Question

A power transformer is used to set up an alternating emf of $220\,V$ to $4.4\,kV$ for transmitting $6.6\,kW$of power. The primary coil has 1000 turns. What is the current rating of secondary? (assume efficiency 100%)
A. $2.5\,A$
B. $1.5\,A$
C. $1.0\,A$
D. $0.75\,A$

Answer 1

Hint: The power input and power output of the transformer will be equal since efficiency is given as 100%. We know that power is the product of voltage and current. So, if we know the value of output power then we can calculate the output current by dividing power output with the output voltage.


Complete step by step answer:
The voltage in the primary of a transformer or the input voltage is given as
${V_i} = 220\,V$
Voltage in the secondary or the output voltage is given as
${V_o} = 4.4\,kV$
$ \Rightarrow {V_o} = 4.4\, \times {10^3}\,V$
Power transmitted by the transformer is given as
${P_i} = 6.6\,kW$
$ \Rightarrow {P_i} = 6.6\, \times {10^3}\,W$
The efficiency is calculated as the ratio of output power to input power.
Since the efficiency of the transformer is given as 100%, we get the power output equal to the power input.
$ \Rightarrow {P_o} = {P_i}$
The number of turns in the primary coil is 1000.
We need to find the current rating of the secondary ${I_o}$ .
We know that power is the product of voltage and current
In equation form we can write it as
$P = VI$
Where, P is the power, V is the voltage and I is the current.
We need to find output current.
So, let us write the equation for power output
$ \Rightarrow {P_o} = {V_o}{I_0}$
$ \Rightarrow {I_o} = \dfrac{{{P_o}}}{{{V_o}}}$
Substituting all the values in this equation we get
$ \Rightarrow {I_o} = \dfrac{{6.6\, \times {{10}^3}\,}}{{4.4\, \times {{10}^3}}}$
$ \Rightarrow {I_o} = 1.5\,A$
This is the current rating of the secondary.
So, the correct answer is option B.

Note: The efficiency of the transformer is given as hundred percent. Since efficiency is the ratio of power output to power input, we will get the power output equal to the power input in this case. But when the efficiency is not hundred percent, we cannot take the output power as the same as input power because there will be some loss in energy in such transformers. The output power will then be less than the input power.