Chapter 10, Problem 1RQ

To determine

The short circuit current.

Explanation of Solution

Given:

The secondary voltage $\left(V\right)$ is $240\text{V}$.

The rating of the transformer $\left(VA\right)$ is $15\text{kVA}$.

The impedance $\left(Z\right)$ is $4.2\%$.

Calculation:

Calculate the full load current $\left(I\right)$ using the relation.

  $\begin{array}{c}I=\frac{VA}{V}\\ =\frac{\left(15\text{kVA}\right)\left(\frac{1000\text{VA}}{1\text{kVA}}\right)}{240\text{V}}\\ =62.5\text{A}\end{array}$

Calculate the short circuit current $\left(I_{sc}\right)$ using the relation.

  $\begin{array}{c}{I}_{sc}=\frac{I}{Z}\\ =\frac{62.5\text{A}}{0.042}\\ =1488.095\text{A}\end{array}$

Thus, the short circuit current in the circuit is $\underset{\_}{1488.095\text{A}}$.