Answer to Question #143234 in Electrical Engineering for NAVANIT KRISHNA

Question

1. A 30 KVA transformer has 300 turns on the primary and 50 turns on the secondary winding. The primary is connected to 1500 V, 50 Hz supply. Find
1) the full load primary and secondary currents
2) the secondary emf
3) maximum flux in the core

Answer 1

1) Since the power is $P_1=1.732\times U_1\times I_1$

$I_1=\frac{P}{1.732\times U_1}=\frac{30\times 10^3}{1.732\times 1500}\approx 11.55$ A.

And the current ratio is equal to the ratio of the number of turns : $\frac {I _ 2} {I _ 1} =\frac {N _ 1} {N _ 2}$ , therefore

$I_2=\frac{N_1\times I_1}{N_2}=\frac{300\times 11.55}{50}=69.3$ A.

2) $\frac{U_1}{U_2}=\frac{N_1}{N_2}$

$U_2=\frac{U_1\times N_2}{N_1}=\frac{1500\times 50}{300}=250$ V.

3) Since magnetic flux is $\Phi =\frac {U _ 1} {4 .44\times 2\pi\times\nu ^ 2}$ $=\frac {1500} {4 .44\times 2\pi\times 50 ^ 2 }\approx 0.043$ Vb.