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Write your responses to the following questions and turn it in. Outline • RCL circuit(s) (L = inductance or coil) – – – – • What activities are, do you feel, most useful to your learning in lectures? • What activities are, do you feel, useless for your learning? • Any suggestions on any activities you want to see during lectures? • What do you like most about the labs? • What do you dislike about the labs? What’s “L?” Ohm’s Law of L? LC circuit RLC circuit(?) Physics of L Side: How can you make L large? • If you can create more B field for a given current, L will be larger • From Faraday’s Law, – VL = L dI/dt, where L is called “self inductance.” – Coil (many loops compared to a single loop) is better. – Putting iron (or ferrite) core will help “trap” B field. • Ferrite (insulator; susceptibility ~ 100) is better since it prevents Eddy currents in the core. • Works at higher frequency (up to GHz!) Responses to sinusoidal signal Solving this Eq. 1K • Ohm’s Law of capacitor A in – I = C dVC /dt = C d/dt (Vin – Vout ) • Ohm’s Law of inductor C 0.1 uF B L 6.8 mH – VL = Vout = L dI/dt. – For sinusoidal, VL = jωL I (ZL ~ jωL) • Block high-freq signal – opposite of capacitor • Combining the two: • Vout = L dI/dt = LC (d/dt)2 (Vin – Vout) • If – Vin = Aejωt and – Vout = Bej(ωt+ϕ ), • The above Eq. can be re-written as – Bej(ωt+ϕ ) = LC (jω) 2 (Aejωt – Bej(ωt+ϕ )) – Vout = L dI/dt = LC (d/dt) 2 (Vin – Vout) 1 More arithmetic • Bej(ωt+ϕ) = LC (jω) 2 (Aejωt – Bej(ωt+ϕ)) or • Bejϕ = – LCω2 (A – Bejϕ) – Single Eq for two unknowns, B and ϕ! – Both imaginary and real parts need to be equal! • Bcosϕ = – LCω2 (A – Bcosϕ) and • Bsinϕ = – LCω2 (– Bsinϕ). More arithmetic 2 • Bcosϕ = – LCω (A – Bcosϕ) and • sinϕ = 0, cosϕ = ±1. (ϕ is either 0 or π: both lead to the same solution!) • B(1 – LCω2 ) = – LCω2 A. • B = – A LCω2/(1 – LCω2) = – A Lω/(1/Cω – Lω) More arithmetic • Bcosϕ = – LCω2 (A – Bcosϕ) and • Bsinϕ = – LCω2 (– Bsinϕ). • The 2nd Eq. holds only when Bsinϕ = 0, or LCω2 = 1. – B = 0 won’t satisfy Eq. 1. – In general LCω 2 ≠ 1. – So sinϕ = 0, cosϕ = ±1 Some interpretation & observations • B = – A Lω/(1/Cω – Lω) • like a voltage divider, but it’s as if the resistance (imaginary) is in the opposite directions for C and L, so the sum = subtraction. • When LCω 2<1 B and A have opposite sign (180° out of phase), and LCω 2 >1, they are in phase. • When 1/Cω ~ Lω (or LCω2 ~1), B will be much larger than A – amplification w/o amplifier! • This is like forced resonant oscillation. – Used in radio (TV) station selection. (C is variable for a typical radio to tune to various stations.) – If C ~ 100 pF, what should L be to tune to 1500 kHz? Dividing by 0? Graphically, minimal resistance 12 LCω 2=1) • When 1/Cω = Lω (or There is no solution – dividing by 0! • In reality, there is always some resistance (long coil has large resistance unless super conductor!) in addition to L and C, so dividing by 0 won’t happen. i.e. • Bejϕ = A jLω / (1/jCω + R + jLω). – When LCω 2=1, Bejϕ = A jLRω (= A jR/Cω) – i.e. ϕ = π /2 (90° shifted: cosine in, – sine out, for example.) 10 Output amplitude RCω0 = 0.1 8 6 Output phase 4 2 0 -5 -4 -3 -2 -1 0 LCω 2 = 1 1 2 3 4 5 Log(ω 2 in unit of 1/L C) 2 More resistance (damping) Strong damping 4 3.5 RCω0 = 0.3 3.5 3 3 Output phase 2.5 2.5 2 Output amplitude 2 Output amplitude Output phase RCω0 = 1.0 1.5 1.5 1 1 0.5 0.5 0 -5 -4 -3 -2 -1 0 0 LCω 2 1 =1 2 3 Log(ω 2 4 5 in unit of 1/L C) -5 -4 -3 -2 -1 0 1 LCω 2 = 1 2 3 4 5 Log(ω 2 in unit of 1/L C) 3