Physics 2 Quiz: Transformers And Mutual Inductance

10 questions · exam conditions

0:00

Transformers And Mutual InductanceQuestion 1 of 10

A long solenoid (solenoid A) has $n_A = 2000 \text{ turns/m}$ , cross-sectional area $\mathcal{A} = 5 \times 10^{-4} \text{ m}^2$ , and length $\ell = 0.50 \text{ m}$ . A short secondary coil B with $N_B = 50$ turns is wound tightly around the center of solenoid A. The permeability of free space is $\mu_0 = 4\pi \times 10^{-7} \text{ T}\cdot\text{m/A}$ .

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?

The first aspect is correct but the second is incorrect:

M_{AB} = M_{BA}

is guaranteed, and computing

M

with B as primary is equally straightforward because the Neumann formula is symmetric and can be applied with equal ease regardless of which coil is designated the primary.

The first aspect is incorrect and the second is correct:

M_{AB} = M_{BA}

holds only for linear, isotropic media with no eddy currents, so it is not universally guaranteed; and computing

M

with B as primary is harder for the stated reason about field non-uniformity.

Both aspects are incorrect:

M_{AB} \neq M_{BA}

in general when the two coils have very different geometries, and computing

M

with B as primary is actually simpler because fewer turns means fewer flux-linkage terms to evaluate.

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.

← Back to quizzes

Physics 2 Quiz

Physics 2 Quiz: Transformers And Mutual Inductance

Practice Transformers And Mutual Inductance in Physics 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Transformers And Mutual Inductance, giving you a quick way to practice the rules, question types, and explanations that matter most for Physics 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

The first aspect is correct but the second is incorrect: $M_{AB} = M_{BA}$ is guaranteed, and computing $M$ with B as primary is equally straightforward because the Neumann formula is symmetric and can be applied with equal ease regardless of which coil is designated the primary.
The first aspect is incorrect and the second is correct: $M_{AB} = M_{BA}$ holds only for linear, isotropic media with no eddy currents, so it is not universally guaranteed; and computing $M$ with B as primary is harder for the stated reason about field non-uniformity.
Both aspects are incorrect: $M_{AB} \neq M_{BA}$ in general when the two coils have very different geometries, and computing $M$ with B as primary is actually simpler because fewer turns means fewer flux-linkage terms to evaluate.
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. (correct answer)

Explanation: 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.

Question 2

A step-down transformer with a turns ratio $N_1 : N_2 = 10 : 1$ is used to supply power to a resistive load. The primary is connected to a $240 \text{ V}$ RMS AC source. The secondary winding has a non-negligible resistance of $r_2 = 1 \ \Omega$ , but all other transformer components are ideal. The external load resistance is $R_L = 4 \ \Omega$ .

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 $r_2$ ?

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 $R_L/r_2 = 4$ . (correct answer)
Power in $R_L$ is $115.2 \text{ W}$ and power in $r_2$ is $0 \text{ W}$ , because in transformer circuit analysis the secondary winding resistance is referred entirely to the primary side, so it does not appear in the secondary power balance and dissipates no power in the secondary.
Power in $R_L$ is $57.6 \text{ W}$ and power in $r_2$ is $57.6 \text{ W}$ , because the same secondary current flows through both $r_2$ and $R_L$ in series, and since the secondary voltage divides equally between the two resistances, each receives half the total secondary power.
Power in $R_L$ is $92.16 \text{ W}$ and power in $r_2$ is $23.04 \text{ W}$ , but the total secondary power $115.2 \text{ W}$ exceeds what an ideal lossless transformer would deliver because the winding resistance $r_2$ creates additional core flux losses that must be supplied by the primary.

Explanation: 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.

Question 3

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 $M$ ?

$M = 60 \text{ mH}$ and $L_{\text{total}} = 250 \text{ mH}$ , because for perfect coupling $M = \sqrt{L_1 L_2}$ and series-aiding total inductance is $L_1 + L_2 + 2M$ . (correct answer)
$M = 60 \text{ mH}$ and $L_{\text{total}} = 190 \text{ mH}$ , because for perfect coupling $M = \sqrt{L_1 L_2}$ , but series-aiding inductance is $L_1 + L_2 + M$ (only one mutual term since the coupling is unidirectional).
$M = 65 \text{ mH}$ and $L_{\text{total}} = 260 \text{ mH}$ , because for perfect coupling $M = (L_1 + L_2)/2$ and the series-aiding total is $L_1 + L_2 + 2M$ .
$M = 60 \text{ mH}$ and $L_{\text{total}} = 130 \text{ mH}$ , because for perfect coupling $M = \sqrt{L_1 L_2}$ , and series-aiding total inductance equals $L_1 + L_2$ with the mutual terms canceling due to symmetry of the shared core.

Explanation: 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.

Question 4

A power company transmits $P = 1 \text{ MW}$ of power over a transmission line with total resistance $R_{\text{line}} = 10 \ \Omega$ . Two scenarios are compared: (A) power is transmitted at $V_A = 10 \text{ kV}$ RMS, and (B) power is transmitted at $V_B = 100 \text{ kV}$ RMS using a step-up transformer at the source. In both cases, the load at the far end receives all power not lost in the line.

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?

The fractional power loss decreases by a factor of 10, and the required step-up turns ratio is $N_1:N_2 = 1:10$ , because doubling the voltage halves the current but the loss scales as $I^2 R$ , giving a factor-of-10 reduction matching the voltage increase ratio.
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 $10^2 = 100$ . (correct answer)
The fractional power loss decreases by a factor of 100, and the required step-up turns ratio is $N_1:N_2 = 10:1$ , because voltage is stepped up by a factor of 10 (requiring more primary turns than secondary turns) and the $I^2R$ loss scales as the square of the voltage ratio.
The fractional power loss decreases by a factor of 10, and the required step-up turns ratio is $N_1:N_2 = 10:1$ , because the transmission current decreases by 10 and loss is linear in current for a resistive line, while a step-up transformer always has more primary turns than secondary turns.

Explanation: 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

Question 5

An autotransformer is constructed from a single tapped winding. The full winding has $N = 1000$ turns and is connected across a $200 \text{ V}$ RMS AC source. A tap is placed at $N_{\text{tap}} = 750$ turns from the bottom of the winding, and the load $R_L = 40 \ \Omega$ is connected between the tap and the bottom terminal (the common terminal). The winding is assumed ideal and the source is connected across the full winding.

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?

Load current is $3.75 \text{ A}$ , and the common section carries $3.75 \text{ A}$ , because in an autotransformer the shared winding section carries the full load current — no current splitting occurs at the tap node since the source and load currents flow in the same direction through the common winding.
Load current is $3.75 \text{ A}$ , and the common section carries $2.8125 \text{ A}$ , because conservation of apparent power gives a source current $I_1 = (V_2/V_1)I_L = (150/200)(3.75) \text{ A}$ , and this source current flows entirely through the common section of the winding below the tap.
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: $I_{\text{common}} = I_L - I_1 = 3.75 - 2.8125 = 0.9375 \text{ A}$ . (correct answer)
Load current is $5.0 \text{ A}$ , and the common section carries $5.0 \text{ A}$ , because the tap at 750 turns out of 1000 provides a voltage of $200 \times (1000/750) \approx 267 \text{ V}$ to the load by transformer action, and the resulting load current flows uniformly through the entire common section.

Explanation: 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.

Question 6

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 $V_2/V_1 = N_2/N_1$ ?

$M$ decreases by a factor of two because $M \propto \mu_r$ , and the turns-ratio voltage relationship also changes because the reduced permeability lowers the flux density, so the secondary voltage becomes proportional to both $\mu_r$ and the turns ratio rather than the turns ratio alone.
$M$ 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. (correct answer)
$M$ is unchanged because mutual inductance for a perfectly coupled transformer depends only on the geometry and number of turns, not on the absolute permeability of the core material, and the turns-ratio voltage relationship is likewise unaffected by the core swap.
$M$ decreases by a factor of four because $M \propto \mu_r^2$ : the primary flux produced per unit current scales with $\mu_r$ , and the fraction of that flux linking the secondary also scales with $\mu_r$ , so both factors multiply to give a quadratic dependence. The turns-ratio voltage relationship remains valid as long as the transformer stays ideal.

Explanation: 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.

Question 7

An ideal transformer has a primary connected to a $120 \text{ V}$ RMS, $60 \text{ Hz}$ source. The secondary is connected to a load that consists of a resistor $R = 10 \ \Omega$ in series with an inductor $L = 26.5 \text{ mH}$ . The turns ratio is $N_1 : N_2 = 1 : 2$ .

What is the RMS current drawn from the primary source? (Use $\omega = 2\pi(60) \approx 377 \text{ rad/s}$ , giving $\omega L \approx 10 \ \Omega$ .)

$I_{1,\text{rms}} \approx 33.9 \text{ A}$ , because the secondary voltage is $240 \text{ V}$ RMS, the secondary impedance magnitude is $|Z_2| = \sqrt{(10)^2+(10)^2} = 10\sqrt{2} \ \Omega$ , giving $I_2 \approx 16.97 \text{ A}$ , and the primary current is $I_1 = (N_2/N_1)I_2 = 2 \times 16.97 \approx 33.9 \text{ A}$ .
$I_{1,\text{rms}} \approx 16.97 \text{ A}$ , because the secondary voltage is $240 \text{ V}$ RMS, the secondary impedance is $10\sqrt{2} \ \Omega$ , and the primary current equals the secondary current directly since the transformer merely steps up voltage while conserving current in a step-up configuration.
$I_{1,\text{rms}} \approx 12.0 \text{ A}$ , because the load impedance $|Z_2| = 10\sqrt{2} \ \Omega$ is first referred to the primary side by dividing by $(N_2/N_1)^2 = 4$ , giving a referred impedance of $\frac{10\sqrt{2}}{4} \ \Omega$ , and the primary current is then $I_1 = V_1/Z_{\text{ref}} = 120/(2.5\sqrt{2}) \approx 33.9 \text{ A}$ .
$I_{1,\text{rms}} \approx 8.49 \text{ A}$ , 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 $I_1 = (N_1/N_2)I_2 = (1/2)(16.97) \approx 8.49 \text{ A}$ . (correct answer)

Explanation: 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.

Question 8

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?

Peak EMF in coil 2 is $10\pi \text{ V}$ ; peak back-EMF in coil 1 due to mutual coupling is $10\pi \text{ V}$ , because mutual inductance is reciprocal ( $M_{12} = M_{21}$ ) and the same value of $M$ governs both interactions symmetrically.
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 $-M\,di_2/dt = 0$ . (correct answer)
Peak EMF in coil 2 is $5\pi \text{ V}$ ; peak back-EMF in coil 1 due to mutual coupling is $5\pi \text{ V}$ , because the total mutual coupling effect $M\,di_1/dt$ is shared equally between the two coils, with each experiencing half the full mutual EMF.
Peak EMF in coil 2 is $10\pi \text{ V}$ ; peak back-EMF in coil 1 due to mutual coupling is $20\pi \text{ V}$ , because when coil 2 is open-circuited its impedance is infinite, causing a voltage-doubling reflection back into coil 1 analogous to an open-circuit transmission line reflection.

Explanation: 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.

Question 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?

The conclusion is incorrect; the total inductance is $L_{\text{opp}} = |L_1 - L_2| - 2M = 100 \text{ mH}$ , because the opposing configuration causes the two self-inductances to partially cancel each other before the mutual terms are subtracted, and this reduced effective self-inductance could in principle reach zero if $M$ were large enough.
The conclusion is correct; the total inductance is $L_{\text{opp}} = L_1 + L_2 - M = 400 \text{ mH}$ , which is positive, because in the series-opposing configuration only one mutual term subtracts — the second mutual term adds due to the asymmetry created by the opposing flux directions.
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. (correct answer)
The conclusion is incorrect; the total inductance is $L_{\text{opp}} = L_1 + L_2 - 2M = 300 \text{ mH}$ , but a coupling coefficient $k < 1$ does not guarantee a positive result — if $M$ were close to $\sqrt{L_1 L_2}$ with $L_1 \gg L_2$ , the series-opposing inductance could become negative for certain asymmetric coil geometries.

Explanation: 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.

Question 10

A transformer is used in a power distribution system. The primary coil has $N_1 = 500$ turns and is connected to an AC source with peak voltage $V_0 = 1000 \text{ V}$ . The secondary coil has $N_2 = 100$ turns and is connected to a purely resistive load $R = 20 \ \Omega$ . Assume the transformer is ideal.

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?

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. (correct answer)
The secondary RMS current increases by a factor of two and the power delivered to the load increases by a factor of four, because the induced EMF in the secondary is proportional to the rate of change of flux, which doubles when frequency doubles, directly raising the secondary voltage and current.
The secondary RMS current decreases and the power delivered to the load decreases, because the increased frequency raises the inductive reactance of the primary coil, reducing the magnetizing current and therefore the effective flux linkage to the secondary — even for an ideal transformer.
The secondary RMS current remains unchanged but the power delivered to the load doubles, because each cycle delivers the same fixed energy to the resistive load, and doubling the frequency means twice as many cycles per second, increasing the time-averaged power by a factor of two.

Explanation: 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.