This set of Network Theory Multiple Choice Questions & Answers (MCQs) focuses on “Series Equalizer”.
1. The value of attenuation D is equal to?
a) log10 (N)
b) 10 log10 (N)
c) 20 log10 (N)
d) 40 log10 (N)
View Answer
Explanation: 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)
View Answer
Explanation: 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
View Answer
Explanation: 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
View Answer
Explanation: 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
View Answer
Explanation: 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)
View Answer
Explanation: 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
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Explanation: 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
View Answer
Explanation: 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
View Answer
Explanation: 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
View Answer
Explanation: 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.
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