Saved Bookmarks
| 1. |
A transformer on a utility pole operates at V_p=8.5 kV on the primary side and supplies electrical energy to a number of nearby houses at V_s=120V , both quantities being rms values. Assume an ideal step-down transformer, a purely resistive load , and a power factor of unity. (a) What is the turns ratio N_p//N_s of the transformer ? |
|
Answer» Solution :The turns ratio `N_p//N_s` is related to the rms primary and secondary voltages `(V_s=V_p N_s//N_p)`. Calculation : We can write as `V_S/V_p = N_S/N_p` `N_p/N_S=V_p/V_S=(8.5xx10^3 V)/(120V)=70.83 approx71.` (b) The average rate of energy consumption (or dissipation) in the houses served by the transformer is 78 kW. What are the rms currents in the primary and secondary of the transformer? For a purely resistive LOAD, the power factor cos `phi` is unity, thus, the average rate at which energy is supplied and dissipated `(P_(avg)=EI=IV)`. Calculations : In the primary circuit, with `V_p`=8.5 kV, `I_p=P_(avg)/V_p = (78xx10^3W)/(8.5xx10^3 V)=9.176A approx 9.2 A` SIMILARLY , in the secondary circuit, `I_S=P_(avg)/V_S=(78xx10^3W)/(120V)`=650 A. You can check that `I_s=I_p(N_p//N_s)` as required by Eq. 31-74. (c) What is the resistive load `R_s` in the secondary circuit? What is the corresponding resistive load `R_p` in the pri mary circuit? One way: We can use V= IR to RELATE the resistive load to the rms VOLTAGE and current. For the secondary circuit, we find `R_S=V_S/I_S=(120V)/(650A)=0.1846Omega approx` 0.18 `Omega` Similarly , for the primary circuit we find `R_p=V_p/I_p =(8.5xx10^3 V)/(9.176A)=926 Omega approx 930 Omega`. Second way: We use the fact that `R_p` equals the equivalent resistive load "seen” from the primary side of the transformer, which is a resistance modified by the turns ratio and given by Eq. 31-76 `(R_(eq)= (N_p//N_s)^2R)` . If we SUBSTITUTE `R_p` for `R_(eq)`and `R_s` for R , that equation yields `R_p=(N_p/N_S)^2 R_S=(70.83)^2 (0.1846Omega)` `=926Omega approx 930 Omega`. |
|