Transformers — Question 5

Question 5

Winding resistances can be referred to the secondary before solving the load

A transformer has ideal ratio $a=N_p/N_s=5.00$, primary winding resistance 2.00 ohms, secondary winding resistance .100 ohm, and a 6.00-ohm resistive load. The primary supply is 120 V rms. Neglect core and leakage effects.

1. Refer primary resistance to the secondary and find load current and terminal voltage.

2. Find copper loss and efficiency.

Question 5 - Solution

Winding resistances can be referred to the secondary before solving the load

(a) Method and calculation

$$R_{\mathrm{eq,s}}=r_s+r_p/a^2=.100+2/25=.180\,\Omega,\quad V_{s,0}=120/5=24.0\,V.$$$$\boxed{I_s=24/6.18=3.88350\,A,\quad V_s=6I_s=23.3010\,V}.$$

(b) Interpretation and further calculation

$$P_{\mathrm{copper}}=I_s^2R_{\mathrm{eq,s}}=2.71468\,W,\quad P_L=I_s^2R_L=90.4892\,W.$$$$\boxed{\eta=\frac{R_L}{R_L+R_{\mathrm{eq,s}}}=6/6.18=.970874}.$$

Physical check

The referred primary copper loss equals the actual $I_p^2r_p$ because $I_p=I_s/a$.