# Test: Inductor Filters & Capacitor Filters

## 20 Questions MCQ Test Analog Electronics | Test: Inductor Filters & Capacitor Filters

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Attempt Test: Inductor Filters & Capacitor Filters | 20 questions in 20 minutes | Mock test for Electrical Engineering (EE) preparation | Free important questions MCQ to study Analog Electronics for Electrical Engineering (EE) Exam | Download free PDF with solutions
QUESTION: 1

### What is the effect of an inductor filter on a multi frequency signal?

Solution:

Presence of inductor usually dampens the AC signal. Due to self induction induces opposing EMF or changes in the current.

QUESTION: 2

### The ripple factor (ϒ) of inductor filter is_________

Solution:

Ripple factor will decrease when L is increased and RL. Inductor has a higher dc resistance. It depends on property of opposing the change of direction of current.

QUESTION: 3

### The inductor filter gives a smooth output because_________

Solution:

The inductor does not allow the ac components to pass through the filter. The main purpose of using an inductor filter is to avoid the ripples. By using this property, the inductor offers an infinite resistance to ac components and gives a smooth output.

QUESTION: 4

A full wave rectifier with a load resistance of 5KΩ uses an inductor filter of 15henry. The peak value of applied voltage is 250V and the frequency is 50 cycles per second. Calculate the dc load current.

Solution:

For a rectifier with an inductor filter,
VDC=2Vm/π, Idc=VDC/RL=2Vm/RLπ
IDC=2*250/(3.14*15*103)=10.6mA.

QUESTION: 5

The output of a rectifier is pulsating because_________

Solution:

For any electronic devices, a steady dc output is required. The filter is used for this purpose. The ac components are removed by using a filter.

QUESTION: 6

A dc voltage of 380V with a peak ripple voltage not exceeding 7V is required to supply a 500Ω load. Find out the inductance required.

Solution:

Given the ripple voltage is 7V. So, 7=1.414VRMS
ϒ=VRMS/VDC=4.95/380=0.0130. ϒ=1/3√2(RL/Lω)
So, L=28.8henry.

QUESTION: 7

The inductor filter should be used when RL is consistently small because_________

Solution:

When RLis infinite, the ripple factor is 0.471. This value is close to that of a rectifier. So, the resistance should be small.

QUESTION: 8

The output voltage VDC for a rectifier with inductor filter is given by_________

Solution:

The inductor with high resistance can cause poor voltage regulation. The choke resistance, the resistance of half of transformer secondary is not negligible.

QUESTION: 9

What causes to decrease the sudden rise in the current for a rectifier?

Solution:

When the output current of a rectifier increases above a certain value, magnetic energy is stored in the inductor. This energy tends to decrease the sudden rise in the current. This also helps to prevent the current to fall down too much.

QUESTION: 10

A full wave rectifier with a load resistance of 5KΩ uses an inductor filter of 15henry. The peak value of applied voltage is 250V and the frequency is 50 cycles per second. Calculate the ripple factor (ϒ).

Solution:

ϒ=IAC/IDC, IAC=2√2Vm/3π(RL2+4ω2L2)1/2
By putting the values, IAC=4.24Ma. VDC=2Vm/π, IDC=VDC/RL=2Vm/RL π
IDC=2*250/(3.14*15*103)=10.6mA. ϒ=4.24/10.6=0.4.

QUESTION: 11

In a shunt capacitor filter, the mechanism that helps the removal of ripples is_________

Solution:

Filtering is frequently done by shunting the load with capacitor. It depends on the fact that a capacitor stores energy when conducting and delivers energy during non conduction. Throughout this process, the ripples are eliminated.

QUESTION: 12

The cut-in point of a capacitor filter is_________

Solution:

The capacitor charges when the diode is in ON state and discharges during the OFF state of the diode. The instant at which the conduction starts is called cut-in point. The instant at which the conduction stops is called cut-out point.

QUESTION: 13

The rectifier current is a short duration pulses which cause the diode to act as a_________

Solution:

The diode permits charge to flow in capacitor when the transformer voltage exceeds the capacitor voltage. It disconnects the power source when the transformer voltage falls below that of a capacitor.

QUESTION: 14

A half wave rectifier, operated from a 50Hz supply uses a 1000µF capacitance connected in parallel to the load of rectifier. What will be the minimum value of load resistance that can be connected across the capacitor if the ripple% not exceeds 5?

Solution:

For a half wave filter,
ϒ=1/2√3 fCRL=1/2√3*50*10-3*RL
RL=103/5√3=115.47Ω.

QUESTION: 15

A 100µF capacitor when used as a filter has 15V ac across it with a load resistor of 2.5KΩ. If the filter is the full wave and supply frequency is 50Hz, what is the percentage of ripple frequency in the output?

Solution:

For a full wave rectifier, ϒ=1/4√3 fCRL
=1/4√3*50*10-3*2.5
=0.01154. So, ripple is 1.154%.

QUESTION: 16

A full wave rectifier uses a capacitor filter with 500µF capacitor and provides a load current of 200mA at 8% ripple. Calculate the dc voltage.

Solution:

The ripple factor ϒ=IL/ 4√3 fCVDC
VDC=200*10-3/ 4√3 *50*500*8
=14.43.

QUESTION: 17

The charge (q) lost by the capacitor during the discharge time for shunt capacitor filter.

Solution:

The ‘T’ is the total non conducting time of capacitor. The charge per unit time will give the current flow.

QUESTION: 18

Which of the following are true about capacitor filter?

Solution:

The rectifier may be full wave or half wave. The capacitors are usually electrolytic even though they are large in size.

QUESTION: 19

The rms ripple voltage (Vrms) of a shunt filter is_________

Solution:

The ripple waveform will be triangular in nature. The rms value of this wave is independent of slopes or lengths of straight lines. It depends only on the peak value.

QUESTION: 20

A shunt capacitor of value 500µF fed rectifier circuit. The dc voltage is 14.43V. The dc current flowing is 200mA. It is operating at a frequency of 50Hz. What will be the value of peak rectified voltage?

Solution:

We know, Vm=Vdc+Idc/4fC
=14.43+ {200/ (200*500)} 103
=14.43+2=16.43V.

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