Saved Bookmarks
| 1. |
The series and parallel circuits shown in figure have the same impedance and the same power factor If R = 3 Omega and X =4Omega find the valuse of R_(1) andX_(1)Also find the impedance power factor . . |
|
Answer» Solution :Impedance: `Z = sqrt(R^(2) + X^(2)) =5 Omega` Power factor: `cos phi = R/(Z) = (3)/(5) implies phi = cos^(-1) (0.6)` `TAN phi = (X)/R = (4)/(3)` `(i_(R1))_(1) = (V_(0))/(R_(1)) , (iX_(1))_(0) = (V_(0))/(X_(1))` `i_(0)= sqrt((i_(R1))_(0)^(2) + (i_(X1))_(0)^(2) ) = sqrt(((V_(0))/(R_(1)))^(2) + ((V_(0))/X_(1)^(2))^(2))` `(V_(0))/(Z) = V_(0)sqrt((1)/(R_(1)^(2)) + (1)/X_(1)^(2))` `Z =(1)/(sqrt((1)/R_(1)^(2) + (1)/(X_(1)^(2))))=5` `tan phi = (i_(X_1))_(0)/((i_(R_1))_(0)) = (V_(0)//X_(1))/(V_(0)//R_(1)) = (R_(1))/(X_(1)) = tan phi = (4)/(3) ` `X_(1) = (3)/(4) R_(1) ` in `(1)/(R_(1)^(2) ) + (1)/((3/(4) R_(1))^(2)) = (1)/(5^(2))` `(1)/(R_(1)^(2)) (1 + (16)/(9)) =(1)/(25) implies (1)/(R_(1)^(2) ) = (9)/((25)^(2))` `R_(1) = (25)/(3) Omega` `X_(1) = (3)/(4) R_(1) = (25)/(5)Omega` .  ` <br> <img src=) .
|
|