. It acts as a short circuit, allowing current to flow freely through the \( 2\ \Omega \) branch.
- Therefore, the active path of the circuit consists only of the \( 4\ \Omega \) resistor and the \( 2\ \Omega \) resistor in series.
- Calculate the equivalent resistance of the circuit:
\[ R_{\text{eq}} = 4\ \Omega + 2\ \Omega = 6\ \Omega \]
- Calculate the total current \( I \):
\[ I = \frac{V}{R_{\text{eq}}} = \frac{12\text{ V}}{6\ \Omega} = 2.00\text{ A} \] Step 4: Final Answer:
The current flowing in the circuit is 2.00A.
