Question

A $220\text{V}$ input is supplied to a transformer. The output circuit draws a current of $2.0\text{A}$ at $440\text{V}$. If the efficiency of the transformer is $80$%, the current drawn by the primary windings of the transformer is

• A. $1\text{A}$
• B. $1.25\text{A}$
• C. $2.5\text{A}$
• D. $5\text{A}$

Answer 1

The correct option is D $5\text{A}$

$V_p=220\text{V};i_p=?;V_s=440\text{V};i_s=2.0\text{A};\eta =80\text{\%}$

Efficiency $\eta =\frac{\text{output power}}{\text{Input power}}$

From the data given in the question, $80\text{\%}=\frac{V_s{i}_s}{V_p{i}_p}$

$\Rightarrow i_p=\frac{V_s{i}_s}{\left(\frac{80}{100}\right)V_p}=\frac{440\times 2\times 100}{80\times 220}=5\text{A}$

Hence, option (a) is the correct answer.