An L - R - C series circuit is constructed using a 175 resistor, a 12.5 F capaci

Question

An L-R-C series circuit is constructed using a 175 resistor, a 12.5F capacitor, and an 8.00-mHinductor, all connected across an ac source having a variable frequency and a voltage amplitude of 25.0 V.

Part A

At what angular frequency will the impedance be smallest?

Part B

What is the impedance at this frequency?

Part C

At the angular frequency in part A, what is the maximum current through the inductor?

Part D

At the angular frequency in part A, find the potential difference across the ac source, the resistor, the capacitor, and the inductor at the instant that the current is equal to one-half its greatest positive value.

Enter your answers numerically separated by commas.

Explanation / Answer

here,

resistance, R= 175 ohms
capacitor, C = 12.5 uF = 12.5*10^-6 F
inductance, L = 8 mH = 0.008 H
Volatge, V = 25 V

Part A
Zmin = Xl - Xc = 0
Xl = Xc
L*w = 1/w*C

Solvign for angular frequency, w = sqrt(1/(LC))
w = sqrt( 1/(0.008*12.5*10^-6) )
w = 3162.278 rad/s

Part B
Z = sqrt(R^2 + ((1/w*L) - (1/wC)) )
Z = sqrt(175^2 + ((1/(3162.278*0.008)) - (1/(3162.278*12.5*10^-6))) )

Z = 174.928 ohms or 175 ohms(rounded off)
Part C

Max Current, I = E/Z = 25/175 = 0.143 A

Part D
I = Imax*Cos(wt)
0.143/2 = 0.143 *Cos(3162.278*t)

Solving for time , t = 0.000331 s

Voltage at inductor
Vl = I*Xl*Cos(wt+pi/2)
Vl = (0.143*1/(3162.278*0.008)) * Cos(3162.278*0.000331 + pi/2)
Vl = -3.13 V

Volatge at capacitor,
Vc = I*Xc*Cos(wt + pi/2)
Vc = (0.143*1/(3162.278*12.5*10^-6)) * Cos(3162.278*0.000331 + pi/2)
Vc = 3.13 V

Volatge at resistance
Vr = I*R*Cos(wt)
Vr = 0.143*175*Cos(3162.278.278 * 0.000331)
Vr = 12.5 V

also Voltage at bettery
V = E*Cos(wt)
V = 25*Cos(3162.278.278 * 0.000331)
V = 12.5 V

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