Q.Which of the following is correct for step-up transformer? (A) V_s < V_p
(B) V_s > V_p
(C) V_s = V_p
(D) V_s << V_p

$N_s/N_p$ (equal to $V_s/V_p$), i.e. less than 1. …

A step-down transformer reduces the secondary voltage; current rises to conserve power.

A transformer works by mutual induction. The turns ratio fixes the voltage ratio:

$\frac{V_s}{V_p}=\frac{N_s}{N_p}$

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A step-up transformer increases voltage, so to conserve power the secondary current must be smaller than the primary current.

For an ideal transformer, power is conserved:

$V_p I_p = V_s I_s$

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For a transformer, $\dfrac{V_p}{V_s}=\dfrac{N_p}{N_s}$.

Given: $N_s = 400$, $N_p = 500$, $V_s = 400\,V$.

Using the transformer turns-ratio relation: …

Using the transformer current relation $I_sN_s=I_pN_p$ with $N_p=410$, $N_s=1230$, $I_p=6$ A gives $I_s=2$ A.

Working

For an ideal transformer, the turns and currents in primary/secondary satisfy

$I_pN_p = I_sN_s \ \Rightarrow\ I_s = I_p\dfrac{N_p}{N_s}$

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For an ideal transformer, $V_s/V_p = N_s/N_p$.

$\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p} \implies V_s = V_p \times \dfrac{N_s}{N_p} = 200 \times \dfrac{1000}{200} = 1000\text{ V}$

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Step-up transformer: voltage up, current down, so secondary current is less than primary.

For an ideal transformer, power is conserved, so $V_p I_p = V_s I_s$, giving

$\frac{I_s}{I_p} = \frac{V_p}{V_s} = \frac{N_p}{N_s}.$ …

Transformation ratio k = V_S/V_P = N_S/N_P.

In an ideal transformer the induced emf is proportional to the number of turns, so the secondary and primary voltages relate to the turns as:

$\frac{V_S}{V_P} = \frac{N_S}{N_P} = k$

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A step-up transformer increases the output voltage: V_s > V_p.

For an ideal transformer the voltage ratio equals the turns ratio: $\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p}$.

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For an ideal transformer, turns ratio equals voltage ratio; rearrange to find the secondary turns.

Formula

$\frac{N_s}{N_p} = \frac{V_s}{V_p}$

Substitute

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For an ideal transformer, $I_pN_p=I_sN_s$; use the given turns ratio and secondary current to find the primary current.

Step 1 — Ideal transformer relation

$\frac{I_p}{I_s} = \frac{N_s}{N_p}$

Step 2 — Substitute

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A step-up transformer has more turns in the secondary, so N2 > N1.

For an ideal transformer the turns ratio equals the voltage ratio:

$\frac{V_s}{V_p} = \frac{N_2}{N_1}.$

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