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Chapter 20: Problem 24

The maximum efficiency of full wave rectifier is (A) \(100 \%\) (B) \(25.20 \%\) (C) \(40.2 \%\) (D) \(81.2 \%\)

Short Answer

Expert verified
The maximum efficiency of a full-wave rectifier is \(50\%\). None of the given options (A, B, C, D) are equal to 50%, indicating an error with the provided options.

Step by step solution

01

1. Understand the concept of efficiency of a full-wave rectifier

The efficiency of a rectifier is the ratio of the output DC power to the input AC power. For a full-wave rectifier, the efficiency can be calculated using the formula: \[\eta = \frac{P_{dc}}{P_{ac}}\]
02

2. Calculate the output DC power

The output DC power for a full-wave rectifier is given by the formula: \[P_{dc} = \frac{1}{2} IV_m\] Here, I is the current, and V_m is the peak output voltage.
03

3. Calculate the input AC power

The input AC power for a full-wave rectifier is given by the formula: \[P_{ac} = IV_m\] Here, I is the average value of the current, and V_m is the peak output voltage.
04

4. Formula for efficiency

Now, we can write the efficiency formula using P_dc and P_ac \[\eta = \frac{P_{dc}}{P_{ac}} = \frac{\frac{1}{2} IV_m}{IV_m}\]
05

5. Calculate the maximum efficiency

We can now simplify the efficiency formula: \[\eta = \frac{1}{2}\] The maximum efficiency of a full-wave rectifier is 50%. None of the given options (A, B, C, D) are equal to 50%. Therefore, there seems to be a mistake with the given options.

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