Mutually coupled circuits and ideal transformer
Mutual inductance, coupling coefficient, dot convention, coupled-coil mesh equations, reflected impedance and the ideal transformer.
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Why it matters
Magnetic coupling is how LVDTs, current transformers, potential transformers, isolation amplifiers and resolvers work. It is also how unwanted crosstalk enters signal cables. To analyse these you need the dot convention, mutual inductance in mesh equations, and the ideal-transformer model for impedance matching and isolation.
Key ideas
Mutual inductance. When the changing current in coil 1 sets up flux that links coil 2, a voltage v2 = M·di1/dt is induced in coil 2. The constant M (in henries) is the mutual inductance, and it is the same in both directions (M12 = M21) for linear magnetic media.
Coefficient of coupling. k = M/√(L1·L2), with 0 ≤ k ≤ 1.
- k = 1 means perfect coupling (no leakage flux).
- Air-cored coils typically have k < 0.5; iron-cored transformers have k close to 1.
Dot convention. Dots mark terminals of the same instantaneous polarity.
- If a current enters the dotted terminal of one coil, the mutual voltage it induces in the other coil is positive at that coil's dotted terminal.
- Equivalently: if both currents enter (or both leave) their dotted terminals, the mutual term has the same sign as the self term (+jωM). Otherwise it is −jωM.
Coupled-coil equations (phasor form).
V1 = jωL1·I1 ± jωM·I2V2 = ±jωM·I1 + jωL2·I2Use mesh analysis with these terms added; nodal analysis is awkward with coupled coils.
Series connection.
- Series aiding:
L = L1 + L2 + 2M. - Series opposing:
L = L1 + L2 − 2M. Measuring both givesM = (L_aid − L_opp)/4, a standard laboratory method.
Parallel connection.
- Aiding:
L = (L1L2 − M²)/(L1 + L2 − 2M). - Opposing:
L = (L1L2 − M²)/(L1 + L2 + 2M).
Energy. The energy stored in coupled coils is w = ½L1i1² + ½L2i2² ± M·i1·i2. Requiring w ≥ 0 for all currents gives M ≤ √(L1L2), so k ≤ 1.
Reflected impedance. With a load ZL on the secondary, the input impedance is Zin = Z1 + (ωM)²/(Z2 + ZL), where Z1 = R1 + jωL1 and Z2 = R2 + jωL2. The sign of M does not matter here. A secondary inductive load reflects as capacitive, and vice versa.
Ideal transformer. Assume k = 1, no losses, and L1, L2 → ∞ with L2/L1 fixed. Then, with turns ratio a = N1/N2:
V1/V2 = N1/N2 = aI1/I2 = N2/N1 = 1/a, both for dot-consistent referencesV1·I1 = V2·I2, so power in equals power outZin = a²·ZL
A real transformer cannot pass DC (it needs changing flux), even though the ideal model's equations would. Transformers provide galvanic isolation and are used to match a load to a source for maximum power transfer.
Linear (real) transformer models. These include leakage inductances, magnetising inductance and winding resistances. They are the subject of electrical machines; in network analysis we use coupled coils or the ideal model.
Formulas
v2 = M·di1/dt;k = M/√(L1·L2).- Series:
L = L1 + L2 ± 2M;M = (L_aid − L_opp)/4(H). - Phasor equations:
V1 = jωL1·I1 ± jωM·I2,V2 = ±jωM·I1 + jωL2·I2. - Reflected impedance:
Zr = (ωM)²/(Z2 + ZL)(Ω). - Energy:
w = ½L1i1² + ½L2i2² ± M·i1·i2(J). - Ideal transformer:
V1/V2 = N1/N2,I1/I2 = N2/N1,Zin = (N1/N2)²·ZL. - Matching:
N1/N2 = √(Rs/RL).
Worked examples
Example 1 (standard): measuring M. Given: two coils in series give 0.8 H when aiding and 0.2 H when opposing. One coil is known to have L1 = 0.3 H. Find M, L2 and k.
M = (0.8 − 0.2)/4 = 0.15 H.L1 + L2 = (0.8 + 0.2)/2 = 0.5 H, soL2 = 0.2 H.k = 0.15/√(0.3 × 0.2) = 0.15/0.2449 = 0.612.
Answer: M = 0.15 H, L2 = 0.2 H, k = 0.61.
Example 2 (GATE level): coupled coils feeding a load. Given: V1 = 100∠0° V (RMS) drives coil 1 with jωL1 = j10 Ω. Coil 2 has jωL2 = j20 Ω and feeds RL = 10 Ω. The mutual reactance is jωM = j8 Ω. I1 enters the dotted terminal of coil 1, and I2 leaves the dotted terminal of coil 2 into the load. Find I1, I2 and the load power.
- Mesh 1:
100 = j10·I1 − j8·I2. - Mesh 2:
0 = −j8·I1 + (10 + j20)·I2. - Reflected impedance:
(ωM)²/(10 + j20) = 64/(10 + j20) = 1.28 − j2.56 Ω. - Input impedance:
Zin = j10 + 1.28 − j2.56 = 1.28 + j7.44 Ω. I1 = 100/(1.28 + j7.44) = 13.25∠−80.2° A.- From mesh 2:
I2 = j8·I1/(10 + j20) = 4.74∠−53.7° A. - Load power:
P = |I2|²·RL = 4.739² × 10 = 224.6 W. - Check:
Re{V1·I1*} = 100 × 13.25 × cos 80.2° = 224.6 W✓. The coils store energy but dissipate none.
Answer: I1 = 13.25∠−80.2° A, I2 = 4.74∠−53.7° A, P = 224.6 W.
Example 3 (impedance matching). Given: an amplifier with Rs = 800 Ω drives an 8 Ω speaker through an ideal transformer.
- For maximum power,
N1/N2 = √(800/8) = 10. - The speaker then appears as
10² × 8 = 800 Ω, matched to the source.
Common mistakes
- Getting the sign of the mutual term wrong. Check whether each current enters or leaves its dot.
- Writing
M = k(L1 + L2)instead ofk·√(L1L2). - Inverting the impedance ratio. The impedance seen at the primary is
(N1/N2)²·ZL, not(N2/N1)²·ZL. - Saying an ideal transformer steps up power. Voltage goes up and current goes down by the same ratio.
- Applying ideal-transformer relations to loosely coupled air-core coils (k much less than 1).
- Ambiguous "1:5" ratios. Always state which is N1 and which is N2.
For GATE IN
Typical questions: equivalent inductance of series or parallel coupled coils; finding M or k from aiding and opposing measurements; mesh equations with dot convention; reflected impedance; ideal-transformer voltage, current and impedance transformation; the turns ratio for maximum power transfer. Practise writing the coupled mesh equations carefully, since sign errors are the main trap.
Quick check
- L1 = 4 H, L2 = 9 H and k = 0.5. What is M?
- What is the series-aiding inductance of the coils in question 1?
- An ideal 5:1 step-down transformer has a 4 Ω load. What resistance is seen at the primary?
- What is the maximum possible value of k?
Answers: 1. 3 H. 2. 4 + 9 + 6 = 19 H. 3. 100 Ω. 4. 1.
Interview questions
All Electrical Circuits interview questionsTry answering each one aloud before you open it.
1.What is mutual inductance in electrical circuits?Concept
Mutual inductance is a measure of the ability of one coil to induce an electromotive force (EMF) in another coil when the current in the first coil changes. It is denoted by the symbol M and is measured in henries (H). The mutual inductance depends on the number of turns in the coils, the area of the coils, the distance between them, and the permeability of the medium between the coils.
2.Explain the working principle of an ideal transformer.Concept
An ideal transformer operates on the principle of electromagnetic induction. It consists of two coils, the primary and the secondary, wound on a common core. When an alternating current flows through the primary coil, it creates a changing magnetic field, which induces a voltage in the secondary coil. The voltage induced in the secondary coil is proportional to the ratio of the number of turns in the secondary coil to the number of turns in the primary coil.
3.How does the turns ratio affect the voltage and current in a transformer?Concept
The turns ratio of a transformer is the ratio of the number of turns in the primary coil to the number of turns in the secondary coil. It determines the voltage transformation between the primary and secondary coils. If the turns ratio is greater than one, the transformer is a step-down transformer, reducing voltage and increasing current. Conversely, if the turns ratio is less than one, it is a step-up transformer, increasing voltage and reducing current.
4.Why is an ideal transformer considered lossless?Concept
An ideal transformer is considered lossless because it assumes no energy losses in the form of heat, magnetic leakage, or resistance. This means that the power input to the primary coil is equal to the power output from the secondary coil. In reality, transformers have some losses, but the ideal transformer model simplifies analysis by ignoring these losses.
5.What happens if the secondary coil of a transformer is open-circuited?Application
If the secondary coil of a transformer is open-circuited, no current flows through it. However, the primary coil still draws a small current known as the magnetizing current, which is necessary to maintain the magnetic field in the core. The voltage across the secondary coil will be at its maximum value, determined by the turns ratio, but no power is transferred to a load.
6.Why are transformers used in power distribution systems?Application
Transformers are used in power distribution systems to efficiently transmit electrical power over long distances. By stepping up the voltage at the generation site, transformers reduce the current for a given power level, minimizing resistive losses in the transmission lines. At the distribution end, transformers step down the voltage to safer levels for consumer use.
7.What is the effect of mutual inductance on coupled circuits?Application
Mutual inductance in coupled circuits causes a change in current in one coil to induce a voltage in the other coil. This can lead to energy transfer between the circuits, which is the basis for transformer operation. In some cases, mutual inductance can cause unwanted coupling, leading to interference or crosstalk between circuits.
8.An ideal transformer has N1:N2 = 1:5 and a primary voltage of 240 V. What is the secondary voltage?Numerical
For an ideal transformer V2/V1 = N2/N1. With N1:N2 = 1:5, V2 = 240 × 5 = 1200 V; it is a step-up transformer. The secondary current is one-fifth of the primary current, so the power is the same on both sides. Always state which side each number of the ratio refers to; '1:5' read the other way gives 48 V.
9.A transformer has a primary current of 10 A and a turns ratio of 10:1. What is the secondary current?Numerical
The secondary current (I_s) can be calculated using the formula I_s = I_p × (N_p / N_s), where I_p is the primary current, N_p is the number of turns in the primary coil, and N_s is the number of turns in the secondary coil. Given I_p = 10 A and the turns ratio N_p / N_s = 10/1, the secondary current I_s = 10 A × (10/1) = 100 A.
10.What are the limitations of the ideal transformer model?Application
The ideal transformer model assumes no losses, perfect coupling, and infinite permeability of the core, which are not achievable in real-world transformers. Real transformers experience core losses (hysteresis and eddy currents), copper losses (resistance in the windings), and leakage inductance (imperfect coupling). These factors lead to efficiency less than 100% and affect the performance of the transformer.
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