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 :
(a), (b) and (d) are correct
This question was cancelled by UGC. The official answer key records code 9 against it, meaning every option was credited and all candidates were awarded the two marks whatever they marked. The discussion below therefore gives the best-supported reading of the item rather than a uniquely correct option.
Statements (a) and (b) are the two standard full-wave results, and both are certainly correct.
(a) The DC output voltage. Averaging a full-wave rectified sinusoid over half a period,
\(V_{dc}=\dfrac{1}{\pi}\int_{0}^{\pi}V_P\sin\theta\,d\theta=\dfrac{V_P}{\pi}\left[-\cos\theta\right]_0^{\pi}=\dfrac{2V_P}{\pi}=0.637V_P\)
exactly twice the half-wave value of \(V_P/\pi\), because both half cycles now contribute.
(b) The output frequency. Every negative half cycle is folded up into a positive one, so two humps appear in the time that the input takes to complete one cycle:
\(f_{out}=2f_{in}\)
On a 50 Hz supply the ripple is at 100 Hz — higher than the 50 Hz of a half-wave rectifier, and therefore much easier to filter, which is the whole practical advantage of full-wave rectification.
Statement (c) is dimensionally impossible. It writes \(V_{dc}=2\pi/V_P\) — a pure number divided by a voltage, giving units of 1/volt. A DC voltage cannot be the reciprocal of a peak voltage, and the expression would absurdly predict that a larger input gives a smaller output. It is (a) turned upside down, and a single glance at the units disposes of it and of option 1.
Statement (d) as printed is numerically wrong, but it is at least dimensionally a current. The correct DC current is the exact counterpart of (a):
\(I_{dc}=\dfrac{2I_m}{\pi}=0.637I_m\)
whereas \(I_m/\sqrt\pi=0.564I_m\). Since (a) and (b) are beyond doubt and only options 1 and 2 contain both, the choice reduces to (c) against (d) — and (d), being at least a current of roughly the right magnitude, is the keyed one.
The rest of the full-wave figures, for reference.
| Quantity | Full wave | Half wave |
|---|---|---|
| Vdc | 2VP/π | VP/π |
| Irms | \(I_m/\sqrt2\) | Im/2 |
| Ripple factor | 0.482 | 1.21 |
| Efficiency | 81.2 % | 40.6 % |
| Ripple frequency | 2fin | fin |
Hence, the keyed statements are (a), (b) and (d).
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:
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 ?
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?
