Question 1

Arithmetic of signed values in circuit calculations: Compute the algebraic sum of the sequence: +2, −3, −6, +4.

Accepted Answer

Introduction / Context: Accurate arithmetic with signed numbers is essential for circuit analysis, including KCL/KVL sums, phasor components, and error budgeting. This quick calculation reinforces careful handling of signs. Given Data / Assumptions: Sequence of signed integers: +2, −3, −6, +4. Standard arithmetic rules apply. No units; pure numeric sum. Concept / Approach: Combine terms stepwise, being consistent with sign operations. Grouping positive and negative contributions can reduce mistakes. This mirrors summing polarities in circuit loops. Step-by-Step Solution: Start: 0 + 2 = 2.Then 2 − 3 = −1.Then −1 − 6 = −7.Finally −7 + 4 = −3. Verification / Alternative check: Group positives and negatives: positives = 2 + 4 = 6; negatives = 3 + 6 = 9; 6 − 9 = −3. Both methods agree. Why Other Options Are Wrong: (a) −6 results from an arithmetic slip; (c) 0 would incorrectly balance terms; (d) +2 is the first partial sum, not the final total. Common Pitfalls: Dropping signs, rushing steps, or failing to regroup terms to check results. Final Answer: -3.

Question 2

Household lamp resistance from nameplate values: Which formula directly gives resistance R when the rated power P and voltage E of a light bulb are known?

Accepted Answer

Introduction / Context: Nameplates list voltage and power ratings. From these, one can estimate the operating resistance of a lamp. This is useful for quick checks and for calculating expected current draw on a circuit. Given Data / Assumptions: E denotes voltage; P denotes power; R denotes resistance; I denotes current. Assume steady-state operation at the rated conditions. Ignore temperature dependence for this idealized calculation. Concept / Approach: Starting from P = E * I and Ohm’s law E = I * R, eliminate I. Substituting I = E / R into P = E * I gives P = E^2 / R. Rearranging yields R = E^2 / P. Step-by-Step Solution: Ohm’s law: E = I * R.Power law: P = E * I.Eliminate I: I = E / R → P = E * (E / R) = E^2 / R.Solve for R: R = E^2 / P. Verification / Alternative check: Dimensional analysis: volts^2 / watts yields ohms, consistent with resistance. Why Other Options Are Wrong: (b) P = P * R is nonsensical; (c) is the unsolved form before isolating R; (d) gives current, not resistance; thus only (a) directly produces R from E and P. Common Pitfalls: Stopping at P = E^2 / R and forgetting to rearrange; misreading E2 as E * 2 instead of E squared. Final Answer: R = E2/P.

Question 3

Parallel resistors — equivalent resistance calculation: Find the total resistance of a parallel network with two resistors: 220 Ohm and 470 Ohm.

Accepted Answer

Introduction / Context: Computing equivalent resistance for parallel resistors is a core skill. It is necessary for predicting current splits, source loading, and power dissipation in practical circuits. Given Data / Assumptions: R1 = 220 Ohm. R2 = 470 Ohm. Ideal components, DC conditions. Concept / Approach: For two resistors in parallel, use R_eq = (R1 * R2) / (R1 + R2), which is an algebraic rearrangement of 1 / R_eq = 1 / R1 + 1 / R2. The equivalent must be less than the smallest individual resistance. Step-by-Step Solution: Compute numerator: R1 * R2 = 220 * 470 = 103,400.Compute denominator: R1 + R2 = 220 + 470 = 690.Divide: R_eq = 103,400 / 690 ≈ 149.855…Round suitably: ≈ 150 ohm. Verification / Alternative check: Using admittances: 1/220 ≈ 0.004545; 1/470 ≈ 0.002128; sum ≈ 0.006673 → R_eq ≈ 1 / 0.006673 ≈ 150 ohm. Both methods agree. Why Other Options Are Wrong: (a) 220 ohm is larger than the correct equivalent; (c) 445 ohm is near a series-like sum mistake; (d) 690 ohm is the series sum, not parallel; (e) is unnecessary as (b) is correct. Common Pitfalls: Forgetting that the equivalent of parallels is less than the smallest branch; arithmetic slips when multiplying and dividing. Final Answer: 150 ohm.

Question 4

Electromechanical energy conversion — generator principle: Which statement best describes the fundamental operating principle of an electric generator?

Accepted Answer

Introduction / Context: Generators convert mechanical energy to electrical energy via electromagnetic induction. Recognizing the correct physical principle is crucial to understanding power generation and motor–generator reciprocity. Given Data / Assumptions: Conventional generator with rotor motion in a magnetic field. No special construction details required. Basic Faraday–Lenz laws apply. Concept / Approach: By Faraday’s law, an electromotive force is induced in a conductor when the magnetic flux linking it changes. A simple way to achieve this change is to move the conductor through a magnetic field, producing an induced current in a closed circuit consistent with Lenz’s law. Step-by-Step Solution: Mechanically rotate or translate a conductor within a magnetic field.Changing flux linkage induces an EMF.If the circuit is closed, a current flows as the generator output. Verification / Alternative check: Flip the scenario for motors: apply current to a conductor in a magnetic field to produce force. This reciprocity affirms the generator principle conversely. Why Other Options Are Wrong: (a) and (c) refer to magnet pole interactions, not induction; (d) is the motor side of electromagnetism (field generation by current) rather than the generator’s cause of induced current. Common Pitfalls: Confusing cause and effect: a generator relies on motion in a field to induce current, not on the field produced by an already flowing current. Final Answer: Current is induced in a conductor that is moving through a magnetic field.

Question 5

Magnetic circuits: which property is directly analogous to electric current? Choose the best comparison used in basic circuit–magnetism analogies.

Accepted Answer

Introduction / Context: In introductory electromagnetics and electrical engineering, magnetic circuits are often taught using a close analogy to electric circuits. This question asks which magnetic quantity most directly corresponds to electric current in that analogy. Understanding the mapping helps learners transfer intuition from Ohm’s law (V = I * R) to its magnetic counterpart (mmf = phi * reluctance). Given Data / Assumptions: The electric–magnetic analogy is conceptual, not a physical identity. Key electric quantities: voltage (V), current (I), resistance (R). Key magnetic quantities: magnetomotive force (mmf), flux (phi), reluctance (R_m), and flux density (B). Concept / Approach: In the standard analogy: voltage maps to magnetomotive force; current maps to magnetic flux (number of lines of force); and resistance maps to reluctance. Flux density (B) is flux per unit area and has no direct 1:1 counterpart with current in this simple analogy. Step-by-Step Solution: 1) Recall electrical law: V = I * R.2) Recall magnetic form: mmf = phi * R_m, where phi is magnetic flux.3) Compare terms: voltage ↔ mmf, current ↔ flux, resistance ↔ reluctance.4) Therefore, the quantity analogous to electric current is the magnetic flux, commonly described as number of lines of force. Verification / Alternative check: If mmf is increased while reluctance is fixed, magnetic flux increases proportionally (just like current increases when voltage is increased across a fixed resistance). This mirrors the behavior of I in V = I * R and validates the mapping phi ↔ I. Why Other Options Are Wrong: Flux density is flux per area and does not play the role of current. Reluctance corresponds to resistance, not current. Magnetomotive force is analogous to voltage, the driver of flux, not the flux itself. Common Pitfalls: Confusing flux (phi) with flux density (B). Learners sometimes think B is the current analog, but it is an areal density of flux, not the flow itself in the magnetic path. Also, remember the analogy is useful for intuition but has limits (e.g., magnetic circuits are not conductive paths of moving charges). Final Answer: Number of lines of force (magnetic flux).

Question 6

Unit conversion checkpoint: what is the equivalent of 1 kV expressed in volts (V)? Select the correct magnitude.

Accepted Answer

Introduction / Context: Accurate unit conversion is a foundational skill in electronics. Engineers frequently move between prefixes such as milli (10^-3), kilo (10^3), and mega (10^6). This question checks the ability to convert kilovolts to volts, which is essential when reading datasheets, sizing components, or programming measurement equipment. Given Data / Assumptions: Prefix “kilo” means 10^3. We are converting 1 kilovolt to volts. No rounding is needed because the conversion is exact. Concept / Approach: Use the SI prefix table: 1 kV = 10^3 V. The general pattern is value_in_base = value_with_prefix * 10^(prefix power). For kilo, multiply by 10^3 to get base units (volts). Step-by-Step Solution: 1) Identify the prefix: k (kilo) = 10^3.2) Apply conversion: 1 kV = 1 * 10^3 V.3) Compute: 1 * 10^3 = 1000.4) Conclude: 1 kV = 1000 V. Verification / Alternative check: Reverse the operation: 1000 V / 10^3 = 1 kV. The forward and reverse conversions are consistent, confirming correctness. Why Other Options Are Wrong: 1.00 V is 10^0 V, not kilovolt. 100 mV equals 0.1 V, which is 10^-1 V. 1000 mV equals 1 V, not 1000 V. “None of the above” is not applicable since 1000 V is correct. Common Pitfalls: Confusing milli with kilo, or assuming k means 1024 (a binary multiple) instead of 1000 in SI electrical contexts. Final Answer: 1000 V

Question 7

Atomic structure refresher: what is the net charge of a neutral copper atom under standard conditions?

Accepted Answer

Introduction / Context: Electronics begins with atoms, charges, and how they balance. A “neutral” atom, by definition, contains equal numbers of positively charged protons and negatively charged electrons, producing no net charge. This question checks core chemistry–physics knowledge relevant to semiconductor behavior and conductor neutrality. Given Data / Assumptions: We are considering an isolated, neutral copper atom. Protons carry +e charge; electrons carry -e charge; neutrons are uncharged. “Neutral” means total positive and negative charges sum to zero. Concept / Approach: Copper’s atomic number is 29, meaning 29 protons. A neutral copper atom has 29 electrons. The charges cancel: 29*(+e) + 29*(-e) = 0. Net charge is zero; this is true for any neutral atom regardless of element. Step-by-Step Solution: 1) Identify the meaning of “neutral”: equal positive and negative charges.2) Copper has Z = 29 protons; a neutral atom therefore has 29 electrons.3) Sum charges: total charge = (+29e) + (-29e) = 0.4) Conclude the net charge is zero. Verification / Alternative check: Ions, not neutral atoms, carry net charge by gaining or losing electrons. For example, Cu^2+ has lost 2 electrons and has net +2e. Since the stem specifies “neutral,” zero is confirmed. Why Other Options Are Wrong: +1 or -1 would indicate ionization by one electron. +4 would indicate a tetravalent ion, which is not what “neutral” means. “None of the above” is unnecessary because 0 is correct. Common Pitfalls: Confusing atomic number (protons) with net charge; thinking that conduction electrons in metals imply atoms carry net charge (they do not in the neutral, bulk sense). Final Answer: 0

Question 8

Transformer calculation: a step-up transformer has a turns ratio of 1:4. If the primary is 115 V (rms), what is the secondary peak voltage?

Accepted Answer

Introduction / Context: Power transformer problems often require two conversions: turns ratio scaling and rms-to-peak conversion. This question integrates both steps for a step-up transformer, from a 115 V rms primary to the secondary peak value. Given Data / Assumptions: Turns ratio Np:Ns = 1:4 (step-up). Primary voltage Vp(rms) = 115 V. Ideal transformer (ignoring losses and regulation for this calculation). Concept / Approach: For an ideal transformer: Vs/Vp = Ns/Np. After getting Vs(rms), convert to peak using Vp(peak) = Vrms * sqrt(2). We apply this to the secondary value. Step-by-Step Solution: 1) Use ratio: Vs(rms) = Vp(rms) * (Ns/Np) = 115 * 4 = 460 V (rms).2) Convert rms to peak: Vpeak = Vrms * sqrt(2) ≈ 460 * 1.414.3) Compute: 460 * 1.414 ≈ 650.44 V.4) Round to the closest option: 651 V. Verification / Alternative check: Check plausibility: a 4x step-up from 115 V yields 460 V rms; a peak near 1.414 times that is expected near 650 V. The result is consistent with sinusoidal assumptions. Why Other Options Are Wrong: 700 V and 707 V suggest misuse of 115 * 1.414 or misapplied ratios. 208 V is far too small and would correspond to a step-down, not step-up. “None of the above” is unnecessary since 651 V matches the correct rounded value. Common Pitfalls: Confusing rms with peak; forgetting to apply the turns ratio before converting; using 2 instead of sqrt(2) for peak conversion. Final Answer: 651 V

Question 9

Filters: which statement best defines the behavior of a low-pass filter in signal processing?

Accepted Answer

Introduction / Context: Filters shape spectra. A low-pass filter (LPF) preserves low-frequency content while attenuating high-frequency components. LPFs are fundamental in audio processing, anti-aliasing before digitization, smoothing sensor outputs, and conditioning power supplies. Given Data / Assumptions: We consider an idealized description of a low-pass filter. “Passes” means minimal attenuation in the passband; “reduces” means attenuation in the stopband. Cutoff frequency marks the transition band between pass and stop regions. Concept / Approach: An LPF transfer function has a magnitude close to 1 (0 dB) at low frequencies and decreases as frequency rises beyond the cutoff (e.g., -20 dB/decade for a first-order RC filter). The phase also changes, typically approaching -90 degrees for a simple RC as frequency moves far beyond the cutoff. Step-by-Step Solution: 1) Identify desired filter type: low-pass.2) Recall its function: preserve low frequencies; attenuate highs.3) Map options: Option A states the canonical definition.4) Select option A. Verification / Alternative check: Examine a first-order RC low-pass where |H(jω)| = 1 / sqrt(1 + (ωRC)^2). As ω increases, magnitude decreases—confirming high-frequency attenuation. Why Other Options Are Wrong: Option B defines a high-pass. Option C is incorrect; filters affect AC signals by design. Option D describes a fixed attenuator, not a selective filter. Common Pitfalls: Confusing cutoff frequency’s exact definition (e.g., -3 dB point) or mixing low-pass with high-pass responses. Final Answer: Passes lower-frequency signals and reduces higher-frequency signals.

Question 10

Transistor fundamentals: among the following, what is one important and common function of transistors in circuits?

Accepted Answer

Introduction / Context: Transistors are the building blocks of modern electronics. While they can be configured for switching, amplification, and more complex functions, their hallmark role in analog electronics is amplifying small signals to larger, usable levels—whether in audio, radio frequency, or sensor interfaces. Given Data / Assumptions: Bipolar junction transistors (BJTs) and field-effect transistors (FETs) both can amplify. Operating in the appropriate region (active for BJTs, saturation/triode or saturation as applicable for FETs) enables linear gain. Other listed functions may be performed by specialized components or specific transistor circuits, but are not the primary general role. Concept / Approach: Amplification means controlling a large power flow with a small signal. In a BJT, a small base current controls a large collector current (gain beta). In a FET, a small gate voltage swing modulates channel conductance, changing drain current. Proper biasing sets a quiescent (Q) point for linear operation. Step-by-Step Solution: 1) Identify the classic transistor role: amplification.2) Map to options: “Amplify weak signals” matches this role directly.3) Confirm others: LEDs emit light, regulators maintain voltage, rectifiers convert AC to DC—these are not generic transistor roles by default.4) Choose option B. Verification / Alternative check: Consider a common-emitter amplifier: voltage gain Av ≈ -gm * Rc (simplified). This depicts how a small input produces a larger output swing across the load. Why Other Options Are Wrong: Light emission is a property of LEDs/laser diodes, not ordinary transistors. Voltage regulation is typically performed by regulators or transistor arrangements but is not the most universal role. Rectification is primarily diode-based. Common Pitfalls: Assuming any function using a transistor equals the fundamental role; context matters—switching and amplification are common, but amplification is the classic answer here. Final Answer: Amplify weak signals.

Question 11

Series RLC at resonance: which relationship holds between inductive reactance and capacitive reactance at the resonant frequency?

Accepted Answer

Introduction / Context: Resonance in a series RLC circuit is a cornerstone concept. At resonance, the reactive effects cancel, minimizing impedance and maximizing current for a given applied voltage. Understanding the equality of reactances is essential for filter design, tuning, and impedance matching. Given Data / Assumptions: Series R–L–C elements with sinusoidal steady-state excitation. Resonant frequency defined by ω0 = 1 / sqrt(L * C). Ideal components (to focus on the core concept). Concept / Approach: At resonance in series RLC, X_L = ω0 * L and X_C = 1 / (ω0 * C) are equal in magnitude and opposite in sign. Their sum is zero, so the net reactive part of impedance is zero, leaving Z ≈ R (minimum magnitude). Step-by-Step Solution: 1) Write reactances: X_L = ωL; X_C = 1/(ωC).2) At resonance: ω0 = 1/sqrt(LC).3) Substitute: X_L(ω0) = ω0L and X_C(ω0) = 1/(ω0C) = ω0L.4) Conclude: X_L equals X_C in magnitude; net reactance is zero. Verification / Alternative check: Measure impedance versus frequency: the minimum occurs at resonance where reactive terms cancel, confirming equality of magnitudes. Why Other Options Are Wrong: Options A and B contradict resonance; option D (difference equals resistance) is not a standard resonance condition; resistance is independent of reactive cancellation. Common Pitfalls: Confusing series with parallel resonance; in series, impedance is minimum at resonance, not maximum. Final Answer: Inductive reactance is equal to the capacitive reactance.

Question 12

Parallel RLC behavior: when operating below its resonant frequency, a parallel RLC circuit most closely behaves like which simpler parallel circuit?

Accepted Answer

Introduction / Context: Recognizing how RLC networks simplify away from resonance is crucial for quick design estimates. For a parallel RLC, behavior hinges on which susceptance (capacitive or inductive) dominates at a given frequency relative to resonance. Given Data / Assumptions: Parallel RLC network driven by a sinusoidal source. Resonant frequency f0 = 1 / (2 * pi * sqrt(L * C)). We consider qualitative behavior sufficiently far below resonance. Concept / Approach: Parallel admittance Y = G + j(B_C - B_L), where B_C = ωC and B_L = 1/(ωL). At low frequency, 1/(ωL) is large while ωC is small, so B_L dominates and B ≈ -1/(ωL), which is inductive. Therefore, the circuit behaves like an RL in parallel, with current lagging voltage and an overall inductive character. Step-by-Step Solution: 1) Write susceptances: B_C = ωC; B_L = 1/(ωL).2) Evaluate at ω < ω0: B_L >> B_C, so net B is negative (inductive).3) Parallel circuit with inductive dominance behaves like a parallel RL.4) Choose option indicating parallel RL behavior. Verification / Alternative check: At high frequency (above resonance), ωC dominates and the behavior flips to capacitive (parallel RC-like). This complementary behavior around resonance confirms the reasoning. Why Other Options Are Wrong: Parallel RC (option B) describes above-resonance behavior, not below. “Purely resistive” is only at exact parallel resonance (ideally) when reactive currents cancel. Option C is unrelated to the below-resonance condition. Common Pitfalls: Accidentally importing series-circuit intuition (where below resonance is capacitive). Parallel and series cases invert the qualitative dominance. Final Answer: Parallel RL circuit (inductive behavior dominates).

Question 13

Frequency effect on capacitive reactance: what happens to Xc when the operating frequency of a capacitive circuit is doubled?

Accepted Answer

Introduction / Context: Designing filters and timing networks requires a precise understanding of reactance behavior with frequency. Capacitive reactance Xc is inversely proportional to frequency. This question asks about the effect of doubling frequency on Xc, a common quick check in analysis. Given Data / Assumptions: Capacitive reactance formula: Xc = 1 / (2 * pi * f * C). Capacitance C is constant. Frequency f changes from f to 2f. Concept / Approach: Since Xc is inversely proportional to f, scaling f by a factor k scales Xc by 1/k. For k = 2, Xc becomes Xc/2. This relationship is independent of the actual numerical values of f and C: only the ratio matters. Step-by-Step Solution: 1) Start with Xc = 1 / (2 * pi * f * C).2) Replace f by 2f: Xc(new) = 1 / (2 * pi * 2f * C) = 1 / (2 * (2 * pi * f * C)).3) Simplify: Xc(new) = (1/2) * [1 / (2 * pi * f * C)] = Xc / 2.4) Conclude that Xc is cut in half. Verification / Alternative check: Use example numbers: If f = 1000 Hz and C = 1 µF, Xc ≈ 1 / (2 * pi * 1000 * 1e-6) ≈ 159.15 ohm. Doubling f to 2000 Hz yields ≈ 79.6 ohm, half the original. Why Other Options Are Wrong: Xc doubling or “no effect” contradicts inverse proportionality. Multiplying by 6.28 confuses constants (2 * pi) with scaling behavior. Common Pitfalls: Mixing up inductive and capacitive reactance trends; for inductors, reactance increases with frequency, but capacitors do the opposite. Final Answer: Cuts the capacitive reactance in half.

Question 14

Transformer step-down calculation: a transformer has a turns ratio of 5:1 (primary:secondary). The primary is 120 V (rms). What is the secondary voltage (rms)?

Accepted Answer

Introduction / Context: Transformers scale AC voltages according to their turns ratio. Quick mental math with ratios is invaluable for power supplies, instrumentation, and isolation applications. This question asks for the secondary rms voltage of a 5:1 step-down transformer driven by 120 Vrms. Given Data / Assumptions: Turns ratio Np:Ns = 5:1 (step-down). Primary voltage Vp(rms) = 120 V. Ideal transformer approximation (neglect losses and regulation). Concept / Approach: For ideal transformers, Vs/Vp = Ns/Np. Because Ns/Np = 1/5, the secondary voltage is one-fifth of the primary rms voltage. Step-by-Step Solution: 1) Compute ratio: Ns/Np = 1/5 = 0.2.2) Apply to voltage: Vs = Vp * (Ns/Np) = 120 * 0.2.3) Calculate: 120 * 0.2 = 24.4) Secondary rms voltage is 24 V. Verification / Alternative check: Reverse-check: If Vs = 24 V, then Vp/Vs = 120/24 = 5, matching the turns ratio 5:1. Why Other Options Are Wrong: 12 V implies a 10:1 step-down; 18 V and 30 V correspond to other ratios; “None of the above” is invalid since 24 V fits exactly. Common Pitfalls: Inverting the ratio (using 5 instead of 1/5) or confusing rms and peak values. The problem explicitly asks for rms, so no sqrt(2) conversion is needed. Final Answer: 24 V

Question 15

Batteries in everyday flashlights (non-rechargeable): which classification correctly describes common flashlight cells used in torches and remotes?

Accepted Answer

Introduction / Context: Household flashlights, TV remotes, toys, and many small devices typically use AA, AAA, C, or D cells. These are almost always non-rechargeable (disposable) cells intended for single use. Identifying their correct classification helps learners distinguish basic battery categories: primary vs secondary and dry vs wet cells. Given Data / Assumptions: “Common flashlight batteries” refers to typical AA/AAA/C/D cylindrical cells found in stores. They are non-rechargeable unless explicitly labeled rechargeable. The classic chemistries are zinc–carbon and alkaline (zinc–manganese dioxide). Concept / Approach: Primary cells are designed for one discharge; secondary cells are rechargeable. Dry cells use a paste or gel electrolyte; wet cells use a liquid electrolyte. AA/AAA alkaline and zinc–carbon batteries use a moist paste electrolyte sealed in a can—hence “dry.” They are also primary because their chemistry is not meant to be reversed repeatedly under normal, safe charging conditions. Step-by-Step Solution: Identify rechargeability: common flashlight cells are non-rechargeable → primary.Identify electrolyte form: paste electrolyte → dry cell.Combine both attributes: primary + dry cell.Select the option that states “primary dry-cell batteries.” Verification / Alternative check: Packaging for alkaline or zinc–carbon cells states “non-rechargeable” and shows the chemistry and construction typical of dry cells. Rechargeable flashlight batteries (e.g., NiMH) are secondary but clearly labeled as such, distinguishing them from the common disposable type. Why Other Options Are Wrong: A: “Secondary wet-cell” implies rechargeable lead–acid–type cells, not used in flashlights. B: “Primary wet-cell” is inconsistent; household cells are not wet-cell types. D: “Secondary dry-cell” would describe rechargeable chemistries like NiMH, which are not the default for common disposables. E: Not applicable since one option is correct. Common Pitfalls: Confusing rechargeable AA cells (secondary, often NiMH) with the far more common disposable alkaline (primary) cells; assuming “dry cell” means literally dry rather than paste electrolyte. Final Answer: Common flashlight batteries are primary dry-cell batteries

Question 16

In semiconductor physics, “holes” in a crystal lattice behave electrically as what kind of charge carriers?

Accepted Answer

Introduction / Context: Understanding how “holes” behave is foundational to semiconductor devices such as diodes, bipolar junction transistors, and integrated circuits. While holes are not physical particles like electrons, their effective behavior in a crystal matters for current flow and device modeling. Given Data / Assumptions: A “hole” is the absence of an electron in the valence band. Materials considered: intrinsic or doped semiconductors (e.g., silicon). We focus on the electrical behavior (effective charge and motion) rather than atomic structure. Concept / Approach: A missing electron in a covalent bond leaves a vacancy that can be filled by a neighboring electron. This movement of vacancies can be modeled as mobile positive charge carriers. In P-type material, holes are the majority carriers and drift in the direction of the applied electric field, opposite to electron motion, producing conventional current. Step-by-Step Solution: Define a hole as the absence of a valence electron.Recognize that when electrons move to fill vacancies, the vacancies appear to move.Model this vacancy as a carrier with positive charge.Select “Positive charges.” Verification / Alternative check: Device equations (e.g., diode current equation, transistor action) treat hole concentration and mobility similarly to electrons but with positive charge sign, confirming the effective positive behavior. Why Other Options Are Wrong: A/B: Atoms or crystals are not carriers; these describe structures, not charge types. C: Holes are not negative; electrons are negative. E: Not applicable since a correct choice exists. Common Pitfalls: Thinking holes are literal particles; forgetting that current direction is defined by positive charge flow (conventional current), which holes represent effectively. Final Answer: Positive charges

Question 17

Nature of alternating current (AC): which statement best describes how current behaves in a typical AC circuit over time?

Accepted Answer

Introduction / Context: Alternating current (AC) underpins electric power distribution and many signal systems. Recognizing how AC differs from direct current (DC) is essential for analyzing circuits, understanding power ratings, and selecting appropriate components like capacitors and inductors. Given Data / Assumptions: Standard AC sources (e.g., sinusoidal mains power or sinusoidal test signals). Focus on current direction and time variation. No special waveforms beyond basic AC behavior. Concept / Approach: In AC, current reverses direction at a regular rate determined by the frequency (e.g., 50 Hz or 60 Hz). The instantaneous current alternates between positive and negative values, causing periodic direction changes in the circuit conductors. This is distinct from DC, where the current flows in a single direction (though its magnitude may vary or remain constant). Step-by-Step Solution: Identify AC: time-varying current that crosses zero.Note that sign reversal implies direction reversal through the load.Recognize periodicity governed by frequency.Select the option stating “changes direction periodically.” Verification / Alternative check: Oscilloscope plots of AC waveforms show positive and negative half-cycles symmetrical about zero, confirming direction reversal every half cycle. Why Other Options Are Wrong: A: Describes pulsating DC (one direction, varying magnitude). B: Describes steady DC. D: “Two directions at the same time” is physically meaningless for a single conductor pair. E: Not applicable; option C is correct. Common Pitfalls: Confusing pulsating DC with AC; assuming RMS values imply unidirectional flow—RMS is a magnitude metric, not direction. Final Answer: The current changes direction periodically

Question 18

Waveform fundamentals: in signal analysis, what is the “period” of a repetitive waveform such as a sine wave or square wave?

Accepted Answer

Introduction / Context: Time-domain descriptions of periodic signals use two key parameters: period (T) and frequency (f). Getting these definitions correct is vital for converting between time and frequency domains and for interpreting oscilloscope measurements and filter behavior. Given Data / Assumptions: We consider a repetitive, periodic waveform (sine, square, triangle, etc.). We want the definition of period T. Standard relationships apply: f = 1 / T. Concept / Approach: The period T is the smallest time interval over which a periodic waveform exactly repeats itself. Equivalently, it is the duration of one complete cycle from any point back to the identical point in phase on the next cycle. Frequency is the reciprocal of period and has units of hertz (cycles per second). Step-by-Step Solution: Identify a full cycle on the waveform (e.g., peak to next peak for a sine wave).Measure the elapsed time between equivalent phase points.That elapsed time is the period T.Select the option defining period as “the time required to complete one full cycle.” Verification / Alternative check: If f = 50 Hz, then T = 1/50 = 0.02 s. Measuring 20 ms between repeating points on an oscilloscope confirms the relationship. Why Other Options Are Wrong: B defines frequency, not period. C describes rise time, a separate specification. D references 0.707 of peak (RMS of a sine), which is an amplitude measure, not time. Common Pitfalls: Confusing frequency with period; using rise time for non-slew-limited signals; mixing amplitude metrics (RMS) with time metrics. Final Answer: the time required to complete one full cycle

Question 19

BJT parameter identification: the (dc) current gain of a bipolar junction transistor is defined as the ratio of collector current to which transistor current?

Accepted Answer

Introduction / Context: Designers often speak of a transistor’s “beta” (β) or h_FE, the dc current gain in common-emitter configuration. Knowing the correct definition is critical when sizing base resistors, estimating load capability, and biasing amplifier stages. Given Data / Assumptions: Device: bipolar junction transistor (BJT), common-emitter usage. Currents: I_C (collector), I_B (base), I_E (emitter). We focus on dc current gain β. Concept / Approach: The dc current gain β is defined as β = I_C / I_B (for a given operating point). The small-signal gain h_fe is a related incremental parameter but conceptually aligned with the ratio of changes in collector and base currents. This definition underpins equations like I_C ≈ β * I_B, with I_E ≈ I_B + I_C. Step-by-Step Solution: Recall β (h_FE) definition: β = I_C / I_B.Identify the denominator: base current I_B.Confirm other currents are not in the definition’s denominator.Choose “base current.” Verification / Alternative check: Datasheets tabulate h_FE or provide graphs of I_C vs I_B at specified V_CE, validating β as the ratio I_C/I_B in common-emitter configuration. Why Other Options Are Wrong: B: I_E is the sum of base and collector currents and is not used in the standard β ratio. C: “Supply current” is ambiguous and not a BJT terminal current. D: Using collector current as both numerator and denominator yields 1, not β. Common Pitfalls: Assuming β is constant across all currents and temperatures; forgetting to check datasheet conditions for h_FE; mixing small-signal h_fe with dc β. Final Answer: base current

Question 20

AC phase relationships: in a purely inductive circuit, how does the current waveform relate in phase to the applied voltage waveform?

Accepted Answer

Introduction / Context: Understanding phase relationships is essential for analyzing reactive AC circuits, power factor correction, and resonance. Inductors and capacitors shift current relative to voltage in opposite ways, which affects real and reactive power flow. Given Data / Assumptions: Purely inductive load (ideal inductor with no resistance). Sinusoidal steady-state excitation. Focus on phase, not magnitude. Concept / Approach: For an ideal inductor, v_L = L * di/dt. A sinusoidal voltage leads the resulting sinusoidal current by 90 degrees; equivalently, current lags voltage by 90 degrees. This stems from the inductor opposing changes in current, causing a delay in current relative to the applied voltage. Step-by-Step Solution: Assume v(t) = V_m * sin(ωt).Then i(t) = (1/ωL) * ∫ v(t) dt = (V_m/(ωL)) * (-cos(ωt)) = (V_m/(ωL)) * sin(ωt − 90°).Therefore, current lags voltage by 90° in an ideal inductor.Select “Current lags the voltage through an inductor.” Verification / Alternative check: Phasor diagrams place the inductor’s impedance at +jX_L; current phasor is 90° behind the voltage across the inductor, confirming the lag. Why Other Options Are Wrong: A: In phase applies to pure resistors. B: “Current leads” describes capacitors, not inductors. D/E: Not applicable because one specific relationship holds for ideal inductors. Common Pitfalls: Forgetting the effect of series resistance (which reduces but does not reverse the lag); confusing capacitor and inductor phase behaviors. Final Answer: Current lags the voltage through an inductor

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