The Transformer Formula relates the voltage, current, and number of turns in the primary and secondary coils of a transformer, expressed as \ ( \frac{V_s}{V_p} = \frac{N_s}{N_p} \). This fundamental concept appears in NCERT Class 12 Physics, Chapter 7 (Alternating Current), and is a high-weightage topic in CBSE board exams, JEE Main, and NEET. This article covers the complete transformer formula, its derivation, a full formula sheet, three solved examples at progressive difficulty levels, CBSE exam tips, common mistakes, and JEE/NEET applications.
Key Transformer Formulas at a Glance
Quick reference for the most important transformer formulas.
- Turns ratio: \( \frac{V_s}{V_p} = \frac{N_s}{N_p} \)
- Current ratio: \( \frac{I_s}{I_p} = \frac{N_p}{N_s} \)
- Ideal transformer power: \( V_p I_p = V_s I_s \)
- Efficiency: \( \eta = \frac{P_{out}}{P_{in}} \times 100\% \)
- Step-up condition: \( N_s > N_p \), so \( V_s > V_p \)
- Step-down condition: \( N_s < N_p \), so \( V_s < V_p \)
- Transformation ratio: \( k = \frac{N_s}{N_p} \)
What is the Transformer Formula?
The Transformer Formula describes the relationship between the voltages and number of coil turns in a transformer. A transformer is an electrical device that transfers electrical energy between two circuits using electromagnetic induction. It operates on alternating current (AC) only.
In NCERT Class 12 Physics, Chapter 7 (Alternating Current), the transformer is introduced as a practical application of Faraday’s law of electromagnetic induction. The core principle is mutual inductance. When an AC voltage is applied to the primary coil, a changing magnetic flux is created in the iron core. This flux links with the secondary coil and induces an EMF in it.
The ratio of the secondary voltage to the primary voltage equals the ratio of the number of turns in the secondary coil to the number of turns in the primary coil. This is the fundamental Transformer Formula. If the secondary has more turns, the voltage is stepped up. If the secondary has fewer turns, the voltage is stepped down. This principle is the backbone of power transmission across India’s national electrical grid.
Transformer Formula — Expression and Variables
The primary Transformer Formula for an ideal transformer is:
\[ \frac{V_s}{V_p} = \frac{N_s}{N_p} \]
Since an ideal transformer conserves power (\( P_{in} = P_{out} \)), we also get the current relationship:
\[ \frac{I_s}{I_p} = \frac{N_p}{N_s} \]
Combining both, the complete ideal transformer relation is:
\[ \frac{V_s}{V_p} = \frac{N_s}{N_p} = \frac{I_p}{I_s} \]
| Symbol | Quantity | SI Unit |
|---|---|---|
| \( V_p \) | Primary (input) voltage | Volt (V) |
| \( V_s \) | Secondary (output) voltage | Volt (V) |
| \( N_p \) | Number of turns in primary coil | Dimensionless |
| \( N_s \) | Number of turns in secondary coil | Dimensionless |
| \( I_p \) | Primary (input) current | Ampere (A) |
| \( I_s \) | Secondary (output) current | Ampere (A) |
| \( k \) | Transformation ratio \( N_s / N_p \) | Dimensionless |
| \( \eta \) | Efficiency of transformer | Percentage (%) |
Derivation of the Transformer Formula
Consider an ideal transformer with a primary coil of \( N_p \) turns and a secondary coil of \( N_s \) turns wound on a common soft iron core.
Step 1: An AC voltage \( V_p \) applied to the primary coil drives a current that creates a changing magnetic flux \( \Phi \) in the core.
Step 2: By Faraday’s law, the EMF induced in the primary coil is \( \varepsilon_p = -N_p \frac{d\Phi}{dt} \).
Step 3: The same flux links the secondary coil. The induced EMF in the secondary is \( \varepsilon_s = -N_s \frac{d\Phi}{dt} \).
Step 4: Dividing the two equations eliminates \( \frac{d\Phi}{dt} \):
\[ \frac{\varepsilon_s}{\varepsilon_p} = \frac{N_s}{N_p} \]
Step 5: For an ideal transformer with negligible resistance, \( \varepsilon_p \approx V_p \) and \( \varepsilon_s \approx V_s \). This gives the standard Transformer Formula: \( \frac{V_s}{V_p} = \frac{N_s}{N_p} \).
Step 6: For an ideal transformer, input power equals output power: \( V_p I_p = V_s I_s \). Therefore, \( \frac{I_s}{I_p} = \frac{V_p}{V_s} = \frac{N_p}{N_s} \).
Complete Transformer Formula Sheet
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Voltage-Turns Ratio | \( \frac{V_s}{V_p} = \frac{N_s}{N_p} \) | V = voltage, N = number of turns | V (Volt) | Class 12, Ch 7 |
| Current-Turns Ratio | \( \frac{I_s}{I_p} = \frac{N_p}{N_s} \) | I = current, N = number of turns | A (Ampere) | Class 12, Ch 7 |
| Ideal Power Equation | \( V_p I_p = V_s I_s \) | V = voltage, I = current | W (Watt) | Class 12, Ch 7 |
| Transformation Ratio | \( k = \frac{N_s}{N_p} \) | k = transformation ratio, N = turns | Dimensionless | Class 12, Ch 7 |
| Efficiency | \( \eta = \frac{V_s I_s}{V_p I_p} \times 100\% \) | \( \eta \) = efficiency, V = voltage, I = current | % | Class 12, Ch 7 |
| Step-Up Condition | \( N_s > N_p \Rightarrow V_s > V_p \) | N = turns, V = voltage | — | Class 12, Ch 7 |
| Step-Down Condition | \( N_s < N_p \Rightarrow V_s < V_p \) | N = turns, V = voltage | — | Class 12, Ch 7 |
| Power Loss in Transmission | \( P_{loss} = I^2 R \) | I = current, R = resistance of wire | W (Watt) | Class 12, Ch 7 |
| Faraday’s Law (EMF) | \( \varepsilon = -N \frac{d\Phi}{dt} \) | N = turns, \( \Phi \) = magnetic flux | V (Volt) | Class 12, Ch 6 |
| Mutual Inductance EMF | \( \varepsilon_s = -M \frac{dI_p}{dt} \) | M = mutual inductance, I = current | V (Volt) | Class 12, Ch 6 |
Transformer Formula — Solved Examples
Example 1 (Class 9-10 Level)
Problem: A transformer has 500 turns in its primary coil and 2500 turns in its secondary coil. If the primary voltage is 220 V, find the secondary voltage.
Given: \( N_p = 500 \), \( N_s = 2500 \), \( V_p = 220 \) V
Step 1: Write the Transformer Formula: \( \frac{V_s}{V_p} = \frac{N_s}{N_p} \)
Step 2: Substitute the known values: \( \frac{V_s}{220} = \frac{2500}{500} \)
Step 3: Simplify the right side: \( \frac{V_s}{220} = 5 \)
Step 4: Solve for \( V_s \): \( V_s = 5 \times 220 = 1100 \) V
Since \( N_s > N_p \), this is a step-up transformer. The output voltage is higher than the input.
Answer
Secondary voltage \( V_s = 1100 \) V
Example 2 (Class 11-12 Level)
Problem: A step-down transformer converts 11000 V to 220 V. The primary coil has 5000 turns. The secondary coil supplies a current of 10 A to an external circuit. Find: (a) the number of turns in the secondary coil, and (b) the current drawn from the primary supply, assuming 100% efficiency.
Given: \( V_p = 11000 \) V, \( V_s = 220 \) V, \( N_p = 5000 \), \( I_s = 10 \) A
Step 1: Find \( N_s \) using the voltage-turns ratio:
\( \frac{N_s}{N_p} = \frac{V_s}{V_p} \Rightarrow N_s = N_p \times \frac{V_s}{V_p} = 5000 \times \frac{220}{11000} = 100 \) turns
Step 2: Find \( I_p \) using the power conservation equation \( V_p I_p = V_s I_s \):
\( I_p = \frac{V_s I_s}{V_p} = \frac{220 \times 10}{11000} = \frac{2200}{11000} = 0.2 \) A
Verification: Current ratio check: \( \frac{I_s}{I_p} = \frac{10}{0.2} = 50 = \frac{N_p}{N_s} = \frac{5000}{100} = 50 \) ✓
Answer
(a) Number of secondary turns \( N_s = 100 \)
(b) Primary current \( I_p = 0.2 \) A
Example 3 (JEE/NEET Level)
Problem: A transformer with efficiency 90% operates from a 200 V, 50 Hz AC supply. It is required to deliver power to a load at 400 V. The primary coil has 1000 turns and draws a current of 5 A. Calculate: (a) the number of turns in the secondary, (b) the output power, and (c) the current delivered to the load.
Given: \( \eta = 90\% = 0.9 \), \( V_p = 200 \) V, \( V_s = 400 \) V, \( N_p = 1000 \), \( I_p = 5 \) A
Step 1: Find \( N_s \) using the voltage-turns ratio:
\( N_s = N_p \times \frac{V_s}{V_p} = 1000 \times \frac{400}{200} = 2000 \) turns
Step 2: Calculate input power:
\( P_{in} = V_p \times I_p = 200 \times 5 = 1000 \) W
Step 3: Calculate output power using efficiency:
\( P_{out} = \eta \times P_{in} = 0.9 \times 1000 = 900 \) W
Step 4: Calculate secondary current from output power:
\( I_s = \frac{P_{out}}{V_s} = \frac{900}{400} = 2.25 \) A
Note: For a real transformer with \( \eta < 100\% \), \( I_s \neq \frac{N_p}{N_s} I_p \). The current ratio formula applies only to ideal transformers. Always use the power equation for efficiency-based problems.
Answer
(a) \( N_s = 2000 \) turns
(b) Output power \( P_{out} = 900 \) W
(c) Secondary current \( I_s = 2.25 \) A
CBSE Exam Tips 2025-26
- Know the 3-part ratio: The combined relation \( \frac{V_s}{V_p} = \frac{N_s}{N_p} = \frac{I_p}{I_s} \) is frequently tested. Write all three parts together in your answer for full marks.
- State assumptions for ideal transformers: In CBSE board exams, always mention that the derivation assumes 100% efficiency, negligible flux leakage, and no resistance in coils. This earns step marks.
- Identify step-up vs. step-down quickly: If \( N_s > N_p \), it is a step-up transformer. If \( N_s < N_p \), it is step-down. State this clearly in your solution.
- Power transmission questions: The 2025-26 CBSE syllabus emphasises why transformers are used in long-distance power transmission. Remember: stepping up voltage reduces current, which reduces \( I^2 R \) losses in transmission lines.
- Use the efficiency formula carefully: For real transformers, \( \eta = \frac{P_{out}}{P_{in}} \times 100\% \). We recommend writing out \( P_{in} \) and \( P_{out} \) separately before applying efficiency.
- 5-mark derivation: The derivation of the transformer formula is a standard 5-mark question in CBSE boards. Practise all six steps from Faraday’s law through to the final ratio. Include a neat labelled diagram of the transformer for an extra mark.
Common Mistakes to Avoid
- Inverting the current ratio: Many students write \( \frac{I_s}{I_p} = \frac{N_s}{N_p} \), which is incorrect. The correct relation is \( \frac{I_s}{I_p} = \frac{N_p}{N_s} \). Current is inversely proportional to the number of turns, not directly proportional.
- Applying the turns ratio to current in real transformers: The current ratio \( I_s / I_p = N_p / N_s \) holds only for ideal (100% efficient) transformers. For real transformers, always use \( P_{out} = \eta \times P_{in} \) to find the output current.
- Using the transformer formula for DC: Transformers work only on AC. They do not work with DC because a steady current produces no changing flux and therefore no induced EMF. Do not apply the Transformer Formula to DC circuits.
- Confusing primary and secondary: The primary coil is connected to the input (supply) voltage. The secondary coil is connected to the output (load). Mixing up \( V_p \) and \( V_s \) is a very common error in numerical problems.
- Forgetting units in efficiency: Efficiency \( \eta \) is dimensionless or expressed as a percentage. Do not attach units like Watts to it. Always express it as a decimal (0.9) when using in equations, and as a percentage (90%) in the final answer.
JEE/NEET Application of the Transformer Formula
In our experience, JEE aspirants encounter the Transformer Formula in approximately 1-2 questions per year in JEE Main, typically in the Alternating Current chapter. NEET also tests this concept at a conceptual and numerical level. Here are the three most common exam-level application patterns:
Pattern 1: Multi-step Efficiency Problems (JEE Main)
JEE Main frequently combines efficiency with the transformer formula. A typical problem gives the input power, efficiency, and one of the output quantities (voltage or current) and asks you to find the other. The key step is to compute \( P_{out} = \eta \times P_{in} \) first. Then use \( P_{out} = V_s I_s \) to find the unknown. Do not use the simple turns ratio for current in these problems.
Pattern 2: Power Transmission Line Problems (JEE Advanced / CBSE)
These problems involve a power plant generating power at a low voltage, a step-up transformer, a long transmission line with resistance \( R \), and a step-down transformer at the consumer end. You must calculate the power loss \( P_{loss} = I_{line}^2 R \), where \( I_{line} \) is the current in the high-voltage transmission line. The Transformer Formula gives you \( I_{line} \) from the step-up transformer’s secondary. This is a classic multi-concept problem that combines the Transformer Formula with \( P = I^2 R \).
Pattern 3: Conceptual MCQs (NEET)
NEET tests conceptual understanding of transformers. Common MCQ themes include: why transformers do not work on DC, the effect of increasing primary turns on secondary voltage, and which type of transformer (step-up or step-down) is used at a power generating station. For NEET, memorise that step-up transformers are used at generating stations (to reduce transmission losses) and step-down transformers are used near consumer areas (to bring voltage to safe levels like 220 V).
FAQs on Transformer Formula
Explore more related Physics formulas on ncertbooks.net. Visit our complete Physics Formulas hub for a full list of Class 12 topics. You may also find these articles useful: Photon Energy Formula for understanding electromagnetic radiation, Heat Transfer Formula for thermal energy concepts, and Wave Power Formula for wave mechanics. For the official NCERT Class 12 Physics syllabus, refer to the NCERT official website.