Question

A transformer with efficiency $80\%$works at $4$kW and $100$V. If the secondary voltage is $200$V, then the primary and secondary currents are respectively.
A. $40{\text{A}}$and$16{\text{A}}$
B. $16{\text{A}}$and$40{\text{A}}$
C. $20{\text{A}}$and$40{\text{A}}$
D. $40{\text{A}}$and$20{\text{A}}$

Answer 1

Here we will use Ohm’s law, along with some other formulas of electric current, voltage and efficiency. Basic percentage rules will also be used.

Complete answer :
Ohm’s Law: According to Ohm’s Law, the current passing through an element is directly proportional to the voltage across the end points of that element provided that element must be conductor of electricity. ${\text{V}}\alpha \;{\text{I}}$.

Efficiency: The efficiency of a transformer is equal to the ratio of output source power to the input source power. It is represented by the symbol ${\text{\eta }}$ where, ${\text{\eta }} = \dfrac{{{{\text{O}}^{\text{p}}}}}{{{{\text{I}}^{\text{p}}}}}$.
The power of the transformer is $4$kW i.e. $4000$W.

Now, the primary voltage and secondary voltage of the transformer are ${V_p} = 100$V and ${V_p} = 200$V respectively.

The relationship between power, voltage and current is given as, ${\text{P}} = {\text{Vi}}$, substituting the given values in this formula we get,

$$40{\text{A}} = {{\text{i}}_p} \\\$$

Now, since we are given the efficiency, therefore, we will substitute values of primary voltage, secondary voltage, primary current and the given efficiency’s value we get,

{\eta} = \dfrac{{{{\text{V}}_s}{{\text{i}}_s}}}{{{{\text{V}}_p}{{\text{i}}_p}}} \\\ \dfrac{{80}}{{100}} = \dfrac{{200{{\text{i}}_{\text{s}}}}}{{4000}} \\\ \dfrac{{80}}{{100}} = \dfrac{{{{\text{i}}_{\text{s}}}}}{{20}} \\\ $$, Simplifying it further we get,

\dfrac{{80 \times 20}}{{100}} = {{\text{i}}{\text{s}}} \\
16{\text{A}} = {{\text{i}}{\text{s}}} \\

Therefore, the primary current in the transformer is $$40{\text{A}}$$and secondary current is $$16{\text{A}}$$. **Hence, option A is the correct option.** **NOTE:** In such types of problems, where voltage and power or related terms are given we must try to apply Ohm’s law. While applying Ohm’s law we should try to alter the formula by replacing either voltage or current with some other formulas.