In a rectifier the ripple factor can be expressed as (a) \(\dfrac{I_{rms}}{I_{dc}}\) (b) \(\sqrt\dfrac{I_{dc}}{I_{rms}}\) (c) \(\sqrt{\left(\dfrac{I_{rms}}{I_{dc}}\right)^{2}-1}\) (d) \(\sqrt{1-\left(\dfrac{I_{rms}}{I_{dc}}\right)^{2}}\) Which one of the above is correct ?
(a) is wrong, but (c) is correct.
The ripple factor of a rectifier is a measure of the effectiveness of a rectifier circuit in converting AC to DC, and it indicates how much AC 'ripple' remains in the DC output of a rectifier. It is defined as the ratio of the root mean square (RMS) value of the alternating component of the waveform to the direct current (DC) component of the waveform.
The formula for the ripple factor is given by:
\[\gamma = \sqrt{\left(\dfrac{I_{rms}}{I_{dc}}\right)^{2} - 1}\]This formula can be derived as follows:
- Define the RMS current: The total RMS current (\(I_{rms}\)) of the output current for a rectifier contains both AC and DC components.
- Define the DC current: Let \(I_{dc}\) be the average or DC current of the output waveform.
- Ripple Factor Formula: The ripple factor \(\gamma\) is given by:
Using the identity:
\[I_{ac}^{2} = I_{rms}^{2} - I_{dc}^{2}\]We thus find:
\[\gamma = \sqrt{\left(\dfrac{I_{rms}}{I_{dc}}\right)^{2} - 1}\]In the given options:
- Option (a) \(\dfrac{I_{rms}}{I_{dc}}\) does not correctly represent the ripple factor, it represents the Form Factor.
- Option (b) \(\sqrt{\dfrac{I_{dc}}{I_{rms}}}\) is incorrect as it doesn't match the definition.
- Option (c) \(\sqrt{\left(\dfrac{I_{rms}}{I_{dc}}\right)^{2}-1}\) matches the correct formula for the ripple factor.
- Option (d) \(\sqrt{1-\left(\dfrac{I_{rms}}{I_{dc}}\right)^{2}}\) is incorrect as it contradicts the relationship dictated by the ripple factor formula.
Thus, the correct option is: (a) is wrong, but (c) is correct.
Similar Questions
Consider the following rectifier circuits :
1. Half-wave rectifier without filter
2. Full-wave rectifier without filter
3. Full-wave rectifier with series inductance filter
4. Full-wave rectifier with capacitance filterThe sequence of these rectifier circuits in decreasing order of the ripple factor is
Assertion (A) : Capacitor input filter gives a high output voltage which drops as the load is increased.
Reason (R) : Capacitor input filters produce more heating in the rectifiers.
For the different values of 'L' while keeping 'RL' as constant, following graph is obtained from the given circuit.
Choose the correct ascending order from the options given below for the different values of ' L ' :
The ratio of peak to peak input ripple voltage to peak to peak ripple output voltage is known as:
Read the statements:
A. The O/P frequency of a half wave rectifier is equal to the input frequency
B. The O/P frequency of a full wave rectifier is double to the input frequency
C. The regulation of an excellent rectifier should be zero.
D. ripple factor in the full wave rectifier is 1.2.
Choose the correct statements from the options given below:
Statements regarding rectifier are :
A. RMS value of current in a half wave rectifier is \(\frac{I_m}{\sqrt{2}}\)
B. PIV of diode in a full wave bridge rectifier is Vm volts.
C. TUF and rectification efficiency for a rectifier are same parametrs.
D. Rectifier converts A.C. to rippled DC not pure D.C.
Choose the correct answer from the options given below:
For a full-wave rectifier, which of the following statements are correct ?
(a) \(V_{dc}=\dfrac{2V_P}{\pi}\)
(b) \(f_{out}=2f_{in}\)
(c) \(V_{dc}=\dfrac{2\pi}{V_P}\)
(d) \(I_{dc}=\dfrac{I_m}{\sqrt{\pi}}\)
Options :
As compared to a full-wave rectifier using two diodes, the four diode bridge rectifier has the dominant advantage of :
In rectifiers :
(a) The efficiency of rectification is given by \(\eta=\dfrac{P_{ac}}{P_{dc}}\)
(b) The AC component on the output side : \(V_{ac}=\sqrt{V_{rms}^{2}-V_{dc}^{2}}\)
(c) Form factor is given by : \(FF=\dfrac{V_{dc}}{V_{rms}}\)
(d) Ripple factor is given by : \(RF=\sqrt{\left(\dfrac{V_{rms}}{V_{dc}}\right)^{2}-1}\)
Of these statements :
A half wave rectifier operates from 50 Hz supply and provides a peak output voltage of 60 V across the 600 Ω resistance. The value of capacitance for the capacitor filter would limit the peak to peak ripple voltage across load is 15 V :
Important Questions from Rectifier Circuits
PIV of a non conducting diode in a bridge rectifier is
The no-load output voltage of an ideal half wave rectifier is _______.
What is the minimum PIV required for a diode used in a full-wave rectifier circuit using a center-tapped transformer with a secondary voltage of 50V-0-50 V (rms)?
One of the following is a negative resistance device
Which of the following is NOT true for a rectifier?
