All the diodes are ideal. The current flowing in `2 Omega` resistor connected between the diodes `D_(1)` and `D_(2)` is

Options

A. `1 A`
B. `2A`
C. `3A`
D. Zero

Answer

To find the current flowing through the 2 Ω resistor connected between the diodes D1 and D2, we can follow these steps:

### Step 1: Analyze the Circuit
First, we need to understand the configuration of the circuit. We have a voltage supply of 10 volts and several resistors and diodes. The resistors are connected in a combination of series and parallel configurations.

### Step 2: Calculate the Total Resistance
Identify the resistors in series and parallel:
- The resistors in the upper part of the circuit are 1 Ω, 2 Ω, and 1 Ω in series.
- The total resistance of these resistors in series is:
\[ R_{\text{upper}} = 1 \, \Omega + 2 \, \Omega + 1 \, \Omega = 4 \, \Omega \]

- The 4 Ω resistor and the 3 Ω resistor are in series with the upper part:
\[ R_{\text{total}} = R_{\text{upper}} + 3 \, \Omega = 4 \, \Omega + 3 \, \Omega = 7 \, \Omega \]

### Step 3: Identify Parallel Resistors
Now, we need to consider the 4 Ω resistor in parallel with another 4 Ω resistor:
- The equivalent resistance of two 4 Ω resistors in parallel is given by:
\[ \frac{1}{R_{\text{parallel}}} = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} \implies R_{\text{parallel}} = 2 \, \Omega \]

### Step 4: Calculate the Total Resistance in the Circuit
Now, we can find the total resistance in the circuit:
- The total resistance now becomes:
\[ R_{\text{total}} = 7 \, \Omega + 2 \, \Omega = 9 \, \Omega \]

### Step 5: Calculate the Total Current
Using Ohm's law (V = IR), we can find the total current flowing through the circuit:
\[ I = \frac{V}{R_{\text{total}}} = \frac{10 \, \text{V}}{9 \, \Omega} \approx 1.11 \, \text{A} \]

### Step 6: Determine Current through the 2 Ω Resistor
Since the 2 Ω resistor is connected between D1 and D2, we need to analyze the current division:
- The current through the 2 Ω resistor will be half of the total current since it is in parallel with another resistor of equal value (assuming the other resistor is also 2 Ω):
\[ I_{2 \, \Omega} = \frac{I}{2} = \frac{1.11 \, \text{A}}{2} \approx 0.56 \, \text{A} \]

### Final Answer
The current flowing through the 2 Ω resistor is approximately 0.56 A.

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