Question

A transformer consisting of 300 turns in the primary and 150 turns in the secondary gives an output power of 2.2kW. If the current in the secondary coil is 10A, then the input voltage and current in the primary coil are:
A. \[220V\]and \[10A\]
B. \[440V\]and \[5A\]
C. \[440V\]and \[20A\]
D.\[220V\]and \[20A\]

Answer 1

Hint: In this question, we need to determine the input voltage and current in the primary coil. In a transformer, the side which is given with inputs is known as the primary side, and the side through which we get output is called the secondary side.

Complete step by step answer:
Number of turns in the primary side \[{N_p} = 300\]
Number of turns in the secondary side \[{N_s} = 150\]
The output power or the secondary side power is \[{P_o} = 2.2kW = 2200W\]
The current in the secondary coil is \[{I_s} = 10A\]
We know that Ohm’s law is given by the formula
\[{P_O} = {V_O}{I_O} - - (i)\]
Hence by using Ohm’s law, we can calculate the output voltage of the transformer
$
{V_O} = \dfrac{{{P_O}}}{{{I_O}}} \\
\implies {V_O} = \dfrac{{2200}}{{10}} \\
\therefore {V_O} = 220V \\
$
Hence the secondary voltage or the output voltage of the transformer is \[ = 220V\]
Now we know that the ratio of the number of turns in the secondary coil to the number of turns in the primary coil is equal to the ratio of the output voltage to the input voltage, hence we can write
\[n = \dfrac{{{N_s}}}{{{N_P}}} = \dfrac{{{V_o}}}{{{V_i}}}\]
Now we substitute the values
\[
\dfrac{{{N_s}}}{{{N_P}}} = \dfrac{{{V_0}}}{{{V_i}}} \\
\dfrac{{150}}{{300}} = \dfrac{{220}}{{{V_i}}} \\
{V_i} = \dfrac{{300}}{{150}} \times 220 \\
 = 220 \times 2 \\
 = 440V \\
 \]
Hence the input voltage or the primary voltage of the transformer is\[ = 440V\]
We also know that the ratio n is equal to the ratio of current in the secondary side to the current on the primary side, so we can write
\[n = \dfrac{{{N_s}}}{{{N_P}}} = \dfrac{{{I_o}}}{{{I_i}}}\]
Hence by substituting the values
$
\dfrac{{{N_s}}}{{{N_P}}} = \dfrac{{{I_o}}}{{{I_i}}} \\
\dfrac{{150}}{{300}} = \dfrac{{10}}{{{I_i}}} \\
\implies {I_i} = \dfrac{{300}}{{150}} \times 10 \\
 = 20A \\
$
So, the current in the primary coil \[ = 20A\]

So, the correct answer is Option C .

Note:
Students must note that if the turn ratio which is the ratio number of turns in the secondary coil to the number of turns in the primary coil is more than 1 then the transformer is a step-up transformer and if the turn’s ratio is less than 1 the transformer is a step-down transformer.