A parallel resonant circuit has a resistance of 2 kΩ and half power frequencies of 86 kHz and 90 kHz.1. The value of capacitor is
BW = ω_{2}  ω_{1} = 2π(90  86)k = 8π krad/s
= 19.89 nF
A parallel resonant circuit has a resistance of 2 kΩ and half power frequencies of 86 kHz and 90 kHz.The value of inductor is
= 0.16 mH
A parallel resonant circuit has a resistance of 2 kΩ and half power frequencies of 86 kHz and 90 kHz.The quality factor is
A parallel resonant circuit has a midband admittance of 25 x 10^{3} S, quality factor of 80 and a resonant frequency of 200 krad/s.The value of R is
Ay midband frequency Z = R, Y = 1/R
A parallel resonant circuit has a midband admittance of 25 x 10^{3} S, quality factor of 80 and a resonant frequency of 200 krad/s.The value of C is
= 10 μF
A parallel RLC circuit has R =1 kΩ and C = 1 μF. Thequality factor at resonance is 200. The value of inductor is
⇒ L = 25 μH
A parallel circuit has R =1 k , C = 50 μF and L = 10mH. The quality factor at resonance is
A series resonant circuit has an inductor L = 10 mH.The resonant frequency ω_{o} = 106 rad/s and bandwidth is BW = 10^{3} rad/s. The value of R and C will be
= 100pF
A series resonant circuit has L = 1 mH and C = 10μF. The required R for the BW 15 9 . Hz is
⇒
For the RLC parallel resonant circuit when R = 8 kΩ , L = 40 mH and C = 0.25 μF, the quality factor Q is
The maximum voltage across capacitor would be
Thevenin equivalent seen by L  C combination
Open Circuit : v_{1} = 0, v∝ = 3V
For the circuit shown in fig. resonant frequency f_{o} is
Applying 1 A at input port V_{1} = 10 V
voltage across 1 A source
Z_{in} = V_{test}
At resonance Im {Z_{in}} = 0
⇒ ⇒
For the circuit shown in fig. the resonant frequency f_{o} is
At resonance Im {Y} = 0
The network function of circuit shown in fig. is
The value of input frequency is required to cause again equal to 1.5. The value is
For any value of ω, R,C gain ≤ 1.
Thus (D) is correct option.
In the circuit of fig. phase shift equal to 45^{0} is required at frequency ω =20 rad/s . The value of R is
phase shift =  tan^{1} ωCR = 450°
ωCR = 1,
20 x 1 x 10^{6} R = 1 ⇒ R = 50KΩ
For the circuit of fig., Vs is the source voltage and the response is the resistance voltage Vo, R = 30Ω and L = 2H. Suppose the input frequency is adjusted until the gain is equal to 0.6. The value of the frequency is
Bode diagram of the network function V_{o} /V_{s }for the circuit of fig. is
20 dB/decade line starting from ω = 5.56 rad/s
20 dB/decade line starting from ω = 16.67 rad/s
Hence 20 dB/decade line for 5.56 < ω < 16.67 parallel to ω axis to ω >16.67
Use Code STAYHOME200 and get INR 200 additional OFF

Use Coupon Code 







