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Physics 1402: Lecture 21 Today’s Agenda Announcements: –Induction, RL circuits Homework 06: due next MondayHomework 06: due next Monday Induction / AC current

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Faraday's Law dS B B v B N S v B S N

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Induction Self-Inductance, RL Circuits L/R V t 0 L X X X X X X X X X

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Recap from the last Chapter: Time dependent flux is generated by change in magnetic field strength due motion of the magnet Note: changing magnetic field can also be produced by time varying current in a nearby loop Faraday's Law of Induction v B N S v B S N B dI/dt Can time varying current in a conductor induce EMF in in that same conductor ?

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Self-Inductance The inductance of an inductor ( a set of coils in some geometry..eg solenoid, toroid) then, like a capacitor, can be calculated from its geometry alone if the device is constructed from conductors and air. If extra material (eg iron core) is added, then we need to add some knowledge of materials as we did for capacitors (dielectrics) and resistors (resistivity) Archetypal inductor is a long solenoid, just as a pair of parallel plates is the archetypal capacitor. r << l SI UNITS for L : Henry

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Calculation l r N turns Long Solenoid: N turns total, radius r, Length l For a single turn, The total flux through solenoid is given by: Inductance of solenoid can then be calculated as: This (as for R and C) depends only on geometry (material)

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RL Circuits At t=0, the switch is closed and the current I starts to flow. Loop rule: Note that this eqn is identical in form to that for the RC circuit with the following substitutions: I a b L I RC: RC RL:

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Lecture 21, ACT 1 At t=0 the switch is thrown from position b to position a in the circuit shown: –What is the value of the current I a long time after the switch is thrown? (a) I = 0 (b) I = / 2R (c) I = 2 / R (a) I = 0 (b) I = / 2R (c) I = 2 / R 1A What is the value of the current I immediately after the switch is thrown? 1B

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RL Circuits To find the current I as a fct of time t, we need to choose an exponential solution which satisfies the boundary condition: We therefore write: The voltage drop across the inductor is given by: a b L II

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RL Circuit ( on) t I 0 R L/R 2L/R VLVL 0 t Current Max = R 63% Max at t=L/R Voltage on L Max = /R 37% Max at t=L/R

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RL Circuits After the switch has been in position a for a long time, redefined to be t=0, it is moved to position b. Loop rule: The appropriate initial condition is: The solution then must have the form: a b L II

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RL Circuit ( off) 0 -- VLVL t L/R 2L/R t I 0 R Current Max = R 37% Max at t=L/R Voltage on L Max = - 37% Max at t=L/R

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t I 0 R L/R 2L/R VLVL 0 t on off 0 -- VLVL t L/R 2L/R t I 0 R

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Review: RC Circuits (Time-varying currents) Discharge capacitor: C initially charged with Q=C Connect switch to b at t=0. Calculate current and charge as function of time. Convert to differential equation for q: C a b ++ -- R I I Loop theorem

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Review: RC Circuits (Time-varying currents) Trial solution: Check that it is a solution: ! Note that this “guess” incorporates the boundary conditions: C a b ++ -- R I I Discharge capacitor:

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Review: RC Circuits (Time-varying currents) Current is found from differentiation: Conclusion: Capacitor discharges exponentially with time constant = RC Current decays from initial max value (= - /R) with same time constant Discharge capacitor: C a b + -- R + I I

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0 - R I t RC 2RC Discharging Capacitor t q 0 CC Current Max = - /R 37% Max at t=RC Charge on C Max = C 37% Max at t=RC

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t q 0 CC RC 2RC I 0 t R Charging Discharging 0 - R I t RC 2RC t q 0 CC

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Energy of an Inductor How much energy is stored in an inductor when a current is flowing through it? Start with loop rule: From this equation, we can identify P L, the rate at which energy is being stored in the inductor: We can integrate this equation to find an expression for U, the energy stored in the inductor when the current = I: a b L I I Multiply this equation by I :

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Where is the Energy Stored? Claim: (without proof) energy is stored in the Magnetic field itself (just as in the Capacitor / Electric field case). To calculate this energy density, consider the uniform field generated by a long solenoid: l r N turns The inductance L is: We can turn this into an energy density by dividing by the volume containing the field: Energy U:

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Mutual Inductance Suppose you have two coils with multiple turns close to each other, as shown in this cross-section We can define mutual inductance M 12 of coil 2 with respect to coil 1 as: Coil 1 Coil 2 B N1N1 N2N2 It can be shown that :

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Inductors in Series What is the combined (equivalent) inductance of two inductors in series, as shown ? a b L2L2 L1L1 a b L eq Note: the induced EMF of two inductors now adds: Since: And:

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Inductors in parallel What is the combined (equivalent) inductance of two inductors in parallel, as shown ? a b L2L2 L1L1 a b L eq Note: the induced EMF between points a and be is the same ! Also, it must be: We can define: And finally:

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LC Circuits Consider the LC and RC series circuits shown: L C C R Suppose that the circuits are formed at t=0 with the capacitor C charged to a value Q. Claim is that there is a qualitative difference in the time development of the currents produced in these two cases. Why?? Consider from point of view of energy! In the RC circuit, any current developed will cause energy to be dissipated in the resistor. In the LC circuit, there is NO mechanism for energy dissipation; energy can be stored both in the capacitor and the inductor!

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RC/LC Circuits RC: current decays exponentially C R i Q -i t 0 0 1 +++ - - - L C LC: current oscillates i 0 t i Q +++ - - -

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LC Oscillations (qualitative) L C ++ - - L C L C ++ -- L C

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LC Oscillations (quantitative) What do we need to do to turn our qualitative knowledge into quantitative knowledge? What is the frequency of the oscillations? L C ++ - -

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LC Oscillations (quantitative) Begin with the loop rule: Guess solution: (just harmonic oscillator!) where: determined from equation , Q 0 determined from initial conditions Procedure: differentiate above form for Q and substitute into loop equation to find . L C ++ - - i Q remember:

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Review: LC Oscillations Guess solution: (just harmonic oscillator!) where: determined from equation , Q 0 determined from initial conditions L C ++ - - i Q which we could have determined from the mass on a spring result:

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Lecture 21, ACT 2 At t=0 the capacitor has charge Q 0 ; the resulting oscillations have frequency 0. The maximum current in the circuit during these oscillations has value I . –What is the relation between 0 and 2, the frequency of oscillations when the initial charge = 2Q 0 ? (a) 2 = 1/2 0 (b) 2 = 0 (c) 2 = 2 0 1A

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Lecture 21, ACT 2 At t=0 the capacitor has charge Q 0 ; the resulting oscillations have frequency 0. The maximum current in the circuit during these oscillations has value I . (a) I = I (b) I = 2 I (c) I = 4 I What is the relation between I and I , the maximum current in the circuit when the initial charge = 2Q 0 ? 1B

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