Analyzing a Series RLC Circuit A series RLC circuit has R = 440 , L-1.30 H, C 4.

Question

Analyzing a Series RLC Circuit A series RLC circuit has R = 440 , L-1.30 H, C 4.0 F. It is connected to an AC source with,f= 60.0 Hz and 7max-150 V SOLVE IT (A) Determine the inductive reactance, the capacitive reactance, and the impedance of the circuit. A series circuit consisting of a resistor, an inductor, and a capacitor connected to an AC source Conceptualize: The circuit of interest in this example is shown in the figure. The current in the combination of the resistor, inductor, and capacitor oscillates at a particular phase angle with respect to the applied voltage Categorize: The circuit is a simple series RLC circuit, so we can use the approach discussed in this section Analyze Find the angular frequency = 2f= 2(60.0 Hz) = 376.8 rad/s Use the following equation to find the inductive reactance: x, = (377 rad/s)( 1.30 H) = 492.1 Use the following equation to find the capacitive reactance CC(377 rad/s)(4.00 x 10-5 F) X-= 1663.13 Use the following equation to find the impedance z = V (440 )2 + (490.09 -663.15 )? Z = 472.81

Explanation / Answer

now frequency = 94 Hz

(a)
z = 472.81
L = 1.30 H
C = 4.0 uF
R = 440 omh

R^2 = z^2 - (XL - Xc)^2

new XL = 2*pi*f*L

= 2*pi*92*1.30

= 751.46

Xc = 1/2*pi*f*C

= 0.5*pi*92*4.0
  
= 432.48

R = sqroot(z^2 - (XL - Xc)^2

= sqroot(472.81^2 - (751.46-432.48)^2)

R = 349 ohm

(b)

theta = tan^-1(XL -Xc)/R

= tan^-1(751.46 - 432.48)/(349)

theta = 42.42 degree

(c)

Imax = Vmax/z

= (150)/(472.81)

= 0.317 A

VR = Imax*R = 139.48 V

VL = Imax*XL = 238.12 V

VC = Imax*XC = 137.09 V

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