In a half wave rectifier output is taken across a `90 ohm` load resistor. If the resistance of diode in forward biased condition is `10 ohm`, the efficiency of rectification of `ac` power into `dc` power is.

Options

A. `40.6 %`
B. `81.2 %`
C. `73.08 %`
D. `36.54 %`

Explanation

To find the efficiency of a half-wave rectifier, we can use the formula for efficiency given by:

\[ \eta = \frac{0.406 \times R_L}{R_F + R_L} \]

where:
- \( R_L \) is the load resistance (90 ohms in this case),
- \( R_F \) is the forward resistance of the diode (10 ohms in this case).

### Step 1: Identify the values
- Load resistance, \( R_L = 90 \, \Omega \)
- Forward resistance of the diode, \( R_F = 10 \, \Omega \)

### Step 2: Substitute the values into the efficiency formula
Now, substituting the values into the efficiency formula:

\[ \eta = \frac{0.406 \times 90}{10 + 90} \]

### Step 3: Calculate the denominator
Calculate \( R_F + R_L \):

\[ R_F + R_L = 10 + 90 = 100 \, \Omega \]

### Step 4: Substitute the denominator back into the formula
Now substitute back into the efficiency formula:

\[ \eta = \frac{0.406 \times 90}{100} \]

### Step 5: Calculate the numerator
Calculate \( 0.406 \times 90 \):

\[ 0.406 \times 90 = 36.54 \]

### Step 6: Final calculation of efficiency
Now, divide by 100:

\[ \eta = \frac{36.54}{100} = 0.3654 \]

### Step 7: Convert to percentage
To express efficiency as a percentage, multiply by 100:

\[ \eta = 0.3654 \times 100 = 36.54\% \]

### Final Answer
The efficiency of rectification of AC power into DC power is **36.54%**.

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