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## Network Theory MCQ Set 1

1. The value of attenuation D is equal to?
a) log10 (N)
b) 10 log10 (N)
c) 20 log10 (N)
d) 40 log10 (N)

Answer: a [Reason:] The value of attenuation D is equal to log10 (N). Attenuation D = log10 (N) where N is input to output power ratio of the load.

2. The value of N in terms of attenuation D is?
a) antilog(D)
b) antilog(D/10)
c) antilog(D/20)
d) antilog(D/40)

Answer: b [Reason:] The value of N in terms of attenuation D is antilog(D/10). N = antilog(D/10) where D is attenuation in decibels.

3. The input to output power ratio of the load (N) is the ratio of the________ to the __________
a) Maximum power delivered to the load when the equalizer is not present, power delivered to the load when equalizer is present
b) Power delivered to the load when equalizer is present, maximum power delivered to the load when the equalizer is not present
c) Maximum power delivered to the load when the equalizer is present, power delivered to the load when equalizer is not present
d) Power delivered to the load when equalizer is not present, maximum power delivered to the load when the equalizer is present

Answer: a [Reason:] The input to output power ratio of the load (N) is the ratio of the maximum power delivered to the load when the equalizer is not present to the power delivered to the load when equalizer is present.

4. The N is defined as?
a) output power/ input power
b) input power/ output power
c) output power at inductor/ input power
d) output power at capacitor/ input power

Answer: b [Reason:] The N is defined as the ratio of input power to the output power. N =Pi/Pl where Pi is input power and Pl is output power.

5. The expression of input power of a series equalizer is?
a) Vmax2/Ro
b) Vmax2/2Ro
c) Vmax2/3Ro
d) Vmax2/4Ro

Answer: d [Reason:] The expression of input power of a series equalizer is Pi=(Vmax/2Ro)2 Ro=Vmax2/4Ro.

6. The expression of current flowing in a series equalizer is?
a) Vmax/√((Ro)2+(X1)2)
b) Vmax/√((2Ro)2+(X1)2 )
c) Vmax/√((2Ro)2+(2X1)2 )
d) Vmax/√((Ro)2+(2X1)2 )

Answer: c [Reason:] When the equalizer is connected, the expression of current flowing in a series equalizer is I1= Vmax/√((2Ro)2+(2X1)2 ) where Vmax is voltage applied to the network and Ro is resistance of the load as well as source and 2X1 is the reactance of the equalizer.

7. What is the power at the load of a series equalizer?
a) [ Vmax2/(Ro2+X12 )]Ro
b) [ Vmax2/(2(Ro2+X12))]Ro
c) [ Vmax2/(3(Ro2+X12))]Ro
d) [ Vmax2/(4(Ro2+X12))]Ro

Answer: d [Reason:] The power at the load of a series equalizer is P=(Vmax/√((2Ro)2+(2X1)2 ))2 Ro =[Vmax2/(4(Ro2+X12))]Ro.

8. Determine the value of N in the series equalizer.
a) 1+ X12/Ro2
b) X12/Ro2
c) 1+ Ro2/X12
d) Ro2/X12

Answer: a [Reason:] The N is defined as the ratio of input power to the output power. N=Pi/Pl =(Vmax2/4Ro)/[Vmax2/(4(Ro2+X12))]Ro=1+X12/Ro2 .

9. The expression of N in a full series equalizer considering Z1 as inductor and Z2 as capacitor is?
a) Ro2/(ωL1)2
b) 1+ Ro2/(ωL1)2
c) (ω2 L12)/Ro2
d) 1+ (ω2 L12)/Ro2

Answer: d [Reason:] The expression of N in a full series equalizer considering Z1 as inductor and Z2 as capacitor is N = 1 + X12/Ro2 = 1+ (ω2 L12)/Ro2 .

10. The expression of N in a full series equalizer considering Z1 as capacitor and Z2 as inductor is?
a) 1+ (ω2 L12)/Ro2
b) (ω2 L12)/Ro2
c) 1+ Ro2/(ωL1)2
d) Ro2/(ωL1)2

Answer: c [Reason:] The expression of N in a full series equalizer considering Z1 as capacitor and Z2 as inductor is N = 1+ Ro2/X22 = 1+Ro2/(ωL1)2 .

## Network Theory MCQ Set 2

1. In the shunt equalizer, the current flowing from the source is?
a) Vmax(2Ro+jX1)/2Ro(Ro+jX1)
b) Vmax(Ro+jX1)/Ro(Ro+jX1)
c) Vmax(Ro+jX1)/2Ro(Ro+jX1)
d) Vmax(2Ro+jX1)/Ro(Ro+jX1)

Answer: a [Reason:] In the shunt equalizer, the current flowing from the source is Is= Vmax/(Ro+(Ro||jX1/2)). On solving, Is= Vmax(2Ro+jX1)/2Ro(Ro+jX1) .

2. What is the load current in terms of source current in the shunt equalizer?
a) Is jX1/(Ro+jX1)
b) Is jX1/(Ro+2jX1)
c) Is jX1/(2Ro+2jX1)
d) Is jX1/(2Ro+jX1)

Answer: d [Reason:] The load current in terms of source current in the shunt equalizer is Il = Is (jX1/2)/(Ro+jX1/2)). On solving, Il= Is jX1/(2Ro+jX1).

3. What is the load current in terms of Vmax in the shunt equalizer?
a) (VmaxjX1)/(Ro(2Ro+jX1))
b) (VmaxjX1)/(2Ro(2Ro+jX1))
c) (VmaxjX1)/(2Ro(Ro+jX1))
d) (VmaxjX1)/(Ro(Ro+jX1))

Answer: c [Reason:] On substituting Is in the load current equation we get the load current in terms of Vmax in the shunt equalizer as Il= (VmaxjX1)/(2Ro(Ro+jX1)).

4. The input power in shunt equalizer is?
a) Vmax2/Ro
b) Vmax2/2Ro
c) Vmax2/3Ro
d) Vmax2/4Ro

Answer: d [Reason:] The expression of input power of a shunt equalizer is Pi=(Vmax/2Ro)2 Ro=Vmax2/4Ro.

5. What is the power at the load of a shunt equalizer?
a) [ (Vmax2 X12)/(Ro(Ro2+X12))].

b) [ (Vmax2 X12)/(2Ro(Ro2+X12))].

c) [ (Vmax2 X12)/(3Ro(Ro2+X12))].

d) [ (Vmax2 X12)/(4Ro(Ro2+X12))].

Answer: d [Reason:] The power at the load of a series equalizer is P=((Vmax jX1)/(2Ro(Ro+jX1)))2 Ro =[(Vmax2 X12)/(4Ro(Ro2+X12))].

6. The value of N in shunt equalizer is?
a) 1+ X12/Ro2
b) X12/Ro2
c) 1+ Ro2/X12
d) Ro2/X12

Answer: c [Reason:] The N is defined as the ratio of input power to the output power. N=Pi/Pl=(Vmax2/4Ro)/( (Vmax2 X12)/4Ro(Ro2+X12 ) )=1+ Ro2/X12 .

7. The propagation constant of a symmetrical T-section and π-section are the same.
a) True
b) False

Answer: a [Reason:] The propagation constant of a symmetrical T-section and π-section are the same.

8. The attenuation is not sharp in the stop band for an m-derived filter.
a) True
b) False

Answer: b [Reason:] The attenuation is sharp in the stop band for an m-derived filter. So the given statement is not true.

9. The bridged-T phase equalizer consists of?
a) Only pure inductors
b) Only pure capacitors
c) Only pure resistors
d) Only pure reactance

Answer: d [Reason:] The bridged-T phase equalizer consists of only pure reactances. So the bridged T-circuit consists of only inductive or capacitive elements not resistive elements.

10. A lattice phase equalizer is a constant equalizer which satisfies the equation?
a) Z1Z2 = Ro
b) Z1 + Z2 = Ro
c) 1/Z1+1/Z2=Ro
d) Z1Z2 = Ro2

Answer: d [Reason:] The lattice phase equalizer consists of only reactive components. So a lattice phase equalizer is a constant equalizer if the following equation is satisfied. Z1Z2 = Ro2.

## Network Theory MCQ Set 3

1. Consider the impedance functionZ(s)=( s2+6s+8)/( s2+3s). Find the value of R1 after performing the first Cauer form.
a) 1
b) 2
c) 3
d) 4

Answer: a [Reason:] To find the first Cauer form, we take the continued fraction expansion by the divide, invert, divide procedure. On performing this we get the first quotient is 1Ω. So, R1 = 1Ω.

2. Find the first reminder obtained by taking the continued fraction expansion in question 1.
a) s + 8
b) 2s + 8
c) 3s + 8
d) 4s + 8

Answer: c [Reason:] The continued fraction expansion is done by the divide, invert, divide procedure. So the first reminder obtained is 3s + 8 .

3. Find the value of R2 in question 1.
a) 4
b) 3
c) 6
d) 9

Answer: d [Reason:] On performing the continued fraction expansion we get the third quotient as 9. So the value of R2 is 9Ω. R2 = 9Ω.

4. The value of C1 in question 1 is?
a) 1/4
b) 1/3
c) 1/2
d) 1

Answer: b [Reason:] The second quotient of the continued fraction expansion is s/3. So the value of C1 is 1/3 F. C1 = 1/3 F.

5. The value of C2 in question 1 is?
a) 1/6
b) 1/12
c) 1/24
d) 1/48

Answer: c [Reason:] We obtain the fourth quotient on performing continued fraction expansion as s/24. So the value of C2 is 1/24 F. C2 = 1/24 F.

6. Consider the impedance function Z(s)=( s2+6s+8)/( s2+3s). Find the value of C1 after performing the second Cauer form.
a) 1/2
b) 3/8
c) 1/4
d) 1/8

Answer: b [Reason:] The second Cauer network can be obtained by arranging the numerator and denominator polynomials of Z(s) in ascending powers of s. After performing the continued fraction expansion, we get C1 = 3/8 F.

7. Find the first reminder obtained by taking the continued fraction expansion in question 6.
a) 10s/3+s2
b) s/3+s2
c) 10s/3+3s2
d) s/3+3s2

Answer: a [Reason:] On performing the continued fraction expansion after arranging the numerator and denominator polynomials of Z(s) in ascending powers of s, the first reminder is 10s/3+s2.

8. The value of R1 in question 6 is?
a) 9/10
b) 10/9
c) 8/9
d) 9/8

Answer: b [Reason:] The second quotient on performing continued fraction expansion is 9/10. This is the value of 1/R1. So the value of R1 is 10/9Ω. R1 = 10/9Ω.

9. The value of C2 in question 6 is?
a) 3
b) 3/10
c) 3/100
d) 3/1000

Answer: c [Reason:] On performing continued fraction expansion, the third quotient is 100/3s. So the value of C2 is 3/100 F. C2 = 3/100 F.

10. The value of R1 in question 6 is?
a) 10
b) 1
c) 100
d) 1000

Answer: a [Reason:] The final quotient is 1/10. So the value of R1 is 1/1/10.So the value of R1 is 10Ω. R1 = 10Ω.

## Network Theory MCQ Set 4

1. The driving point impedance of a one-port reactive network is given by Z(s)=5(s2+4)(s2+25)/s(s2+16) . After taking the partial fractions, find the coefficient of 1/s.
a) 25/4
b) 50/4
c) 100/4
d) 125/4

Answer: d [Reason:] Since there is an extra term in the numerator compared to the denominator and also an s term in the denominator, the two poles exists at 0 and infinity. Therefore the network consists of first element and last element. By taking the partial fraction expansion of Z(s), we obtain A=5(s2+4)(s2+25)/((s2+16) ) |s=0 =(5×4×25)/16=125/4.

2. In question 1, find the coefficient of (s + j4).
a) 135/4
b) 145/4
c) 155/4
d) 165/4

Answer: a [Reason:] On taking the partial fraction expansion, B=5 (s2+4)(s2+25)/s(s-j4) |s=-j4 = 135/8.

3. The value of H from Z (s) in question 1 is?
a) 3
b) 4
c) 5
d) 6

Answer: c [Reason:] By inspection, the value of H is 5. So H = 5.

4. The value of C0 from the information provided in question 1 is?
a) 1/125
b) 4/125
c) 2/125
d) 3/125

Answer: b [Reason:] The coefficient of 1/s is 125/4. And C0=1/P0 = 1/(125/4)=4/125 Farad.

5. The value of L in question 1 is?
a) 2
b) 3
c) 4
d) 5

Answer: d [Reason:] We know L = H = 5 H.

6. The value of C2 from the information provided in question 1 is?
a) 4/270
b) 8/270
c) 12/270
d) 16/270

Answer: b [Reason:] We know C2= 1/2P2 = 8/(2×135)=8/270 F.

7. The value of L2 from the information provided in question 1 is?
a) 135/60
b) 135/62
c) 135/64
d) 135/66

Answer: c [Reason:] We know L2= 2P2n2 = (2×135)/(16×8)=135/64 H.

8. For performing second Foster form, after splitting the Z (S) given in question 1 into partial fractions, the coefficient of s/((s2+4)) is?
a) 1/35
b) 2/35
c) 3/35
d) 4/35

Answer: b [Reason:] The value of A is (1/5)(s(s2+16)))/(s-j2)(s2+25) at s=-j2 On solving we get the value of A as 2/35. So the coefficient of s/((s2+4)) is 2/35.

9. In question 8, the coefficient of s/((s2+4) ) is?
a) 4/35
b) 3/35
c) 2/35
d) 1/35

Answer: c [Reason:] The coefficient of s/((s2+4) ) is B=(1/5)((s(s2+16) ))/(s-j5)(s2+4) at s=-j5. On solving we get B = 2/35.

10. Determine the value of L1 by performing second foster form for question 1.
a) 35/4
b) 35/3
c) 35/2
d) 35

Answer: a [Reason:] P1 = 2/35. We know L1= 1/2P1 = 35/4 H.

## Network Theory MCQ Set 5

1. Consider the impedance function; Z(s)=((s+4)(s+8))/((s+2)(s+6)) . Find the value of R1 after converting into first Cauer form.
a) 1
b) 2
c) 3
d) 4

Answer: a [Reason:] To find out the first Cauer form, we have to take the continued fraction expansion of Z (s). On solving, we get the first quotient as 1. So the value of R1 as 1Ω.

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

Answer: c [Reason:] On taking the continued fraction expansion of Z (s), the second quotient is s/4. So the value of L2 is ¼ H. L2 = ¼ H.

3. Find the value of R2 in question 1.
a) 1/4
b) 2/4
c) 3/4
d) 4/4

Answer: c [Reason:] We get the third quotient on performing continued fraction expansion of Z (s) as 4/3 and is the value of 1/R2. So the value of R2 is ¾Ω . R2 = ¾Ω .

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

Answer: b [Reason:] The fourth quotient obtained is 3s/4. And this is the value of sL3. So the value of L3 is ¾ H. L3 = ¾ H.

5. Find the value of R3 in question 1.
a) 4
b) 3
c) 2
d) 1

Answer: b [Reason:] The value of 1/R3 (fourth quotient) obtained by continued fraction expansion 1/3. So the value of R3 is 3Ω.

6. Consider the impedance function; Z(s)=(2s2+8s+6)/( s2+8s+12). Find the value of R1 after converting into second Cauer form.
a) 1
b) 3/4
c) 1/2
d) 1/4

Answer: c [Reason:] To find out the second Cauer form, we have to write the impedance function in ascending powers and by taking the continued fraction expansion of Z (s). On solving, the first quotient obtained is 1/2. So the value of R1 as ½ Ω.

7. Find the value of L1 in question 6.
a) 1/3
b) 2/3
c) 3/3
d) 4/3

Answer: a [Reason:] The second quotient obtained is 3/s and this is the value of 1/sL1. So the value of L1 is 1/3 H. L1 = 1/3 H.

8. Find the value of R2 in question 6.
a) 6/7
b) 7/6
c) 7/8
d) 8/7

Answer: d [Reason:] On taking the continued fraction expansion, we get third quotient as 8/7. So the value of R2 is 8/7Ω. R2 = 8/7Ω.

9. Find the value of L2 in question 6.
a) 5/50
b) 10
c) 5/49
d) 49/5

Answer: c [Reason:] We obtain the fourth quotient i.e., 1/sL2 as 49/5s. So the value of L2 is 5/49 H. L2 = 5/49H.

10. Find the value of R3 in question 6.
a) 1/5
b) 14/5
c) 5/14
d) 5

Answer: b [Reason:] The value of R3 is 14/5Ω as the fifth quotient obtained is 5/14. R3 = 5/14Ω.

## Network Theory MCQ Set 6

1. Consider a function Z(s)=5(s+1)(s+4)/(s+3)(s+5) . Find the value of R1 after performing the first form of Foster method.
a) 1/3
b) 2/3
c) 3/3
d) 4/3

Answer: d [Reason:] After splitting the given function into partial according to the properties of first form of Foster method, we get Z(s)=4/3+(5/3)(s/(s+3))+2s/(s+5). So, R1 = 4/3Ω.

2. The value of R1 in the question 1 is?
a) 4/3
b) 5/3
c) 3/5
d) 3/4

Answer: b [Reason:] On taking the partial fractions, we get P1 as 5/3 and we know R1=R1. So the value of R1 is 5/3Ω. R1 = 5/3Ω.

3. The value of L1 in the question 1 is?
a) 5/9
b) 9/5
c) 4/9
d) 9/4

Answer: a [Reason:] We know R1=(3)(L1) and as R1 is 5/3Ω. So the value of L1 is 5/9 H. L1 = 5/9 H.

4. The value of R2 in the question 1 is?
a) 1
b) 2
c) 3
d) 4

Answer: b [Reason:] We obtain P2 as 2. R2=P2. So the value of R2 is 2Ω. R2 = 2Ω.

5. The value of L2 in the question 1 is?
a) 4/5
b) 3/5
c) 2/5
d) 1/5

Answer: c [Reason:] As R2=(5)(L2) So the value of L2 is 2/5 H. L2 = 2/5 H.

6. Consider the admittance function, Y(s)=((2s2+16s+30))/( s2+6s+8). Determine the value of L1 after performing the second form of Foster method.
a) 1/3
b) 2/3
c) 3/3
d) 4/3

Answer: a [Reason:] After splitting the given function into partial according to the properties of first form of Foster method, we get Y(s)=2+3/(s+2)+1/(s+4). We know L1=1/P1 and as P1 = 3, the value of L1 is 1/3H. L1 = 1/3H.

7. The value of R1 in the question 6 is?
a) 4/3
b) 3/3
c) 2/3
d) 1/3

Answer: c [Reason:] R1 = 2/P1 and as P1 is 3, the value of R1 is 2/3Ω. R1 = 2/3Ω.

8. The value of R2 in the question 6 is?
a) 1
b) 2
c) 3
d) 4

Answer: d [Reason:] On taking the partial fractions we get P2 as 1. And we know R2 = 4/P2. So the value of R2 is 4Ω. R2 = 4Ω.

9. The value of L2 in the question 6 is?
a) 4
b) 1
c) 2
d) 3

Answer: b [Reason:] We got P2 = 1 And L2=1/P2 So the value of L2 is 1H. L2= 1H.

10. The value of R in the question 6 is?
a) 3
b) 1
c) 2
d) 4