Question

A step-down transformer reduces the voltage of a transmission line from $2200V$ to $220V$. The power delivered by it is $880W$ and its efficiency is $88\%$. The input current is

• A. $4.56mA$
• B. $0.045A$
• C. $0.45A$
• D. $4.65A$

Answer 1

The correct option is C

$0.45A$

Step 1: Given data

Input voltage, $V_1=2200V$

Output voltage, $V_2=220V$

Output power, $P_2=880W$

Efficiency, $\eta =88\%$

Step 2: Finding the input current.

We know that,

Efficiency, $\eta =\frac{P_2}{P_1}$

Where, $P_2$ and $P_1$ is the output and input power respectively.

Now,

$88\%=\frac{880}{P_1}$

$P_1=\frac{880}{88\%}$

$V_1{I}_1=\frac{880\times 100}{88}$ $\left[\because power=V\times I\right]$

$2200\times I_1=\frac{880\times 100}{88}$

$I_1=\frac{880\times 100}{88\times 2200}$

$I_1=\frac{880\times 100}{88\times 2200}$

$I_1=0.45A$

Hence, option C is correct. The input current is $0.45A$.