Find the efficiency of a half wave rectifier with [tex]$\frac{r_f}{R_L}=4$[/tex]
A. [tex]$1.0125 \%$[/tex]
B. [tex]$8.9125 \%$[/tex]
C. [tex]$8.10 \%$[/tex]
D. [tex]$9.9125 \%$[/tex]
E. None of the above.
Answer (1)
The efficiency of a half-wave rectifier is given by η = π 2 4 ( 1 + R L r f ) .
Substitute the given value R L r f = 4 into the efficiency formula: η = π 2 4 ( 1 + 4 ) = π 2 20 .
Calculate the numerical value of the efficiency: η = π 2 20 ≈ 2.0264 .
Express the efficiency as a percentage: η ≈ 20.264% . The answer is N o n eo f t h e ab o v e .
Explanation
Problem Analysis We are given a half-wave rectifier with the ratio of forward resistance to load resistance, R L r f = 4 . Our goal is to find the efficiency of this rectifier.
Efficiency Formula The efficiency of a half-wave rectifier is given by the formula: η = P a c P d c = V r m s 2 / ( R L + r f ) V d c 2 / R L For a half-wave rectifier, the DC voltage is V d c = π V m and the RMS voltage is V r m s = 2 V m , where V m is the peak voltage.
Substituting Values Substituting these values into the efficiency formula, we get: η = ( V m /2 ) 2 / ( R L + r f ) ( V m / π ) 2 / R L = π 2 R L 4 ( R L + r f ) Simplifying the expression, we have: η = π 2 4 ( 1 + R L r f )
Calculating Efficiency We are given that R L r f = 4 . Substituting this value into the efficiency formula, we get: η = π 2 4 ( 1 + 4 ) = π 2 20 Calculating the numerical value of the efficiency: η = π 2 20 ≈ 2.0264
Final Answer Expressing the efficiency as a percentage: η ≈ 2.0264 × 100% = 20.264% Comparing the calculated efficiency with the given options, we see that none of the options match our calculated value. Therefore, the correct answer is (e) None of the above.
Examples
Half-wave rectifiers are commonly used in low-power applications such as battery chargers and simple DC power supplies. Understanding the efficiency of a rectifier is crucial in designing power electronic circuits. For example, if you are designing a battery charger and need to ensure that at least 10W of power is delivered to the battery, knowing the efficiency of the rectifier will help you determine the required AC power input. By calculating the efficiency, engineers can optimize the design to minimize power loss and improve overall system performance. This ensures that the charger operates effectively and conserves energy.
