Saved Bookmarks
| 1. |
Consider the LCR circuit shown in Fig. Find the net current I and the phse of i. show that `i= (upsilon)/(Z)` . Find the impedence Z for this circuit. |
|
Answer» In Fig. let i. be the total current form the source. It is divided into two parts : `i_(1)` through R and `i_(2)` through series combination of C and L, such that `i= i_(1) + i_(2)`. As `R i_(1) = upsilon_(m) sin omega t, i_(1) = (upsilon_(m) sin omega t)/(R )` If `q_(2)` is charge on C at time t, then for series combination of C and L, `(q_(2))/(C ) + (dq_(2)^(2))/(dt^(2)) = upsilon_(m) omega t` Let `q_(2) = q_(m) sin (omega t + phi)` `:. (dq_(2))/(dt) = q_(m) omega cos (omega t + phi)` and `(d^(2) q_(2))/(dt) = - q_(m) omega^(2) sin (omega t + phi)` Putting in (ii), we get `q_(m) [(1)/(C ) = L omega^(2)] sin (omega t + phi) - upsilon_(m) sin omega t` If `phi = 0` and `((1)/(C ) - L omega^(2)) gt 0`, then `q_(m) = (upsilon_(m))/(((1)/(C ) - L omega^(2)))` From (iii), `i_(2) = (dq_(2))/(dt) = omega q_(m) cos (omega t + phi)` , using (iv), `i_(2) = (omega upsilon_(m) cos (omega t + phi))/((1)/(C ) - L omega^(2))` Taking `phi = 0`, `i_(2) = (upsilon_(m) cos (omega t))/(((1)/(omega C) - L omega))` From (i) and (v), we find that `i_(1)` and `i_(2)` are out of phase. Now, `i_(1) + i_(2) = (upsilon_(m) sin omega t)/(R ) + (upsilon_(m) cos omega t)/(((1)/(omega C) - L omega))` Put `(upsilon_(m))/(R ) = A = C cos phi` and `(upsilon_(m))/(((1)/(omega C) - L omega)) = B = C sin phi` `:. i_(1) + i_(2) = C cos phi sin omeg t + C sin phi cos omega t = C sin (omega t = phi)` Where `C = sqrt(A^(2) + B^(2))` and `phi = tan^(-1) ((B)/(A))` `C = [(upsilon_(m)^(2))/(R^(2)) + (upsilon_(m)^(2))/((1)/(omega C) - L omega)^(2) ]^(1//2)` and `phi (tan^(-1)) (R )/(((1)/(omega C) - L omega))` Hence, `i= i_(1) + i_(2) = [(upsilon_(m)^(2))/(R^(2)) + (upsilon_(m)^(2))/(((1)/(omega C) - L omega)^(2))]^(1//2) sin (omega t + phi)` or `(i)/(upsilon_(m)) = (1)/(Z) = [(1)/(R^(2)) = (1)/(((1)/(omega C) - L omega)^(2))]^(1//2)` This is the expression for impedance Z of the circuit. |
|