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.
Both (A) and (R) are true, but (R) is not correct explanation of (A).
Both statements are true, and both follow from the same underlying behaviour — but the reason is a parallel consequence, not the cause of the assertion. Option 2.
Why the output starts high. With no load, the capacitor charges to the peak of the rectified waveform and stays there, so
\(V_{DC}=V_{m}\)
rather than the \(2V_{m}/\pi=0.636V_{m}\) of an unfiltered full-wave rectifier — a 57 % improvement obtained from one component.
Why it falls under load. Between peaks the capacitor is the only source, and the load discharges it:
\(V_{DC}\approx V_{m}-\dfrac{I_{DC}}{4fC}\)
so the voltage falls linearly with load current, and the ripple grows with it. The regulation is poor precisely because the output rests on charge stored between peaks rather than on the waveform itself.
Why the rectifiers run hot — the reason's own explanation. The diode conducts only during the short interval near each peak when the supply exceeds the capacitor voltage — a conduction angle of perhaps 20° instead of 180°. All the charge the load will draw over the whole cycle must be replaced in that brief window, so the peak current is many times the average:
\(I_{peak}\gg I_{DC}\)
and since dissipation depends on \(I^{2}\), the heating is far worse than the average current suggests. This is why capacitor-input supplies need diodes with a high surge rating and why a surge-limiting resistor is often fitted.
| Capacitor input | Choke input | |
|---|---|---|
| Output voltage | High — near Vm | Lower — near 0.636 Vm |
| Regulation | Poor | Good |
| Peak diode current | Very high | Modest |
| Best for | Light loads | Heavy loads |
Why R does not explain A. Both the falling output and the diode heating are consequences of the short conduction angle, but the heating does not cause the voltage to fall. The voltage falls because charge drains from the capacitor between peaks — a fact about the load and the capacitor, in which the diodes' dissipation plays no part. The two statements are siblings, not parent and child.
Hence, both are true, but (R) is not the correct explanation of (A).
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
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 ?
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?
