# Electronic Devices and Circuits Questions and Answers – Inductor Filters

This set of Electronic Devices and Circuits Multiple Choice Questions & Answers (MCQs) focuses on “Inductor Filters”.

1. What is the effect of an inductor filter on a multi frequency signal?
a) Dampens the AC signal
b) Dampens the DC signal
c) To reduce ripples
d) To change the current

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

2. The ripple factor (ϒ) of inductor filter is_________
a) ϒ = RZ3/√2ωL
b) ϒ = RZ/3√2ωL
c) ϒ = RZ3√2/ωL
d) ϒ = RZ3/√2ωL

Explanation: 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.

3. The inductor filter gives a smooth output because_________
a) It offers infinite resistance to ac components
b) It offers infinite resistance to dc components
c) Pulsating dc signal is allowed
d) The ac signal is amplified

Explanation: 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.

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.
a) 0.7mA
b) 17mA
c) 10.6mA
d) 20mA

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

5. The output of a rectifier is pulsating because_________
a) It has a pulse variations
b) It gives a dc output
c) It contains both dc and ac components
d) It gives only ac components

Explanation: 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.
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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.
a) 10.8henry
b) 30.7henry
c) 28.8henry
d) 15.4henry

Explanation: 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.

7. The inductor filter should be used when RL is consistently small because_________
a) Effective filtering takes place when load current is high
b) Effective filtering takes place when load current is low
c) Current lags behind voltage

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

8. The output voltage VDC for a rectifier with inductor filter is given by_________
a) (2Vm/π)-IDCR
b) (2Vm/π)+IDCR
c) (2Vmπ)-IDCR
d) (2Vmπ)+IDCR

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

9. What causes to decrease the sudden rise in the current for a rectifier?
a) the electrical energy
b) The ripple factor
c) The magnetic energy
d) Infinite resistance

Explanation: 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.

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).
a) 0.1
b) 0.6
c) 0.5
d) 0.4

Explanation: 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.

Sanfoundry Global Education & Learning Series – Electronic Devices and Circuits.

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