Question

A 50 Hz, single phase transformer draws a short-circuit current of 30 A at 0.2 pf lag when connected to 16 V, 50 Hz source. The short-circuit current when same transformer is energized from 16 V, 25 Hz source is ______A.

• A. 56.73

Answer 1

The correct option is A 56.73
$\text{Power factor when it is energized from 16 V, 50 Hz}$

$\frac{r_e}{z_e}=\frac{r_e}{\sqrt{r_e^2+x_e^2}}=0.2$

$\text{and current,}{I}_{sc}=\frac{16}{\sqrt{r_e^2+x_e^2}}=30$

$\sqrt{r_e^2+x_e^2}=0.533\Omega$

$r_e=0.2\times 0.533\Omega =0.1066\Omega$

$x_e=\sqrt{(0.533{)}^2-(0.1066{)}^2}=0.5222\Omega$

$x_e\text{at 25 Hz is}$ $x_e=\frac{25}{50}\times 0.5222=0.2611\Omega$

$I_{sc}\text{at 25 Hz}$ $=\frac{16}{\sqrt{(0.1066{)}^2+(0.2611{)}^2}}=\frac{16}{0.2820}=56.73A$