Question

The secondary coil of an ideal step-down transformer is delivering 500-watt power at 12.5 A current. If the ratio of turns in the primary to the secondary is $5:1$, then the current flowing in the primary coil will be:

Answer 1

We know that power in a transformer is constant on both sides it can be expressed as $\dfrac{{{N_P}}}{{{N_S}}} = \dfrac{{{I_S}}}{{{I_P}}}$. We will find the value of ${N_S}$ ${N_P}$ and put it in the expression to find the value of current in the primary coil of the transformer.

Complete step by step answer:
It is given in the question that the secondary coil of an ideal step-down transformer is delivering 500-watt power at 12.5 A current, it is also mentioned that the ratio of turns in the primary to the secondary is $5:1$. Then we have to find the current flowing in the primary coil. We know that power in a transformer is constant on both sides i.e., primary and secondary sides. It means VI is also constant on both sides. The constant power in the transformer can be represented $\dfrac{{{N_P}}}{{{N_S}}} = \dfrac{{{I_S}}}{{{I_P}}}$.
The value of ${N_P}{I_S} = 500$ watt. Also, the value of ${I_S} = 12.5A$. From here we get the value of ${N_P} = \dfrac{{500}}{{12.5}} = 40$
Now we know that ratio of $\dfrac{{{N_P}}}{{{N_S}}} = \dfrac{5}{1}$
So, on putting the value of ${N_P} = 40$ in the above equation we get-
$\dfrac{{40}}{{{N_S}}} = \dfrac{5}{1}$
${N_S} = 40$
Now putting the value of ${N_S} = 40$, ${N_P} = 40$, ${I_S} = 12.5A$ in $\dfrac{{{N_P}}}{{{N_S}}} = \dfrac{{{I_S}}}{{{I_P}}}$, we get-
$\dfrac{{40}}{8} = \dfrac{{12.5}}{{{I_P}}}$
$\dfrac{{12.5}}{{{I_P}}} = 5$
${I_P} = \dfrac{{12.5}}{5}$
${I_P} = 2.4A$.

Thus, the current through the primary coil will be ${I_P} = 2.4A$.

Additional information:
A transformer has two coils, the first one is the primary coil and the second one is the secondary coil. AN alternating current flow in the primary coil and this induces a voltage in the secondary coil. The primary coil is a part of a circuit with a battery whereas the secondary coil is connected to an ammeter.

Note:
One can make a mistake in finding the value of ${N_S}$, ${N_P}$ this will result in the formation of the wrong answer. This mistake generally occurs because of repeated substitution calculations to find them but if we do all the substitution carefully this mistake can be avoided.