FAD1022 Quick Quiz 2026 — Inductance & Transformers

Classroom quick quiz on mutual inductance, self-inductance, and transformer principles. 11 questions covering conceptual understanding and quantitative problem solving.

Q1 — Mutual Inductance and Current

A pair of solenoids having mutual inductance $M$ are shown below. Current of a certain magnitude is flowing through the inner solenoid. What is the effect of decreasing the magnitude of current flowing in the inner solenoid?

  • A. $M$ decrease
  • B. $M$ remain the same
  • C. $M$ increase

Answer: B — $M$ remains the same.

Mutual inductance depends only on geometry (number of turns, area, length, separation, core material) and not on the current flowing through either coil. Decreasing current reduces the induced emf, but $M$ itself is unchanged.

Q2 — Mutual Inductance and Distance

Two coils kept a distance $d$ apart are shown below. Which of these best describes the effect of decreasing $d$ on the mutual inductance $M$ of the coils?

  • A. $M$ increases
  • B. $M$ remains the same
  • C. $M$ decreases

Answer: A — $M$ increases.

Decreasing the separation distance increases the magnetic flux linkage between the coils, thereby increasing mutual inductance.

Q3 — Reciprocal Induced EMF

Two concentric circular loops are shown below. When current increases by $\Delta i_1 = 2.0\ \text{A}$ in the smaller loop, an emf of $\varepsilon_2 = 6.0\ \text{V}$ is induced in the bigger loop. Find the emf induced in the smaller loop when current increases by $\Delta i_2 = 0.25\ \text{A}$ in the bigger loop in the same duration.

Solution:

From reciprocity of mutual inductance: $$\frac{\varepsilon_2}{\Delta i_1} = \frac{\varepsilon_1}{\Delta i_2}$$

$$\varepsilon_1 = \varepsilon_2 \frac{\Delta i_2}{\Delta i_1} = 6.0 \times \frac{0.25}{2.0} = \boxed{0.75\ \text{V}}$$

Q4 — Mutual Inductance from Flux

Two coils 1 and 2 having 25 and 50 turns respectively are placed near each other. On passing a current $i_1 = 2.0\ \text{A}$ through coil 1, a flux of $\phi_2 = 1.2\ \text{Wb}$ passes through each turn of coil 2. Find the value of mutual inductance of coil 2 with respect to coil 1.

Solution:

$$M_{21} = \frac{N_2 \phi_2}{i_1} = \frac{50 \times 1.2}{2.0} = \boxed{30\ \text{H}}$$

Q5 — Average Induced EMF in Coaxial Solenoids

A solenoid is placed inside another solenoid. The mutual inductance of the two solenoids is $15\ \text{H}$. The current in the inner solenoid changes from $5.0\ \text{A}$ to $9.0\ \text{A}$ in $5.0\ \text{s}$. Find the magnitude of average emf induced in the outer solenoid in this period.

Solution:

$$|\varepsilon_{\text{outer}}| = M \left|\frac{\Delta I}{\Delta t}\right| = 15 \times \frac{9.0 - 5.0}{5.0} = 15 \times 0.8 = \boxed{12\ \text{V}}$$

Q6 — Transformer Turns and Current

The ______ the number of turn of primary coil than secondary, the ______ the value of current in secondary coil.

  • A. Higher & higher
  • B. Higher & lower
  • C. Lower & higher
  • D. Lower & lower

Answer: C — Lower & higher.

For an ideal transformer: $\displaystyle \frac{I_s}{I_p} = \frac{N_p}{N_s}$. If the primary has fewer turns than the secondary ($N_p < N_s$), it is a step-up transformer, and the secondary current is higher than the primary current.

Q7 — Inductor Behaviour in RL Circuit

Which of these best describes the effect of the inductor in the circuit after the switch is closed?

  • a) Behaves like a connecting wire and does not affect the flow of current.
  • b) Initially opposes the flow of current, later behaves like a connecting wire
  • c) Permanently blocks the flow of current in the circuit.
  • d) Provides constant opposition to the flow of current.

Answer: b — Initially opposes the flow of current, later behaves like a connecting wire.

At $t = 0$, the inductor generates a back emf equal and opposite to the applied voltage, preventing instantaneous current rise. As $t \to \infty$, $dI/dt \to 0$, so the back emf vanishes and the inductor acts as a short circuit (connecting wire) with only the wire resistance remaining.

Q8 — Maximum Self-Induced EMF

Choose the situation in which the magnitude of EMF generated by the inductor would be maximum.

  • a) A constant small current is flowing in the circuit.
  • b) A constant large current is flowing in the circuit.
  • c) Current is decreasing slowly in the circuit.
  • d) Current is increasing rapidly in the circuit.

Answer: d — Current is increasing rapidly in the circuit.

$\displaystyle |\varepsilon| = L\left|\frac{dI}{dt}\right|$. The induced emf magnitude depends on the rate of change of current, not the current itself. A rapidly increasing current gives the largest $|dI/dt|$ and therefore the maximum emf.

Q9 — Eddy Current Loss Location

Eddy current losses occur in which part of the transformer?

  • a) The primary winding.
  • b) The secondary winding.
  • c) The external load.
  • d) The iron core

Answer: d — The iron core.

The time-varying magnetic field induces circulating currents (eddy currents) within the conducting iron core itself. These currents dissipate energy as heat due to the core's resistance.

Q10 — Iron Loss (Core Loss) Cause

Which of the following is primarily responsible for 'Iron Loss' (Core Loss) in a transformer?

  • a) Radiation of lights from the coil.
  • b) Friction in the moving parts of the transformer.
  • c) Eddy current and hysteresis in the magnetic core.
  • d) Resistance in the primary and secondary copper windings.

Answer: c — Eddy current and hysteresis in the magnetic core.

Iron/core loss is the sum of hysteresis loss (energy dissipated during cyclic magnetization reversal) and eddy current loss (I²R heating from induced currents in the core). Copper losses (d) are separate and occur in the windings, not the core.

Q11 — Transformer Efficiency and Heat Loss

A transformer has an efficiency of 90%. If the output power is 900 W, what is the power lost as heat?

  • a) 10 W
  • b) 90 W
  • c) 100 W
  • d) 810 W

Solution:

$$\eta = \frac{P_{\text{out}}}{P_{\text{in}}} \Rightarrow 0.90 = \frac{900}{P_{\text{in}}} \Rightarrow P_{\text{in}} = 1000\ \text{W}$$

$$P_{\text{loss}} = P_{\text{in}} - P_{\text{out}} = 1000 - 900 = \boxed{100\ \text{W}}$$

Answer: c — 100 W.

Quiz Summary

Q Topic Type
1 Mutual inductance independence from current Conceptual
2 Mutual inductance dependence on distance Conceptual
3 Reciprocal induced emf Calculation
4 Mutual inductance from flux and turns Calculation
5 Average emf in coupled solenoids Calculation
6 Transformer turns–current relationship Conceptual
7 Inductor transient behaviour in RL circuit Conceptual
8 Self-induced emf maximization Conceptual
9 Eddy current loss location Conceptual
10 Iron loss mechanisms Conceptual
11 Transformer efficiency and power loss Calculation

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