Network Theory Multiple Choice Questions 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
Answer: B
Clarification: 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
Answer: A
Clarification: 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
Answer: C
Clarification: 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
Answer: B
Clarification: 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
Answer: C
Clarification: 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.
Answer: B
Clarification: 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
Answer: A
Clarification: 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
Answer: D
Clarification: 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
Answer: B
Clarification: 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
Answer: D
Clarification: If in a circuit, if the Q value is decreased then bandwidth increases and the bandwidth do not decrease.
