Question

A series RLC circuit consists of a 58.0 ? resistor, a 2.50 mH inductor, and a 450 nF capacitor. It is connected to a 3.0 kHz oscillator with a peak voltage of 5.60 V. What is the instantaneous emf E when i = I ? What is the instantaneous emf E when i = 0A and is decreasing? What is the instantaneous emf E when i = - I ?

Answer 1

Given Data

f = 3 kHz = 3000 Hz

l = 2.5 mH = 2.5*10^-3 H

c = 450 nF = 450*10^-9 F

V = 5.6 v

R = 58 ohms

Xl = 2? fl = 2?*3000*2.5*10^-3 = 47.12 Ohms
Xc = 1/2? fc =1/(2?*3000*450*10^-9) = 117.89 Ohms
X = Xl - Xc = - 70.772 Ohms
Z = 58 Ohms / cos (arc tan 70.772/58) = 91.5 Ohms
I peak = 5.6Vp/91.5 Ohms = 61.2 mA

Part A;
I never reaches 1 Amp.

Therefore the value of E can not be evaluated at I = 1.

E = 0

Part B
Phase angle = arc tan 70.772/58 = 50.66 degrees with (I) leading (E).
Instantaneous value of E @ (180deg - 50.46deg )

                                = (5.6V)*(sin 129.54 deg)

                                = 4.33 Volts

Part C

I never reaches 1 Amp.

Therefore the value of E can not be evaluated at I = 1.

E = 0