If the resistance in parallel with a parallel resonant circuit is reduced, the bandwidth
Answer: Option D
Explanation:
A 12 Ω resistor, a 40 μF capacitor, and an 8 mH coil are in series across an ac source. The resonant frequency is
Answer: Option B
At resonance Xc = XL. i.e 4 1/(2*3.14*C*F) = 2*3.14*L*F. F = 1/(2*3.14*sqrt(L*C)). By substituting values we get F = 281 Hz.
A 15 Ω resistor, a 220 μH coil, and a 60 pF capacitor are in series across an ac source. What is the bandwidth of the circuit?
BW = f/Q. Q = (1/R)(L/C)^1/2. = 127.65. f = 1/2*3.14*(L/C)^1/2. = 1385265.971 f/Q = 10852 = Answer.
In a series RLC circuit that is operating above the resonant frequency, the current
Answer: Option A
Explanation
At resonant condition. XL = XC @ resonant frequency. Resonant frequency more if XL>XC and less if XL<XC. Here it is more and it means XL is high and hence L is high (XL = 360FL). L always takes lags the current. Hence the current lags the applied voltage.
The term choke is more commonly used in connection with an inductor used to block off high frequencies.
Inductor will oppose sudden changes in current and choke is made up of copper winding which will balance the voltages. So, for sudden raise in voltage will be oppose by choke and it will bring the system back 2 healthy.
