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**Hint:**If the output voltage increases in proportion then, we will increase the turns on the secondary coil. If the secondary coil has more turns then the primary coil then the output voltage is bigger than the input voltage. We call this a step up transformer.

**Complete step by step answer:**

A high-voltage electricity takes a transformer with a small amount of current and it changes into a very low-voltage electricity with any outsized current, or we can say that the other way around. That is, intensify or step down the voltage during a circuit.

Again we discuss how the transformer turns low-voltage electricity into high-voltage electricity while dropping the present. A transformer changes high-voltage electricity into low-voltage electricity.

If we modify the amount of turns within the coils we change the induced $emf$. This allows us to vary (transform) the voltage from the first to the secondary winding.

The Turns Rule is:

$\dfrac{{{N_S}}}{{{V_P}}} = \dfrac{{{V_S}}}{{{V_P}}}$

Where,

${N_S} = $ Number of activates the secondary winding

${N_P} = $ Number of activates the first coil

${V_S} = $ Voltage across the secondary coil

${V_P} = $ Voltage across the primary coil

So if the number of activates of the first coil is quite high on the secondary winding, the output voltage is going to be lesser than the input voltage. This is called a step down transformer.

In this case, the turn’s ratio is $5:4$

$\dfrac{{{N_P}}}{{{N_S}}} = \dfrac{{{V_P}}}{{{V_S}}};$

That is, $\dfrac{5}{4} = \dfrac{{220}}{X}.$

Taking cross multiplication we get,

$5X = 220 \times 4$

Let us divide \[5\] on both sides we get,

$X = \dfrac{{220 \times 4}}{5}$

On simplification we get,

$ = 176V$

The ratio of the voltages is $5:4$

It is a known incontrovertible fact that voltage and current are inversely proportional then the ratio of the first current to the secondary current is $4:5$

Hence, the ratio of the first current to the secondary current is $4:5$

**So the correct answer is option (A).**

**Additional information:**

In other words, \[\dfrac{{{I_1}}}{{{I_2}}} = \dfrac{{{V_2}}}{{{V_1}}}.\]for example, if the present and drop through the secondary winding is ${\text{3amps}}$ and ${\text{10volts}}$, and therefore the drop through the first coil is 5 volts, then the current through the primary coil is $\dfrac{{{{10}}}}{{{5}}}{{ \times 3 = 6 amps}}$. So, the secondary takes less voltage and more current.

**Note:**A Current Transformer (CT) is employed to live the present of another circuit. CTs are used worldwide to watch high-voltage lines across national power grids. A CT is meant to supply an AC in its secondary coil that's proportional to the present that it's measuring in its primary.

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