A transformer having efficiency of $75\%$ is working on $220V$ and $4.4kW$ power supply. If the current in the secondary coil is $5A$. What will be the voltage across the secondary coil and the current in the primary coil?
A) ${{V}_{s}}=220V,{{i}_{p}}=20A$
B) ${{V}_{s}}=660V,{{i}_{p}}=15A$
C) ${{V}_{s}}=660V,{{i}_{p}}=20A$
D) ${{V}_{s}}=220V,{{i}_{p}}=15A$

Hint: The power of the transformer is always the same for both primary and secondary coil, and source of voltage are always connected with the primary coil.


Formula Used:
$P={{v}_{1}}{{i}_{1}}={{v}_{2}}{{i}_{2}}$
${{v}_{1}}=$ Voltage across primary coil
${{i}_{1}}=$ Current in primary coil
${{v}_{2}}=$ Voltage across secondary coil
${{i}_{2}}=$ Current in secondary coil

Complete step by step answer:
Transformer is a device based on the principle of mutual induction, which is used for converting large alternating current at low voltage into small current at high voltage and vice-versa. The transformers which convert low voltage into higher ones are called ‘step up’ transformers while those which convert high voltage into lower ones are called ‘step down’ transformers.
There are two coils inside the transformer named primary coil and secondary coil. The voltage source is always connected to the primary coil.
Power in secondary coil = power in primary coil
$P={{v}_{1}}{{i}_{1}}={{v}_{2}}{{i}_{2}}$
Transformer efficiency$=75\%=\dfrac{75}{100}=\dfrac{3}{4}$
It means that it convert $\dfrac{3}{4}$ part of voltage and $\dfrac{3}{4}$ part of current
Given
$\begin{align}
  & P=4.4kW \\
 & P=4.4\times {{10}^{3}}W \\
 & {{v}_{1}}=220V \\
 & {{i}_{2}}=5A \\
\end{align}$
$\begin{align}
  & P={{v}_{2}}{{i}_{2}} \\
 & {{v}_{2}}=\dfrac{P}{{{i}_{2}}}=\dfrac{4.4\times {{10}^{3}}}{5} \\
\end{align}$
${{v}_{2}}=0.88\times {{10}^{3}}=880V$
But efficiency $=\dfrac{3}{4}$
Then
 $\begin{align}
  & {{v}_{2}}=\dfrac{3}{4}\times 880 \\
 & {{v}_{2}}=660V \\
\end{align}$
$\begin{align}
  & P={{v}_{1}}{{i}_{1}} \\
 & {{i}_{1}}=\dfrac{P}{{{v}_{1}}}=\dfrac{4.4\times {{10}^{3}}}{220} \\
 & {{i}_{1}}=\dfrac{4400}{220} \\
 & {{i}_{1}}=20A \\
\end{align}$
$\therefore$ the correct option is C.

Note: When we find out the ${{v}_{2}}$, maximum times we use simply $P={{v}_{2}}{{i}_{2}}$ not multiplying with efficiency but we should multiply with efficiency in this case.