W9_parallel_resonance.eps

Week 9: Alternating Current Circuits 311

R

Z

φ

χL− χC

Figure 119: The impedance diagram for theLRCcircuit.

Z=

√

R^2 + (χL−χC)^2 (678)

so that

I 0 =

V 0

Z

(679)

and

φ= tan−^1

(

χL−χC

R

)

(680)

• The Parallel LRC Circuit:
  The parallel LRC circuit is actually muchsimplerthan the series as far as understanding the
  solution is concerned. This is because thesamevoltage dropV 0 sin(ωt) occurs across allthree
  components, and so we can just write down the currents througheach component using the
  elementary single-component rules above:

IR = V^0

R

sin(ωt) (681)

IL =

V 0

χL

sin(ωt−π/2) (682)

IC =

V 0

χC

sin(ωt+π/2) (683)

Note well that we use the rules we derived where the current through the inductor isπ/ 2 behind

the voltage (which is thereforeπ/ 2 aheadof the current) and vice versa for the capacitor. To

find the total current provided by the voltage, we simply add thesethree currents according to

Kirchhoff’s junction rule. Of course, we are adding three trig functions with different relative

phases, so we once again must accomplish this with suitable phasors:

Itot = V^0

R

sin(ωt) +V^0

χL

sin(ωt−π/2) +V^0

χC

sin(ωt+π/2)

=

V 0

Z

sin(ωt−φ)

= I 0 sin(ωt−φ) (684)

In this expression, a bit of contemplation should convince you that the impedanceZfor this

circuit is given by the entirely reasonable:

1

Z

=

√(

1

R^2

+ (

1

χC

−

1

χL

)^2

)

(685)

which we recognize as the phasor equivalent of the familiar rule for reciprocal addition of

resistances in parallel, and:

φ = tan−^1

( 1

χC−

1

χL

1

R

)

= tan−^1

(

RC(ω^2 −ω^20 )

ω

)

(686)

for the phase.