MIT 6.002 Circuits and Electronics, Spring 2007
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
so help us too...brother
dear i've only started but not solved yet!!!
I'll tell help as soon as possible....
Lab is easy dude. just need an hour more to complete.
i know u r a smart guy @ chiragsala :)....
In the space provided below write an algebraic expression in terms of the device parameters for the bandwidth Δω of the impedance Z looking into this tank circuit. _________ The antenna tank of a Graymark 536 Radio Kit has an inductance L≈0.65mH. The resistance of the inductor RL≈4.0Ω. The equivalent resistance across the capacitor is RC≈490.0kΩ. The capacitor is variable, for tuning. If we tune to a station at f=980.0kHz what is the capacitance, in picoFarads, of the tuning capacitor? __________ What is the bandwidth, in kHz, of the tank at f=980.0kHz? _________ If next we tune to a station at f=1600.0kHz what is the capacitance, in picoFarads, of the tuning capacitor? _________ What is the bandwidth, in kHz, of the tank at f=1600.0kHz? ______________ So it is apparent that this is not the only circuit in the radio that selects the desired station from stations on adjacent channels.
Please Help Based off of the value of Q you determined earlier, you could also estimate the value of the element removed from the circuit! Enter the capacitance in nF or the inductance in mH, depending on which element you determined was removed from the circuit.
@Nurali pls repeat your question.. I can't understand.
H11P1 a) 1/(RC*C) + RL/L b) (thats the formula = 1/(2*pi*f)^(2)*L) c) (thats the formula = 1/(RC*C) + RL/L) d) (thats the formula = 1/(2*pi*f)^(2)*L) e) c) (thats the formula = 1/(RC*C) + RL/L) H11P2 a) (The Formula = vm*0.1) b) Cs / Rp c) Cs / Rp d) 0.1*VM(t) H11P3 a)C b)R c)L d ? e) L f) 1000 g) 1 h ?
Sorry buddy was late because of Diwali preparation. By the way the answers for LAB 11 are: 1)0.001592 2)1.59E-7 3)0.0356 4)10 Still if you liked please give best response.