Question

A series LCR circuit is made by taking R = 100 Omega, L = (2)/(pi) and C = (100)/(pi) mu F. The series combination is connected across an a.c. source of 220 V, 50 Hz. Calculate impedance of the circuit and peak value of current flowing n the circuit. What is power factor of the circuit ? Compare it with one at resonance frequency.

Answer 1

Answer:

Explanation:

Impedence of LCR circuit is given by

$\boxed{R''=\sqrt{R^2+{|X_l-X_c |}^2}}$

$R''=\sqrt{(R)^2+{|(2\pi fL) -(\frac{1}{2\pi Cf} ) |}^2}$

$R''=\sqrt{(100)^2+{|(2\pi (50)(2/\pi) -(\frac{10^6}{2\pi (100/\pi)50} ) |}^2}\\\\R''=\sqrt{10000+{|200 -100 |}^2}\\\\\boxed{R''=100\sqrt{2}}$

Peak value of current is related to impedance and peak voltage as:

$\boxed{V_0=I_0 R}$

∴$I_0=\frac{220}{100\sqrt{2}} \\\\=> \boxed{I_0=1.55A}$

Power factor is given by $PF=cos(\phi)$

where $cos(\phi) = \frac{R}{Impedence}=\frac{100}{100\sqrt{2}}=\boxed{1/\sqrt{2} }$