250+ TOP MCQs on CLC Filter and Answers

Posted on by engineerComments Off on 250+ TOP MCQs on CLC Filter and Answers

Electronic Devices and Circuits Multiple Choice Questions on “CLC Filter”.

1. What is the number of capacitors and inductors used in a CLC filter?
A. 1, 2 respectively
B. 2, 1 respectively
C. 1, 1 respectively
D. 2, 2 respectively
Answer: B
Clarification: A very smooth output can be obtained by a filter consisting of one inductor and two capacitors connected across each other. They are arranged in the form of letter ‘pi’. So, these are also called as pi filters.

2. Major part of the filtering is done by the first capacitor in a CLC filter because _________
A. The capacitor offers a very low reactance to the ripple frequency
B. The capacitor offers a very high reactance to the ripple frequency
C. The inductor offers a very low reactance to the ripple frequency
D. The inductor offers a very high reactance to the ripple frequency
Answer: A
Clarification: The CLC filters are used when high voltage and low ripple frequency is needed than L section filters. The capacitor in a CLC filter offers very low reactance to the ripple frequency. So, maximum of the filtering is done by the first capacitor across the L section part.

3. At f=50Hz, the ripple factor of CLC filter is_________
A. ϒ=5700RL / (LC1C2)
B. ϒ=5700/ (LC1C2RL)
C. ϒ=5700LC1/ (C2RL)
D. ϒ=5700C1C2/ (LRL)
Answer: B
Clarification: The ripple factor of a rectifier is the measure of disturbances produced in the output. It’s the effectiveness of a power supply filter to reduce the ripple voltage. The ratio of ripple voltage to DC output voltage is ripple factor which is 5700 / (LC1C2RL) at 50Hz.

4. A single phase full wave rectifier makes use of pi section filter with 10µF capacitors and a choke of 10henry. The secondary voltage is 280V and the load current is 100mA. Determine the dc output voltage when f=50Hz.
A. 345V
B. 521V
C. 243V
D. 346V
Answer: D
Clarification: Given, VRMS=280V
So, V¬m = 1.414*280=396V.
From theory of capacitor filter, VDC = Vm –IDC/4fC=396-0.1/ (4*50*10*10-6)=346V.

5. For a given CLC filter, the operating frequency is 50Hz and 10µF capacitors useC. The load resistance is 3460Ω with an inductance of 10henry. Calculate the ripple factor.
A. 0.165%
B. 0.142%
C. 0.178%
D. 0.321%
Answer: A
Clarification: We have, ϒ=5700 / (LC1C2RL)
=5700 / (10*100*10-12*3460)
=0.165%.

6. The inductor is placed in the L section filter because_________
A. It offers zero resistance to DC component
B. It offers infinite resistance to DC component
C. It bypasses the DC component
D. It bypasses the AC component
Answer: A
Clarification: The inductor offers high reactance to ac component and zero resistance to dc component. So, it blocks the ac component which cannot be bypassed by the capacitors.

7. The voltage in case of a full wave rectifier in a CLC filter is_________
A. Vϒ = IDC/2fC
B. Vϒ = IDC fC
C. Vϒ = IDC/fC
D. Vϒ = 2IDCfC
Answer: A
Clarification: T he filter circuit is a combination of capacitors and inductors. The RMS value depends on the peak value of charging and discharging magnitude, VPEAK.

8. The advantages of a pi-filter is_________
A. low output voltage
B. low PIV
C. low ripple factor
D. high voltage regulation
Answer: C
Clarification: Due to the use of two capacitors with an inductor, an improved filtering action is provideC. This leads to decrement in ripple factor. A low ripple factor signifies regulated and ripple free DC voltage.

9. What is the relation between time constant and load resistance?
A. They don’t depend on each other
B. They are directly proportional
C. They are inversely proportional
D. Cannot be predicted
Answer: C
Clarification: If the load resistance value is large, the discharge time constant will be of a high value. Thus the capacitors time to discharge will get over soon. This lowers the amount of ripples in the output and increases the output voltage. If the load resistance is small, the discharge time constant will be more with decrease in output voltage.

10. The output waveform of CLC filter is superimposed by a waveform referred to as_________
A. Square wave
B. Triangular wave
C. Saw tooth wave
D. Sine wave
Answer: C
Clarification: Since the rectifier conducts current only in the forward direction, any energy discharged by the capacitor will flow into the loaC. This result in a DC voltage upon which is superimposed a waveform referred to as a saw tooth wave.