(2) what values of r and c make the impedance between c- d equal the impedance between a-b for a source frequency of ω = 5 rad/s? [hint: it is probably easier to work with admittance values to get ycd = yab.]

R= 5/2 and L= 3/2 H

Explanation:

Note: check attached file/picture for the diagrams and written solution.

Knowing that Z(c)= -j and Z(L)=3j.

Y(RL)= 1/2-j. Also, R+Z(L) =Z-j.

Step one(1): find Yab and Y(L) from;

Yab = 2/5 +j / 5 and Y(L)=-j/3.

Yab= 2/5+ (1/5-1/3)

Yab= 2/5 - j(2/15)----------------------------------------------------------------------------(2).

Worthy of note is that; Y(RC) = (2+j)/5 and that Y(L) = -j/3.

Step two(2): find R using Ycd formula

Therefore, in order to find R, we have to use the formula below;

Ycd= 1/R + 1/jwL --------------------------------------------------------------------------------(2).

Equating equation (1) and equation (2) together. From the hint; Ycd=Yab.

Hence, 1/R = 2/5; R = 5/2.

Step three(3): finding L from equation (2).

Equating equation (1) and equation (2) together. From the hint; Ycd=Yab.

1/jwL = -j(2/15); our w(Omega)= 5 rad per seconds.

Therefore, L= 10/15= 3/2 H.

v=ir = 3x15 =45 v