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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).

1 Answer

3 votes

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,
$I=(Vs)/(Z)$

Z = 40 + j40 - j100

= 46 - j 60 Ω

=
$72.1 \angle -56.3^\circ$

So,
$I=(Vs)/(Z)$


$=(600 \angle 20^\circ)/(72.1 \angle -56.3^\circ)$


$=8.32\ \angle 76.3^\circ$

Therefore,


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

A 40Ω resistor, a 5 mH inductor, and a 1.25μF capacitor are connected in series. The-example-1
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User Watson
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