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Lecture37

Lecture37 - 11/16 Lecture Addendum to previous...

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Unformatted text preview: 11/16 Lecture Addendum to previous lecture: Resistance of very long wire (Length l, cross ­sectional area A) ρl R = (ρ is the resistivity). A € € € € € E Recall the “local” form of Ohm’s Law: J = ρ Capacitance of parallel ­plate capacitor (plates of area A, distance l between them, A >> l2) € εK A C = o E (KE is the dielectric constant). l Eo E= ( E is the electric field in a medium with dielectric constant KE, and E o is KE the electric field in a vacuum). € In a vacuum, KE = 1. In a medium that is not a vacuum, KE > 1. € Inductance of very long solenoid (N loops, area A of each loop, length l of solenoid) µoK M N 2 A 2 L= l Relative permeability KM: B = K M Bo where B is the magnetic field in a medium and Bo is the magnetic field in a vacuum. KM > 1 for paramagnetic materials€ € KM < 1 for diamagnetic materials KM = 1 for vacuum The formulas above are for ideal circuit elements. Circuit elements – General case: Resistance: R = ρ[ f ( geometry )] Capacitance: C = ε oK E [ f ( geometry )] € Inductance: L = µoK M [ f ( geometry )] € The quantity f(geometry) is a function of the geometry of the circuit element. In the resistance formula, f(geometry) has units of 1/length. In the other two formulas, € f(geometry) has units of length. RC circuits – review Charging: WV (dissipated)R + (stored)C Discharging: (stored)C (dissipated)R RL circuits – review Growth of current: WV (dissipated)R + (stored)L Decay of current: (stored)L (dissipated)R As you can see, the RC and RL circuits are similar. LC circuit  ­Different from RC and RL: (stored)C (stored)L (stored)C … Loop equation for LC circuit: − q dI − L = 0 C dt € € € € € € d 2q 1 This can be re ­written as: 2 + q = 0 dt LC € Note: this equation looks similar to that for a mass on a spring: € ma = −kx d2x k + x = 0 dt 2 m Recall that the solution to this equation was harmonic oscillations (sines & cosines). d2x In general, the solution to the differential equation of the form 2 + w 2 x = 0 is sines dt and cosines. € In the case of the LC circuit equation, the solution can be written as: 1 q( t ) = A cos(ωt ) + B sin(ωt ) ; ω = LC An equivalent way of writing the solution is: q( t ) = A cos(ωt + φ€ o) Here, A is the amplitude, and φo is the initial phase. These two ways of writing the solution both have two unknowns. The unknowns (A and B or A and φo) are determined by the initial conditions of the problem. In this case, the initial conditions give A = qo and φo = 0. Therefore, we can write: q( t ) = qo cos(ωt ) dq I( t ) = = −ωq sin(ωt ) dt The energy stored in the capacitor is given by: 1 q2 12 UC = = qo cos 2 (ωt ) 2 C 2C € € The energy stored in the inductor is given by: 1 L2 1 U L = LI 2 = qo w 2 sin 2 (ωt ) = q sin 2 (ωt ) 2 2 2C o The graphs of energy stored versus time for the capacitor and inductor are shown below. The two energies are oscillating in time with the same amplitude but different phases. This is an ideal case (no resistance). RLC circuit  ­More realistic case  ­If resistance is small enough, we still expect oscillations Note: consider the following current through a resistor: € Assume VA > VB. Then, ΔVAB = VB − VA = − IR . This is the convention we are using to determine the signs in the loop equation. dI q Loop equation for RLC circuit: −IR − L − = 0 € dt C d 2q R dq q This can be re ­written as: 2 + + = 0 dt L dt LC € Look for a solution of the form: q( t ) = A( t ) cos(ω ' t + φ ) € If you plug this solution into the differential equation above, you get a differential −( R ) t equation for A(t). If you solve this equation, you get A( t ) = Aoe 2 L . € 1 R2 In this case, w ' = − 2 and the initial conditions give Ao = qo and φ o = 0 LC 4 L € Therefore, the equation for q(t) can be written as: € € € Ⱥȹ Ⱥ 2 ȹ 1 R −( R ) t Ⱥ q( t ) = qoe 2 L cosȺȹ ȹ LC − 4 L2 ȹ t + φ o Ⱥ ȹ Ⱥȹ Ⱥ Ⱥ Ⱥ This solution is known as damped oscillation. The charge q still oscillates in time (as in the LC circuit), but the amplitude of these oscillations decreases in time according −( R ) t to the exponential decay factor qoe 2 L . This exponential decay factor is known as the envelope of the oscillations. 1 R2 − 2 , so if R is too large, you can LC 4 L get the square root of a negative number. Therefore, the solution q(t) above includes 4L 1 R2 the implicit condition that . If this condition > 2 , which simplifies to R 2 < C LC 4 L € is satisfied (as in the solution q(t) above), the system is said to be underdamped. € 4L € If the condition R 2 > is satisfied, the system is overdamped. In this case, the C resistance is too high for there to be oscillations in the charge, so the charge will decay in one of the two ways shown in the graph below: One concern with the solution above: w ' = € € To solve the differential equation in this case, try a solution of the form: q( t ) = A exp( − β1t ) + B exp( −β2 t ) (Note: exp(x) = e x ) If the solution above is plugged into the differential equation for the LRC circuit, we get the following result: € ȹ ȹ R q Ⱥ ȹ ȹ R R 1 ȹ ȹ R 1 ȹ ȹ Ⱥ q( t ) = o Ⱥexpȹ −ȹ + − t ȹ + expȹ −ȹ − − ȹ ȹ t ȹ Ⱥ ȹ 2 L 2 Ⱥ ȹ ȹ 2 L 4 L2 LC Ⱥ ȹ 4 L2 LC Ⱥ ȹ Ⱥ ȹ Ⱥ ȹ ȹ Ⱥ Ⱥ Ⱥ € ...
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