Question

Show that the phase angle (argument) of the impedance between A and B in the following circuit is zero only at a single frequency, namely f (27LC)-1 L R. C A

Answer 1

We found this as -

We have Z_eq implies R + jwL + 1/jwL R + jwL middot jwC + 1/jwC R + 1 - w^2 LC/jwC (j^2 implies -1) Therefore Z_eq implies R jw C + 1 - w^2 LC/jwC 1 - w^2 LC + R jw C/jwC \r\n Now, by rationalization, multiply & divide by -jwC as - Z_eq implies (R jw C + (1 - w^2 LC)) - jwC/jwC times -jwC -R j^2 w^2 C^2 - jwC (1 - w^2 LC)/-j^2 w^2 C^2 & j^2 implies -1, Z_eq implies R w^2 C^2/w^2 C^2 -j(1 - w^2 LC/w C) Z_eq implies R - j/wc (1 - w^2 LC) phase angle is zero when imaginary part is zero. Therefore By equating imaginary part to zero as: - -j/wc (1 - w^2 LC) implies 0 1 - w^2 LC implies 0, w^2 LC implies 1 Therefore w implies 1/squareroot LC & 2 pi f implies 1/squareroot LC Therefore f implies 1/2 pi squareroot LC