Question

A power transformer is used to step up an alternating e.m.f. of $220V$ to $11kV$ to transmit $4.4kW$ of power. If the primary coil has $1000$ turns, what is the current rating of the secondary? Assume $100\%$ efficiency for the transformer.
(A) $2A$
(B) $1.7A$
(C) $0.4A$
(D) $0.8A$

Answer 1

Since the efficiency is $100\%$, the power before step up will be equal to the power after. So we are given the power in the first case and the power in the second case can be written as the product of the current and the voltage. Therefore, we can find the current from that equation.
Formula used: In this solution we will be using the following formula,
$\Rightarrow P = VI$ where $P$ is the power and $V$ is the voltage and $I$ is the current.

Complete step by step answer:
In this question, we are told that the given transformer has an efficiency of $100\%$. This means that during the process of step up, no power is lost in any form. So, the power before and after the step up will be the same. Therefore we can write
$\Rightarrow {P_1} = {P_2}$
Now the power is given by the product of the current and the voltage. So,
$\Rightarrow {P_2} = {V_2}{I_2}$
The power before is given as,
$\Rightarrow {P_1} = 4.4kW = 4.4 \times {10^3}W$
So on substituting the values we get,
$\Rightarrow 4.4 \times {10^3}W = {V_2}{I_2}$
After the step up, the voltage is ${V_2} = 11kV = 11 \times {10^3}V$
So substituting this value,
$\Rightarrow 4.4 \times {10^3} = 11 \times {10^3}{I_2}$
So we can get the current as,
$\Rightarrow {I_2} = \dfrac{{4.4 \times {{10}^3}}}{{11 \times {{10}^3}}}$
On doing this calculation we get the current in the secondary as,
$\Rightarrow {I_2} = 0.4A$
Therefore, the correct answer is option (C).

Note:
Transformer’s are devices that are used to transfer the power from one circuit to another at constant frequency by altering the voltage according to our requirement. In practice the transformers can’t have an efficiency of $100\%$ as there is always some power loss due to heating or other causes. The transformers work on the principle of Faraday’s electromagnetic induction.