An L - R - C series circuit consists of a 40.0 ? resistor, a 20.0 ? F capacitor,

Question

An L-R-C series circuit consists of a 40.0?resistor, a 20.0?F capacitor, a 3.50mH inductor, and an ac voltage source of voltage amplitude 65.0V operating at 1050Hz.

(A) Find the current amplitude across the inductor, the resistor, and the capacitor

(B) Find the voltage amplitudes across the inductor, the resistor, and the capacitor VL, VR,VC

(C) Find new current amplitude across the inductor, the resistor, and the capacitor.

(D) Find new voltage amplitudes across the inductor, the resistor, and the capacitor.

Explanation / Answer

impedence Z^2 = R^2 + (xL-xc)^2

xL = wL = 2pifL = 2*3.14* 1050 *0.0035 = 23 ohms

xC= 1/wC = 1/(2pifC) = 1/2*3.14* 1050 *20*10^-6 =7.6 ohms

so Z^2 = 40^2 + (23-7.6)^2

Z = 42.86 ohms

total current i = V/Z = 65/42.86 = 1.5165 Amps

A. currnt i across inductor = V/XL = 65/23 = 2.82 Amps

current i across resistor = V/R = 65/40 = 1.625 Amps

current i across capacitor = V/xc = 65/7.6 = 8.55 Amps

B. V across Inductor VL = i XL = 1.516 * 23 = 34.88 volts

VR = iR = 1.516*40 = 60.64 volts

Vc = iXC = 1.516*7.6 = 11.526 volts

c. i across inductor for above volatge 34.88 Volts is i = 34.88/23 = 1.51 Amps

i across R is V/R = 60.64/40 = 1.516 Amps

i across C is V/XC = 11.526/7.6 = 1.516 Amps

d.if there is specfic condtion

new voltages will be same as that of B.

i.e

V across Inductor VL = i XL = 1.516 * 23 = 34.88 volts

VR = iR = 1.516*40 = 60.64 volts

Vc = iXC = 1.516*7.6 = 11.526 volts

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