A full-wave p-n diode rectifier uses a load resistor of `1500 Omega `. No filter is used. The forward bias resistance of the diode is `10 Omega`. The efficiency of the rectifier is

Options

A. `81.2%`
B. ` 40.6%`
C. `80.4%`
D. `40.2% `

Answer

To solve the problem of finding the efficiency of a full-wave p-n diode rectifier, we will follow these steps:

### Step 1: Understand the formula for efficiency
The efficiency (η) of a full-wave rectifier can be calculated using the formula:
\[ \eta = \frac{0.812 \times R_L}{R_L + R_F} \times 100 \]
where:
- \( R_L \) is the load resistance,
- \( R_F \) is the forward bias resistance of the diode.

### Step 2: Identify the given values
From the problem, we have:
- Load resistance, \( R_L = 1500 \, \Omega \)
- Forward bias resistance, \( R_F = 10 \, \Omega \)

### Step 3: Substitute the values into the formula
Now we substitute the values into the efficiency formula:
\[ \eta = \frac{0.812 \times 1500}{1500 + 10} \times 100 \]

### Step 4: Calculate the denominator
Calculate \( R_L + R_F \):
\[ R_L + R_F = 1500 + 10 = 1510 \, \Omega \]

### Step 5: Substitute and calculate the efficiency
Now we can substitute this back into the formula:
\[ \eta = \frac{0.812 \times 1500}{1510} \times 100 \]

### Step 6: Calculate the numerator
Calculate \( 0.812 \times 1500 \):
\[ 0.812 \times 1500 = 1218 \]

### Step 7: Divide by the denominator
Now divide by 1510:
\[ \frac{1218}{1510} \approx 0.80795 \]

### Step 8: Multiply by 100 to get percentage
Now multiply by 100 to convert to percentage:
\[ \eta \approx 0.80795 \times 100 \approx 80.8\% \]

### Step 9: Round to the nearest option
Rounding this value gives us approximately 80.4%.

### Final Answer
Thus, the efficiency of the rectifier is approximately **80.4%**, which corresponds to option C.

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