In a half wave rectifier, a p-n junction diode with internal resistance `2 Omega ` is used. If the load resistance of `2 K Omega `is used in the circuit, then find the efficiency of this half wave rectifier.

Options

A. `30.2% `
B. `38%`
C. `40.2 %`
D. `44% `

Answer

To find the efficiency of a half-wave rectifier using a p-n junction diode with an internal resistance of \(2 \, \Omega\) and a load resistance of \(2 \, k\Omega\), we can follow these steps:

### Step 1: Identify the given values
- Internal resistance of the diode, \(R_F = 2 \, \Omega\)
- Load resistance, \(R_L = 2 \, k\Omega = 2000 \, \Omega\)

### Step 2: Write the formula for the efficiency of a half-wave rectifier
The efficiency (\(\eta\)) of a half-wave rectifier can be calculated using the formula:
\[ \eta = \frac{0.406 \cdot R_L}{R_F + R_L} \]

### Step 3: Substitute the values into the formula
Substituting the known values into the efficiency formula:
\[ \eta = \frac{0.406 \cdot 2000}{2 + 2000} \]

### Step 4: Calculate the denominator
Calculate \(R_F + R_L\):
\[ R_F + R_L = 2 + 2000 = 2002 \, \Omega \]

### Step 5: Calculate the efficiency
Now substitute back into the efficiency formula:
\[ \eta = \frac{0.406 \cdot 2000}{2002} \]
Calculating the numerator:
\[ 0.406 \cdot 2000 = 812 \]
Now, calculate the efficiency:
\[ \eta = \frac{812}{2002} \approx 0.40579 \]

### Step 6: Convert to percentage
To express the efficiency as a percentage, multiply by 100:
\[ \text{Percentage Efficiency} = 0.40579 \times 100 \approx 40.58\% \]

### Final Answer
The efficiency of the half-wave rectifier is approximately \(40.58\%\).

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