In a series LCR circuit having L = 30 mH, R = 8 Ω and the resonant frequency is 50 Hz. The quality factor of the circuit is: The quality factor of the circuit is: View Solution

The resonance of a series LCR circuit occurs when the inductive and capacitive reactances are equal in magnitude but cancel each other because they are 180 degrees apart in phase. In LCR circuit, the impedance is given by:

Write its value in LCR circuit. View Solution. Q2. The values of L and R in L C R circuit are 1 m H and 10 ...

The figure show a series LCR circuit with L = 5.0 H, C = 80 μ F, R = 40 Ω connected to variable frequency 240 V source. Calculate (i)The angular frequency of the source which driver the circuit at resonance (ii) The current at the frequency. (iii)The rms potential drop across the capacitor at …

In series LCR circuit in which R=3 ohm.,L=25mH, and C=800 micro F a sinusoidal a.c voltage of peak value 250v is applied.Find (i)frequency at which resonance occurs. (ii)current in the circuit at resonant condition.

LCR circuit has L = 10 m H, R = 3 Ω and C = 1 μ F connected in series to a source of 15 cos ω t V. Calculate the current amplitude and the average power dissipated per cycle at a frequency that is 10 % lower than the resonance frequency.

In a series LCR circuit connected to an a.c. source of voltage v = v m s i n ω t, use phasor diagram to derive an expression for the current in the circuit. Hence, obtain the expression for the power dissipated in the circuit. Show that power dissipated at resonance is maximum.

Q-factor: In LCR Circuit, the ratio of resonance frequency to the difference of its neighbouring frequencies so that their corresponding current is 1 / √ 2 times of the peak value, is called Q-factor of the circuit. Formula: Q = 1 R √ L C Conditions for the large value of Q factor: (i) Value of L C should be large. (ii) Value of R should be ...

A resonance circuit having inductance and resistance 2 × 10 − 4 H and 6.28 Ω respectively oscillates at 10 MHz frequency. The value of quality factor of this resonator is [ π = 3.14 ] View Solution

The figure shows basic LCR series circuit where a voltage V s is applied across RLC series circuit. We will solve this circuit by using vector method. Vector drawn for resistance is along the X-axis because current and voltage are in phase in case of a purely resistive circuit and magnitude will be R.