Network Theory MCQ Set 1
1. The value of one decibel is equal to?
a) 0.115 N
b) 0.125 N
c) 0.135 N
d) 0.145 N
Answer
Answer: a [Reason:] The value of one decibel is equal to 0.115 N. One decibel = 0.115 N where N is the number of nepers and N = loge(V1/V2).
2. A filter which passes without attenuation all frequencies up to the cut-off frequency fc and attenuates all other frequencies greater than fc is called?
a) high pass filter
b) low pass filter
c) band elimination filter
d) band pass filter
Answer
Answer: b [Reason:] A filter is called a low pass filter if it passes all frequencies up to the cut-off frequency fc without attenuation and attenuates all other frequencies greater than fc. This transmits currents of all frequencies from zero up to the cut-off frequency.
3. A filter which attenuates all frequencies below a designated cut-off frequency fc and passes all other frequencies greater than fc is called?
a) band elimination filter
b) band pass filter
c) low pass filter
d) high pass filter
Answer
Answer: d [Reason:] A filter is called high pass filter if attenuates all frequencies below a designated cut-off frequency fc and passes all other frequencies greater than fc. Thus the pass band of this filter is the frequency range above fc and the stop band is the frequency range below fc.
4. A filter that passes frequencies between two designated cut-off frequencies and attenuates all other frequencies is called?
a) high pass filter
b) band elimination filter
c) band pass filter
d) low pass filter
Answer
Answer: c [Reason:] A band pass filter passes frequencies between two designated cut-off frequencies and attenuates all other frequencies. A band pass filter has two cut-off frequencies and will have the pass band f2-f1; f1 is the lower cut-off frequency, f2 is the upper cut-off frequency.
5. A filter that passes all frequencies lying outside a certain range, while it attenuates all frequencies between the two designated frequencies is called?
a) low pass filter
b) high pass filter
c) band elimination filter
d) band pass filter
Answer
Answer: c [Reason:] A band elimination filter passes all frequencies lying outside a certain range, while it attenuates all frequencies between the two designated frequencies. It is also referred to as band stop filter.
6. The expression of the characteristic impedance of a symmetrical T-section is?
a) ZOT=√(Z12/4-Z1Z2)
b) ZOT=√(Z12/4+Z1)
c) ZOT=√(Z12/4+Z2)
d) ZOT=√(Z12/4+Z1Z2)
Answer
Answer: d [Reason:] For a T-section, the value of input impedance when it is terminated in Zo is
Zin=(Z1/2)+(Z2((Z1/2)+Zo))/((Z1/2)+Z2+Zo) and Zin=Zo. On solving, the expression of the characteristic impedance of a symmetrical T-section is ZOT=√(Z12/4+Z1Z2).
7. The expression of the open circuit impedance Zoc is?
a) Zoc=Z1/2+Z2
b) Zoc=Z2/2+Z2
c) Zoc=Z1/2+Z1
d) Zoc=Z1/2-Z2
Answer
Answer: a [Reason:] On open circuiting the port 2 of T-section, we get the expression of the open circuit impedance Zoc as Zoc=Z1/2+Z2.
8. The expression of short circuit impedance Zsc is?
a) Zsc=(Z12-4Z1Z2)/(2Z1-4Z2)
b) Zsc=(Z12+4Z1Z2)/(2Z1+4Z2)
c) Zsc=(Z12-4Z1Z2)/(2Z1+4Z2)
d) Zsc=(Z12+4Z1Z2)/(2Z1-4Z2)
Answer
Answer: b [Reason:] On short circuiting the port 2 of T-section, we get the expression of short circuit impedance Zsc as Zsc=(Z1/2)+((Z1/2)xZ2)/((Z1/2)+Z2). On solving we get Zsc=(Z12+4Z1Z2)/(2Z1+4Z2).
9. The relation between ZOT, Zoc, Zsc is?
a) ZOT=√ZocZsc
b) Zoc=√(ZOT Zsc)
c) Zsc=√(ZOT Zoc)
d) Zoc=√(ZOT Zoc)
Answer
Answer: a [Reason:] Zoc=Z1/2+Z2 and Zsc=(Z12+4Z1Z2)/(2Z1+4Z2) => ZocxZsc=Z1Z2+Z12/4 =Zo2T. The relation between ZOT, Zoc, Zsc is ZOT=√ZocZsc.
10. The value of sinhϒ/2 in terms of Z1 and Z2 is?
a) sinhϒ/2=√(4Z1/Z2)
b) sinhϒ/2=√(Z1/Z2)
c) sinhϒ/2=√(Z1/4Z2)
d) sinhϒ/2=√(2Z1/Z2)
Answer
Answer: c [Reason:] sinhϒ/2=√((1/2(coshϒ-1)/(1/2(1+Z1/2Z2-1))). The value of sinhϒ/2 in terms of Z1 and Z2 is sinhϒ/2=√(Z1/4Z2).
Network Theory MCQ Set 2
1. The denominator polynomial in a transfer function may not have any missing terms between the highest and the lowest degree, unless?
a) all odd terms are missing
b) all even terms are missing
c) all even or odd terms are missing
d) all even and odd terms are missing
Answer
Answer: c [Reason:] All the quotients in the polynomial P(s) are positive. The denominator polynomial in a transfer function may not have any missing terms between the highest and the lowest degree, unless all even or odd terms are missing.For example P(s) = s3+3s is Hurwitz because all quotient terms are positive and all even terms are missing.
2. The roots of the odd and even parts of a Hurwitz polynomial P (s) lie on ____________
a) right half of s plane
b) left half of s-plane
c) on jω axis
d) on σ axis
Answer
Answer: c [Reason:] The roots of the odd and even parts of a Hurwitz polynomial P (s) lie on jω axis not on right half of s plane or on left half of s-plane.
3. If the polynomial P (s) is either even or odd, then the roots of P (s) lie on __________
a) on σ axis
b) on jω axis
c) left half of s-plane
d) right half of s plane
Answer
Answer: b [Reason:] If the polynomial P (s) is either even or odd, then the roots of P (s) lie on jω axis not on right half of s plane or on left half of s-plane.
4. If the ratio of the polynomial P (s) and its derivative gives a continued fraction expansion with ________ coefficients, then the polynomial P (s) is Hurwitz.
a) all negative
b) all positive
c) positive or negative
d) positive and negative
Answer
Answer: b [Reason:] If the ratio of the polynomial P (s) and its derivative P‘(s) gives a continued fraction expansion with all positive coefficients, then the polynomial P (s) is Hurwitz. If all the quotients in the continued fraction expansion are positive, the polynomial P(s) is positive.
5. Consider the polynomial P(s)=s4+3s2+2. The given polynomial P (s) is Hurwitz.
a) True
b) False
Answer
Answer: a [Reason:] P(s)=s4+3s2+2 => P‘ (s)=4s3+6s
After doing the continued fraction expansion, we get all the quotients as positive. So, the polynomial P (s) is Hurwitz.
6. When s is real, the driving point impedance function is _________ function and the driving point admittance function is _________ function.
a) real, complex
b) real, real
c) complex, real
d) complex, complex
Answer
Answer: b [Reason:] When s is real, the driving point impedance function is real function and the driving point admittance function is real function because the quotients of the polynomials P(s) and Q(s) are real. When Z(s) is determined from the impedances of the individual branches, the quotients are obtained by adding together, multiplying or dividing the branch parameters which are real.
7. The poles and zeros of driving point impedance function and driving point admittance function lie on?
a) left half of s-plane only
b) right half of s-plane only
c) left half of s-plane or on imaginary axis
d) right half of s-plane or on imaginary axis
Answer
Answer: c [Reason:] The poles and zeros of driving point impedance function and driving point admittance function lie on left half of s-plane or on imaginary axis of the s-plane..
8. For real roots of sk, all the quotients of s in s2+ωk2 of the polynomial P (s) are __________
a) negative
b) non-negative
c) positive
d) non-positive
Answer
Answer: b [Reason:] For real roots of sk, all the quotients of s in s2+ωk2 of the polynomial P (s) are non-negative. So by multiplying all factors in P(s) we find that all quotients are positive.
9. The real parts of the driving point function Z (s) and Y (s) are?
a) positive and zero
b) positive
c) zero
d) positive or zero
Answer
Answer: d [Reason:] The real parts of the driving point impedance function Z (s) and driving point admittance function Y (s) are positive or zero.
10. For the complex zeros to appear in conjugate pairs the poles of the network function are ____ and zeros of the network function are ____________
a) complex, complex
b) complex, real
c) real, real
d) real, complex
Answer
Answer: c [Reason:] P(s) and Q(s) are real when s is real. So the poles of the network function are real and zeros of the network function are real, the complex zeros to appear in conjugate pairs.
Network Theory MCQ Set 3
1. A network is said to be symmetrical if the relation between A and D is?
a) A = D
b) A = 2 D
c) A = 3 D
d) A = 4 D
Answer
Answer: a [Reason:] We know V1=AV2-BI2 and I1=CV2-DI2. If the network is symmetrical, then the relation between A and D is A = D.
2. The relation between Z11 and Z12 if the network is symmetrical is?
a) Z11 = 2 Z12
b) Z11 = Z12
c) Z11 = 3 Z12
d) Z11 = 4 Z12
Answer
Answer: b [Reason:] For a network to be symmetrical A=D. So the relation between Z11 and Z12 for the network is symmetrical is Z11 = Z12.
3. The relation between Z12 and Z11 and B and C parameters if the network is symmetrical is?
a) Z11 = Z12 = B/C
b) Z11 = Z12 = C/B
c) Z11 = Z12 =√(B/C)
d) Z11 = Z12 = √(C/B)
Answer
Answer: c [Reason:] For symmetrical network, A=D. On substituting this we get the relation between Z12 and Z11 and B and C parameters if the network is symmetrical is
Z11 = Z12 =√(B/C).
4. Determine the transmission parameter A in the circuit shown below.
a) 3/4
b) 4/3
c) 5/6
d) 6/5
Answer
Answer: d [Reason:] We know V1=AV2-BI2 and I1=CV2-DI2. A=(V1/V2) |I2=0. On solving we get the transmission parameter A as A = 6/5.
5. Determine the transmission parameter B in the circuit shown in question 4.
a) 17/5
b) 5/17
c) 13/5
d) 5/13
Answer
Answer: a [Reason:] The transmission parameter B is -V1/I2 |V2=0. On short cicuiting the port 2, from the circuit we get -I2 = (5/17)V1 => -V1/I2 = 17/5. On substituting we get B = 17/5.
6. Determine the transmission parameter C in the circuit shown in question 4.
a) 2/5
b) 1/5
c) 4/5
d) 3/5
Answer
Answer: b [Reason:] The transmission parameter C is I1/V2 |I2=0. This parameter is obtained by open circuiting the port 2. So we get V2 = 5I1 => I1/V2 = 1/5. On substituting, we get C = 1/5.
7. Determine the transmission parameter D in the circuit shown in question 4.
a) 3/5
b) 4/5
c) 7/5
d) 2/5
Answer
Answer: c [Reason:] The transmission parameter D is -I1/I2 |V2=0. This is obtained by short circuiting the port 2. We get I1 = (7/17)V1 and -I2 = (5/17)V1. On solving, we get -I1/I2 = 7/5. On substituting we get D = 7/5.
8. The value of Z11 in the circuit shown in question 4 is?
a) 1.8
b) 2.8
c) 3.8
d) 4.8
Answer
Answer: c [Reason:] The relation between Z11 and ABCD parameters is Z11=√(AB/CD). We know A = 6/5, B = 17/5, C = 1/5, D = 7/5. On substituting, Z11 = √((6/5×17/5)/(1/5×7/5))=3.8Ω.
9. The value of Z12 in the circuit shown in question 4 is?
a) 1.1
b) 2.2
c) 3.3
d) 4.4
Answer
Answer: d [Reason:] The relation between Z12 and ABCD parameters is Z12=√(BD/AC). We got B = 17/5, D = 7/5, A = 6/5, C = 1/5. On substituting Z12=√(BD/AC) = √((17/5×7/5)/(6/5×1/5))=4.4Ω.
10. Determine the value of Ø in the circuit shown in question 4.
a) 0.25
b) 0.5
c) 0.75
d) 1
Answer
Answer: c [Reason:] Ø is called image transfer constant and it is also used to describe reciprocal networks and this parameter is obtained from the voltage and current ratios. We know Ø= tanh-1√(BC/AD) = tanh-1√(17/42)=0.75.
Network Theory MCQ Set 4
1. The impedances Z1 and Z2are said to be inverse if?
a) Z1Z2 = R0
b) Z1 + Z2 = R0
c) 1/Z1+1/Z2=R0
d) Z1Z2 = R02
Answer
Answer: d [Reason:] The impedances Z1 and Z2 are said to be inverse if the geometric mean of the two impedances is a real number.
2. An inverse network may be obtained by?
a) Converting each series branch into another series branch
b) Converting each series branch into another parallel branch
c) Converting each parallel branch into another series branch
d) None of the mentioned
Answer
Answer: c [Reason:] An inverse network may be obtained by converting each parallel branch into another series branch and vice-versa and not by converting each series branch into another series branch and not by converting each series branch into another parallel branch.
3. An inverse network may be obtained by converting each resistance element R into a corresponding resistive element of value?
a) R02/R
b) R/R02
c) R0/R
d) R/R0
Answer
Answer: a [Reason:] To obtain the inverse network we have to convert each resistance element R into a corresponding resistive element of value R02/R.
4. An inverse network may be obtained by converting each inductance L into a capacitance of value?
a) L/R0
b) L/R02
c) R0/L
d) R02/L
Answer
Answer: b [Reason:] An inverse network may be obtained by converting Each inductance L should be converted into a capacitance of value L/R02 to obtain the inverse network.
5. An inverse network may be obtained by converting each capacitance C into an inductance of value?
a) CR02
b) CR0
c) R02/C
d) C/R02
Answer
Answer: a [Reason:] An inverse network is obtained by converting each capacitance C into an inductance of value CR02 where R0 is resistance.
6. Consider the network shown below. Find the value of capacitance C1‘ after converting the inductance L1 into a capacitance.
a) R02/L1
b) R0/L1
c) L1/R02
d) L1/R0
Answer
Answer: c [Reason:] An inverse network may be obtained by converting Each inductance L should be converted into a capacitance of value L/R02 to obtain the inverse network. The value of capacitance C1‘ after converting the inductance into a capacitance is L1/R02. C1’ = L1/R02.
7. In the network showed in question 6, find the value of inductance L1‘ after converting the capacitance into an inductance.
a) C1/R02
b) R02/C1
c) C1R0
d) C1R02
Answer
Answer: d [Reason:] An inverse network is obtained by converting each capacitance C into an inductance of value CR02 where R0 is resistance. The value of inductance L1‘ after converting the capacitance into an inductance is L1‘ = C1R02.
8. From the network showed in question 6, find the value of resistance R1‘ after converting the resistance R1.
a) R1/R0
b) R0/R1
c) R1/R02
d) R02/R1
Answer
Answer: d [Reason:] To obtain the inverse network we have to convert each resistance element R into a corresponding resistive element of value R02/R. The value of resistance R1‘ after converting R1 is R1‘ = R02/R1.
9. The value of the capacitance C2‘ after converting the inductor into the C2‘ in the network showed in question 6.
a) L2/R02
b) L2/R0
c) R02/L2
d) R0/L2
Answer
Answer: a [Reason:] An inverse network may be obtained by converting Each inductance L should be converted into a capacitance of value L/R02 to obtain the inverse network. The value of the capacitance C2‘ after converting the inductor into the capacitance is
C2‘ = L2/R02.
10. The value of the inductor L2‘ after converting the capacitor into the L2‘ in the network showed in question 6.
a) R02/C2
b) C2R02
c) C2R0
d) R02/C2
Answer
Answer: b [Reason:] An inverse network is obtained by converting each capacitance C into an inductance of value CR02 where R0 is resistance. The value of the inductor L2‘ after converting the capacitor into the inductance is L2‘ = C2R02.
Network Theory MCQ Set 5
1.The relation between ZoT and ZoT‘ in the circuits shown below.
a) ZoT = ZoT‘
b) ZoT = 2 ZoT‘
c) ZoT = 3 ZoT‘
d) ZoT = 4 ZoT‘
Answer
Answer: a [Reason:] The relation between ZoT and ZoT’ is ZoT = ZoT’ where ZoT’ is the characteristic impedance of the modified (m-derived) T-network.
2. The value of Z2’ in terms of Z1, Z2 from the circuits shown in question 1 is?
a) Z2‘=Z2/4 m (1-m2 )+Z2/m
b) Z2‘=Z1/4 m (1-m2 )+Z1/m
c) Z2‘=Z2/4 m (1-m2 )+Z1/m
d) Z2‘=Z1/4 m (1-m2 )+Z2/m
Answer
Answer: d [Reason:] As ZoT = ZoT’, √(Z12/4+Z1Z2)=√(m2 Z12/4+m Z2‘). On solving, Z2‘=Z1/(4 m (1-m2))+Z2/m.
3. The relation between Zoπ and Zoπ’ in the circuits shown below is?
a) Zoπ = 2 Zoπ’
b) Zoπ = 4 Zoπ’
c) Zoπ = Zoπ’
d) Zoπ = 3 Zoπ’
Answer
Answer: c [Reason:] The characteristic impedances of the prototype and its modified sections have to be equal for matching. The relation between Zoπ and Zoπ‘ is Zoπ = Zoπ’.
4. The value of Z1‘ in terms of Z1, Z2 from the circuits shown in question 3 is?
a) Z1‘=(m Z2(Z2 4 m)/(1-m2 ))/m Z1(Z2 4 m/(1-m2 ))
b) Z1‘=(m Z1(Z2 4 m)/(1-m2 ))/m Z2(Z2 4 m/(1-m2 ))
c) Z1‘=(m Z1(Z2 4 m)/(1-m2 ))/m Z1(Z2 4 m/(1-m2 ))
d) Z1‘=(m Z1(Z2 4 m)/(1-m2 ))/m Z1(Z1 4 m/(1-m2 ))
Answer
Answer: c [Reason:] As Zoπ = Zoπ’, √(Z1Z2/(1+Z1/4 Z2))=√(((Z1‘ Z2)/m)/(1+(Z1‘)/(4 Z2/m))). On solving, Z1‘=(m Z1(Z2 4 m)/(1-m2 ))/m Z1(Z2 4 m/(1-m2 )) .
5. The value of resonant frequency in the m-derived low pass filter is?
a) fr=1/(√(LC(1+m2 ) ))
b) fr=1/(√(πLC(1+m2 ) ))
c) fr=1/(√(LC(1-m2 ) ))
d) fr=1/(√(πLC(1-m2 ) ))
Answer
Answer: d [Reason:] ωr2 = 1/(LC(1-m2)). So the value of resonant frequency in the m-derived low pass filter is fr=1/√(πLC(1-m2 ) ).
6. The cut-off frequency of the low pass filter is?
a) 1/√LC
b) 1/(π√LC)
c) 1/√L
d) 1/(π√L)
Answer
Answer: b [Reason:] To determine the cut-off frequency of the low pass filter we place m = 0. So fc=1/(π√LC).
7. The resonant frequency of m-derived low pass filter in terms of the cut-off frequency of low pass filter is?
a) fc/√(1-m2 )
b) fc/√(1+m2 )
c) fc/(π√(1-m2 ))
d) fc/(π√(1+m2 ))
Answer
Answer: a [Reason:] If a sharp cut-off is desired, the frequency at infinity should be near to fc. The resonant frequency of m-derived low pass filter in terms of the cut-off frequency of low pass filter is fr=fc/√(1-m2).
8. The expression of m of the m-derived low pass filter is?
a) m=√(1+(fc/fr)2 )
b) m=√(1+(fc/f)2)
c) m=√(1-(fc/fr)2 )
d) m=√(1-(fc/f)2 )
Answer
Answer: c [Reason:] As fr=fc/√(1-m2). The expression of m of the m-derived low pass filter is m=√(1-(fc/fr)2 ).
9. Given a m-derived low pass filter has cut-off frequency 1 kHz, design impedance of 400Ω and the resonant frequency of 1100 Hz. Find the value of k.
a) 400
b) 1000
c) 1100
d) 2100
Answer
Answer: a [Reason:] The value of k is equal to the design impedance. Given design impedance is 400Ω. So, k = 400.
10. The value of m from the information provided in question 9.
a) 0.216
b) 0.316
c) 0.416
d) 0.516
Answer
Answer: c [Reason:] m=√(1-(fc/fr)2) fc = 1000, fr = 1100. On substituting m=√(1-(1000/1100)2 )=0.416.
