Question

A single phase, 50-KVA, 2400-240-volt, 60 Hz distribution transformer has the following parameters: •

Resistance of the 2400-volt winding Ri=0.75 ohm
Resistance of the 240-volt winding R=0.0075 ohm
Leakage reactance of the 2400-volt winding X:=1 ohm
Leakage reactance of the 240-volt winding X2=0.01 ohm
Core loss resistance on the 2400 side Rc=733.5 ohms
Magnetizing reactance on the 2400 side X=4890 ohms
a. Draw the equivalent circuit referred to the high voltage side and referred to the low- voltage side. Label the impedances numerically.
b. The transformer is used as a step-down transformer at the load end of a feeder having impedance of (0.5+j2.0) ohms. Determine the voltage V. at the ending end of the feeder when the transformer delivers rated load at rated secondary voltage and 0.8 laggin power factor. Neglect the "exciting current" l. of the transformer (this is the current into the parallel Rc/jXm combination), which is another way of saying that you should use what we called in class the "approximate circuit #2).

Answer 1

B) voltage at the sending end of the feeder = 2483.66 v

Step-by-step explanation:

attached below is the the equivalent circuits and the remaining solution for option A

B) voltage = 2400 v

I =
`(50*10^3)/(2400)` = 20.83 A

calculate voltage at sending end ( Vs )

Vs = 2400 + 20.83 ∠ -cos^-1 (0.8) ( 0.75*2 + 0.5 + j 2 + j2 )

hence Vs = 2483.66 ∠ 0.961

therefore voltage at the sending end = 2483.66 v