Question

An ideal transformer converts 220V a.c. to 3.3kV a.c. to transmit a power of 4.4 kW. If primary coil has 600 turns, then alternating current in secondary coil is
$\begin{array}{l}
{\rm{A}}{\rm{. }}\dfrac{1}{3}A\\
{\rm{B}}{\rm{. }}\dfrac{4}{3}A\\
{\rm{C}}{\rm{. }}\dfrac{5}{3}A\\
{\rm{D}}{\rm{. }}\dfrac{7}{3}A
\end{array}$

Answer 1

Hint: If we consider that the transformer has 100$\% $ efficiency then the input power and the output power of a transformer are equal. The power of a transformer is equal to the product of voltage in the given coil and the current flowing through that coil.

Formula used:
The efficiency of a transformer is given as
$\eta = \dfrac{{{\rm{Output power}}}}{{{\rm{Input power}}}}$
The input power is given as
${P_I} = {V_P}{I_P}$
Here ${V_P}$ is the voltage through the primary coil while ${I_P}$ is the current flowing through the primary coil.
The output power is given as
${P_O} = {V_S}{I_S}$
Here ${V_S}$ is the voltage through the secondary coil while ${I_S}$ is the current flowing through the secondary coil.

Complete step-by-step answer:
We are given an ideal transformer. The input voltage which is the voltage in the primary coil of the transformer is given as
${V_P} = 220V$
The output voltage which is the voltage in the secondary coil of the transformer is given as
${V_S} = 3.3kV = 3300V$
The input power which is the power transmitted by the transformer is given as
${P_I} = 4.4kW = 4400W$
Therefore, we can calculate the current through the primary coil by using the input power formula in the following way.
${I_P} = \dfrac{{{P_I}}}{{{V_P}}} = \dfrac{{4400}}{{220}} = 20A$
For an ideal transformer, the efficiency is 100$\% $, therefore, input power is equal to the output power of the transformer. Therefore, we can calculate the current through the secondary coil in the following way.
$\begin{array}{l}
{P_I} = {P_O}\\
{V_P}{I_P} = {V_S}{I_S}\\
 \Rightarrow {I_P} = \dfrac{{{V_S}}}{{{V_P}}}{I_S}\\
 \Rightarrow {I_P} = \dfrac{{220}}{{3300}} \times 20 = \dfrac{4}{3}A = 1.33A
\end{array}$
Hence, the correct answer to the question is option B.

Note: Since we are given an ideal transformer, we have considered the efficiency 100$\% $ for the transformer which is possible only in the ideal situations. But in actual practise, a transformer cannot achieve efficiency equal to 100$\% $.