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AC Circuits

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E&M Topic

  • LRC circuit and alternating current

DataStudio File

  • 81 AC Circuit.ds

Equipment List

Introduction

The purpose of this activity is to study AC circuits with a resistor, a capacitor, and inductor. You will be able to do that by examining the current through the circuit as a function of the frequency of the applied voltage. Determine what happens to the resistance and the current when we change the parameters of voltage and frequency. Use the 'OUTPUT' feature of the PASCO 750 Interface to apply voltage to the circuit. Use the voltage sensor and DataStudio to measure the voltage across the resistor in the circuit as the frequency of the voltage is changed. Also, investigate the phase relationship between the applied voltage and the resistor voltage as you vary the frequency.

Background

An AC circuit consists of circuit elements and a power source that provides alternating voltage Δv. This time-varying voltage from the source is described by
( 1 )
Δv = ΔVmax sin ωt 
where
ΔVmax 
 is the maximum output voltage of the source, or the voltage amplitude. In resistors, if
ΔVmax 
 is the emf supplied by the generator, the Kirchhoff's loop rule gives
( 2 )
Δv − iR · R = 0
where iR =
Δv
R
 =
ΔVmax
R
 = Imax · sin(ωt)
where Imax =
ΔVmax
R
and ω = 2 · π · f = 2 ·
π
f
 
So, the instantaneous voltage across the resistor is:
( 3 )
ΔvR = iR · R = Imax · R · sin(ωt). 
It means that for such a current, the current and voltage are in phase: both of them are zero at the same instant, both of them pass through their maximum values at the same instant. For capacitors, the Kirhhoffs's loop rule gives us:
( 4 )
Δv −
q
C
 = 0. 
Substituting
ΔVmax · sin(ωt) 
 for Δv and rearranging gives:
( 5 )
q = C · ΔVmax · sin(ωt). 
So, the instantaneous voltage across the resistor is:
( 6 )
iC =
dq
dt
 = ω · C · ΔVmax · cos(ωt). 
And for inductors, again, Kirchhoff's loop rule gives us
( 7 )
Δv − L ·
diL
dt
 = 0. 
Substituting
ΔVmax · sin(ωt) 
 for Δv and rearranging gives:
( 8 )
Δv = L ·
diL
dt
 = ΔVmax · sin(ωt). 
And solving for diL gives:
( 9 )
diL =
ΔVmax
L
 · sin(ωt) · dt. 
Integrating this expression gives the instantaneous current iL in the inductor as a function of time:
( 10 )
iL =
ΔVmax
ωL
 · sin
ωt −
π
2
. 
The instantaneous current iL in the inductor and the instantaneous voltage
ΔvL 
 across the inductor are out of the phase by
π
2
 rad = 90°.

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