If I have a series LC circuit with 2 inductors, and if the inductors have different inductance values, what effect will they have on the resonance of the circuit?
At resonance; Xl Xc (inductive reactance capacitive reactance) but Xc 1/ (2*pi*f*C) and Xl 2*pi*f*L where C is capacitance, L is inductance and f as frequency at resonance therefore fr (resonant frequency) 1 / (2*pi*sqrtLC) since series inductors, the total inductance adds up the total inductive reactance increases and inductance increases so the resonant frequency will become lower
Resonance 1/(2 *pi * sqrt (L*C)) The larger they are, the lower the resonance.
The differential equation on your circuit is: Lq'' + Rq' + q/C 12sin(10t).(a million) in case you assume that sin(10t) 0 for t0 you will discover answer making use of one sided Laplace rework. I even have assumed that sin(10t) is valid for -inf t inf . the final answer of non-homogeneous differential equation (a million) consists of the sum of the particular (any) answer of the equation (a million) and the final answer of the homogeneous equation (2) under: Lq'' + Rq' + q/C 0.(2) particular answer of (a million) is assumed as q Qsin(10t + fi). in case you plug it in the answer is: q -0.05cos(10t).(3) the final answer of (2) is: q R*(e^at)*sin(bt + fi),.(4) the place: a -R/2L -6 and b sqrt(4L/C-R^2)/2L 8. Plugging (4) into (2) you get R 0.051/0.8, fi 0.923. ultimately your answer is the sum of (3) and (4): q - 0.05cos(10t) + (0.051/0.8)*(e^(-6t))*sin(8t + 0.923). you could verify that qo 0.001, and io q'(0) 0. answer (2) is so stated as table sure answer. hence LC is in resonance at frequency omega 10 and their serial impedance is 0. hence the table sure modern-day is E/R 12/24 0.5 A, which you get via fact the 1st by-manufactured from q. answer (4) is the quick answer brought about by making use of preliminary cost q0.
