Question

A step-down transformer having an efficiency of 75% supplies 3-ampere current at 120 volts. Calculate the current in its primary coil, if it is operated at 440 volts.
A. 2
B. 1
C. 3
D. 4

Answer 1

Hint: The efficiency of a transformer is the ratio of output power to the input power of the transformer. And, power is the product of current and voltage. Thus this problem can be solved using only these concepts.

Formula used:
\[\eta =\dfrac{{{P}_{O}}}{{{P}_{I}}}\]
\[P=V\times I\]

Complete step-by-step solution:
From given, we have,
The efficiency of the transformer, \[\eta =75%\]
The output current of the transformer, \[{{I}_{S}}=3A\]
The output voltage of the transformer, \[{{V}_{O}}=120V\]
The input voltage of the transformer, \[{{V}_{I}}=440V\]
The efficiency of a transformer defines its working ability. Higher the ratio of output power to the input power greater the efficiency of that transformer.
Firstly, compute the output power of the transformer
The output power is the product of the secondary current and the output voltage.
Thus, the formula for calculating the output power of the transformer is,
\[{{P}_{O}}={{V}_{O}}\times {{I}_{S}}\]
Substitute the values in the above equation.
\[\begin{align}
  & {{P}_{O}}=120\times 3 \\
 &\Rightarrow {{P}_{O}}=360 W \\
\end{align} \]
The efficiency of the transformer is given by,
\[\eta =\dfrac{{{P}_{O}}}{{{P}_{I}}}\]
Substitute the given values in the above equation.
\[\begin{align}
  & 75\%=\dfrac{{{P}_{O}}}{{{P}_{I}}} \\
 &\Rightarrow \dfrac{75}{100}=\dfrac{360}{{{P}_{I}}} \\
\end{align}\]
Rearrange the terms to find the value of the input power of the transformer
\[\begin{align}
  & {{P}_{I}}=\dfrac{360\times 100}{75} \\
 &\Rightarrow {{P}_{I}}=480\,W \\
\end{align}\]
Therefore, the value of the input power of the transformer is 480 W.
Now compute the value of the primary current of the transformer.
The input power is the product of the primary current and the input voltage.
Thus, the formula for calculating the primary current of the transformer is,
\[{{P}_{I}}={{V}_{I}}\times {{I}_{P}}\]
Substitute the obtained values in the above equation.
\[480=440\times {{I}_{P}}\]
Rearrange the terms to find the value of the primary current of the transformer
\[\begin{align}
  & {{I}_{P}}=\dfrac{480}{440} \\
 &\Rightarrow {{I}_{P}}=1.09\,A \\
\end{align}\]
As the value of current in the primary coil of the transformer equals nearly 1 A, thus the option (B) is correct.

Note: The things to be on your finger-tips for further information on solving these types of problems are: The primary current refers to the current that flows through the primary coil and is also called as the input current whereas, the secondary current refers to the current that flows through the secondary coil and is also called as the output current.