EE101: Resonance in RLC circuits

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1 EE11: Resonance in RLC circuits M. B. Patil Deartment of Electrical Engineering Indian Institute of Technology Bombay

2 I V R V L V C I = I m = R + jωl + 1/jωC = Im θ, where R + j(ωl 1/ωC) R 2 + (ωl 1/ωC) 2, θ = tan 1» ωl 1/ωC R.

3 I V R V L V C I = I m = R + jωl + 1/jωC = Im θ, where R + j(ωl 1/ωC) R 2 + (ωl 1/ωC) 2, * As ω is varied, both I m and θ change. θ = tan 1» ωl 1/ωC R.

4 I V R V L V C I = I m = R + jωl + 1/jωC = Im θ, where R + j(ωl 1/ωC) R 2 + (ωl 1/ωC) 2, θ = tan 1» ωl 1/ωC R. * As ω is varied, both I m and θ change. * When ωl = 1/ωC, I m reaches its maximum value, m = /R, and θ becomes, i.e., the current I is in hase with the alied voltage.

5 I V R V L V C I = I m = R + jωl + 1/jωC = Im θ, where R + j(ωl 1/ωC) R 2 + (ωl 1/ωC) 2, θ = tan 1» ωl 1/ωC R. * As ω is varied, both I m and θ change. * When ωl = 1/ωC, I m reaches its maximum value, m = /R, and θ becomes, i.e., the current I is in hase with the alied voltage. * The above condition is called resonance, and the corresonding frequency is called the resonance frequency (ω ). ω = 1/ LC

6 VR VL I Vm VC I m = R 2 + (ωl 1/ωC) 2,» ωl 1/ωC θ = tan 1. R

7 f VR VL I Vm VC Im (A).1 R = 1 Ω L = 1 mh C = 1 µf Frequency (Hz) 1 5 I m = R 2 + (ωl 1/ωC) 2,» ωl 1/ωC θ = tan 1. R * As ω deviates from ω, I m decreases.

8 f VR VL I Vm VC Im (A).1 R = 1 Ω L = 1 mh C = 1 µf Frequency (Hz) 1 5 I m = R 2 + (ωl 1/ωC) 2,» ωl 1/ωC θ = tan 1. R * As ω deviates from ω, I m decreases. * As ω, the term 1/ωC dominates, and θ π/2.

9 f f I VR Vm VL VC Im (A).1 R = 1 Ω L = 1 mh C = 1 µf θ (degrees) Frequency (Hz) Frequency (Hz) 1 5 I m = R 2 + (ωl 1/ωC) 2,» ωl 1/ωC θ = tan 1. R * As ω deviates from ω, I m decreases. * As ω, the term 1/ωC dominates, and θ π/2. * As ω, the term ωl dominates, and θ π/2.

10 f f I VR Vm VL VC Im (A).1 R = 1 Ω L = 1 mh C = 1 µf θ (degrees) Frequency (Hz) Frequency (Hz) 1 5 I m = R 2 + (ωl 1/ωC) 2,» ωl 1/ωC θ = tan 1. R * As ω deviates from ω, I m decreases. * As ω, the term 1/ωC dominates, and θ π/2. * As ω, the term ωl dominates, and θ π/2. (SEQUEL file: ee11 reso rlc 1.sqroj)

11 I V R V L V C m m / 2 ω 1 ω ω 2 ω

12 I V R V L V C m m / 2 ω 1 ω ω 2 ω * The maximum ower that can be absorbed by the resistor is P max = 1 2 max (Im )2 R = /R.

13 I V R V L V C m m / 2 ω 1 ω ω 2 ω * The maximum ower that can be absorbed by the resistor is P max = 1 2 max (Im )2 R = /R. * Define ω 1 and ω 2 (see figure) as frequencies at which I m = Im max / 2, i.e., the ower absorbed by R is P max/2.

14 I V R V L V C m m / 2 ω 1 ω ω 2 ω * The maximum ower that can be absorbed by the resistor is P max = 1 2 max (Im )2 R = /R. * Define ω 1 and ω 2 (see figure) as frequencies at which I m = Im max / 2, i.e., the ower absorbed by R is P max/2. * The bandwidth of a resonant circuit is defined as B = ω 2 ω 1, and the quality factor as Q = ω /B. Quality is a measure of the sharness of the I m versus frequency relationshi.

15 I m = R 2 + (ωl 1/ωC) 2. m m / 2 For ω = ω, I m = m = /R. For ω = ω 1 or ω = ω 2, I m = m / 2. ω 1 ω ω 2 ω

16 I m = R 2 + (ωl 1/ωC) 2. m m / 2 For ω = ω, I m = m = /R. For ω = ω 1 or ω = ω 2, I m = m / 2. 1 «Vm = for ω = ω 1,2. 2 R R 2 + (ωl 1/ωC) 2 ω 1 ω ω 2 ω

17 I m = R 2 + (ωl 1/ωC) 2. m m / 2 For ω = ω, I m = m = /R. For ω = ω 1 or ω = ω 2, I m = m / 2. 1 «Vm = for ω = ω 1,2. 2 R R 2 + (ωl 1/ωC) 2 ω 1 ω ω 2 ω 2 R 2 = R 2 + (ωl 1/ωC) 2 R = ±(ωl 1/ωC).

18 I m = R 2 + (ωl 1/ωC) 2. For ω = ω, I m = m = /R. For ω = ω 1 or ω = ω 2, I m = m / 2. m m / 2 1 «Vm = for ω = ω 1,2. 2 R R 2 + (ωl 1/ωC) 2 ω 1 ω ω 2 ω 2 R 2 = R 2 + (ωl 1/ωC) 2 R = ±(ωl 1/ωC). Solving for ω (and discarding negative solutions), we get ω 1,2 = R 2L + s R 2L «2 + 1 LC.

19 I m = R 2 + (ωl 1/ωC) 2. For ω = ω, I m = m = /R. For ω = ω 1 or ω = ω 2, I m = m / 2. m m / 2 1 «Vm = for ω = ω 1,2. 2 R R 2 + (ωl 1/ωC) 2 ω 1 ω ω 2 ω 2 R 2 = R 2 + (ωl 1/ωC) 2 R = ±(ωl 1/ωC). Solving for ω (and discarding negative solutions), we get ω 1,2 = R 2L + s R 2L «2 + 1 LC. * Bandwidth B = ω 2 ω 1 = R/L.

20 I m = R 2 + (ωl 1/ωC) 2. For ω = ω, I m = m = /R. For ω = ω 1 or ω = ω 2, I m = m / 2. m m / 2 1 «Vm = for ω = ω 1,2. 2 R R 2 + (ωl 1/ωC) 2 ω 1 ω ω 2 ω 2 R 2 = R 2 + (ωl 1/ωC) 2 R = ±(ωl 1/ωC). Solving for ω (and discarding negative solutions), we get ω 1,2 = R 2L + s R 2L «2 + 1 LC. * Bandwidth B = ω 2 ω 1 = R/L. * Quality Q = ω /B = ω L/R.

21 I m = R 2 + (ωl 1/ωC) 2. For ω = ω, I m = m = /R. For ω = ω 1 or ω = ω 2, I m = m / 2. m m / 2 1 «Vm = for ω = ω 1,2. 2 R R 2 + (ωl 1/ωC) 2 ω 1 ω ω 2 ω 2 R 2 = R 2 + (ωl 1/ωC) 2 R = ±(ωl 1/ωC). Solving for ω (and discarding negative solutions), we get ω 1,2 = R 2L + s R 2L «2 + 1 LC. * Bandwidth B = ω 2 ω 1 = R/L. * Quality Q = ω /B = ω L/R. * Show that, at resonance (i.e., ω = ω ), V L = V C = Q.

22 I m = R 2 + (ωl 1/ωC) 2. For ω = ω, I m = m = /R. For ω = ω 1 or ω = ω 2, I m = m / 2. m m / 2 1 «Vm = for ω = ω 1,2. 2 R R 2 + (ωl 1/ωC) 2 ω 1 ω ω 2 ω 2 R 2 = R 2 + (ωl 1/ωC) 2 R = ±(ωl 1/ωC). Solving for ω (and discarding negative solutions), we get ω 1,2 = R 2L + s R 2L «2 + 1 LC. * Bandwidth B = ω 2 ω 1 = R/L. * Quality Q = ω /B = ω L/R. * Show that, at resonance (i.e., ω = ω ), V L = V C = Q. * Show that ω = ω 1 ω 2.

23 9 VR VL.1 I = Im θ L = 1 mh C = 1 µf R = 1 Ω VC Vm R = 2 Ω Im (A) θ (degrees) R = 1 Ω R = 2 Ω Frequency (Hz) Frequency (Hz) 1 5 As R is increased,

24 9 VR VL.1 I = Im θ L = 1 mh C = 1 µf R = 1 Ω VC Vm R = 2 Ω Im (A) θ (degrees) R = 1 Ω R = 2 Ω Frequency (Hz) Frequency (Hz) 1 5 As R is increased, * The quality factor Q = ω L/R decreases, i.e., I m versus ω curve becomes broader.

25 9 VR VL.1 I = Im θ L = 1 mh C = 1 µf R = 1 Ω VC Vm R = 2 Ω Im (A) θ (degrees) R = 1 Ω R = 2 Ω Frequency (Hz) Frequency (Hz) 1 5 As R is increased, * The quality factor Q = ω L/R decreases, i.e., I m versus ω curve becomes broader. * The maximum current (at ω = ω ) decreases (since m = /R).

26 9 VR VL.1 I = Im θ L = 1 mh C = 1 µf R = 1 Ω VC Vm R = 2 Ω Im (A) θ (degrees) R = 1 Ω R = 2 Ω Frequency (Hz) Frequency (Hz) 1 5 As R is increased, * The quality factor Q = ω L/R decreases, i.e., I m versus ω curve becomes broader. * The maximum current (at ω = ω ) decreases (since m = /R). * The resonance frequency (ω = 1/ LC) is not affected.

27 I V R V L V C I = I m = R + jωl + 1/jωC = Im θ, where R + j(ωl 1/ωC) R 2 + (ωl 1/ωC) 2, θ = tan 1» ωl 1/ωC R.

28 I V R V L V C I = I m = R + jωl + 1/jωC = Im θ, where R + j(ωl 1/ωC) R 2 + (ωl 1/ωC) 2, θ = tan 1» ωl 1/ωC R * For ω < ω, ωl < 1/ωC, the net imedance is caacitive, and the current leads the alied voltage..

29 I V R V L V C I = I m = R + jωl + 1/jωC = Im θ, where R + j(ωl 1/ωC) R 2 + (ωl 1/ωC) 2, θ = tan 1» ωl 1/ωC R * For ω < ω, ωl < 1/ωC, the net imedance is caacitive, and the current leads the alied voltage. * For ω = ω, ωl = 1/ωC, the net imedance is urely resistive, and the current is in hase with the alied voltage..

30 I V R V L V C I = I m = R + jωl + 1/jωC = Im θ, where R + j(ωl 1/ωC) R 2 + (ωl 1/ωC) 2, θ = tan 1» ωl 1/ωC R * For ω < ω, ωl < 1/ωC, the net imedance is caacitive, and the current leads the alied voltage. * For ω = ω, ωl = 1/ωC, the net imedance is urely resistive, and the current is in hase with the alied voltage. * For ω > ω, ωl > 1/ωC, the net imedance is inductive, and the current lags the alied voltage..

31 I V R V L V C I = I m = R + jωl + 1/jωC = Im θ, where R + j(ωl 1/ωC) R 2 + (ωl 1/ωC) 2, θ = tan 1» ωl 1/ωC R * For ω < ω, ωl < 1/ωC, the net imedance is caacitive, and the current leads the alied voltage. * For ω = ω, ωl = 1/ωC, the net imedance is urely resistive, and the current is in hase with the alied voltage. * For ω > ω, ωl > 1/ωC, the net imedance is inductive, and the current lags the alied voltage. * Let us look at an examle (next slide)..

32 1.1 f =4.3 khz 1.1 i 1.1 f =5 khz f V s R = 1 Ω L = 1 mh C = 1 µf f =5.9 khz Time (µsec) V s (V) (left axis) i (A) (right axis)

33 : hasor diagrams VR VL I Vs VC R = 1 Ω L = 1 mh C = 1 µf 4 3 VL 2 1 VL VL Im(V) VR Vs Vs, VR VR Vs 1 VL VC VL 2 VC VR VR 3 4 f = 4.3 khz VC f = f 5 khz Re(V) Re(V) Re(V) f = 5.9 khz

34 Resonance in arallel RLC circuits I R I L I C I m V I m = Y V, where Y = G + jωc + 1/jωL (G = 1/R). I m V = G + jωc + 1/jωL = I m Vm θ, where G + j(ωc 1/ωL) = I m G 2 + (ωc 1/ωL) 2, θ = tan 1» ωc 1/ωL G.

35 Resonance in arallel RLC circuits I R I L I C I m V I m = Y V, where Y = G + jωc + 1/jωL (G = 1/R). I m V = G + jωc + 1/jωL = I m Vm θ, where G + j(ωc 1/ωL) = I m G 2 + (ωc 1/ωL) 2, * As ω is varied, both and θ change. θ = tan 1» ωc 1/ωL G.

36 Resonance in arallel RLC circuits I R I L I C I m V I m = Y V, where Y = G + jωc + 1/jωL (G = 1/R). I m V = G + jωc + 1/jωL = I m Vm θ, where G + j(ωc 1/ωL) = I m G 2 + (ωc 1/ωL) 2, θ = tan 1» ωc 1/ωL G * As ω is varied, both and θ change. * When ωc = 1/ωL, reaches its maximum value, ax m = I m/g = I mr, and θ becomes, i.e., the voltage V is in hase with the source current..

37 Resonance in arallel RLC circuits I R I L I C I m V I m = Y V, where Y = G + jωc + 1/jωL (G = 1/R). I m V = G + jωc + 1/jωL = I m Vm θ, where G + j(ωc 1/ωL) = I m G 2 + (ωc 1/ωL) 2, θ = tan 1» ωc 1/ωL G * As ω is varied, both and θ change. * When ωc = 1/ωL, reaches its maximum value, ax m = I m/g = I mr, and θ becomes, i.e., the voltage V is in hase with the source current. * The above condition is called resonance, and the corresonding frequency is called the resonance frequency (ω ). ω = 1/ LC.

38 Resonance in arallel RLC circuits Series RLC circuit: I m = Parallel RLC circuit: = R 2 + (ωl 1/ωC) 2, I m G 2 + (ωc 1/ωL) 2,» ωl 1/ωC θ = tan 1. R θ = tan 1» ωc 1/ωL G.

39 Resonance in arallel RLC circuits Series RLC circuit: I m = Parallel RLC circuit: = R 2 + (ωl 1/ωC) 2, I m G 2 + (ωc 1/ωL) 2,» ωl 1/ωC θ = tan 1. R θ = tan 1» ωc 1/ωL G * The two situations are identical if we make the following substitutions: I V, R 1/R, L C..

40 Resonance in arallel RLC circuits Series RLC circuit: I m = Parallel RLC circuit: = R 2 + (ωl 1/ωC) 2, I m G 2 + (ωc 1/ωL) 2,» ωl 1/ωC θ = tan 1. R θ = tan 1» ωc 1/ωL G * 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 arallel RLC circuits..

41 Resonance in arallel RLC circuits Series RLC circuit: I m = Parallel RLC circuit: = R 2 + (ωl 1/ωC) 2, I m G 2 + (ωc 1/ωL) 2,» ωl 1/ωC θ = tan 1. R θ = tan 1» ωc 1/ωL G * 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 arallel RLC circuits. s * Show that ω 1,2 = 1 «1 2 2RC RC LC Bandwidth B = 1/RC..

42 Resonance in arallel RLC circuits Series RLC circuit: I m = Parallel RLC circuit: = R 2 + (ωl 1/ωC) 2, I m G 2 + (ωc 1/ωL) 2,» ωl 1/ωC θ = tan 1. R θ = tan 1» ωc 1/ωL G * 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 arallel RLC circuits. s * Show that ω 1,2 = 1 «1 2 2RC RC LC Bandwidth B = 1/RC. * Show that, at resonance (i.e., ω = ω ), I L = I C = Q I m..

43 Resonance in arallel RLC circuits Series RLC circuit: I m = Parallel RLC circuit: = R 2 + (ωl 1/ωC) 2, I m G 2 + (ωc 1/ωL) 2,» ωl 1/ωC θ = tan 1. R θ = tan 1» ωc 1/ωL G * 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 arallel RLC circuits. s * Show that ω 1,2 = 1 «1 2 2RC RC LC Bandwidth B = 1/RC. * Show that, at resonance (i.e., ω = ω ), I L = I C = Q I m. * Show that ω = ω 1 ω 2..

44 Resonance in arallel RLC circuits: home work I R I L I m V R = 2 kω L = 4 mh I C I m = 5 ma C =.25 µf * Calculate ω, f, B, Q. * Calculate I R, I L, I C at ω = ω, ω 1, ω 2. * Verify grahically that I R + I L + I C = I s in each case. * Plot the ower absorbed by R as a function of frequency for f /1 < f < 1 f.

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