PowerPoint PresentationLecture OutlineSelf-InductanceSlide 4Slide 5Slide 6Lecture 19, CQSlide 8Slide 9RL CircuitsLecture 18, ACT 2Slide 12RL Circuit (e on)Slide 14RL Circuit (e off)e on e offLecture 18, ACT 3Energy of an InductorWhere is the Energy Stored?Slide 20Physics 1304: Lecture 15, Pg 1Self Inductance and RL CircuitsB(t)~iLIBLdI dt( / )IRabLIPhysics 1304: Lecture 15, Pg 2Lecture Lecture OutlineOutlineConcept of Self-InductanceDefinition of Self-InductanceCalculation of Self-Inductance for Simple CasesRL CircuitsEnergy in Magnetic FieldText Reference: Chapter 32Physics 1304: Lecture 15, Pg 3Self-Self-InductanceInductanceX X X X X X X • emf induced in loop opposing initial emf•Self-Induction: the act of a changing current through a loop inducing an opposing current in that same loop. Just as a variable current can induce an EMF on another coil, it can induce an EMF on itself (back EMF).X X X X X X X Consider the loop at the right.• magnetic field produced in the area enclosed by the loop. • flux through loop changes•switch closed Þ current starts to flow in the loop.Physics 1304: Lecture 15, Pg 4We can put solenoids in a circuit:Self-InductanceSelf-InductanceRabLIBecause of the back EMF the current in this circuit does not instantaneouslyreach steady state. This is due to the presence of the solenoid L.Flux through the whole solenoid is Flux throughone loop }Recall that for a solenoid}N0BAn Ai 0B nior, N Li Thus, 20N AN il where, 20N ALlFaradays Law then takes form, BdNdtdiLdtPhysics 1304: Lecture 15, Pg 5Self-Self-InductanceInductanceThe 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 << llrN turnsdA- - - - -+ + + +d ALdI dt ( / )LIBCQVALR0CCPhysics 1304: Lecture 15, Pg 6Self-InductanceSelf-InductanceLdI dt( / )LIBUnits of Inductance are Henrys:[ ]:[ ]VsHA[ ]HPhysics 1304: Lecture 15, Pg 7Lecture 19, Lecture 19, CQCQConsider the two inductors shown: Inductor 1 has length l, N total turns and has inductance L1. Inductor 2 has length 2l, 2N total turns and has inductance L2. What is the relation between L1 and L2?(a) L2 < L1(b) L2 = L1(c) L2 > L1lrN turnsr2lr2N turnsPhysics 1304: Lecture 15, Pg 8Lecture 19, Lecture 19, CQCQConsider the two inductors shown: Inductor 1 has length l, N total turns and has inductance L1. Inductor 2 has length 2l, 2N total turns and has inductance L2. What is the relation between L1 and L2?(a) L2 < L1(b) L2 = L1(c) L2 > L1lrN turnsr2lr2N turns• To determine the self-inductance L, we need to determine the flux B which passes through the coils when a current I flows: L B / I.• To calculate the flux, we first need to calculate the magnetic field B produced by the current: B = 0(N/l)I• ie the B field is proportional to the number of turns per unit length.• Therefore, B1 = B2. But does that mean L1 = L2?• To calculate L, we need to calculate the flux.• Since B1 = B2 , the flux through any given turn is the same in each inductor.• However, there are twice as many turns in inductor 2; therefore the flux through inductor 2 is twice as much as the flux through inductor 1!!! Therefore, L2 = 2L1.Physics 1304: Lecture 15, Pg 9Physics 1304: Lecture 15, Pg 10RL CircuitsRL 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:RIabLIIR LdIdt 0RC:0εdtdqRCqRCRCRLLRRCRL: R L1CRIQ Physics 1304: Lecture 15, Pg 11Lecture 18, ACT 2Lecture 18, ACT 21) 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) I0 = 0(b) I = / 2R(c) I = 2 / R2) What is the value of the current I0 immediately after the switch is thrown?• A long time after the switch is thrown, the current approaches an asymptotic value. ie as t , dI/dt 0.• As dI/dt 0 , the voltage across the inductor 0. I = / 2R.• Immediately after the switch is thrown, the rate of change of current is as large as it can be (we’ve been assuming it was !)• The inductor limits dI/dt to be initially equal to / L. ie the voltage across the inductor = ; the current then must be 0!abRLIIRPhysics 1304: Lecture 15, Pg 12RL CircuitsRL CircuitsTo find the current I as a fcn 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:RL = LRRabLI IdIdtt( ) 0I tR( ) IReRt L 1//Rt LLdIV L edt Physics 1304: Lecture 15, Pg 13RL Circuit (RL Circuit ( on)on) CurrentMax = R63% Max at t=L/RIReRt L 1/L/RtI2L/R0RVL0t Voltage on LMax = /R37% Max at t=L/RV LdIdteLRt L /Sketch curves !Physics 1304: Lecture 15, Pg 14RL CircuitsRL CircuitsAfter 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:RabLI IIR LdIdt 0I tR( ) 0IReRt L/V LdIdteLRt L /Physics 1304: Lecture 15, Pg 15RL Circuit (RL Circuit ( off)off) 0-VLtL/Rt2L/RI0R CurrentMax = R37% Max at t=L/RIReRt L/ Voltage on LMax = -37% Max at t=L/RV LdIdteLRt L /Sketch curves !Physics 1304: Lecture 15, Pg 16 on on off off t0-It0RL/Rt2L/R0RI0tL/R2L/RIReRt L/V LdIdteLRt L / IReRt L 1/V LdIdteLRt L
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