EE101: Resonance in RLC circuitssequel/ee101/ee101_resonance.pdf · 2015-06-30 · Resonance in...

EE101: Resonance in RLC circuits

M. B. [email protected]

www.ee.iitb.ac.in/~sequel

Department of Electrical EngineeringIndian Institute of Technology Bombay

M. B. Patil, IIT Bombay

Resonance in series RLC circuits

Vm 6 0

I =Vm ∠ 0

R + jωL + 1/jωC=

R + j(ωL− 1/ωC)≡ Im ∠ θ , where

Im =Vmp

R2 + (ωL− 1/ωC)2, θ = − tan−1

»ωL− 1/ωC

* As ω is varied, both Im and θ change.

* When ωL = 1/ωC , Im reaches its maximum value, Imaxm = Vm/R, and

θ becomes 0, i.e., the current I is in phase with the applied voltage.

* The above condition is called “resonance,” and the corresponding frequency iscalled the “resonance frequency” (ω0).

ω0 = 1/√

Vm 6 0

I =Vm ∠ 0

R + jωL + 1/jωC=

Im =Vmp

R2 + (ωL− 1/ωC)2, θ = − tan−1

»ωL− 1/ωC

ω0 = 1/√

Vm 6 0

I =Vm ∠ 0

R + jωL + 1/jωC=

Im =Vmp

R2 + (ωL− 1/ωC)2, θ = − tan−1

»ωL− 1/ωC

ω0 = 1/√

Vm 6 0

I =Vm ∠ 0

R + jωL + 1/jωC=

Im =Vmp

R2 + (ωL− 1/ωC)2, θ = − tan−1

»ωL− 1/ωC

ω0 = 1/√

Vm 6 0

Frequency (Hz)

105104103102

L = 1mHC = 1µF

R = 10Ω

Frequency (Hz)

105104103102

Im =Vmp

R2 + (ωL− 1/ωC)2, θ = − tan−1

»ωL− 1/ωC

* As ω deviates from ω0, Im decreases.

* As ω → 0, the term 1/ωC dominates, and θ → π/2.

* As ω →∞, the term ωL dominates, and θ → −π/2.

(SEQUEL file: ee101 reso rlc 1.sqproj)

Vm 6 0

Frequency (Hz)

105104103102

L = 1mHC = 1µF

R = 10Ω

Frequency (Hz)

105104103102

Im =Vmp

R2 + (ωL− 1/ωC)2, θ = − tan−1

»ωL− 1/ωC

Vm 6 0

Frequency (Hz)

105104103102

L = 1mHC = 1µF

R = 10Ω

Frequency (Hz)

105104103102

Im =Vmp

R2 + (ωL− 1/ωC)2, θ = − tan−1

»ωL− 1/ωC

Vm 6 0

Frequency (Hz)

105104103102

L = 1mHC = 1µF

R = 10Ω

Frequency (Hz)

105104103102

Im =Vmp

R2 + (ωL− 1/ωC)2, θ = − tan−1

»ωL− 1/ωC

Vm 6 0

Frequency (Hz)

105104103102

L = 1mHC = 1µF

R = 10Ω

Frequency (Hz)

105104103102

Im =Vmp

R2 + (ωL− 1/ωC)2, θ = − tan−1

»ωL− 1/ωC

Vm 6 0

Imaxm /

ω2ω0ω1

* The maximum power that can be absorbed by the resistor is

Pmax =1

2(Imax

m )2 R =1

* Define ω1 and ω2 (see figure) as frequencies at which Im = Imaxm /√

2, i.e., thepower absorbed by R is Pmax/2.

* The bandwidth of a resonant circuit is defined as B = ω2 − ω1, and the qualityfactor as Q = ω0/B. Quality is a measure of the sharpness of the Im versusfrequency relationship.

Vm 6 0

Imaxm /

ω2ω0ω1

Pmax =1

2(Imax

m )2 R =1

Vm 6 0

Imaxm /

ω2ω0ω1

Pmax =1

2(Imax

m )2 R =1

Vm 6 0

Imaxm /

ω2ω0ω1

Pmax =1

2(Imax

m )2 R =1

Im =Vmp

R2 + (ωL− 1/ωC)2.

For ω = ω0, Im = Imaxm = Vm/R .

For ω = ω1 or ω = ω2, Im = Imaxm /

√2 .

Imaxm /

ω2ω0ω1

⇒ 1√2

VmpR2 + (ωL− 1/ωC)2

for ω = ω1,2 .

2 R2 = R2 + (ωL− 1/ωC)2 → R = ±(ωL− 1/ωC) .

Solving for ω (and discarding negative solutions), we get

ω1,2 = ∓ R

* Bandwidth B = ω2 − ω1 = R/L .

* Quality Q = ω0/B = ω0L/R .

* Show that, at resonance (i.e., ω = ω0), |VL| = |VC | = Q Vm .

* Show that ω0 =√ω1ω2 .

Im =Vmp

R2 + (ωL− 1/ωC)2.

√2 .

Imaxm /

ω2ω0ω1

⇒ 1√2

for ω = ω1,2 .

2 R2 = R2 + (ωL− 1/ωC)2 → R = ±(ωL− 1/ωC) .

ω1,2 = ∓ R

Im =Vmp

R2 + (ωL− 1/ωC)2.

√2 .

Imaxm /

ω2ω0ω1

⇒ 1√2

for ω = ω1,2 .

2 R2 = R2 + (ωL− 1/ωC)2 → R = ±(ωL− 1/ωC) .

ω1,2 = ∓ R

Im =Vmp

R2 + (ωL− 1/ωC)2.

√2 .

Imaxm /

ω2ω0ω1

⇒ 1√2

for ω = ω1,2 .

2 R2 = R2 + (ωL− 1/ωC)2 → R = ±(ωL− 1/ωC) .

ω1,2 = ∓ R

Im =Vmp

R2 + (ωL− 1/ωC)2.

√2 .

Imaxm /

ω2ω0ω1

⇒ 1√2

for ω = ω1,2 .

2 R2 = R2 + (ωL− 1/ωC)2 → R = ±(ωL− 1/ωC) .

ω1,2 = ∓ R

Im =Vmp

R2 + (ωL− 1/ωC)2.

√2 .

Imaxm /

ω2ω0ω1

⇒ 1√2

for ω = ω1,2 .

2 R2 = R2 + (ωL− 1/ωC)2 → R = ±(ωL− 1/ωC) .

ω1,2 = ∓ R

Im =Vmp

R2 + (ωL− 1/ωC)2.

√2 .

Imaxm /

ω2ω0ω1

⇒ 1√2

for ω = ω1,2 .

2 R2 = R2 + (ωL− 1/ωC)2 → R = ±(ωL− 1/ωC) .

ω1,2 = ∓ R

Im =Vmp

R2 + (ωL− 1/ωC)2.

√2 .

Imaxm /

ω2ω0ω1

⇒ 1√2

for ω = ω1,2 .

2 R2 = R2 + (ωL− 1/ωC)2 → R = ±(ωL− 1/ωC) .

ω1,2 = ∓ R

Frequency (Hz)Frequency (Hz)

I = Im 6 θ

Vm 6 00

105105 104104 103103 102102

R = 20Ω

R = 10Ω

R = 20Ω

R = 10Ω

L = 1mHC = 1 µF

As R is increased,

* The quality factor Q = ω0L/R decreases, i.e., Im versus ω curve becomesbroader.

* The maximum current (at ω = ω0) decreases (since Imaxm = Vm/R).

* The resonance frequency (ω0 = 1/√

LC) is not affected.

I = Im 6 θ

Vm 6 00

105105 104104 103103 102102

R = 20Ω

R = 10Ω

R = 20Ω

R = 10Ω

L = 1mHC = 1 µF

As R is increased,

I = Im 6 θ

Vm 6 00

105105 104104 103103 102102

R = 20Ω

R = 10Ω

R = 20Ω

R = 10Ω

L = 1mHC = 1 µF

As R is increased,

I = Im 6 θ

Vm 6 00

105105 104104 103103 102102

R = 20Ω

R = 10Ω

R = 20Ω

R = 10Ω

L = 1mHC = 1 µF

As R is increased,

Vm 6 0

I =Vm ∠ 0

R + jωL + 1/jωC=

Im =Vmp

R2 + (ωL− 1/ωC)2, θ = − tan−1

»ωL− 1/ωC

* For ω < ω0, ωL < 1/ωC , the net impedance is capacitive, and the current leadsthe applied voltage.

* For ω = ω0, ωL = 1/ωC , the net impedance is purely resistive, and the currentis in phase with the applied voltage.

* For ω > ω0, ωL > 1/ωC , the net impedance is inductive, and the current lagsthe applied voltage.

* Let us look at an example (next slide).

Vm 6 0

I =Vm ∠ 0

R + jωL + 1/jωC=

Im =Vmp

R2 + (ωL− 1/ωC)2, θ = − tan−1

»ωL− 1/ωC

Vm 6 0

I =Vm ∠ 0

R + jωL + 1/jωC=

Im =Vmp

R2 + (ωL− 1/ωC)2, θ = − tan−1

»ωL− 1/ωC

Vm 6 0

I =Vm ∠ 0

R + jωL + 1/jωC=

Im =Vmp

R2 + (ωL− 1/ωC)2, θ = − tan−1

»ωL− 1/ωC

Vm 6 0

I =Vm ∠ 0

R + jωL + 1/jωC=

Im =Vmp

R2 + (ωL− 1/ωC)2, θ = − tan−1

»ωL− 1/ωC

0 100 200 300 400

−0.1

Time (µsec)

i (A) (right axis)

Vs (V) (left axis)

f =4.3 kHz

f =5 kHz ≃ f0

f =5.9 kHzR = 10Ω

L = 1 mH

C = 1 µF

Resonance in series RLC circuits: phasor diagrams

−40 1 2−1 0 1 2−1 0 1 2−1

C = 1 µF

R = 10Ω

L = 1 mH

Re(V) Re(V) Re(V)

Vs, VR

f = f0 ≃ 5 kHzf = 4.3 kHz f = 5.9 kHz

Resonance in parallel RLC circuits

IL ICIR

VIm 6 0

Im ∠ 0 = Y V, where Y = G + jωC + 1/jωL (G = 1/R) .

V =Im ∠ 0

G + jωC + 1/jωL=

G + j(ωC − 1/ωL)≡ Vm ∠ θ , where

Vm =Imp

G2 + (ωC − 1/ωL)2, θ = − tan−1

»ωC − 1/ωL

* As ω is varied, both Vm and θ change.

* When ωC = 1/ωL, Vm reaches its maximum value, Vmaxm = Im/G = ImR, and

θ becomes 0, i.e., the voltage V is in phase with the source current.

ω0 = 1/√

IL ICIR

VIm 6 0

V =Im ∠ 0

G + jωC + 1/jωL=

Vm =Imp

G2 + (ωC − 1/ωL)2, θ = − tan−1

»ωC − 1/ωL

ω0 = 1/√

IL ICIR

VIm 6 0

V =Im ∠ 0

G + jωC + 1/jωL=

Vm =Imp

G2 + (ωC − 1/ωL)2, θ = − tan−1

»ωC − 1/ωL

ω0 = 1/√

IL ICIR

VIm 6 0

V =Im ∠ 0

G + jωC + 1/jωL=

Vm =Imp

G2 + (ωC − 1/ωL)2, θ = − tan−1

»ωC − 1/ωL

ω0 = 1/√

Series RLC circuit: Im =Vmp

R2 + (ωL− 1/ωC)2, θ = − tan−1

»ωL− 1/ωC

Parallel RLC circuit: Vm =Imp

G2 + (ωC − 1/ωL)2, θ = − tan−1

»ωC − 1/ωL

* The two situations are identical if we make the following substitutions:I↔ V,R ↔ 1/R,L↔ C .

* Thus, our results for series RLC circuits can be easily extended to parallel RLCcircuits.

* Show that ω1,2 = ∓ 1

LC⇒Bandwidth B = 1/RC .

* Show that, at resonance (i.e., ω = ω0), |IL| = |IC | = Q Im .

R2 + (ωL− 1/ωC)2, θ = − tan−1

»ωL− 1/ωC

G2 + (ωC − 1/ωL)2, θ = − tan−1

»ωC − 1/ωL

R2 + (ωL− 1/ωC)2, θ = − tan−1

»ωL− 1/ωC

G2 + (ωC − 1/ωL)2, θ = − tan−1

»ωC − 1/ωL

R2 + (ωL− 1/ωC)2, θ = − tan−1

»ωL− 1/ωC

G2 + (ωC − 1/ωL)2, θ = − tan−1

»ωC − 1/ωL

R2 + (ωL− 1/ωC)2, θ = − tan−1

»ωL− 1/ωC

G2 + (ωC − 1/ωL)2, θ = − tan−1

»ωC − 1/ωL

R2 + (ωL− 1/ωC)2, θ = − tan−1

»ωL− 1/ωC

G2 + (ωC − 1/ωL)2, θ = − tan−1

»ωC − 1/ωL

Resonance in parallel RLC circuits: home work

IL ICIR

VIm 6 0 R = 2 kΩ

L = 40 mH

C = 0.25 µF

Im = 50 mA

* Calculate ω0, f0, B, Q.

* Calculate IR , IL, IC at ω = ω0, ω1, ω2.

* Verify graphically that IR + IL + IC = Is in each case.

* Plot the power absorbed by R as a function of frequency for f0/10 < f < 10 f0.

EE101: Resonance in RLC circuitssequel/ee101/ee101_resonance.pdf · 2015-06-30 · Resonance in...

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