Transformers & Mutual Inductance - Physics 2

1. What is the RMS current drawn from the primary source? (Use $\omega = 2\pi(60) \approx 377 \text{ rad/s}$, giving $\omega L \approx 10 \ \Omega$.)
   Answer: $I_{1,\text{rms}} \approx 8.49 \text{ A}$, because the secondary voltage is $240 \text{ V}$ RMS, the secondary impedance is $|Z_2| = \sqrt{(10)^2+(10)^2} = 10\sqrt{2} \ \Omega$, giving $I_2 \approx 16.97 \text{ A}$, and the correct transformer current relation $I_1 N_1 = I_2 N_2$ gives $I_1 = (N_1/N_2)I_2 = (1/2)(16.97) \approx 8.49 \text{ A}$. Transformer problems test two key relationships: the voltage ratio and the current ratio. These are not the same ratio — they go in opposite directions. For a turns ratio $N_1 : N_2$, voltage steps up by $N_2/N_1$, while current steps down by the same factor (because an ideal transformer conserves power: $V_1 I_1 = V_2 I_2$). Here's the correct approach. The secondary voltage is $V_2 = V_1 \cdot (N_2/N_1) = 120 \times 2 = 240 \text{ V RMS}$. The secondary load has $R = 10\ \Omega$ and $X_L = \omega L \approx 10\ \Omega$, so $|Z_2| = \sqrt{10^2 + 10^2} = 10\sqrt{2}\ \Omega$. The secondary current is $I_2 = 240/(10\sqrt{2}) \approx 16.97 \text{ A}$. Now apply the transformer current relation: $I_1 N_1 = I_2 N_2$, which gives $I_1 = I_2 \cdot (N_2/N_1) \cdot (N_1/N_1)$... more simply, $I_1 = (N_2/N_1) \cdot I_2$ is wrong — the correct form is $I_1 = (N_2/N_1) \cdot I_2$ reversed: $I_1 = (1/2)(16.97) \approx 8.49 \text{ A}$. This is answer D. Choice A incorrectly multiplies $I_2$ by 2 (the voltage ratio) instead of dividing — confusing the voltage and current ratios. Choice B claims current is unchanged in a step-up transformer, which violates power conservation entirely. Choice C correctly sets up the referred-impedance method but then computes $I_1 = V_1/Z_\text{ref}$ and gets 33.9 A, which is actually the same error as A — it divided the impedance by 4 correctly, but $120/(10\sqrt{2}/4) = 120 \times 4/(10\sqrt{2}) = 33.9$ — an arithmetic inconsistency with its own stated logic. Remember: In a step-up transformer, primary current is larger than secondary current — the source supplies more current at lower voltage. Always use $I_1/I_2 = N_2/N_1$, not the voltage ratio.
2. A student claims that the mutual inductance $M$ computed using coil B as the 'primary' (flux from B linking A) must equal $M$ computed using solenoid A as the 'primary' (flux from A linking B), but also argues that the numerical computation is much harder when B is treated as the primary because coil B's field is non-uniform inside A. Which of the following best evaluates both aspects of this claim?
   Answer: Both aspects are correct: $M_{AB} = M_{BA}$ is guaranteed by the Neumann formula (a fundamental theorem), and computing $M$ with B as primary is indeed significantly harder because the short coil B produces a highly non-uniform field that does not uniformly link all turns of the long solenoid A. Whenever you encounter questions about mutual inductance, two key ideas should come to mind: the reciprocity theorem and the practical asymmetry of computation. The reciprocity theorem, derived from the Neumann formula, states that $M_{AB} = M_{BA}$ universally — it follows from the mathematical symmetry of the double line integral $M = \frac{\mu_0}{4\pi} \oint \oint \frac{d\vec{\ell}_A \cdot d\vec{\ell}_B}{r}$, which treats both coils identically. This is not a conditional result; it holds regardless of geometry, coil size, or turns count. So the student's first claim is rock solid. The second claim is equally valid. When solenoid A is the primary, its field inside is beautifully uniform: $B_A = \mu_0 n_A I_A$, so the flux through each turn of coil B is simply $\Phi = B_A \cdot \mathcal{A}$, giving $M = \mu_0 n_A N_B \mathcal{A} = (4\pi \times 10^{-7})(2000)(50)(5\times10^{-4}) \approx 6.28 \times 10^{-5} \text{ H}$. Clean and simple. But if coil B is the primary, its short-coil field spreads non-uniformly over the entire length of solenoid A, requiring a complicated integral — much harder numerically. Both aspects of the student's claim are correct, making D the answer. A is wrong because it dismisses the computational difficulty — yes, the Neumann formula is symmetric in principle, but applying it numerically is far easier from one direction than the other. B is wrong because it incorrectly limits reciprocity to special media; the theorem is universal. C is wrong on both counts — reciprocity holds generally, and fewer turns does not make the non-uniform field integral simpler. Your study tip: always default to the long solenoid as primary when computing $M$ — uniform field means trivial flux calculation. Reciprocity then guarantees the answer applies both ways.
3. What is the RMS power dissipated in the external load $R_L$, and how does it compare to the power dissipated in the secondary winding resistance $r_2$?
   Answer: Power in $R_L$ is $92.16 \text{ W}$ and power in $r_2$ is $23.04 \text{ W}$, so the load receives four times the power lost in the secondary winding resistance, consistent with the resistance ratio $R_L/r_2 = 4$. When a transformer has a resistive secondary winding, you must treat the secondary circuit as a simple series loop: the induced secondary EMF drives current through both the winding resistance $r_2$ and the load $R_L$ in series. Start by finding the secondary voltage using the turns ratio: $V_2 = V_1 \times \frac{N_2}{N_1} = 240 \times \frac{1}{10} = 24 \text{ V RMS}$. This 24 V is the open-circuit secondary EMF — the voltage the ideal transformer delivers before internal losses. The secondary current is then $I_2 = \frac{V_2}{r_2 + R_L} = \frac{24}{1 + 4} = 4.8 \text{ A RMS}$. Power in the load: $P_{R_L} = I_2^2 \cdot R_L = (4.8)^2 \times 4 = 92.16 \text{ W}$. Power lost in the winding: $P_{r_2} = I_2^2 \cdot r_2 = (4.8)^2 \times 1 = 23.04 \text{ W}$. Since the same current flows through both, power divides proportionally to resistance, giving a 4:1 ratio — exactly $R_L / r_2 = 4$. This confirms A is correct. B is wrong because "referring" $r_2$ to the primary is a circuit analysis technique for simplification — it doesn't mean $r_2$ physically stops dissipating power. The heat in $r_2$ is real regardless of which side of the circuit you model it on. C is wrong because $r_2 \neq R_L$; equal power would require equal resistances. The voltage doesn't split evenly here — it splits as 4.8 V vs. 19.2 V. D is wrong because winding resistance is purely ohmic; it does not create "core flux losses" or cause the primary to supply extra energy beyond what the secondary circuit actually consumes. Study tip: Whenever you see transformer problems with non-ideal windings, immediately redraw the secondary as a plain series circuit with $V_2$, $r_2$, and $R_L$ — then apply Ohm's law and $P = I^2 R$ normally.
4. Two ideal inductors with self-inductances $L_1 = 40 \text{ mH}$ and $L_2 = 90 \text{ mH}$ are wound on the same toroidal core such that all flux from one coil links the other. They are connected in series-aiding (fluxes add). Which of the following correctly gives the total inductance of the series combination and the mutual inductance $M$?
   Answer: $M = 60 \text{ mH}$ and $L_{\text{total}} = 250 \text{ mH}$, because for perfect coupling $M = \sqrt{L_1 L_2}$ and series-aiding total inductance is $L_1 + L_2 + 2M$. When two inductors share a magnetic core with perfect coupling, two key formulas govern their behavior. First, the mutual inductance is $M = k\sqrt{L_1 L_2}$, where $k$ is the coupling coefficient. "All flux links the other coil" means perfect coupling, so $k = 1$, giving $M = \sqrt{(40)(90)} = \sqrt{3600} = 60 \text{ mH}$. Second, when inductors are connected series-aiding (currents enter the dotted terminals, so fluxes reinforce), the total inductance is $L_{\text{total}} = L_1 + L_2 + 2M$. The factor of 2 arises because mutual coupling adds voltage to both inductors simultaneously — inductor 1 sees extra EMF from inductor 2's changing current, and inductor 2 sees extra EMF from inductor 1's changing current. This gives $L_{\text{total}} = 40 + 90 + 2(60) = 250 \text{ mH}$, confirming answer A is correct. Answer B gets $M$ right but uses $L_1 + L_2 + M$ — a classic error of counting only one mutual term. Mutual coupling is bilateral; both coils experience the effect, so you must count it twice. Answer C uses the wrong formula for $M$, incorrectly averaging the self-inductances instead of taking their geometric mean, which inflates $M$ to 65 mH. Answer D correctly computes $M$ but then ignores the mutual terms entirely, as if the coils were magnetically isolated — the whole point of mutual inductance is that it changes the total. Remember the series-aiding formula as $L_{\text{total}} = L_1 + L_2 + 2M$ (aiding adds, opposing subtracts) and $M = k\sqrt{L_1 L_2}$ for coupling — these two formulas together solve nearly every coupled-inductor problem.
5. Two coils, 1 and 2, are placed near each other. The mutual inductance of the pair is $M = 50 \text{ mH}$. Coil 1 carries a current that varies as $i_1(t) = I_0 \sin(\omega t)$ where $I_0 = 2 \text{ A}$ and $\omega = 100\pi \text{ rad/s}$. Coil 2 is open-circuited (no current flows in it). What is the peak magnitude of the EMF induced in coil 2, and what is the peak magnitude of the back-EMF induced in coil 1 due to mutual inductance with coil 2?
   Answer: Peak EMF in coil 2 is $10\pi \text{ V}$; peak back-EMF in coil 1 due to mutual coupling is zero, because coil 2 is open-circuited so $i_2 = 0$ for all time, meaning $di_2/dt = 0$, and the mutually-induced EMF in coil 1 is $-M\,di_2/dt = 0$. Mutual inductance questions require you to carefully track which coil's current is changing and apply the correct form of Faraday's law to each coil independently. The EMF induced in coil 2 by coil 1's changing current is $\mathcal{E}_2 = -M\frac{di_1}{dt}$. Since $i_1 = I_0\sin(\omega t)$, we get $\frac{di_1}{dt} = I_0\omega\cos(\omega t)$, so the peak EMF in coil 2 is $MI_0\omega = (0.050)(2)(100\pi) = 10\pi \text{ V}$. Now, the back-EMF in coil 1 due to mutual coupling is $\mathcal{E}_{1,\text{mutual}} = -M\frac{di_2}{dt}$. Because coil 2 is open-circuited, no current can flow: $i_2 = 0$ for all time, so $di_2/dt = 0$, and this mutual back-EMF is exactly zero. Answer B captures both results correctly. Answer A is tempting because the reciprocity relation $M_{12} = M_{21}$ is real — but reciprocity tells you the same M governs both coils, not that both experience the same EMF regardless of current. The EMF in coil 1 depends on $di_2/dt$, not $di_1/dt$. Answer C incorrectly imagines the mutual EMF being "split" between the coils — there is no such sharing principle. Each coil gets its own Faraday expression. Answer D imports a transmission-line concept (open-circuit voltage doubling) into a lumped-circuit context where it simply doesn't apply; no such reflection mechanism exists here. Your strategy: always write out both mutual EMF equations, $\mathcal{E}_2 = -M\frac{di_1}{dt}$ and $\mathcal{E}_1 = -M\frac{di_2}{dt}$, separately. Then substitute what you actually know about each current.
6. What is the RMS current delivered to the load and the RMS current flowing through the common (shared) section of the winding (the lower 750 turns), and how does the current in the common section differ from the load current?
   Answer: Load current is $3.75 \text{ A}$, and the common section carries $0.9375 \text{ A}$, because applying KCL at the tap node shows that the common section current equals the difference between the load current and the source input current: $I_{\text{common}} = I_L - I_1 = 3.75 - 2.8125 = 0.9375 \text{ A}$. Whenever you see an autotransformer problem, think of it as a KCL problem at the tap node — current doesn't just pass through; it splits and recombines in ways that differ from a conventional two-winding transformer. Start with the tap voltage. Since the winding is uniform, the voltage across the lower 750 turns is $V_2 = 200 \times \frac{750}{1000} = 150 \text{ V}$. The load current is then $I_L = \frac{V_2}{R_L} = \frac{150}{40} = 3.75 \text{ A}$. Now apply conservation of apparent power (ideal transformer) to find the source current: $I_1 = \frac{V_2}{V_1} I_L = \frac{150}{200} \times 3.75 = 2.8125 \text{ A}$. At the tap node, KCL requires that the currents balance: the source delivers $I_1$ downward through the full winding, and the load draws $I_L$ from the tap. The common (lower) section must carry the difference: $I_{\text{common}} = I_L - I_1 = 3.75 - 2.8125 = 0.9375 \text{ A}$. This is answer C. A is wrong because it ignores KCL entirely — in an autotransformer, source and load currents both flow through the common section but in opposite directions, so they partially cancel, not add. B correctly finds $I_1$ and $I_L$ but then incorrectly assumes $I_1$ is what flows through the common section, confusing where the source current goes relative to the tap node. D inverts the turns ratio, applying it as a step-up when the tap clearly gives a lower fraction of the source voltage. The key study tip: always draw the tap node and write KCL explicitly — $I_{\text{common}} = I_L - I_1$ is the defining relationship of the shared winding section in a step-down autotransformer.
7. By what factor does the fractional power loss (power lost in the line divided by total power transmitted) decrease when switching from scenario A to scenario B, and what turns ratio $N_1 : N_2$ is needed at the step-up transformer if the generator produces $10 \text{ kV}$ RMS?
   Answer: The fractional power loss decreases by a factor of 100, and the required step-up turns ratio is $N_1:N_2 = 1:10$, because increasing the voltage by a factor of 10 reduces the current by a factor of 10, and since $P_{\text{loss}} \propto I^2$, the loss decreases by a factor of $10^2 = 100$. Whenever you see a question about high-voltage power transmission, anchor your thinking to two key relationships: $P = IV$ and $P_{\text{loss}} = I^2 R$. These two together reveal why voltage matters so much. For a fixed transmitted power $P$, the current in the line is $I = P/V$. In scenario A, $I_A = 1\text{ MW}/10\text{ kV} = 100\text{ A}$, giving a line loss of $P_{\text{loss},A} = (100)^2 \times 10 = 100\text{ kW}$, or a 10% fractional loss. In scenario B, $I_B = 1\text{ MW}/100\text{ kV} = 10\text{ A}$, so $P_{\text{loss},B} = (10)^2 \times 10 = 1\text{ kW}$, a 0.1% fractional loss. The ratio is $100\text{ kW}/1\text{ kW} = 100$. Because voltage increased by a factor of 10 and loss scales as $I^2 \propto 1/V^2$, the fractional loss drops by $10^2 = 100$. To step from $10\text{ kV}$ up to $100\text{ kV}$, you need $N_1:N_2 = 1:10$ — fewer primary turns than secondary turns. This confirms B is correct. Choice A gets the turns ratio right but claims only a factor-of-10 reduction, confusing $I^2R$ scaling with linear $IR$ scaling. Choice C correctly identifies the factor-of-100 reduction but inverts the turns ratio — a step-*up* transformer needs more secondary turns, not more primary. Choice D makes both errors: it assumes loss is linear in current and inverts the transformer ratio. Study tip: Always remember the "double penalty" of low-voltage transmission — higher current hits you twice through $I^2R$. When voltage increases by factor $n$, power loss drops by $n^2$.
8. A transformer manufacturer specifies a core material with relative permeability $\mu_r = 5000$. An engineer proposes replacing it with a material of $\mu_r = 2500$ while keeping all winding geometries identical. Assuming the transformer remains ideal (no core losses, complete flux linkage), which of the following correctly predicts the effect on the mutual inductance $M$ and the turns-ratio voltage relationship $V_2/V_1 = N_2/N_1$?
   Answer: $M$ decreases by a factor of two because $M \propto \mu_r$, but the turns-ratio voltage relationship $V_2/V_1 = N_2/N_1$ remains valid because it follows from Faraday's law applied to a shared flux — the ratio $N_2 d\Phi/dt$ to $N_1 d\Phi/dt$ is independent of the absolute permeability as long as coupling is perfect. Transformer problems like this require you to separate two distinct questions: how does core permeability affect inductance values? and does the fundamental turns-ratio voltage law still hold? Keep those two questions independent in your mind. For a toroidal or ideal core geometry, mutual inductance scales as $M = k\sqrt{L_1 L_2}$, where each self-inductance goes as $L \propto \mu_r N^2 A / \ell$. Since both $L_1$ and $L_2$ scale linearly with $\mu_r$, mutual inductance follows $M \propto \mu_r$. Cutting $\mu_r$ in half — from 5000 to 2500 — therefore cuts $M$ by exactly a factor of two. That confirms the first part of B. The turns-ratio voltage relationship $V_2/V_1 = N_2/N_1$ comes directly from Faraday's law: $V_1 = N_1 \frac{d\Phi}{dt}$ and $V_2 = N_2 \frac{d\Phi}{dt}$. When you divide them, $d\Phi/dt$ cancels completely. The ratio depends only on the winding turns, not on how large $\Phi$ actually is. So as long as coupling remains perfect (ideal transformer assumption), the voltage ratio is preserved regardless of permeability. B captures both truths correctly. A is wrong because it incorrectly claims the voltage ratio changes with $\mu_r$ — the $d\Phi/dt$ cancellation disproves this. C is wrong because it denies that $M$ depends on permeability at all, ignoring that $L \propto \mu_r$. D is wrong because it argues a quadratic dependence $M \propto \mu_r^2$; the "flux linking the secondary" is the same flux as produced by the primary — it doesn't scale with $\mu_r$ a second time. Study tip: On transformer questions, always split your analysis — inductance magnitudes depend on core properties, but voltage ratios in an ideal transformer are purely geometric (turns counts). These are governed by different physics.
9. Two ideal inductors with self-inductances $L_1 = 100 \text{ mH}$ and $L_2 = 400 \text{ mH}$ have mutual inductance $M = 100 \text{ mH}$. They are connected in series-opposing (fluxes partially cancel). A student calculates the coupling coefficient as $k = M/\sqrt{L_1 L_2} = 0.5$ and then claims that because $k < 1$, the series-opposing total inductance must be positive (greater than zero). Is the student's conclusion correct, and what is the actual series-opposing total inductance?
   Answer: The conclusion is correct; the total inductance is $L_{\text{opp}} = L_1 + L_2 - 2M = 300 \text{ mH}$, which is indeed positive, consistent with $k = 0.5 < 1$ preventing complete cancellation of the self-inductance terms. When two inductors are connected in series, the total inductance depends on whether their magnetic fluxes aid or oppose each other. For series-aiding, you add the mutual terms: $L_{\text{aid}} = L_1 + L_2 + 2M$. For series-opposing, both mutual terms subtract: $L_{\text{opp}} = L_1 + L_2 - 2M$. The factor of 2 appears because each inductor's flux cuts through the other — there are two cross-coupling contributions, both reversed in the opposing case. Plugging in the given values: $L_{\text{opp}} = 100 + 400 - 2(100) = 300 \text{ mH}$. This is positive, and the student's conclusion (that it must be positive) happens to be correct — but for the right reason. The coupling coefficient $k = M/\sqrt{L_1 L_2} = 100/\sqrt{100 \times 400} = 0.5$ guarantees $2M < 2\sqrt{L_1 L_2}$, and by the AM-GM inequality, $L_1 + L_2 \geq 2\sqrt{L_1 L_2} > 2M$, so the result is always non-negative when $k \leq 1$. This confirms C is correct. A is wrong because it invents a formula using $|L_1 - L_2|$ that has no basis in circuit theory — you always sum both self-inductances, never subtract them from each other. B is wrong because it uses only a single mutual term ($-M$) instead of $-2M$. Both cross-coupling interactions reverse sign in the opposing configuration, not just one. D is a tempting distractor — it gets the arithmetic right (300 mH) but falsely claims $k < 1$ doesn't guarantee a positive result. In fact, it mathematically does, for any passive inductor pair. Study tip: Always remember the $2M$ factor in series inductor formulas — dropping one of the mutual terms is the most common error on coupled-inductor problems.
10. If the AC source frequency is doubled while keeping the peak voltage $V_0$ constant, which of the following correctly describes what happens to the secondary RMS current and the power delivered to the load?
    Answer: The secondary RMS current remains unchanged and the power delivered to the load remains unchanged, because for an ideal transformer neither quantity depends on frequency — only on the turns ratio and applied voltage. The flux amplitude adjusts inversely with frequency so that the product $\omega \Phi_0$ remains constant. Whenever you see a transformer question involving frequency changes, anchor yourself to one key equation: the transformer turns ratio $\frac{V_2}{V_1} = \frac{N_2}{N_1}$. For an ideal transformer, this relationship depends only on the turns ratio — not on frequency. Here's the deeper physics: the primary voltage is fixed at $V_0 = 1000\text{ V}$, so the secondary voltage is always $V_2 = V_1 \cdot \frac{N_2}{N_1} = 1000 \cdot \frac{100}{500} = 200\text{ V}$, regardless of frequency. What actually adjusts internally is the flux amplitude $\Phi_0$: since $V_1 = N_1 \omega \Phi_0$, doubling $\omega$ cuts $\Phi_0$ in half, keeping the product $\omega \Phi_0$ — and therefore the induced voltage — constant. With $V_2$ unchanged, the secondary RMS current $I_2 = \frac{V_{2,\text{rms}}}{R}$ stays the same, and so does the power $P = \frac{V_{2,\text{rms}}^2}{R}$. Answer A is correct. Answer B misunderstands what "rate of change of flux" means in this context. Doubling frequency also halves flux amplitude, so the EMF — which depends on $\omega \Phi_0$ — doesn't change. Answer C describes behavior relevant to a real transformer with significant primary inductance, but for an ideal transformer, there is no meaningful "magnetizing current" limiting flux linkage — the primary voltage fully determines the secondary voltage. Answer D confuses energy-per-cycle with power; if each cycle delivers the same energy but you have twice as many cycles, total power is unchanged because energy-per-cycle is halved when voltage is constant. Your takeaway: for ideal transformers, frequency is irrelevant — only the turns ratio and applied voltage matter. When frequency changes, flux amplitude compensates silently.

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