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Unformatted text preview: steps of 40Hz. f(Hz) V R (mV) 7460 318 7500 338 7540 343 7580 390 7620 410 7660 433 7700 444 7740 445 7780 432 7820 412 7860 392 7900 367 7940 323 7980 314 Resonant frequency from the table and graph below is 7740Hz at which voltage is maximum of 445mV. VI(3) R L =1.317KΩ Quality factor for V(3) ∆ f for V(3) = 160Hz f R for V(3) = 1360 Thus, Q from graph = f R /∆ f Thus Q=8.5 Now Q from Equation: Q= 2π f R L/R+ R L Therefore Q = 5.6 Quality factor for V(3’) ∆ f for V(3) = 610Hz f R for V(3) = 7740Hz Thus, Q from graph = f R /∆ f Thus Q=12.68 Now Q from Equation: Q= 2π f R L/R+ R L Therefore Q =32.05 New Question μ pearmeability = L’ / L where L’ is for inductor with iron core and L is for inductor without iron core. Now L is given by: Thus, L’ = 0.105 & L = 247.366 Thus, μ = 4.24e4 Hm1...
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This note was uploaded on 10/20/2011 for the course EE 257 taught by Professor staff during the Spring '11 term at SUNY Buffalo.
 Spring '11
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