Physics Circle home Student Login

Home›Chapters›Alternating Current PYQs›Resonance, Bandwidth & Q Factor›Question 95

NEET 2021 · Resonance, Bandwidth & Q Factor

Resonance, Bandwidth & Q Factor question from NEET 2021

Q95NEET 2021Medium

Resonance, Bandwidth & Q Factor

An inductor of inductance L, a capacitor of capacitance C and a resistor of resistance 'R' are connected in series to an ac source of potential difference 'V' volts as shown in figure. Potential difference across L, C and R is 40V, 10V and 40V, respectively. The amplitude of current flowing through LCR series circuit is 102\sqrt{2} A. The impedance of the circuit is:

Figure

Show answer and solution

Answer: (a)

  1. V=VR2+(VL−VC)2=402+302=50 VV = \sqrt{V_R^2 + (V_L - V_C)^2} = \sqrt{40^2 + 30^2} = 50\text{ V} (rms).
  2. Irms=I02=1022=10 AI_{rms} = \frac{I_0}{\sqrt{2}} = \frac{10\sqrt{2}}{\sqrt{2}} = 10\text{ A}
  3. Z=VIrms=5010=5 ΩZ = \frac{V}{I_{rms}} = \frac{50}{10} = 5\ \Omega
  4. So the answer is (a) 5 Ω5\ \Omega.
Report a mistake

More questions on Resonance, Bandwidth & Q Factor