Question

Exercise 6 An AC circuit is composed of a serial connection of: a resistor with resistance R-50 a coil with inductance L-0.3 H and a capacitor with capacitance C. I 5·E The circuit is connected to an AC voltage source with ar 50V,600 ㎐ 1. Determine the impedance and the effective electric current in the circuit. 2. Determine the phase shift between the voltage and the current. Conclude. 3. Find the appropriate frequency that make the circuit pure resistance circuit.

Answer 1

Given:

R = 50 ohm

L = 0.3 H

C = 15 x 10^-6 F

V = 50 V

f = 600 Hz

here,

$\omega = 2 \pi f$

$\omega = 2 \times 3.14 \times 600$

$\omega = 3768 rad/s$

Now find the X_C and X_L

_{$X_c = \frac{1}{\omega C}$}

_{$X_c = \frac{1}{3768 \times 15 \times 10^{-6}}$}

_{$X_c = 17.7$}

_and

$X_L = \omega L$

$X_L =3768 \times0.3$

X_L = 1130.4

A)

Impedance of LCR circuit is given by

$Z = \sqrt{R^2 + (X_L^2 - X_C^2)}$

$Z = \sqrt{50^2 + (1130.4^2 - 17.7^2)}$

$Z = \sqrt{2500 + 1277490.87}$

$Z = 1131.36$

and the value of electric current is found out by using

$I = \frac {V}{Z}$

$I = \frac {50}{1131.36}$

I = 0.04419 A

B)the phase of the LCR circuit is find out by using

$\phi = tan^{-1} \frac{X_L - X_C}{R}$

$\phi = tan^{-1} \frac{1130.4 - 17.7}{50}$

$\phi = 87.42$

C)

when the current and the emf are in the same phase i.e $\phi = 0$ then LCR circuit behave like an pure resistive circuit.

in this case,

$X_L = X_C$

$\omega L = \frac {1}{\omega C}$

$\omega = \sqrt{}\frac {1}{LC}$

$\omega = \sqrt{}\frac {1}{0.3 \times 15 \times 10^{-6}}$

$\omega = 471.4 rad/s$