Network Theory Questions and Answers – Advanced Problems on Filters – 2

This set of Network Theory Technical Interview Questions & Answers focuses on “Advanced Problems on Filters – 2”.

1. In a parallel RL circuit, 10 A current enters into the resistor R and 15 A current enters into the Inductor L. The total current I is _____________
a) 15.5 A
b) 25.05 A
c) 18.03 A
d) 30.15 A
View Answer

Answer: c
Explanation: Currents in resistance and inductance are out of phase by 90°.
Hence, I = \(I_1^2 + I_2^2\)
Or, I = [102 + 152]0.5
Or, I = \(\sqrt{100+225} = \sqrt{325}\)
= 18.03 A.

2. In the circuit given below, a dc circuit fed by a current source. With respect to terminals AB, Thevenin’s voltage and Thevenin’s resistance are ____________
Thevenin’s voltage & Thevenin’s resistance are VTH = 7.14 V, RTH = 2.14 Ω for dc circuit
a) VTH = 5 V, RTH = 0.75 Ω
b) VTH = 7.14 V, RTH = 2.14 Ω
c) VTH = 7.5 V, RTH = 2.5 Ω
d) VTH = 5 V, RTH = 1 Ω
View Answer

Answer: b
Explanation: VTH = \(\frac{5 X 10}{35}\) X 5 = 7.14 V
Also, RTH = \(\frac{5 X 15}{35}\) = 2.14 Ω.

3. In the circuit given below, the value of R is ____________
The value of R is 6 Ω in given circuit
a) 12 Ω
b) 6 Ω
c) 3 Ω
d) 1.5 Ω
View Answer

Answer: b
Explanation: The resistance of parallel combination is given by,
Req = \(\frac{40}{3}\) – 10 = 3.33 Ω
Or, \(\frac{1}{3.33} = \frac{1}{12} + \frac{1}{15} + \frac{1}{15}\)
Or, R = 6 Ω.
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4. In the circuit given below, M = 1. The resonant frequency is _______________
The resonant frequency is 14.53 Hz in given circuit
a) 145.3 Hz
b) 0.1453 Hz
c) 1.453 Hz
d) 14.53 Hz
View Answer

Answer: d
Explanation: IEQ = L1 + L2 + 2M
LEQ = 5 + 5 + 2 × 1
= 10 + 2 = 12 H
∴ FO = \(\frac{1}{2π\sqrt{LC}}\)
= \(\frac{1}{2π\sqrt{12 × 0.1}}\)
= 0.1453 Hz.

5. A thermistor is used for ____________
a) Over voltage protection
b) Temperature alarm circuit
c) Automatic light control
d) Automatic sensor
View Answer

Answer: b
Explanation: The resistance of a thermistor decreases with increases in temperature. It is used to monitor the hot spot temperature of electric machines.

6. Two network functions are given below.
H1 = \(\frac{1}{s^2 + 2s + 1}\), H2 = \(\frac{1}{s^3+3s^2+3s+1}\)
The frequency response indicates that the filter is ___________
a) Low-pass
b) Band-pass
c) High-pass
d) Band reject
View Answer

Answer: a
Explanation: | H1 |2 = \(\frac{1}{[(jω)^2 + 2jω + 1] [(jω)^2 + 2 jω + 1]}\)
Similarly, | H2 |2 = \(\frac{1}{1 + ω^6}\)
Therefore at ω = 0, 1 and ∞, we have | H |2 = 1, \(\frac{1}{2}\) and 0 respectively.
Hence, the filter is a Low-pass filter.

7. A particular band-pass function has a network function as H(s) = \(\frac{s}{s^2+2s+1}\) then, its quality factor Q is ___________
a) \(\frac{3}{4}\)
b) \(\frac{1}{2}\)
c) \(\frac{3}{2}\)
d) \(\frac{1}{4}\)
View Answer

Answer: d
Explanation: H(s) = \(\frac{Ks}{s^2+as+b}\)
Then, quality factor is given as \(\frac{\sqrt{b}}{a}\)
Here, b = 1, a = 2, K = 1
∴ Q = \(\frac{1}{2}\) = 0.5.
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8. The filter which passes all frequencies above fc by attenuating significantly, all frequencies above fc is _______________
a) Low-pass
b) High-pass
c) Band-pass
d) Band-stop
View Answer

Answer: b
Explanation: A high-pass filter is one which passes all frequencies above fc by attenuating significantly, all frequencies below fc.

9. For a Band Pass Filter, the slope of the filter is given as 80dB/decade. The order of the Band Pass Filter is __________
a) 10
b) 8
c) 4
d) 6
View Answer

Answer: b
Explanation: The Bode plot is a logarithmic plot which helps in fitting a large scale of values into a small scale by the application of logarithm. Plotting the slope 80dB/decade on the Bode plot, we get n=8. Hence the order of the Band Pass Filter is 8.
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10. In which of the filter circuits given below, will the bandwidth be equal to the critical frequency?
a) Low-pass
b) High-pass
c) Band-pass
d) Band-stop
View Answer

Answer: a
Explanation: Bandwidth can be calculated by considering,
Largest positive value – Smallest Positive Value
Here, in case of the Low-pass filter only, the largest positive value will of course be the critical frequency, beyond which frequencies have to be blocked. Hence, the bandwidth in a Low-pass filter equals the critical frequency.

11. For providing a Roll-off greater than 60dB/decade/pole, filters with which characteristics are useful?
a) Butterworth
b) Chebyshev
c) Bessel
d) Butterworth & Bessel
View Answer

Answer: b
Explanation: Roll off is a term commonly refers to the steepness of the transmission function with respect to the frequency. For a Chebyshev filter, the Roll-off value greater than 60. This characteristic feature is useful when a rapid roll-off is required because it provides a Roll-off rate more than 60. On the other hand, both Butterworth and Bessel have the Roll-off rate less than or equal to 60 dB/decade/pole.

12. A Low-pass filter circuit has a cut-off frequency of 50 kHz. The bandwidth of the filter is ______________
a) 24.6 kHz
b) 50 kHz
c) 61.5 kHz
d) 36.9 kHz
View Answer

Answer: b
Explanation: The bandwidth is defined as the highest cut-off frequency to the lowest cut-off frequency. Here the lowest cut-off frequency is Zero.
For a Low-pass filter therefore, Cut-off Frequency = Bandwidth of the filter
∴ Bandwidth = 50 kHz.

13. Given a system function H(s) = \(\frac{1}{s+10}\). Let us consider a signal sin 10t. Then the steady state response is ___________
a) 1
b) 2
c) 0
d) 5
View Answer

Answer: c
Explanation: H(s) = \(\frac{V(s)}{J(s)}\)
= \(\frac{1}{s+10}\)
V(s) = \(\frac{1}{s+10}\) . J(s)
J(s) = L (sin 10t) = \(\frac{10}{s^2+100}\)
V (s) = \(\frac{1}{s^2+100} . \frac{10}{s+10}\)
VSS = lims→0⁡ sV(s)
= 0.

14. Current I = – 1 + \(\sqrt{2}\) (sin (ωt + 60°)) A is passed through three meters. The respective readings will be?
a) \(\sqrt{2}\) A, \(\sqrt{2}\) A and 1 A
b) 1 A, \(\sqrt{2}\) A and 1 A
c) – 1 A, \(\sqrt{2}\) A and \(\sqrt{2}\) A
d) -1 A, 1 A and 1 A
View Answer

Answer: c
Explanation: We know that a PMMC instrument reads only DC value and since it is a centre zero type, so it will give – 1.
So, rms = \(\sqrt{1^2 + (\frac{\sqrt{2}}{\sqrt{2}})^2} = \sqrt{2}\) A
Moving iron also reads rms value, so its reading will also be \(\sqrt{2}\) A.

15. For a Band Pass Filter, the slope of the filter is given as 40dB/decade. The order of the Band Pass Filter is __________
a) 2
b) 3
c) 4
d) 6
View Answer

Answer: c
Explanation: The Bode plot is a logarithmic plot which helps in fitting a large scale of values into a small scale by the application of logarithm. Plotting the slope 40dB/decade on the Bode plot, we get n=4. Hence order of the Band Pass Filter is 4.

Sanfoundry Global Education & Learning Series – Network Theory.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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