Oscillating Currents
• Ch.30: Induced E Fields: Faraday’s Law • Ch.30: RL Circuits • Ch.31: Oscillations and AC Circuits Review: Inductance
• If the current through a coil of wire changes, there is an induced emf proportional to the rate of change of the current. •Define the proportionality constant to be the inductance L : di εεε === −−−L dt
• SI unit of inductance is the henry (H). LC Circuit Oscillations Suppose we try to discharge a capacitor, using an inductor instead of a resistor:
At time t=0 the capacitor has maximum charge and the current is zero.
Later, current is increasing and capacitor’s charge is decreasing Oscillations (cont’d)
What happens when q=0? Does I=0 also? No, because inductor does not allow sudden changes.
In fact, q = 0 means i = maximum!
So now, charge starts to build up on C again, but in the opposite direction! Textbook Figure 31 1
Energy is moving back and forth between C,L
1 2 1 2 UL === UB === 2 Li UC === UE === 2 q / C Textbook Figure 31 1 Mechanical Analogy • Looks like SHM (Ch. 15) Mass on spring. • Variable q is like x, distortion of spring. • Then i=dq/dt , like v=dx/dt , velocity of mass.
By analogy with SHM, we can guess that q === Q cos(ωωω t) dq i ======−−−ωωωQ sin(ωωω t) dt Look at Guessed Solution dq q === Q cos(ωωω t) i ======−−−ωωωQ sin(ωωω t) dt q
i Mathematical description of oscillations
Note essential terminology: amplitude, phase, frequency, period, angular frequency. You MUST know what these words mean! If necessary review Chapters 10, 15. Get Equation by Loop Rule If we go with the current as shown, the loop rule gives: q di −−− −−− L === 0 C dt
2 Now replace i by d q 1 dq/dt to get: === −−− q dt 2 LC Guess Satisfies Equation!
Start with q === Q cos(ωωω t) d 2q 1 === −−− q dq 2 so that === −−−ωωω Q sin(ωωω t) dt LC dt And taking one more derivative gives us d 2q === −−−ωωω 2Q cos(ωωω t) === −−−ωωω 2q dt 2 So the solution is correct, provided our angular frequency satisfies 1 ωωω2 === LC LC Circuit Example Given an inductor with L = 8.0 mH and a capacitor with C = 2.0 nF, having initial charge q(0) = 50 nC and initial current i(0) = 0.
(a) What is the frequency of oscillations (in Hz)? (b) What is the maximum current in the inductor? (c) What is the capacitor’s charge at t = 30 s? Example: Part (a) L = 8 mH, C = 2 nF, q(0) = 50 nC, i(0) = 0
(a) What is the frequency of the oscillations?
1 1 rad ωωω======5.2 ×××105 LC 8 ×××10−−−3 ××× 2 ×××10−−−9 s
ωωω 5.2 ×××105 rad / s f ======4 ×××104 === 40 kHz 2πππ .6 28 rad / cycle Example: Part (b) ( L = 8 mH, C = 2 nF, q(0) = 50 nC, i(0) = 0 ) (b) What is the maximum current? Q 2 (50 ×××10−−−9 )2 E ======.6 25 ×××10−−−7 J 2C 2 ××× 0.2 ×××10−−−9 1 2E But E === LI 2 so I 2 === 2 L 2E 2 ××× .6 25 ×××10−−−7 So I ======12 5. mA L 8 ×××10−−−3 Example: Part (c) ( L = 8 mH, C = 2 nF, q(0) = 50 nC, i(0) = 0 ) (c) What is the charge at t = 30 s? q(t) === Q cos(ωωωt) === (50nC)cos( 5.2 ×××105 ××× 30 ×××10−−−6 ) === (50nC)cos( 5.7 rad) === (50nC)cos(430°°°) === 50 ××× .0 347 === 17 nC AC Voltage Sources • For an AC circuit, we need an alternating emf , or AC power supply. • This is characterized by its amplitude and its frequency .
ε = ε m sinωt
amplitude angular frequency Notation for oscillating functions Note that the textbook uses lower-case letters for oscillating time dependent voltages and currents, with upper-case letters for the corresponding amplitudes .
v === V sinωωωt i === I sinωωωt AC Currents and Voltages
ε = ε m sinωt = vR = VR sinωt
Ohm’s Law gives: iR = vR / R = (ε m / R)sinωt
So the AC current is: iR = I R sinωt So the amplitudes are related by: VR = I R R AC Voltage Current Relations • First apply an alternating emf to a resistor, a capacitor, and an inductor separately , before dealing with them all at once. • For C, L, define reactance X analogous to resistance R for resistor. Measured in ohms.
VR === I R R
VC === IC XC
VL === I L X L Summary for R, C, L Separately “ELI the ICE man”
VR === I R R {v and i are in phase R R
1 VC === IC XC with XC === { i leads v by 90 °. ωωω C C C
V === I X with X === ωωω L { L L L L vL leads iL by 90 °. Q.31 1 Which of the following is true about the phase relation between the current and the voltage for an inductor?
(1) The current is in phase with the voltage. (2) The current is ahead of the voltage by 90º. (3) The current is behind the voltage by 90º. (4) They are out of phase by 180º. Q.31 1 Which of the following is true about the phase relation between the current and the voltage for an inductor? Remember ELI the ICE man!
Inductor: voltage leads current. (1) The current is in phase with the voltage. (2) The current is ahead of the voltage by 90º. (3) The current is behind the voltage by 90º. (4) They are out of phase by 180º. What is the phase relation between the Q.31 1 current and the voltage for an inductor?
Let i === I sin(ωωωt) Proof: di Drop is v === L === LIωωωcos(ωωωt) dt i ∝∝∝ sin(ωωωt)
v hits peak before i v ∝∝∝ cos(ωωωt) Q.31 2
An inductor L carries a current with I === 0.3 A amplitude I at angular frequency ωωω. ωωω=== 200 rad / s What is the amplitude V of the voltage across this inductor? L === .03 H
)1( VL === 5.0 V )2( VL === 0.2 V )3( VL === 0.6 V
)4( VL === 12V )5( VL === 18 V )6( VL === 600 V Q.31 2
The reactance is: I === 0.3 A X L === ωωωL === 200 ××× .03 === 0.6 ωωω=== 200 rad / s Thus the voltage amplitude is: L === .03 H
VL === I L X L === 0.3 A ××× 0.6 === 18 V
)1( VL === 5.0 V )2( VL === 0.2 V )3( VL === 0.6 V
)4( VL === 12V )5( VL === 18 V )6( VL === 600 V Impedance
The “AC Ohm’s Law” is: εm = IZ where Z is called the impedance.
Obviously, for a resistor, Z = R for a capacitor and for an inductor Z = X C Z = X L
But if you have a combination of circuit elements, Z is more complicated. Impedance and Phase Angle General problem: if we are given
ε (t) = εm sinω t can we find i(t)? We can always write (ti ) = I sin(ω t −φ )
By definition of impedance I = εm / Z Given εεε and ωωω, find Z and φφφ. m ?? Series RLC Circuit 1. The currents are all equal. 2. The voltage drops add up to the applied emf as a function of time: εεε === vR +++ vC +++ vL BUT: Because of the phase differences , the amplitudes do not add: εεε m ≠≠≠ VR +++VC +++VL So the impedance is not just a sum: Z ≠≠≠ R +++ XC +++ X L Results for series circuits
εεε(t) === εεεm sinωωωt (ti ) === I sin(((ωωω t −−−φφφ )))
It turns out that for this particular circuit
2 2 Z === R +++ ( X L −−− XC ) Z X L −−− XC tanφφφ === X L −−− XC R φφφ
R Series Circuit Example
Given L = 50 mH, C = 60 F, εm=120V, f=60Hz, and I=4.0A. (a) What is the impedance? (b) What is the resistance R?
Solution to (a) is easy: εεε 120 Z === m ======30 I 4 Example (part b) (a) Z = 30
L=50 mH, C = 60 F, εm=120V, f=60Hz, I=4.0A (b) What is the resistance R? ωωω === 2πππ f === 2 ××× .3 14 ××× 60 === 377 rad / s −−−3 X L === ωωωL === 377 ××× 50 ×××10 === 18 8. 1 1 X ======44 2. C ωωω C 377 ××× 60 ×××10−−−6
XC −−− X L === 44 2. −−− 18 8. === 25 4.
2 2 2 2 R === Z −−− ( X L −−− XC ) === 30 −−− 25 4. === 16 AC Circuits
• Ch.30: Faraday’s Law • Ch.30: Inductors and RL Circuits: • Ch.31: AC Circuits Review: Inductance
• If the current through a coil of wire changes, there is an induced emf proportional to the rate of change of the current. •Define the proportionality constant to be the inductance L : di εεε === −−−L dt
• SI unit of inductance is the henry (H). Review: LC Circuits q === Q cos(ωωω t) dq i ======−−−ωωωQ sin(ωωω t) dt d 2q 1 Loop rule gives differential equation: === −−− q dt 2 LC Solution if frequency is correct: 1 ωωω 2 === LC Natural frequency of oscillations! Review: AC Voltage Sources • For an AC circuit, we need an alternating emf , or AC power supply. • This is characterized by its amplitude and its frequency .
ε = ε m sinωt
amplitude angular frequency Review: R, C, L Separately “ELI the ICE man”
VR === I R R {v and i are in phase R R
1 VC === IC XC with XC === { i leads v by 90 °. ωωω C C C
V === I X with X === ωωω L { L L L L vL leads iL by 90 °. Review: Simple series circuit
εεε(t) === εεεm sinωωωt (ti ) === I sin(((ωωω t −−−φφφ )))
It turns out that for this particular circuit
2 2 Z === R +++ ( X L −−− XC ) Z X L −−− XC tanφφφ === X L −−− XC R φφφ
R Series Circuit Example
Given L = 50 mH, C = 60 F, εm=120V, f=60Hz, and I=4.0A. (a) What is the impedance? (b) What is the resistance R?
Solution to (a) is easy: εεε 120 Z === m ======30 I 4 Example (part b) (a) Z = 30
L=50 mH, C = 60 F, εm=120V, f=60Hz, I=4.0A (b) What is the resistance R? ωωω === 2πππ f === 2 ××× .3 14 ××× 60 === 377 rad / s −−−3 X L === ωωω L === 377 ××× 50 ×××10 === 18 8. 1 1 X ======44 2. C ωωω C 377 ××× 60 ×××10−−−6
XC −−− X L === 44 2. −−− 18 8. === 25 4.
2 2 2 2 R === Z −−− ( X L −−− XC ) === 30 −−− 25 4. === 16 Q.31 3 A certain series RLC circuit is driven by an applied emf with angular frequency 20 radians per second. If the maximum charge on the capacitor is 0.03 coulomb, what is the maximum current in the circuit?
(1) 6 A (2) 1.5 A (3) 0.6 A (4) 0.15 A (5) Not enough information Q.31 3 A certain series RLC circuit is driven by an applied emf with angular frequency 20 radians per second. If the maximum charge on the capacitor is 0.03 coulomb, what is the maximum current in the circuit? dq q === Q sin((()(ωωωt))) i ======ωωωQ cos((()(ωωωt))) dt I === ωωωQ === 20 ××× .0 03 === 6.0 A
(1) 6 A (2) 1.5 A (3) 0.6 A (4) 0.15 A (5) Not enough information Q.31 4
In a series RLC circuit, find the resistance, given the impedance and phase constant:
Z = 500 φφφ = 60° R = ?
(1) 400 (2) 350 (3) 300 (4) 250 In an RLC circuit: Q.31 4 Z = 500 φφφ = 60° R = ?
Z X L −−− XC φφφ R === Z cosφφφ R 1 cos((()(60°°°)))(=== sin((()30°°°)))=== 2 R === Z 2/ === 250
(1) 400 (2) 350 (3) 300 (4) 250 Resonance For a given series RLC circuit, what applied frequency will give the biggest current? εεε I === m 2 2 R +++ ( X L −−− XC ) I is biggest when denominator is smallest, which is when reactances cancel:
1 2 1 ωωωL === Which gives ωωω === ωωωC LC Same as natural frequency for oscillations! Resonance Plot: I vs ωωω
Peak at ωωω === /1 LC Becomes sharper as R → 0 Power in AC Circuits
To know how much power will be consumed in an AC circuit we need more than the impedance Z; we also need the phase angle φφφ. AC Power What is the power provided by an AC source? • Only interested in the time average! • Time average power to L and C is zero.
• So P ave (from source) = P ave (to resistor).
Pave === vRiR === (VR sinωωωt)(I R sinωωωt) 1 === V I sin2 ωωωt === V I R R 2 R R RMS Values
1 V === v2 === V 2 sin2 ωωωt === V === V / 2 RMS 2
2 Likewise I RMS === i === I / 2
So for a resistor: V I 1 P === V I ======VI ave RMS RMS 2 2 2 Power and Phase Angle
Power to resistor 2 2 P === iR R === I R2 Rsin (((ωωωt −−−φφφ)))
2 1 2 P === I R sin2 === I R avg R avg 2 R
But we want result in terms of impedance and phase angle: R === Z cosφφφ 1 So P === I 2 Z cosφφφ avg 2 Power factor What about more complicated circuits? Method of Phasors
•Useful mathematical trick for working with oscillating functions having different phase relationships. •We’ll use it to derive the known result for the series circuit, to show how it works. Good for general case.
iR === I R sin(((ωωω t)))
Length of phasor I R =
iR = Vertical component of phasor Series RLC Circuit 1. The currents are all equal. 2. The voltage drops add up to the applied emf: ε = vR + vC + vL Because of the phase differences, the amplitudes do not add, but in the phasor diagram, this means that the vertical components do add, so we get the right answer if we add the phasors as vectors. Phasors for Series RLC Circuit
εεε === vR +++ vC +++ vL === εεε m sinωωωt
iR === iC === iL === I sin(((ωωωt −−−φφφ)))
i I
(((ωωω t −−−φφφ ))) Adding the Voltage Phasors
εεε === vR +++ vC +++ vL === εεε m sinωωωt I ε Adding the Voltage Phasors
εεε === vR +++ vC +++ vL === εεε m sinωωωt
2 2 εεε m === VR +++ (VL −−−VC )
V −−−V tanφφφ === L C VR Impedance and Phase Angle V −−−V 2 2 tanφφφ === L C εεε m === VR +++ (VL −−−VC ) VR Divide by the common current amplitude I:
εεε m === IZ , VR === IR, VL === IX L , VC === IXC
2 2 Z === R +++ ( X L −−− XC ) X −−− X tanφφφ === L C R