Question

In an ideal transformer, the voltage is stepped down from $11KV$ to $220V$ . If the primary current be $20A$ the current in the secondary should be
(A) $5KA$
(B) $1KA$
(C) $0.5KA$
(D) $0.1KA$

Answer 1

In an ideal transformer there will be no losses between the two coils of the transformer (primary and secondary transformers). Thus the power ratio between the power supplied by the primary coil $({P_p})$ to the power supplied by the secondary coil $({P_s})$ is unity, ${P_p}:{P_s} = 1:1$ .

Formulas used: We will be using the formula $P = VI\cos \phi$ where $P$ is the power of the transformer coil, $V$ is the voltage of the transformer coil, $I$ is the current passing through the transformer coil , and $\cos \phi$ is the power factor of the transformer coil(which is the ratio between true power to the apparent power)

Complete Answer:
Basically, a transformer is a simple electrical device that consists of two coils wound around a conductor that works using the principle of electromagnetic induction. The alternating current on the primary winding of the transformers induces a varying magnetic flux on the transformer core which in turn induces an electromotive force on the secondary winding. And a secondary current is thus produced due to the electromagnetic induction by the primary winding.
In an ideal transformer there will be no losses at all, which means it has high core magnetic permeability, and winding inductances. Also, it should have zero net EMF. And hence the word “ideal”.
Now when there are no losses, the power supplied to the primary coil will be supplied to the secondary coil without any losses. Thus we can say that ${P_p} = {P_s}$ where ${P_p}$ is the power supplied by the primary winding and ${P_s}$ is the power supplied by the secondary winding.
We know that the power of a transformer is given by, $P = VI\cos \phi$ . Since it is an ideal transformer and it has no losses, the true power and apparent power will be the same. Thus, the power factor of the transformer, $\cos \phi = 1$ .
Thus the power is now dependent on the voltage $V$ and current $I$ . Since it is a transformer, the current introduced in it will be alternating current and hence voltage and current will be alternating. Although the power produced by the primary coil and the power produced by the secondary coil will be the same, the voltage and current for both coils may vary.
$\Rightarrow {P_p} = {P_s}$
${V_p}{I_p} = {V_s}{I_s}$
This means that the product of voltage and current is always constant.
$\Rightarrow VI = const$
Thus, we can say that ${V_1}{I_1} = {V_2}{I_2}$
Substituting the values from the problem ${V_1} = 110KV$ , ${V_2} = 220V$ , ${I_1} = 20A$
$\Rightarrow (11000) \times 20 = (220){I_2}$
Solving for ${I_2}$
${I_2} = \dfrac{{11000 \times 20}}{{220}}$
$\Rightarrow {I_2} = 1000V = 1KA$
Thus, the current required to maintain the power of the transformer constant is ${I_2} = 1KA$ .
Hence the correct answer for option B.

Note:
Although the power of the transformer is considered to be constant the values of $V$ and $I$ can be subjected to changes because $V \propto {I^2}$ and hence if the value of $V$ changes then the value of $I$ changes itself such that $P$ remains constant.