Question

1.) What is the value of the admittance Y₂ in the path through which I₂ flows? Present your answer in polar form.

2.)Now consider the current path for I₁. The expression for the admittance Y₁ will be of the form  Y1=[A+jXC]/[B+XC^2]
where X_C = 1/ωC is the capacitive reactance.What are the values of the coefficients A and B in this expression?

3.) Using the result of Part 4, develop an expression in terms of the capacitive reactance X_C which will be solved to obtain the necessary capacitance values. This expression should be of the form  D*XC^2+E*XC+F=0. Determine the numerical values of the D, E, and F coefficients.

4.) Solve the quadratic equation obtained in Part 5 for the two values of reactance X_C which accomplish the resonance condition. Enter the minimum and maximum values in the order indicated in the answer spaces.

5.) Now obtain the two capacitance values C₁ and C₂ which would accomplish the resonance condition.
Considering the current phasor relationships, can you explain why there are two possible values for the capacitance necessary to achieve the resonance condition?

Answer 1

• 4- W-9 rad/sec at jw (11) - 9+99 -0.0) (-84.8 2) xc wc - ac 2515). Y = - 1 2- jxc 2+ S 2-1 6 2 - 24 h/ac 4+tach y = 24 JXc - Atjx e ut x2 B+ x2 where, A=2, B=4 5 A resonance, Imaginary part of Y, +42 = 0 Prig £0.01 E84 80 + 2+ 3x cho I 44x, 2d -0.01 + x = 0 utx txo - 100x = 0

X 100 Xc+ 4 =0 x = 99.95 , x = 0.062 - tox - a (99.951 - 1 - wer 960.04 - 899.556 - 1.112 mp, 2.778 F: - 4) sering quadratie equation, xu = 99.952 ixe2 = 0.042 Xeri - o.o4 л, хмел 49. Сл 3 dxet Exet F=o- corporing ¢ DEL, EN-100, Rzy 2 At resonance phasol diagrams & As it has two parallel branches of RL and RC, it can have two different capacitance values to get resonance condition Condition.