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authorpriyanka2015-06-24 15:03:17 +0530
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+clc
+clear
+
+//input
+kva=10;//kVA rating of the transformer
+vp=400;//voltage on primary side in volts
+vs=230;//voltage on secondary side in volts
+//short circuit test
+ppd1=18;//primary p.d. in volts
+ip1=25;//primary current in amperes
+inp1=120;//power input in watts
+//short circuit test
+ppd2=400;//primary p.d. in volts
+ip2=0.5;//primary current in amperes
+inp2=70;//power input in watts
+
+//calculations
+zp=ppd1/ip1;//equivalent primary impedance in ohms
+rp=inp1/(ip1^2);//equivalent resistance in ohms
+xp=((zp^2)-(rp^2))^0.5;//equivalent leakage reactance in ohms
+r0=(vp^2)/(1000*inp2);//resistance of parallel circuit
+phi=sin(acos(inp2/(vp*ip2)));//sine of power factor
+im=ip2*phi;//magnetizing current in amperes
+x0=vp/im;//reactance in ohms
+
+//output
+mprintf('the equivalent circuit parameters are \n Rp=%3.3f ohms \n Xp=%3.3f ohms \n r0=%3.3f kilo ohms \n x0=%3.1f ohms',rp,xp,r0,x0)