The number of turns in the primary of an ideal transformer is 400 and that in the secondary is 2000. If the ou

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Q. The number of turns in the primary of an ideal transformer is 400 and that in the secondary is 2000. If the output power from the secondary at 1000 V is 10 kW then calculate the voltage and current in the primary coil.

Tamil Nadu DgeTamil Nadu HSC (DGE) Board 2019Subjective· 2mImportance★★★★★

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Concept understanding — Transformer Turns Ratio

What the Turns Ratio Really Means

Imagine you have two coils of wire wrapped around a shared iron core. One coil (the primary) has a certain number of loops — say 100 turns. The other (the secondary) has a different number — say 200 turns. When you feed AC voltage into the primary, it creates a changing magnetic field that cuts through both coils. That field induces a voltage in the secondary.

The key insight: each turn of wire gets the same "share" of the magnetic push. If one turn in the primary gets 1 volt induced across it, then every turn in the secondary also gets 1 volt. So if the secondary has twice as many turns, it gets twice the total voltage.

That's the whole idea in a nutshell. The turns ratio directly determines how voltage transforms from primary to secondary.

The Precise Relation

The voltage across each coil is proportional to the number of turns:

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

Where $V_p$ and $V_s$ are the primary and secondary voltages, and $N_p$ and $N_s$ are the number of turns on each coil.

\frac{V_s}{V_p} = \frac{N_s}{N_p} = \frac{I_p}{I_s}

The current relation follows from conservation of energy. An ideal transformer doesn't create or destroy power — it just transforms it. So the power in equals the power out:

V_p I_p = V_s I_s

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

Watch out

The current ratio is the inverse of the turns ratio. If you step up voltage (more secondary turns), the secondary current drops proportionally. Many students flip this by accident.

What This Tells You

If $N_s > N_p$ , you have a step-up transformer — secondary voltage is higher, secondary current is lower. If $N_s < N_p$ , it's a step-down transformer — secondary voltage is lower, secondary current is higher.

The turns ratio itself is often written as $a = \frac{N_p}{N_s}$ (primary over secondary). Then:

V_s = \frac{V_p}{a}, \quad I_s = a I_p

A Concrete Example

A transformer has 500 turns on the primary and 100 turns on the secondary. The primary is connected to 230 V AC.

Turns ratio $a = \frac{500}{100} = 5$ . This is a step-down transformer.

Secondary voltage: $V_s = \frac{230}{5} = 46$ V. …

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Using the turns ratio for an ideal transformer, the primary voltage is $200\,\text{V}$ , and using power conservation, the primary current is $50\,\text{A}$ .

For an ideal transformer, the ratio of secondary to primary voltage equals the ratio of secondary to primary turns:

$\dfrac{V_2}{V_1} = \dfrac{N_2}{N_1} \quad\Rightarrow\quad V_1 = V_2\cdot\dfrac{N_1}{N_2}$

Here $N_1 = 400$ , $N_2 = 2000$ , and the secondary voltage $V_2 = 1000\,\text{V}$ . Substituting,

$V_1 = 1000 \times \dfrac{400}{2000} = 1000 \times 0.2 = 200\,\text{V}$

An ideal transformer has no losses, so the power delivered to the primary equals the power delivered by the secondary to its load:

$P_1 = P_2 = 10\,\text{kW} = 10000\,\text{W}$

The primary current is then …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set V11 mark
$Q.For a step-down transformer, the ratio of primary current to secondary current is __________. Fill in the blank choosing the appropriate answer from the bracket: (photons, diffraction, polarity, monopoles, greater than unity, less than unity)$
›Reveal solutionSolution

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

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CBSE 2026Set A1 markMCQ
$Q. Which quantity decreases in step-down transformer? (A) Current (B) Voltage (C) Frequency (D) Power$
›Reveal solutionSolution

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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CBSE 2025Set D1 markMCQ
$Q. In a step-up transformer, the value of current in secondary compared to the primary coil is (A) equal (B) less (C) more (D) no relation between the two$
›Reveal solutionSolution

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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CBSE 2025Set ANNUAL1 markMCQ
$Q. A transformer has 400 secondary turns and 500 primary turns. If the secondary voltage is 400 V, the primary voltage is:$
(a) 500V
(b) 400V
(c) 1000V
(d) 800V.
›Reveal solutionSolution

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: …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. In a transformer, the number of turns in the primary and the secondary are 410 and 1230 respectively. If the current in primary is 6 A, then that in the secondary coil is :$
(a) 12 A
(b) 2 A
(c) 1 A
(d) 18 A
›Reveal solutionSolution

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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CBSE 2024Set ANNUAL1 mark
$Q. In an ideal transformer, number of turns in the primary and secondary are 200 and 1000 respectively. If the input voltage is 200 V, what will be its output voltage ?$
›Reveal solutionSolution

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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CBSE 2023Set F1 markMCQ
$Q. In step-up transformer, the value of current in secondary coil compared to primary coil is (A) equal (B) less (C) more (D) none of these$
›Reveal solutionSolution

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

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

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CBSE 2022Set I1 markMCQ
$Q. Transformation ratio in transformer is (A) V_S/V_P (B) V_P/V_S (C) 1/(V_S\cdotV_P) (D) 1/(I_S\cdotI_P)$
›Reveal solutionSolution

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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CBSE 2022Set I1 markMCQ
$Q. Which of the following is correct for step-up transformer? (A) V_s < V_p (B) V_s > V_p (C) V_s = V_p (D) V_s << V_p$
›Reveal solutionSolution

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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CBSE 2020Set NC1 mark
$Q. A power transmission line feeds input power at 2300 V to a step-down transformer with its primary windings having 4000 turns. What should be the number of turns in the secondary in order to get an output power at 230 V?$
›Reveal solutionSolution

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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CBSE 2019Set ANNUAL1 markMCQ
$Q. A step-up transformer operates on a 230 V line and supplies current of 2 A to a load. The ratio of the primary and secondary windings is 1:25. The current in the primary coil is$
(a) 12.5 A
(b) 15 A
(c) 25 A
(d) 50 A
›Reveal solutionSolution

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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CBSE 2019Set ANNUAL1 markMCQ
$Q. In a step - up transformer, no. of turns in primary and secondary coils are N1 and N2, then -$
(a) N1 > N2
(b) N2 > N1
(c) N1 = N2
(d) N1 = 0
›Reveal solutionSolution

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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Q.The number of turns in the primary of an ideal transformer is 400 and that in the secondary is 2000. If the output power from the secondary at 1000 V is 10 kW then calculate the voltage and current in the primary coil.

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