A resistor R = 150 ohm, a capacitor C = 12.0 mu F, and an inductor L = 0.6H are

Question

A resistor R = 150 ohm, a capacitor C = 12.0 mu F, and an inductor L = 0.6H are connected in series with an AC emf source epsilon (t) = epsilon_0 sin(omega t)with epsilon_0 = 50.0V and omega = 300s^-1. What is the impedance of the circuit What is the current amplitude i_0 What are the voltage amplitudes across the resistor, V_0R, the capacitor, V_0C, and the inductor, V_0L What is the phase angle closed integral of the source voltage with respect to the current Does the current lag behind the voltage, or otherwise Give a reason for your answer. What is the average power dissipated in the circuit

Explanation / Answer

(a)
R = 150
C = 12.0 * 10^-6 F
L = 0.6 H
= 300 rad/s


E = Eo * sin(*t)
Impedance of the circut , Z = sqrt(R^2 + (XL - XC)^2)

XL = L
XL = 300 * 0.6 = 180

XC = 1/C
XC = 1/(300 * 12.0 * 10^-6)
XC = 277.8

Z = sqrt(150^2 + (277.8 - 180)^2)
Z = 179.1

(b)
Io = Eo/z
Io = 50.0/179.1
Io = 0.279 Amp
Current Amplitude, Io = 0.279 Amp

(c)
Voltage Amplitude across -

Resitor,
Vr = Io *R
Vr = 0.279 * 150 V
Vr = 41.85 Volt

Capacitor,
Vc = Io * XC
Vc = 0.279 * 277.8
Vc = 77.51 Volt

Inductor,
VL = Io * XL
VL = 0.279 * 180
VL = 50.22 Volt

(d)
= cos^-1(R/Z)
= cos^-1(150/179.1)
=  33.2
Phase Angle, =  33.1

As the Circut is Capacitive, Current Leads the Voltage

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