Question

A step-down transformer with an efficiency of $80\%$ is used on a 1000 V line to deliver 10 A at 100 V at the secondary coil. The Current drawn from the line is

• A. 1.5 A
• B. 2 A
• C. 3 A
• D. 1.25 A
• E. 1 A

Answer 1

Answer: (D)

1.25 A

The efficiency of the transformer is $e=\frac{poweroutput}{powerinput}=\frac{V_2{I}_2}{V_1{I}_1}$

So, $\frac{80}{100}=\frac{100\left(10\right)}{\left(1000\right)I_1}$

or $I_1=1.25A$