Correct answer: always less than that of the smallest capacitor

For capacitors connected in series, the total capacitance is:

\[\frac{1}{C_{\text{total}}}=\frac{1}{C_1}+\frac{1}{C_2}+\cdots\]

Because reciprocals are added, \(C_{\text{total}}\) must be smaller than the smallest capacitor in the series.

• It cannot be greater than the largest capacitor.
• Adding capacitances applies to parallel, not series.
• Averaging capacitances is not a valid method.

Therefore, the total capacitance of capacitors in series is always less than the smallest capacitor.

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Correct answer: A — withstand a greater voltage than a single capacitor can withstand

When electrolytic capacitors are connected in series, the applied voltage is shared across each capacitor. This allows the combination to handle a higher total voltage than any single capacitor in the string could tolerate on its own. This technique is used when the supply voltage exceeds the voltage rating of available capacitors — for example, combining two 200 V capacitors in series to work in a 350 V supply. Equalising resistors are often placed across each capacitor to ensure the voltage divides evenly.

• B — increase the total capacity: Series connection reduces total capacitance (like resistors in parallel — the reciprocals add). Parallel connection is used to increase total capacitance.
• C — reduce the ripple voltage further: Ripple reduction depends on capacitance value and load current, not on series vs parallel connection. Series connection actually reduces capacitance, which would worsen ripple filtering.
• D — resonate the filter circuit: Resonance involves a specific combination of inductance and capacitance in LC filters; simply connecting capacitors in series does not create a resonant filter circuit.

Therefore, series connection of filter capacitors is used specifically to share the voltage across each device, allowing the combination to safely withstand a higher supply voltage than a single capacitor could manage alone.

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Correct answer: microfarads

Capacitance is measured in farads (F), and in practical circuits commonly in:

• microfarads (\(\mu\)F)

• nanofarads (nF)

• picofarads (pF)

• Microvolts measure voltage.

• Millihenrys measure inductance.

• Megohms measure resistance.

Therefore, a capacitor is identified by values in microfarads.

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Correct answer: bringing the metal plates closer together

The capacitance of a parallel-plate capacitor is given by:

\[ C = \frac{\varepsilon A}{d} \]

where:

• \(C\) is the capacitance
• \(\varepsilon\) is the permittivity of the dielectric
• \(A\) is the plate area
• \(d\) is the distance between the plates

Capacitance is inversely proportional to the plate spacing:

\[ C \propto \frac{1}{d} \]

So, reducing the distance between the plates increases the capacitance.

• Making the plates smaller reduces \(A\), lowering capacitance.
• Moving the plates apart increases \(d\), lowering capacitance.
• Touching the plates together would short-circuit them.

Therefore, the capacitance may be increased by bringing the metal plates closer together.

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Correct answer: dielectric

The material between the plates of a capacitor is called the dielectric.

The dielectric:

• separates the conductive plates

• increases the capacitance by its permittivity

• prevents direct current flow between the plates

• A semiconductor is a different type of material used in devices like diodes and transistors.

• A resistor limits current, not separates capacitor plates.

• Lamination refers to layered materials, typically in transformers.

Therefore, the material is the dielectric.

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Correct answer: A — 45 picofarad

When capacitors are wired in parallel, their capacitances add directly. This is the opposite of resistors, where parallel combinations reduce the total value. Each capacitor shares the same voltage, so their charge-storage abilities simply sum together.

\[ C_{\text{total}} = C_1 + C_2 + C_3 \]

Substituting the values:

\[ C_{\text{total}} = 15 + 15 + 15 = 45\ \mathrm{pF} \]

• B. 18 picofarad — Incorrect; this does not correspond to any standard parallel or series formula for three equal 15 pF capacitors.
• C. 12 picofarad — Incorrect; this is also not a result of any standard combination formula for these values.
• D. 5 picofarad — Incorrect; this would be the result if the capacitors were in series (15/3 = 5 pF), but the question specifies parallel.

Therefore, three 15 pF capacitors in parallel combine to give a total capacitance of 45 pF.

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Correct answer: reactance

Capacitors and inductors oppose alternating current due to their energy storage properties.

This opposition is called reactance:

• Capacitive reactance (\(X_C\)) decreases with frequency

• Inductive reactance (\(X_L\)) increases with frequency

• Resistance applies to both AC and DC without frequency dependence.

• Resonance is a specific condition where reactances cancel.

• Conductance is the inverse of resistance.

Therefore, the correct term is reactance.

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Correct answer: frequency decreases

The reactance of a capacitor is given by:

\[ X_C = \frac{1}{2\pi f C} \]

where:

• \(X_C\) is capacitive reactance (ohms)
• \(f\) is frequency (Hz)
• \(C\) is capacitance (farads)

From the formula, reactance is inversely proportional to frequency. As the frequency decreases, the denominator becomes smaller, so the reactance becomes larger.

• frequency increases causes reactance to decrease, not increase.
• applied voltage increases does not change reactance, reactance depends only on frequency and capacitance.
• applied voltage decreases also has no effect on reactance.

Therefore, the reactance of a capacitor increases as the frequency decreases.

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Correct answer: A — frequency increases

Inductive reactance \(X_L\) is the opposition an inductor presents to alternating current. It rises proportionally with frequency because a faster-changing current induces a greater back-EMF in the coil, making it harder for current to flow.

The governing formula is:

\[ X_L = 2\pi f L \]

Where:

• X_L = inductive reactance (ohms)
• f = frequency (Hz)
• L = inductance (henries)

Worked example: A 10 mH inductor at 1 kHz:

\[ X_L = 2\pi \times 1000 \times 0.01 = 62.8\ \Omega \]

At 2 kHz the reactance doubles to 125.6 Ω, confirming the direct relationship with frequency.

• B. frequency decreases — From X_L = 2πfL, lowering f reduces X_L; reactance falls, not rises.
• C. applied voltage increases — Voltage does not appear in the reactance formula; X_L depends only on frequency and inductance.
• D. applied voltage decreases — Similarly, reducing voltage has no effect on the reactance of an inductor.

Therefore, inductive reactance increases directly with frequency, making higher frequencies increasingly difficult to pass through an inductor.

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Correct answer: B — increase

Inductance is determined by how strongly a coil can store energy in a magnetic field. When current flows through a coil, each turn contributes to the total magnetic flux. Adding more turns increases the total flux linkage, which directly raises the inductance. The relationship is governed by:

\[ L = \frac{\mu_0 \mu_r N^2 A}{l} \]

Where:

• L = inductance (henries)
• N = number of turns
• A = cross-sectional area of the core (m²)
• l = length of the coil (m)
• μ₀ = permeability of free space
• μᵣ = relative permeability of the core material

Because N is squared in the formula, doubling the number of turns increases inductance by a factor of four.

• A — decrease: More turns increases flux linkage; inductance rises, it does not fall.
• C — remain unchanged: Inductance is directly dependent on the number of turns; changing N always changes L.
• D — become resistive: Adding turns adds a small amount of DC resistance, but inductance is not converted into resistance — the component remains inductive.

Therefore, increasing the number of turns on an inductor increases its inductance, with inductance rising proportionally to the square of the number of turns.

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Correct answer: henry

Inductance is the property of a component that opposes changes in current.

The unit of inductance is the henry (H).

• The farad is the unit of capacitance.
• The ohm is the unit of resistance.
• Reactance is a property measured in ohms, not a unit itself.

Therefore, the unit of inductance is the henry.

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Correct answer: C — 40 uH

When inductors are connected in series (with no mutual coupling between them), their inductances add directly, just as resistors in series add together.

\[ L_{\text{total}} = L_1 + L_2 \]

Given two 20 µH inductors:

\[ L_{\text{total}} = 20\ \mu\text{H} + 20\ \mu\text{H} = 40\ \mu\text{H} \]

• A — 10 µH: This would result from placing the two inductors in parallel using the product-over-sum rule, and even then the parallel result for two equal inductors is half of one value (10 µH), not a series result.
• B — 20 µH: This is simply the value of one inductor unchanged; no combination rule produces this result for two identical inductors in series.
• D — 80 µH: This would be double the correct answer; there is no standard series formula that multiplies inductances.

Therefore, two 20 µH inductors connected in series produce a total inductance of 40 µH.

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Correct answer: A — 10 uH

When inductors are connected in parallel (with no mutual coupling), they combine in the same way as resistors in parallel. For two equal inductors, the total inductance is simply half the value of one inductor.

\[ L_{total} = \frac{L_1 \times L_2}{L_1 + L_2} \]

Given two 20 µH inductors:

\[ L_{total} = \frac{20 \times 20}{20 + 20} = \frac{400}{40} = 10\ \mu\mathrm{H} \]

• B — 20 µH: This would be correct if the inductors were in parallel but only one were present, or if the question were asking for the value of a single inductor. Parallel connection always reduces the total below either individual value.
• C — 40 µH: This is the result of adding the two inductors in series (20 + 20 = 40 µH), not parallel.
• D — 80 µH: This has no basis in standard inductor combination rules; it doubles the series value and is incorrect.

Therefore, two 20 µH inductors connected in parallel give a total inductance of 10 µH.

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Correct answer: A — windings are wound on a closed ring of magnetic material

A toroid is a doughnut-shaped (annular) core, usually made from ferrite or powdered iron. Wire is wound around this closed ring so that the magnetic flux circulates entirely within the core material. Because the flux path is fully contained inside the ring, very little magnetic field escapes — giving a toroidal inductor excellent self-shielding and low interference with neighbouring components.

• B — windings are air-spaced: An air-spaced coil uses no core material at all. This describes a standard air-core inductor, not a toroid.
• C — windings are wound on a ferrite rod: A ferrite rod (or ferrite-slug) core is a straight, open-ended former. The flux path is not closed, so field leaks from both ends — this is a rod-core inductor, not a toroid.
• D — the inductor is enclosed in a magnetic shield: An external magnetic shield is a separate housing placed around a component to contain stray fields. A toroid achieves self-shielding by geometry, not by an enclosure.

Therefore, a toroidal inductor is specifically defined by its windings being wound on a closed ring of magnetic material, which gives it a fully contained flux path and inherently low external field leakage.

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Correct answer: B — 23 volt

A transformer steps voltage up or down in direct proportion to the turns ratio. With more turns on the primary than the secondary, this is a step-down transformer and the output voltage will be lower than the input.

\[ \frac{V_s}{V_p} = \frac{N_s}{N_p} \]

Given:

• \(V_p = 230\ \mathrm{V}\)
• \(N_p = 100\ \mathrm{turns}\)
• \(N_s = 10\ \mathrm{turns}\)

\[ V_s = V_p \times \frac{N_s}{N_p} = 230 \times \frac{10}{100} = 23\ \mathrm{V} \]

• A — 10 volt: This would result from dividing by the number of secondary turns alone, ignoring the primary voltage entirely.
• C — 110 volt: This is close to half of 230 V and does not follow from the turns ratio; it may be confused with a US mains voltage.
• D — 2300 volt: This would be the result of multiplying rather than dividing by the turns ratio (10:1), which would describe a step-up transformer with the windings reversed.

Therefore, applying the transformer turns ratio to 230 V with a 100:10 (10:1) step-down ratio gives a secondary voltage of 23 V.

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Correct answer: minimum

For a series LC circuit, the inductive reactance \(X_L\) and capacitive reactance \(X_C\) are:

\[ X_L = 2\pi f L \]

\[ X_C = \frac{1}{2\pi f C} \]

At the resonant frequency, these reactances are equal in magnitude and opposite in sign:

\[ X_L = X_C \]

They cancel each other, so the total reactive component becomes zero. The remaining impedance is only the small resistive losses in the circuit, making the total impedance minimum.

• maximum applies to a parallel resonant circuit, not a series circuit.
• totally reactive is incorrect because the reactive components cancel at resonance.
• totally inductive is incorrect because there is no net reactance at resonance.

Therefore, at resonance a series LC circuit has minimum impedance.

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Correct answer: maximum

For a parallel LC circuit, resonance occurs when the inductive reactance equals the capacitive reactance:

\[ X_L = X_C \]

At this frequency, the currents flowing through the inductor and capacitor are equal in magnitude but opposite in phase, effectively cancelling each other. As a result, very little current is drawn from the source.

This causes the circuit to present a very high impedance at resonance.

• A minimum impedance occurs in a series resonant circuit, not a parallel one.
• At resonance, the reactive effects cancel, so the impedance is not totally reactive.
• It is not totally inductive, since the inductive and capacitive effects balance.

Therefore, the impedance at resonance in a parallel LC circuit is maximum.

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Correct answer: decrease to half

The resonant frequency of an LC circuit is given by:

\[ f = \frac{1}{2\pi\sqrt{LC}} \]

If the capacitance is increased by a factor of 4:

\[ f_{\text{new}} = \frac{1}{2\pi\sqrt{L(4C)}} = \frac{1}{2\pi \cdot 2\sqrt{LC}} = \frac{1}{2} \times \frac{1}{2\pi\sqrt{LC}} \]

So the new resonant frequency is:

\[ f_{\text{new}} = \frac{f}{2} \]

Exam tip:

Resonant frequency varies with the inverse square root of \(L\) or \(C\):

\[ f \propto \frac{1}{\sqrt{C}} \]

So if capacitance increases by 4:

\[ \sqrt{4} = 2 \Rightarrow f \text{ decreases by } 2 \]

Memory aid:

More C or L → lower f, by the square root

• Increasing by four times would require reducing \(LC\) by a factor of 16.
• Doubling would require reducing \(LC\) by a factor of 4.
• Decreasing to one quarter would require increasing \(LC\) by a factor of 16.

Therefore, the resonant frequency will decrease to half.

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Correct answer: increase by a factor of two

The resonant frequency of an LC circuit is given by:

\[ f = \frac{1}{2\pi\sqrt{LC}} \]

If the inductance \(L\) is decreased by a factor of four, the new inductance is:

\[ L_{new} = \frac{L}{4} \]

Substituting into the formula:

\[ f_{new} = \frac{1}{2\pi\sqrt{(L/4)C}} \]

\[ f_{new} = \frac{1}{2\pi\left(\frac{1}{2}\sqrt{LC}\right)} = 2f \]

So reducing the inductance by four causes the square root term to halve, which makes the frequency double.

• increase by a factor of four would require the inductance to decrease by a factor of sixteen.
• decrease by a factor of two is the opposite of what happens.
• decrease by a factor of four is also incorrect because reducing inductance raises frequency.

Therefore, decreasing the inductance by a factor of four causes the resonant frequency to increase by a factor of two.

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Correct answer: is highly selective

The quality factor \(Q\) of a resonant circuit is a measure of how sharply it responds to its resonant frequency.

A high \(Q\) circuit has:

• low energy loss
• narrow bandwidth
• sharp tuning

Since bandwidth is related to \(Q\) by:

\[ Q = \frac{f_0}{\text{Bandwidth}} \]

a high \(Q\) results in a small bandwidth, making the circuit highly selective.

• Quiescent current is unrelated to \(Q\).
• A high \(Q\) circuit has a narrow, not wide, bandwidth.
• Inductance value alone does not determine \(Q\).

Therefore, a high \(Q\) resonant circuit is highly selective.

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