Back to Directory
NEET PHYSICSMedium

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. The potential difference across L, C and R is 40 V, 10 V and 40 V, respectively. The amplitude of the current flowing through the LCR series circuit is 10210\sqrt{2} A. The impedance of the circuit will be:

A

4 Ω\Omega

B

5 Ω\Omega

C

42 Ω4\sqrt{2} \ \Omega

D

52 Ω5\sqrt{2} \ \Omega

Step-by-Step Solution

To find the impedance (ZZ), we first determine the total RMS voltage (VV) of the source. In a series LCR circuit, this is given by the phasor sum: V=VR2+(VLVC)2V = \sqrt{V_R^2 + (V_L - V_C)^2}. Substituting the given values, V=402+(4010)2=1600+900=50V = \sqrt{40^2 + (40 - 10)^2} = \sqrt{1600 + 900} = 50 V . The RMS current (II) is calculated from the peak current amplitude (imi_m) using the relationship I=im2I = \frac{i_m}{\sqrt{2}}. Given im=102i_m = 10\sqrt{2} A, I=1022=10I = \frac{10\sqrt{2}}{\sqrt{2}} = 10 A . The impedance is then the ratio of the RMS voltage to the RMS current: Z=VI=50 V10 A=5 ΩZ = \frac{V}{I} = \frac{50 \text{ V}}{10 \text{ A}} = 5 \ \Omega .

Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started