28 Common Questions of a Simple Alternating Current Circuit
- Category: High School
- Written by fisikastudycenter
physics.fisikastudycenter.com,- Learning ac current circuit in 28 common questions of a single ac current circuit problem. This topic discussed at grade 12 high school.
Problem
Given an ac circuit contents a resistor (R), an inductor (L) and a capacitor (C) in series combination with voltage source of V = 120 sin (125t) volt.
Questions
- a) angular frequency of the ac source
- b) frequency of the ac source
- c) period of ac source
- d) voltage amplitude of ac source
- e) effective voltage of ac source
- f) peak to peak voltage of ac source
- g) inductive reactance of the inductor
- h) capacitive reactance of the capacitor
- i) circuit impedance
- j) amplitude of circuit current
- k) root-mean-square current of circuit
- l) voltage across d-e
- m) voltageacross e-f
- n) voltage across f-g
- o) voltage across d-f
- p) voltage across e-g
- q) voltage across d-g
- r) circuit power factor
- s) phase angle
- t) power dissipated in the circuit
- u) Circuit properties whether resistive, inductive or capacitive
- v) instantaneous value of voltage when t = (π/150) sec
- w) current equation of ac source
- x) instantaneous value of current of ac source when t = (0,016 π)
- y) average voltage
- z) current average
- aa) current and voltage in phasor
- bb) resistance, inductive reactance and capacitive reactance in phasor
Answers
a) Expression of sinusoidal voltage
where V is the instataneous value at time t, Vmax is the amplitude (maximum value), ω is the angular frequency. So then ω = 125 rad/s
b) angular frequency calculation :
c) period :
d) amplitude of voltage:
e) efective voltage or root-mean-square voltage :
f) peak to peak voltage :
g) inductive reactance :
h) capacitive reactance :
i) circuit impedance :
j) amplitude of current (maximum value) :
k) root-man-square current:
l) voltage across d-e :
We assume the voltage as the efective value, and the current is effective value Ief
m) voltage across e-f :
n) voltage across f-g:
o) voltage across d-f :
where VR , VL and VC are the voltage acorss R, the voltage across L and the voltage across C respectively.
There is no capacitor on d-f, so then the equation above becomes:
p) voltage across e-g :
q) voltage across d-g:
r) power factor (pf) :
pf = cos φ = R/Z = 8/10 = 0.8
or use anaother relationship
pf = cos φ = VR/V
s) angle phase between I and V ::
t) power dissipated in the circuit :
u) Circuit properties whether resistive, inductive or capacitive
To define the properties of circuit if :
XL > XC → inductive circuit
XC > XL → capacitive circuit
XL = XC → resistive circuit
So then the circuit is capacitive
v) instantaneous value of V at t = ( π/150) sec:
w) current equation of ac source :
For V = Vmax sin ω t
the current equation is:
I = Imax sin (ωt + φ) → capacitive circuit
I = Imax sin (ωt − φ) → inductive circuit
The circuit is capacitive so then use he first equation
x) instantaneous value of current at t = (0,016 π) sec :
y) average voltage :
z) average current :
aa) phasor diagram
bb) phasor diagram of R, XL and XC