This set of Network Theory Multiple Choice Questions & Answers (MCQs) focuses on “Parallel Resonance”.
1. For the circuit shown below, determine its resonant frequency.
a) 6.12
b) 7.12
c) 8.12
d) 9.12
View Answer
Explanation: The resonant frequency of the circuit is fr = 1/(2π√LC). Given L = 5H and C = 100uf. On substituting the given values in the equation we get resonant frequency = 1/(2π√(5×100×10-6)) = 7.12 Hz.
2. Find the quality factor of the following circuit.
a) 2.24
b) 3.34
c) 4.44
d) 5.54
View Answer
Explanation: The quality factor of the circuit is Q = XL/R = 2πfrL/R. Given f = 7.12 Hz and L = 5H and R = 100. On substituting the given values in the equation we get the quality factor = (6.28×7.12×5)/100 = 2.24.
3. Find the bandwidth of the circuit shown below.
a) 1
b) 2
c) 3
d) 4
View Answer
Explanation: The bandwidth of the circuit is BW = fr/Q. we obtained fr = 7.12 Hz and Q = 2.24. On substituting the given values in the equation we get the bandwidth = 7.12/2.24 = 3.178Hz.
4. The magnification in resonance considering the voltage across inductor is?
a) V/VL
b) VL/V
c) V x VL
d) VL
View Answer
Explanation: The ratio of voltage across inductor to the voltage applied at resonance can be defined as magnification. The magnification in resonance considering the voltage across inductor is Q = VL/V.
5. Considering the voltage across the capacitor, the magnification in resonance is?
a) VC
b) V x VC
c) VC/V
d) V/VC
View Answer
Explanation: The ratio of voltage across capacitor to the voltage applied at resonance can be defined as magnification. Considering the voltage across the capacitor, the magnification in resonance is Q = VC/V.
6. The value of ωr in parallel resonant circuit is?
a) 1/(2√LC)
b) 1/√LC
c) 1/(π√LC)
d) 1/(2π√LC)
View Answer
Explanation: Basically parallel resonance occurs when XL = XL. The frequency at which the resonance occurs is called the resonant frequency. The value of ωr in parallel resonant circuit is ωr = 1/√LC.
7. The expression of resonant frequency for parallel resonant circuit is?
a) 1/(2π√LC)
b) 1/(π√LC)
c) 1/(2√LC)
d) 1/√LC
View Answer
Explanation: The condition for resonance occurs when XL = XL. The expression of resonant frequency for parallel resonant circuit is fr = 1/(2π√LC).
8. Find the resonant frequency in the ideal parallel LC circuit shown in the figure.
a) 7.118
b) 71.18
c) 711.8
d) 7118
View Answer
Explanation: The expression for resonant frequency is fr = 1/(2π√LC). Given L = 50mH and C = 0.01uF. On substituting the given values in the equation we get the resonant frequency = 1/(2π√(50×10-3)×0.01×10-6))=7117.6 Hz.
9. If the value of Q of the circuit is high, then its effect on bandwidth is?
a) large bandwidth
b) small bandwidth
c) no effect on bandwidth
d) first increases and then decreases
View Answer
Explanation: If the value of Q of the circuit is high, then small bandwidth because bandwidth is inversely proportional to the quality factor.
10. If in a circuit, if Q value is decreased then it will cause?
a) small bandwidth
b) no effect on bandwidth
c) first increases and then decreases
d) large bandwidth
View Answer
Explanation: If in a circuit, if the Q value is decreased then bandwidth increases and the bandwidth do not decrease.
Sanfoundry Global Education & Learning Series – Network Theory.
To practice all areas of Network Theory, here is complete set of 1000+ Multiple Choice Questions and Answers.
