Question

A 40Ω resistor, a 5 mH inductor, and a 1.25μF capacitor are connected in series. The series-connected elements are energized by a sinusoidal voltage source whose voltage is 600 cos(8000t + 20°) V.

Required:
a. Draw the frequency-domain equivalent circuit.
b. Reference the current in the direction of the voltage rise across the source, and find the phasor current.
c. Find the steady-state expression for i(t).

Answer 1

Solution :

Given  

From the voltage expression, w = 8000 rad/s

Inductive reactance, XL = jwL = j 40 Ω

Capacitive reactance, XC = -j/wC

                                    = - j 100 Ω

Now, $I=\frac{Vs}{Z}$

Z = 40 + j40 - j100

  = 46 - j 60 Ω

   = $72.1 \angle -56.3^\circ$

So, $I=\frac{Vs}{Z}$

      $=\frac{600 \angle 20^\circ}{72.1 \angle -56.3^\circ}$

      $=8.32\ \angle 76.3^\circ$

Therefore,

$i(t) = 8.32 \ \cos(8000t+76.3^\circ)$