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## Network Theory MCQ Set 1

1. The driving point impedance of an LC network is given by Z(s)=(2s5+12s3+16s)/( s4+4s2+3). By taking the continued fraction expansion using first Cauer form, find the value of L1.
a) s
b) 2s
c) 3s
d) 4s

Answer: b [Reason:] The first Cauer form of the network is obtained by taking the continued fraction expansion of given Z(s). And we get he first quotient as 2s. So, L1 = 2s.

2. Find the first reminder obtained by taking the continued fraction expansion in question 1.
a) 4s3+10s
b) 12s3+10s
c) 4s3+16s
d) 12s3+16s

Answer: a [Reason:] On taking the continued fraction expansion, the first reminder obtained is 4s3+10s.

3. Find the value of C2 in question 1.
a) 1
b) 1/2
c) 1/3
d) 1/4

Answer: d [Reason:] The second quotient obtained on taking the continued fraction expansion is s/4 and this is the value of sC2. So the value of C2 = 1/4.

4. Find the value of L3 in question 1.
a) 8
b) 8/3
c) 8/5
d) 8/7

Answer: b [Reason:] By taking the continued fraction expansion, the third quotient is 8s/3. sL3 = 8s/3. So L3 = 8/3H.

5. Find the value of C4 in question 1.
a) 1/2
b) 1/4
c) 3/4
d) 1

Answer: c [Reason:] We get the fourth quotient as 3s/4. So sC4 = 3s/4. C4 = 3/4F.

6. Find the value of L5 in question 1.
a) 2
b) 2/5
c) 2/7
d) 2/3

Answer: d [Reason:] On taking the continued fraction expansion fifth quotient is 2s/3. sL5 = 2s/3 So L5 = 2/3H.

7. The driving point impedance of an LC network is given by Z(s)=(s4+4s2+3)/(s3+2s) . By taking the continued fraction expansion using second Cauer form, find the value of C1.
a) 2/3
b) 2/2
c) 1/2
d) 4/2

Answer: a [Reason:] To obtain the second Cauer form, we have to arrange the numerator and the denominator of given Z(s) in ascending powers of s before starting the continued fraction expansion. By taking the continued fraction expansion we get the first quotient as 3/2s. So 1/sC1 = 3/2s C1 = 2/3F..

8. Find the value of L2 in question 7.
a) 1/5
b) 2/5
c) 3/5
d) 5/4

Answer: d [Reason:] On taking the continued fraction expansion the second quotient is 4/5s. 1/sL2 = 4/5s So L2 = 5/4H.

9. Find the value of C3 in question 7.
a) 25/s
b) 2/25s
c) 25/3s
d) 25/4s

Answer: b [Reason:] The third quotient is 25/2s. 1/sC3 = 25/2s. C3 = 2/25F.

10. Find the value of L4 in question 7.
a) 5
b) 2/5
c) 3/5
d) 4/5

Answer: a [Reason:] We obtain the fourth quotient as 1/5s. 1/sL4 = 1/5s L4 = 5H.

## Network Theory MCQ Set 2

1. The attenuation in dB in terms of input power (P1) and output power (P2) is?
a) log10 (P1/P2)
b)10 log10 (P1/P2)
c) log10 (P2/P1)
d) 10 log10 (P2/P1)

Answer: b [Reason:] The increase or decrease in power due to insertion or substitution of a new element in a network can be conveniently expressed in decibels or in nepers. The attenuation in dB in terms of input power (P1) and output power (P2) is Attenuation in dB = 10 log10 (P1/P2).

2. If V1 is the voltage at port 1 and V2 is the voltage at port 2, then the attenuation in dB is?
a) 20 log10 (V1/V2)
b) 10 log10 (V1/V2)
c) 20 log10 (V2/V1)
d) 10 log10 (V2/V1)

Answer: a [Reason:] If V1 is the voltage at port 1 and V2 is the voltage at port 2, then the attenuation in dB is Attenuation in dB =20 log10 (V1/V2) where V1 is the voltage at port 1 and V2 is the voltage at port 2.

3. What is the attenuation in dB assuming I1 is the input current and I2 is the output current leaving the port?
a) 10 log10 (I1/I2)
b) 10 log10 (I2/I1)
c) 20 log10 (I2/I1)
d) 20 log10 (I1/I2)

Answer: d [Reason:] Assuming I1 is the input current and I2 is the output current leaving the port, the attenuation in dB is Attenuation in dB =20 log10 (I1/I2) where I1 is the input current and I2 is the output current leaving the port.

4. The value of one decibel is equal to?
a) log10 (N)
b) 10 log10 (N)
c) 20 log10 (N)
d) 40 log10 (N)

Answer: c [Reason:] The value of one decibel is equal to 20 log10 (N). One decibel = 20 log10 (N) where N is the attenuation.

5. The value of N in dB is?
a) N= anti log (dB)
b) N= anti log(dB/10)
c) N=anti log(dB/20)
d) N=anti log(dB/40)

Answer: c [Reason:] The value of N in dB can be expressed as N=anti log(dB/20).

6. In the circuit shown below, find the value of I1/I2.

a) (R1-R2+R0)/R2
b) (R1+R2+R0)/R2
c) (R1-R2-R0)/R2
d) (R1+R2-R0)/R2

Answer: b [Reason:] R2(I1-I2)=I2(R1+R0) => I2(R2+R1+R0)I1R2. On solving, I1/I2=(R1+R2+R1)/R2.

7. Determine the value of N in the circuit shown in question 6.
a) (R1+R2-R0)/R2
b) (R1-R2-R0)/R2
c) (R1+R2+R0)/R2
d) (R1-R2+R0)/R2

Answer: c [Reason:] N = I1/I2. We got I1/I2=(R1+R2+R1)/R2. So on substituting we get N = (R1+R2+R0)/R2.

8. The value of the characteristic impedance R0 in terms of R1 and R2 and R0 in the circuit shown in question 6 is?
a) R1+R2(R1+R0)/(R1+R0+R2)
b) R1+ R2(R1+R0)/(R1+R0+R2)
c) R2+ R2(R1+R0)/(R1+R0+R2)
d) R0+R2(R1+R2)/(R1+R0+R2)

Answer: b [Reason:] The value of the characteristic impedance R0 in terms of R1 and R2 and R0 when it is terminated in a load of R0 is R0=R1+ R2(R1+R0)/(R1+R0+R2).

9. Determine the value of R1 in terms of R0 and N in the circuit shown in question 6 is?
a) R1= R0(N-1)/(N+1)
b) R1= R0(N+1)/(N+1)
c) R1= R0(N-1)/(N-1)
d) R1= R0(N+1)/(N-1)

Answer: a [Reason:] R0 = R1+(R1+R0)/N. On solving, the value of R1 in terms of R0 and N is R1= R0(N-1)/(N+1).

10. Determine the value of R2 in terms of R0 and N in the circuit shown in question 6 is?
a) R2= NR0/(N2-1)
b) R2= 2 NR0/(N2-1)
c) R2= 3 NR0/(N2-1)
d) R2= 4 NR0/(N2-1)

Answer: b [Reason:] NR2 = R1+R0+R2. On substituting the value of R1, we get the value of R2 in terms of R0 and N as R2= 2 NR0/(N2-1).

## Network Theory MCQ Set 3

1. Based on the location of zeros and poles, a reactive one-port can have ____________ types of frequency response.
a) 1
b) 2
c) 3
d) 4

Answer: d [Reason:] A reactive one-port can have four types of frequency response based on the location of zeros and poles. (i) frequency response with two external poles (ii) frequency response with two external zeros (iii) frequency response with an external zero ω = 0 and an external poles at ω = ∞ (iv) frequency response with an external zero ω = ∞ and an external poles at ω = 0.

2. A driving point impedance with poles at ω = 0, ω = ∞ must have ___________ term in the denominator polynomial.
a) s
b) s+1
c) s+2
d) s+3

Answer: a [Reason:] As there is a pole at ω = 0, (s-jω)=s. Poles are written in the denominator of Z(s). So there will be s term in the denominator polynomial in a driving point impedance function Z(s).

3. A driving point impedance with poles at ω = 0, ω = ∞ must have excess ___________ term in the numerator polynomial.
a) s1n1
b) s1n2
c) s2n2
d) s2n1

Answer: c [Reason:] The driving point impedance of the one-port is infinite, and it will not pass either direct current or alternating current of an infinitely high frequency.

4. A driving point impedance with zeros at ω = 0, ω = ∞ must have ___________ term in the numerator polynomial.
a) s+3
b) s+2
c) s+1
d) s

Answer: d [Reason:] As there is a zero at ω = 0, (s-jω)=s. Zeros are written in the numerator of Z(s). So there will be s term in the numerator polynomial in a driving point impedance function Z(s).

5. A driving point impedance with zeros at ω = 0, ω = ∞ must have an excess ___________ term in the denominator polynomial.
a) s2n1
b) s2n2
c) s1n2
d) s1n1

Answer: b [Reason:] The driving point impedance of the one-port is zero, and it will pass both direct current and an alternating current of an infinitely high frequency.

6. A driving point impedance with zero at ω = 0 and pole at ω = ∞ must have ___________ term in the numerator polynomial.
a) s+1
b) s
c) s+3
d) s+2

Answer: b [Reason:] As ω = 0, (s-jω)=s. The numerator of Z(s) contains poles and denominator contains zeros. So there will be s term in the numerator polynomial.

7. A driving point impedance with zero at ω = 0 and pole at ω = ∞ must have ___________ term in the numerator polynomial.
a) s1n1
b) s2n1
c) s1n2
d) s2n2

Answer: d [Reason:] If a pole is at ω = ∞, there will be an equal number of s2n2 type terms in the numerator polynomial and the denominator polynomial.

8. A driving point impedance with zero at ω = 0 and pole at ω = ∞ must have ___________ term in the denominator polynomial.
a) s2n2
b) s1n1
c) s2n1
d) s1n2

Answer: a [Reason:] If there is a zero at ω = 0 and pole at ω = ∞ , the one-port will pass direct current and block the alternating current of an infinitely high frequency.

9. A driving point impedance with pole at ω = 0 and zero at ω = ∞ must have ___________ term in the denominator polynomial.
a) s
b) s+3
c) s+1
d) s+2

Answer: a [Reason:] s-jω = (s-j(0)) = s. As pole is at ω = 0, there will be s term in the denominator polynomial.

10. A driving point impedance with pole at ω = 0 and zero at ω = ∞ must have ____________ term in the numerator and denominator.
a) s1n2
b) s2n2
c) s1n1
d) s2n1

Answer: b [Reason:] If a pole at ω = 0 and zero at ω = ∞, the one-port will block the direct current and pass the alternating current of an infinitely high frequency.

## Network Theory MCQ Set 4

1. A network either T or π, is said to be of the constant-k type if Z1 and Z2 of the network satisfy the relation?
a) Z1Z2 = k
b) Z1Z2 = k2
c) Z1Z2 = k3
d) Z1Z2 = k4

Answer: b [Reason:] Z1,Z2 are inverse if their product is a constant, independent of frequency, k is real constant, that is the resistance. k is often termed as design impedance or nominal impedance of the constant k-filter.

2. In the circuit shown below, find the value of Z1.

a) jωL
b) 2 jωL
c) jωL/2
d) 4 jωL

Answer: a [Reason:] The constant k, T or π type filter is also known as the prototype because other more complex networks can be derived from it. From the given figure, the value of Z1 is jωL.

3. In the circuit shown in the question 2, find the value of Z2.
a) jωC
b) 2 jωC
c) 1/jωC
d) 1/2 jωC

Answer: c [Reason:] From the prototype T section and prototype π section shown in figures, we get the value of Z2 is 1/jωC.

4. The value of Z1Z2 in the circuit shown in the question 2 is?
a) L/C
b) C/L
c) 1/LC
d) LC

Answer: a [Reason:] Z1 = jωL and Z2 = 1/jωC. So the product Z1Z2 is jωL x 1/jωC = L/C.

5. Determine the value of k in the circuit shown in the question 2.
a) √LC
b) √((L/C) )
c) √((C/L) )
d) √((1/CL) )

Answer: b [Reason:] We got Z1Z2 = L/C. And we know Z1Z2= k2. So k2 = L/C. So the value of k is √(L/C).

6. The cut-off frequency of the constant k-low pass filter is?
a) 1/√LC
b) 1/(π√LC)
c) √LC
d) π√LC

Answer: b [Reason:] Z1/4Z2 = 0. Z1 = jωL and Z2 = 1/jωC. On solving the cut-off frequency of the constant k-low pass filter is fc= 1/(π√LC).

7. The value of α in the pass band of constant k-low pass filter is?
a) 2 cosh-1⁡(fc/f)
b) cosh-1⁡⁡(fc/f)
c) cosh-1⁡(f/fc)
d) 2 cosh-1⁡(f/fc)

Answer: d [Reason:] The value of α in the pass band of constant k-low pass filter is α= 2 cosh-1⁡(f/fc).

8. The value of β in the attenuation band of constant k-low pass filter is?
a) 0
b) π
c) π/2
d) π/4

Answer: b [Reason:] We know that in the attenuation band, Z1/4Z2 < -1 i.e., f/fc < 1. So the value of β in the pass band of constant k-low pass filter is β= π.

9. The value of α in the attenuation band of constant k-low pass filter is?
a) α=2 cosh-1⁡(fc/f)
b) α=cosh-1⁡(f/fc)
c) α=2 cosh-1⁡(f/fc)
d) α=cosh-1⁡(fc/f)

Answer: c [Reason:] α = 2 cosh-1[Z1/4Z2] and Z1/4Z2 = f/fc. On substituting we get α = 2 cosh-1⁡(f/fc).

10. The value of α in the pass band of constant k-low pass filter is?
a) π
b) π/4
c) π/2
d) 0

Answer: d [Reason:] We know that in the pass band, the condition is -1 < Z1/4Z2 < 0. So α= π.

## Network Theory MCQ Set 5

1. The relation between α, β, ϒ is?
a) α = ϒ + jβ
b) ϒ = α + jβ
c) β = ϒ + jα
d) α = β + jϒ

Answer: b [Reason:] We know that the propogation constant is a complex function and the real part of the complex propogation constant is a measure of the change in magnitude of the current or voltage in the network known as attenuation constant and imaginary part is a measure of the difference in phase between the input and output currents or voltages known as phase shift constant. ϒ = α + jβ.

2. If Z1, Z2 are same type of reactance, then |Z1/4 Z2| is real, then the value of α is?
a) α = sinh-1⁡√( Z1/4 Z2)
b) α = sinh-1⁡⁡√( Z1/Z2)
c) α = sinh-1⁡√( 4 Z1/Z2)
d) α = sinh-1⁡⁡√( Z1/2 Z2)

Answer: a [Reason:] Z1, Z2 are same type of reactance and |Z1/4 Z2| is real. |Z1/4 Z2| > 0. The value of α is α = sinh-1⁡√(Z1/4 Z2).

3. If Z1, Z2 are same type of reactance, then |Z1/4 Z2| is real, then?
a) |Z1/4 Z2|=0
b) |Z1/4 Z2| < 0
c) |Z1/4 Z2| > 0
d) | Z1/4 Z2|>= 0

Answer: c [Reason:] If Z1 and Z2 are same type of reactances, then √(Z1/4 Z2) should be always positive implies that |Z1/4 Z2|>0.

4. Which of the following expression is true if Z1, Z2 are same type of reactance?
a) sinh⁡α/2 sin⁡β/2=0
b) coshα/2 sin⁡β/2=0
c) coshα/2 cos⁡β/2=0
d) sinhα/2 cos⁡β/2=0

Answer: d [Reason:] If Z1, Z2 are same type of reactance, then the real part of sinhϒ/2 = sinhα/2 cos⁡β/2 + jcoshα/2 sin⁡β/2 should be zero. So sinhα/2 cos⁡β/2=0.

5. Which of the following expression is true if Z1, Z2 are same type of reactance?
a) sinhα/2 cos⁡β/2=x
b) coshα/2 cos⁡β/2=0
c) coshα/2 sin⁡β/2=x
d) sinhα/2 sin⁡β/2=0

Answer: c [Reason:] If Z1, Z2 are same type of reactance, then the imaginary part of sinhϒ/2 = sinhα/2 cos⁡β/2 + jcoshα/2 sin⁡β/2 should be some value. So coshα/2 sin⁡β/2=x.

6. The value of α if Z1, Z2 are same type of reactance?
a) 0
b) π/2
c) π
d) 2π

Answer: a [Reason:] As sinhα/2 cos⁡β/2=0 and coshα/2 sin⁡β/2=x, the value of α if Z1, Z2 are same type of reactance is α= 0.

7. The value of β if Z1, Z2 are same type of reactance?
a) 2π
b) π
c) π/2
d) 0

Answer: b [Reason:] The value of β if Z1, Z2 are same type of reactances, then sinhα/2 cos⁡β/2=0 and coshα/2 sin⁡β/2=x. So the value of β is β= π.

8. If Z1, Z2 are same type of reactance, and if α = 0, then the value of β is?
a) β=2 sin-1⁡(√(Z1/4 Z2))
b) β=2 sin-1⁡(√(4 Z1/Z2))
c) β=2 sin-1⁡(√(4 Z1/Z2))
d) β=2 sin-1⁡(√(Z1/Z2))

Answer: a [Reason:] If α = 0, sin β/2 = x(√(Z1/4 Z2). But sine can have a maximum value of 1. Therefore the above solution is valid only for Z1/4 Z2, and having a maximum value of unity. It indicates the condition of pass band with zero attenuation and follows the condition as -1 < Z1/4 Z2 <= 0. So β=2 sin-1⁡(√(Z1/4 Z2)).

9. If the value of β is π, and Z1, Z2 are same type of reactance, then the value of β is?
a) α=2 cosh-1⁡√(Z1/2 Z2)
b) α=2 cosh-1⁡√(Z1/Z2)
c) α=2 cosh-1⁡√(4 Z1/Z2)
d) α=2 cosh-1⁡√(Z1/4 Z2)

Answer: d [Reason:] If the value of β is π, cos β/2 = 0. So sin β/2 = ±1; cosh α/2 = x = √(Z1/4 Z2). This solution is valid for negative Z1/4 Z2 and having magnitude greater than or equal to unity. -α &lt= Z1/2 Z2 <= -1. α=2 cosh-1⁡√(Z1/4 Z2).

10. The relation between Z, Z1, Z2, ZoT is?
a) ZoT = Z1Z2/Z
b) Z = Z1Z2/ZoT
c) ZoT = Z1Z1/Z
d) ZoT = Z2Z2/Z