Question 1

Resistor color code validation (5-band): A 2% 55 kΩ fixed resistor would be coded as green, green, black, red, red. Evaluate this statement for correctness.

Accepted Answer

Introduction / Context: Five-band resistor color codes use the first three bands for significant digits, the fourth for the multiplier, and the fifth for tolerance. Verifying codes quickly is essential in troubleshooting and assembly. Given Data / Assumptions: Nominal value: 55 kΩ = 55,000 Ω. Tolerance: 2% (typically red). Five-band scheme: digit-digit-digit, multiplier, tolerance. Concept / Approach: Translate the proposed bands into numbers. Green = 5, black = 0, red multiplier = 10^2, red tolerance = 2%. The first three bands must represent 550 (5, 5, 0). With multiplier 10^2, the value is 550 * 100 = 55,000 Ω = 55 kΩ. Step-by-Step Solution: Band 1: green → 5.Band 2: green → 5.Band 3: black → 0 (three-digit format → 550).Band 4: red multiplier → 10^2 = 100.Band 5: red tolerance → 2%. Verification / Alternative check: Compute nominal: 550 * 100 = 55,000 Ω. Compare to target: 55 kΩ. Tolerance color matches 2%, confirming correctness. Why Other Options Are Wrong: Incorrect: Conflicts with the direct calculation from standardized color assignments. Common Pitfalls: Confusing 4-band with 5-band rules, misplacing the third digit, or misreading red as 10^3 (which is orange). Also, reading the resistor in reverse can swap tolerance and first digit. Final Answer: Correct

Question 2

Ohm’s law check: With a 10 Ω resistor across 20 V, what current flows? Assess the statement “4 A will flow.”

Accepted Answer

Introduction / Context: Ohm’s law is the foundational relationship in basic circuits: V = I * R. Verifying numeric statements quickly is a frequent task during lab work and exams. Here we test a simple current computation. Given Data / Assumptions: Resistor R = 10 Ω. Applied voltage V = 20 V (DC). Assume ideal source and resistor with no temperature effects. Concept / Approach: Use I = V / R. Substitution yields the true current. Compare the calculated current to the claimed 4 A. Step-by-Step Solution: I = V / R.I = 20 / 10.I = 2 A (not 4 A).Therefore, the statement “4 A will flow” is incorrect. Verification / Alternative check: Cross-check with power: P = V * I = 20 * 2 = 40 W, and also P = I^2 * R = 2^2 * 10 = 40 W. Consistent results confirm 2 A. Why Other Options Are Wrong: Correct: Would imply 4 A, which violates Ohm’s law with the given values. Common Pitfalls: Arithmetic slips (dividing 20 by 5 instead of 10). Confusing series/parallel equivalents that change R. Ignoring that doubling current would require either doubling voltage or halving resistance. Final Answer: Incorrect

Question 3

Variable resistors (rheostats and potentiometers): The resistive track can be manufactured with a linear or a tapered (audio/log) law. Evaluate the statement.

Accepted Answer

Introduction / Context: Potentiometers and rheostats are used for volume controls, sensor calibration, and bias adjustments. Their mechanical angle-to-resistance profile, called taper, determines how resistance changes with rotation or slider position. Given Data / Assumptions: Two common tapers: linear and logarithmic (audio). Applications differ: instrumentation versus audio perception. Construction may be carbon film, cermet, or wirewound. Concept / Approach: A linear taper produces resistance proportional to travel, ideal for measurement and control systems needing uniform response. A tapered (log/audio) profile matches human loudness perception; equal angular increments produce perceived equal loudness steps. Other specialty tapers (reverse log, S-curve) exist but are less common. Step-by-Step Solution: Define linear: R changes uniformly with knob angle.Define tapered: R follows a non-linear law (e.g., log) suited to psychoacoustics.Map to use cases: linear for instruments; log for volume controls.Therefore, the statement that tracks can be linear or tapered is correct. Verification / Alternative check: Datasheets identify parts as “A” (log/audio) or “B” (linear), confirming the industry practice of offering both tapers. Why Other Options Are Wrong: Incorrect: Conflicts with ubiquitous availability and labeling of linear and audio tapers. Common Pitfalls: Assuming taper letters are universal across regions (some standards swap A/B). Confusing taper with total resistance value or power rating. Final Answer: Correct

Question 4

Fixed resistors today: Are carbon composition resistors the most common and least expensive type of fixed-value resistor? Assess the statement for modern practice.

Accepted Answer

Introduction / Context: Resistor technologies have evolved. While carbon composition parts were once widespread, modern electronics overwhelmingly uses film and thick-film chip resistors for cost, stability, and availability. This question asks you to evaluate a statement that reflects an older era. Given Data / Assumptions: Contemporary mass production favors SMD thick-film and metal-film resistors. Selection criteria include tolerance, temperature coefficient, noise, surge behavior, and cost. Carbon composition remains niche (e.g., surge/impulse applications) and is relatively costly. Concept / Approach: Compare technologies: carbon composition has higher noise, wider tolerances, and poorer long-term stability than metal-film or thick-film. Large-scale manufacturing of chip resistors drove prices down, making them the most common and economical for general use. Therefore, the statement that carbon composition are the most common and least expensive is not aligned with current industry practice. Step-by-Step Solution: Identify mainstream technology: thick-film SMD and metal-film through-hole.Evaluate cost and availability: these dominate catalogs and reels.Consider performance: better stability and tolerance than carbon composition.Conclusion: the statement is incorrect for modern electronics. Verification / Alternative check: Distributor inventories show far greater selection and lower pricing for film-based resistors; carbon composition is limited and comparatively expensive per unit. Why Other Options Are Wrong: Correct: Would imply a market reality that no longer exists broadly. Common Pitfalls: Assuming historical textbooks reflect current sourcing. Ignoring that some specialty surge applications still specify carbon composition, which does not make them “most common.” Final Answer: Incorrect

Question 5

In electrical insulation and materials engineering, the term for the maximum electric stress a material can withstand without dielectric rupture or flashover is often confused. State whether the following statement is accurate: “The maximum potentia…

Accepted Answer

Introduction / Context: Engineers frequently encounter two closely related terms when discussing insulation limits: breakdown voltage and dielectric strength. Although they are connected, they are not identical. Confusing these terms can lead to poor component selection, over-stressed designs, and avoidable failures in high-voltage power supplies, cables, and capacitors. Given Data / Assumptions: The statement claims a definition for the “maximum potential a material can withstand without rupture.” We are discussing intrinsic properties of insulating media (solid, liquid, gas) versus device- or geometry-specific ratings. No particular thickness or geometry is specified in the statement. Concept / Approach: Dielectric strength is a material property expressed as an electric field value (for example, kV/mm). It indicates the maximum field the material can withstand without breakdown. Breakdown voltage is a system or specimen value that depends on geometry, thickness, surface condition, and environment. For a given sample of thickness t, the breakdown voltage V_bd ≈ (dielectric strength) * t, assuming uniform fields and ideal conditions. Thus, calling the general “maximum potential a material can withstand” a breakdown voltage ignores the crucial dependence on thickness and geometry and is not strictly correct as a general materials definition. Step-by-Step Solution: Identify what the statement defines: a general maximum potential “a material” can withstand.Recognize that materials are characterized by dielectric strength (field), not a universal voltage.Note that breakdown voltage applies to a specific specimen dimension; doubling thickness doubles voltage withstand approximately.Therefore, the statement is not accurate as a general definition. Verification / Alternative check: Standards data sheets list dielectric strength for materials (e.g., 20 kV/mm for a film). To get breakdown voltage for a 0.5 mm sheet, you multiply: V_bd ≈ 20 kV/mm * 0.5 mm = 10 kV. Change thickness and the voltage changes, proving the distinction. Why Other Options Are Wrong: Correct: Incorrect because it confuses a geometry-dependent rating (breakdown voltage) with a material property (dielectric strength). Only true for gases / Only true above 1 kV: The distinction holds for solids, liquids, and gases across voltage ranges. Depends only on temperature: Temperature affects limits but does not change the definition. Common Pitfalls: Using “breakdown voltage” as a universal material constant; ignoring thickness, edges, humidity, and contamination, which all impact actual breakdown voltage of a specimen. Final Answer: Incorrect

Question 6

In electronics prototyping practice, a solderless breadboard (also called a protoboard) is commonly used to assemble and test experimental circuits without permanent soldering. Assess the statement: “A protoboard is used to make experimental circuit…

Accepted Answer

Introduction / Context: Solderless breadboards, often called protoboards, are ubiquitous in labs and classrooms. They allow rapid iteration, reuse of components, and safe exploration of circuit ideas before committing to printed circuit boards. Understanding their intended role helps learners select the right construction method for a given task. Given Data / Assumptions: We are evaluating the typical use case of a protoboard. The context is general-purpose analog and digital electronics experiments. No high-frequency or high-power edge cases are assumed. Concept / Approach: A protoboard provides spring contacts arranged in rows and columns to quickly interconnect ICs, resistors, capacitors, and wires. Because connections are temporary and solder-free, the platform is ideal for experimental circuits, debugging, and short-term demonstrations. For production, vibration-heavy environments, or radio-frequency applications, other construction methods are preferred, but the statement’s general claim remains true. Step-by-Step Solution: Identify the function of a protoboard: rapid, solderless interconnection.Match that function to “experimental circuits”: prototyping and testing fit perfectly.Conclude that the statement aligns with common and intended usage.Therefore the statement is correct. Verification / Alternative check: Standard lab kits and educational courses specify protoboards for initial builds. Once a design stabilizes, it is migrated to perfboard, stripboard, or a PCB—confirming the experimental focus of protoboards. Why Other Options Are Wrong: Incorrect: Conflicts with widespread practice. Only correct for analog / below 1 kHz / depends on PCB substrate: Breadboards are used for both analog and digital, within reasonable frequency limits; PCB substrate is unrelated to a protoboard’s purpose. Common Pitfalls: Trying to operate at very high frequencies on a protoboard (excessive stray capacitance/inductance), or drawing high currents through the thin spring contacts—these are limitations, not contradictions of the primary use case. Final Answer: Correct

Question 7

Instrument capability clarification: Evaluate the statement: “A digital voltmeter can be used to measure resistance.” (Assume a device marketed strictly as a digital voltmeter.)

Accepted Answer

Introduction / Context: Test instruments are often named for their primary function, and that name matters. A digital voltmeter measures voltage, whereas a digital multimeter (DMM) measures multiple quantities such as voltage, current, and resistance. Understanding the distinction prevents misuse and incorrect expectations in the lab. Given Data / Assumptions: The instrument in question is a digital voltmeter (DVM), not a multimeter. The statement claims resistance measurement capability for a DVM. We assume no add-on fixtures or external sources are used. Concept / Approach: Measuring resistance requires injecting a known current through the unknown resistor and measuring the resulting voltage (or applying a known voltage and measuring current). A DMM provides an internal source and measurement path for this; a DVM typically does not. Therefore, a dedicated DVM lacks the native circuitry to perform ohmic measurements unless it includes a resistance function, in which case it is by definition a multimeter or a specialized ohmmeter—not a pure voltmeter. Step-by-Step Solution: Define instrument scope: DVM = voltage only.Define requirement for resistance: needs internal source and computation R = V/I.Note absence of source path in a pure DVM.Conclude that the statement is incorrect unless the device is actually a DMM. Verification / Alternative check: Product datasheets label meters with resistance ranges (Ω) as multimeters. Pure voltmeters list DCV/ACV ranges only—no Ω function present—confirming the functional distinction. Why Other Options Are Wrong: Correct: Would only be true if the device were a multimeter, not a dedicated voltmeter. External current source: With elaborate setups a DVM can participate, but that is not what the statement implies. Only above 10 kΩ / analog meters only: These are unrelated and inaccurate limitations. Common Pitfalls: Using “DVM” as a generic synonym for DMM; assuming any digital meter can measure all quantities. Always check the meter’s front panel and specifications. Final Answer: Incorrect

Question 8

Reliability of variable resistors (potentiometers and rheostats): Consider the maintenance issue: “Variable resistors often develop problems with the wiper making reliable contact with the resistive track.” Evaluate this statement.

Accepted Answer

Introduction / Context: Variable resistors—potentiometers and rheostats—use a sliding wiper that contacts a resistive element. Over time, environmental contaminants, wear, and oxidation can degrade that contact, leading to noisy or intermittent operation. Recognizing this common failure mode is helpful in troubleshooting audio equipment, controls, and instrumentation. Given Data / Assumptions: Devices considered: carbon composition, cermet, conductive plastic, and wirewound potentiometers. Operating in typical lab or field environments. No special hermetic sealing unless specified. Concept / Approach: The wiper-to-track interface is a mechanical contact subject to friction and contamination. Dust, smoke, and moisture can infiltrate the housing, while mechanical wear creates debris. Oxidation films increase contact resistance and create scratching or “crackling” noise in audio paths. Even in wirewound types with robust windings, the wiper junction can degrade. Hence wiper contact issues are a well-known reliability concern for variable resistors. Step-by-Step Solution: Identify the mechanical interface: metal wiper to resistive element.List degradation mechanisms: wear, oxidation, contamination, and mechanical shock.Relate symptoms: intermittent resistance, noise when adjusted, sudden open circuits.Conclude that the statement describing wiper contact problems is correct. Verification / Alternative check: Maintenance practices include contact cleaning sprays, exercising the pot, or replacing with sealed or digital (encoder plus DAC) alternatives—evidence that wiper contact reliability is a known issue manufacturers mitigate in datasheets and application notes. Why Other Options Are Wrong: Incorrect: Contradicts common field experience. Only true for wirewound / above 1 A / sealed enclosures: Contact problems can occur across types and at small signal levels; sealing helps but does not eliminate wear. Common Pitfalls: Assuming that increasing contact pressure always helps (it can accelerate wear), and overlooking that DC across the wiper can create micro-arcing and residue, worsening noise. Final Answer: Correct

Question 9

Power dissipation calculation (Ohm’s law and power law): A 12 V source supplies 3 A to a resistor. Evaluate the statement: “The power dissipated by the resistor is 4 W.”

Accepted Answer

Introduction / Context: Accurate power calculations are critical for component selection and thermal design. Overheating resistors can fail catastrophically. This question checks quick application of the fundamental relation between voltage, current, and power in a purely resistive DC circuit. Given Data / Assumptions: DC source voltage V = 12 V. Load current I = 3 A. Resistive load; no reactive elements are implied. Concept / Approach: For DC resistive circuits, power dissipated by a load is P = V * I. Alternative forms using resistance are P = I^2 * R and P = V^2 / R, but those require R. With both V and I known, the direct product gives the correct result immediately. Step-by-Step Solution: Use the power identity: P = V * I.Substitute values: P = 12 * 3.Compute: P = 36 W.Compare to the claim “4 W”: the claim is incorrect by a factor of 9. Verification / Alternative check: Determine R from Ohm’s law: R = V / I = 12 / 3 = 4 Ω. Then compute P with P = I^2 * R = 3^2 * 4 = 9 * 4 = 36 W. Both methods agree, confirming the result. Why Other Options Are Wrong: Correct / correct only if R = 1.33 Ω: Neither makes the 4 W figure right for this V and I. Reactive load / indeterminate with phase angle: In DC with a resistor, there is no phase; P is unambiguously V * I. Common Pitfalls: Dropping a zero or misreading units; confusing wattage rating (how much a resistor can handle) with power actually dissipated; mixing up AC reactive cases with straightforward DC loads. Final Answer: Incorrect

Question 10

Component failure modes — fixed resistors: Assess the statement: “Fixed-value resistors can internally short-circuit.”

Accepted Answer

Introduction / Context: Understanding typical and atypical failure modes of passive components helps technicians diagnose faults efficiently. While open-circuit failures are more common for resistors, short-circuit failures, though rarer, can occur and carry serious risk for downstream circuitry. Given Data / Assumptions: Resistor types considered: carbon film, metal film, thick/thin film, wirewound, carbon composition. Operating in realistic environments with possible surge, overheating, or contamination. We are discussing failure possibilities, not probabilities. Concept / Approach: Resistors can fail open due to element cracking or fusing. Shorts can result from carbonization of encapsulant after severe overheating, melted or collapsed spacing between turns in wirewound devices, conductive residue bridging terminations, or moisture ingress creating low-resistance paths. Though statistically less frequent than opens, shorts are a legitimate failure mode acknowledged by manufacturers and failure analysis reports. Step-by-Step Solution: Identify common open-circuit mechanisms (element fracture, burn-out).List short-circuit mechanisms (carbonized paths, melted insulation, conductive contamination).Relate to specific constructions (wirewound turn-to-turn shorts; film resistors with cracked coatings and carbon tracks).Conclude that internal shorts are possible → the statement is correct. Verification / Alternative check: Field reliability data and failure-analysis photos often show charred bodies with measurable low resistance across leads after over-stress—a practical demonstration of short failure modes. Why Other Options Are Wrong: Incorrect: Ignores documented short mechanisms. True only for certain types or ratings: While mechanisms vary, shorts are not exclusive to a single type or power rating. Depends solely on humidity: Moisture is one factor, not the only cause. Common Pitfalls: Assuming a resistor “can only go open.” This bias can lead to missed diagnoses when a supply rail is unexpectedly pulled low by a failed resistor. Final Answer: Correct

Question 11

Terminology clarity — “load resistance”: Evaluate the statement: “A load resistance is the resistance of the load.”

Accepted Answer

Introduction / Context: Electronics relies on shared vocabulary. The term “load” refers to the device or network that consumes power from a source. Understanding that “load resistance” is simply the equivalent resistance presented by that load helps with circuit analysis, voltage divider design, and matching considerations. Given Data / Assumptions: We use the word “load” in its standard circuit sense. We consider both DC and AC contexts but focus on the resistance concept. If the load is not purely resistive, an equivalent resistance or impedance may be used depending on frequency. Concept / Approach: For resistive loads, the load can be modeled by a single resistance R_load. For complex loads in AC, the generalized quantity is impedance Z_load, which includes resistance and reactance. When someone says “load resistance,” they either mean the pure resistive case or the resistive part of the impedance. In plain language, it is indeed the resistance of the load as “seen” by the source. Step-by-Step Solution: Define load: the element(s) drawing power from the source.Define load resistance: the equivalent resistance of that load under the operating condition.Note that in AC, we may instead speak of load impedance; resistance is the real component.Thus, the statement is an accurate plain-language definition. Verification / Alternative check: Basic circuit textbooks define R_load in divider equations such as V_out = V_in * (R_load / (R_source + R_load)), illustrating the use of “load resistance” as the resistance attributable to the load connection. Why Other Options Are Wrong: Incorrect: Contradicts standard terminology. Only for AC/DC: The definition holds regardless of frequency; only the modeling approach changes. Depends on power factor: Power factor relates to AC impedance angle, not the definition of resistance. Common Pitfalls: Mixing “resistance” and “impedance.” While related, impedance includes reactance; still, calling a purely resistive load’s opposition “load resistance” is fully correct. Final Answer: Correct

Question 12

Ohm’s law relationship (clarified): For a fixed applied voltage, assess the statement: “Resistance is inversely proportional to current.”

Accepted Answer

Introduction / Context: Ohm’s law relates voltage, current, and resistance in linear (ohmic) conductors. Many statements derived from Ohm’s law are correct only under specific conditions. Here the phrase “for a fixed applied voltage” is the crucial qualifier that makes the inverse proportion statement meaningful. Given Data / Assumptions: Linear, ohmic behavior is assumed (R constant over the operating region). The applied voltage is held constant. Temperature effects and non-linear devices are ignored for this basic relationship check. Concept / Approach: Ohm’s law is V = I * R. With V fixed, I = V / R. This shows I is inversely proportional to R. Equivalently, solving for R gives R = V / I, implying R is inversely proportional to I at constant V. This inverse relationship underpins behaviors such as the decrease in current when a series resistance is increased while the supply voltage is unchanged. Step-by-Step Solution: Start with V = I * R.Hold V constant; rearrange: I = V / R.Observe inverse proportionality: as R increases, I decreases proportionally.Therefore, the statement is correct under the stated condition. Verification / Alternative check: Example: With V = 10 V, R1 = 1 kΩ → I1 = 10 mA. If R2 = 2 kΩ, I2 = 5 mA. Doubling R halves I, confirming inverse proportionality at fixed V. Why Other Options Are Wrong: Incorrect: Would contradict algebra from Ohm’s law. Only for non-ohmic devices / above 10 V / indeterminate: These caveats are irrelevant or misleading; the relationship is general for ohmic devices at any voltage where linearity holds. Common Pitfalls: Forgetting to specify the held variable; without “fixed voltage,” proportionality statements can be ambiguous. Also, some components (lamps, semiconductors) are non-ohmic—Ohm’s law may apply instantaneously but with R not constant. Final Answer: Correct

Question 13

Tolerance interpretation — resistor value range: A resistor marked 9.9 kΩ, tolerance ±10%, may have an actual value in what range? Evaluate the statement: “between 8.91 kΩ and 10.89 kΩ.”

Accepted Answer

Introduction / Context: Resistor tolerances define the permissible deviation from the nominal value at the time of manufacture under specified conditions. Being able to compute tolerance bands quickly is essential when checking component bins, assembling filters, and ensuring bias networks fall within design margins. Given Data / Assumptions: Nominal resistance R_nom = 9.9 kΩ. Tolerance = ±10% (i.e., ±0.10 of nominal). Room-temperature initial tolerance is implied; long-term drift and temperature coefficient are separate specifications. Concept / Approach: For a tolerance of ±T, the minimum and maximum initial values are R_min = R_nom * (1 − T) and R_max = R_nom * (1 + T). With T = 0.10, compute the endpoints and compare to the claim. This is a straightforward percentage calculation based on the nominal value. Step-by-Step Solution: Compute lower bound: R_min = 9.9 kΩ * 0.90 = 8.91 kΩ.Compute upper bound: R_max = 9.9 kΩ * 1.10 = 10.89 kΩ.Compare with the statement range: exactly matches 8.91 kΩ to 10.89 kΩ.Therefore, the statement is correct. Verification / Alternative check: Express deviation in absolute terms: 10% of 9.9 kΩ is 0.99 kΩ. Bounds are 9.9 − 0.99 = 8.91 kΩ and 9.9 + 0.99 = 10.89 kΩ—same result by a different arithmetic path. Why Other Options Are Wrong: Incorrect: Conflicts with basic percentage math. Only for 5% / only at 25 °C / needs power rating: Tolerance is specified independently of power rating; temperature and aging are separate specs. Common Pitfalls: Confusing tolerance (initial manufacturing spread) with temperature coefficient (change with temperature) or with derating due to power dissipation. Also, mixing kΩ and Ω in calculations can create decimal errors. Final Answer: Correct

Question 14

Ohmmeter principle of operation: Evaluate the statement: “An ohmmeter uses an internal battery to measure the resistance of a resistor.”

Accepted Answer

Introduction / Context: Resistance cannot be measured directly without applying a stimulus. Both analog and digital ohmmeters contain an internal source to drive a small current through the device under test, then infer resistance from the measured voltage or current. Recognizing this principle explains common cautions such as “never measure resistance on a live circuit.” Given Data / Assumptions: Device: ohmmeter function, whether stand-alone or inside a digital multimeter. Unknown resistor is passive and disconnected from external power. Standard two-wire measurement method is used. Concept / Approach: An ohmmeter sources a known current I_test (or applies a known voltage V_test) from its internal battery. It then measures the corresponding voltage V_DUT across the resistor or the current through it and computes R using R = V_DUT / I_test or R = V_test / I_DUT. The internal battery is essential for this stimulus. Consequently, if the battery is weak, resistance readings drift or the meter indicates “low battery.” Step-by-Step Solution: Recognize measurement method: apply stimulus using internal source.Relate to formula: R = V / I, with one quantity fixed/known by the meter.Understand why live circuits must be avoided: external voltages corrupt readings and may damage the meter.Therefore, the statement about the internal battery is correct. Verification / Alternative check: User manuals specify open-circuit test voltages and test currents for various ranges (for example, 0.3 V test on sensitive ranges). This confirms the use of an internal source derived from the meter’s battery. Why Other Options Are Wrong: Incorrect: Conflicts with the fundamental measurement method. Only true for analog / only above 1 kΩ / depends on lead capacitance: Modern digital multimeters also use internal sources; lead capacitance is irrelevant to static resistance measurement. Common Pitfalls: Attempting to measure resistance in-circuit with power applied (backfeeds the meter); measuring components in parallel without isolating them, which lowers the apparent resistance. Final Answer: Correct

Question 15

In conductor sizing practice (using circular mils and feet), is it correct to express electrical resistivity in the customary unit Ω·cmil/ft (ohm–circular-mil per foot)? Provide the most appropriate evaluation of this unit statement.

Accepted Answer

Introduction / Context: Resistivity describes how strongly a material opposes current flow. In SI it is commonly given in Ω·m. However, electricians and wire tables in North America often use circular mils (cmil) and feet (ft) to size conductors. This question checks whether Ω·cmil/ft is a legitimate customary unit for resistivity in that context. Given Data / Assumptions: Conductor cross-section sometimes specified in circular mils (1 circular mil = diameter in mils squared). Length measured in feet. We are discussing material property ρ and the design constant used in wire tables. Concept / Approach: Ohm’s law for a uniform conductor is R = ρ * (L / A). If A is expressed in circular mils and L in feet, then ρ must carry units that cancel A and L to produce ohms. Therefore, a consistent customary unit is Ω·cmil/ft. In many cable tables, a material constant k is tabulated in Ω·cmil/ft at a stated temperature (for copper near 20 °C, k is about 10.4 Ω·cmil/ft). Designers compute R by R = k * L / A, with L in feet and A in circular mils. Step-by-Step Solution: Start from R = ρ * L / A.Choose L in ft and A in cmil.Require ρ to have units Ω·cmil/ft so that R is in ohms.Conclude that Ω·cmil/ft is a valid customary resistivity unit. Verification / Alternative check: Compare with SI: if A is in m^2 and L in m, then ρ is Ω·m. Unit systems change the numeric value but not the physics; the formula remains R = ρ * L / A. Why Other Options Are Wrong: Ω/ft^2 (option b) would make R dimensionless when used in R = ρ * L / A; it is inconsistent. “Only Ω·m” (option c) ignores accepted industry practice in AWG/cmil systems. cmil/Ω·ft (option d) inverts units and would not yield ohms from R = ρ * L / A. Common Pitfalls: Confusing resistivity (material property) with resistance (component property), and mixing SI with customary units without converting. Also, writing the order as cmil·Ω/ft or Ω·cmil/ft is equivalent multiplicatively, but clarity favors Ω·cmil/ft. Final Answer: Valid — Ω·cmil/ft is a correct customary unit for resistivity used with circular-mil and feet measurements.

Question 16

Mechanical–electrical equivalence: Evaluate the statement “One horsepower equals 746 watts.” Choose the most accurate assessment for engineering calculations.

Accepted Answer

Introduction / Context: Power conversions between imperial and SI units are common in motors, generators, and mechanical drives. Horsepower (hp) is frequently converted to watts (W) to compare specifications or compute energy consumption. Given Data / Assumptions: Using mechanical (imperial) horsepower. Steady power; no efficiency correction included. Standard engineering approximation is acceptable. Concept / Approach: By definition, 1 mechanical horsepower ≈ 550 ft·lbf/s. Converting to SI: 1 ft = 0.3048 m and 1 lbf = 4.4482216 N, so 1 hp ≈ 550 * 4.4482216 * 0.3048 W ≈ 745.7 W. In practice, engineers round to 746 W for quick calculations. Electrical horsepower used by some utilities is also set to 746 W for convenience. Step-by-Step Solution: Start with 1 hp = 550 ft·lbf/s.Convert ft·lbf to N·m: 1 ft·lbf = 1.3558179 N·m.Compute: 550 * 1.3558179 W ≈ 745.7 W.Round to 746 W for engineering use. Verification / Alternative check: Check motor nameplates: power is often given in hp and kW, with 1 hp listed near 0.746 kW. This cross-verifies the standard equivalence. Why Other Options Are Wrong: Exactly 1000 W (option b) confuses horsepower with kilowatt. ≈ 550 W (option c) confuses force–distance units with power directly. “Unit of energy” (option d) is incorrect; horsepower is a rate (power), analogous to watts. Common Pitfalls: Mixing metric horsepower (PS ≈ 735.5 W) with mechanical horsepower; both are close but not identical. Always check which definition applies. Final Answer: Correct — 1 hp ≈ 746 W.

Question 17

Component terminology check: Is a rheostat considered a type of variable resistor used to adjust current or resistance?

Accepted Answer

Introduction / Context: Rheostats and potentiometers are common adjustable resistive components. Understanding their definitions and typical connections avoids specification and wiring errors in labs and industry. Given Data / Assumptions: Rheostat: high-power adjustable resistor, commonly wired as a 2-terminal device. Potentiometer: 3-terminal resistor with a wiper; can be used as voltage divider or variable resistor. Goal: determine whether a rheostat counts as a variable resistor. Concept / Approach: A variable resistor is any resistor whose resistance can be adjusted. A potentiometer is a variable resistor with three terminals; when two terminals are used (one end and the wiper), it functions as a rheostat. Many rheostats are physically similar to potentiometers but designed for higher current/power. Therefore, a rheostat is indeed a type of variable resistor. Step-by-Step Solution: Define variable resistor: a component with adjustable resistance.Define rheostat: 2-terminal adjustable resistor (often one end plus wiper of a potentiometer), typically higher power.Conclude: a rheostat is a variable resistor. Verification / Alternative check: Handbooks specify wiring a potentiometer as a rheostat by tying one end to the wiper, leaving two terminals to vary series resistance in a circuit. Why Other Options Are Wrong: Fixed high-power resistor (option b) lacks adjustability. Inductor (option c) is a different passive component. “Only a voltage divider” (option d) ignores the common 2-terminal rheostat configuration. Common Pitfalls: Assuming “potentiometer” and “rheostat” are interchangeable without noting connection method and power rating. Always check the application and required wattage. Final Answer: Correct — a rheostat is a variable resistor.

Question 18

Power rating adequacy: A 10 Ω resistor carries 500 mA continuously. Is selecting a 2 W resistor adequate for safe operation?

Accepted Answer

Introduction / Context: Selecting an appropriate resistor wattage rating is critical for reliability and safety. The rating must exceed the expected power dissipation, typically with margin to account for temperature rise, airflow, enclosure, and tolerance. Given Data / Assumptions: Resistance R = 10 Ω. Current I = 0.5 A (500 mA), steady-state. Continuous operation; ambient conditions not specified (assume typical lab/room temperature). Concept / Approach: Power in a resistor is computed using P = I^2 * R = V^2 / R = V * I. After calculating the actual dissipation, choose a resistor with a power rating comfortably above that value. Many designers target 1.5× to 2× margin depending on environment and derating guidelines. Step-by-Step Solution: Compute power: P = I^2 * R = (0.5 A)^2 * 10 Ω = 0.25 * 10 = 2.5 W.Compare with proposed rating: 2 W < 2.5 W, so the part would be overstressed.Apply margin: choose ≥ 3 W (preferably 5 W if enclosure is warm or airflow is limited). Verification / Alternative check: Using P = V * I: first compute V = I * R = 0.5 * 10 = 5 V. Then P = 5 V * 0.5 A = 2.5 W, confirming the result. Why Other Options Are Wrong: “Adequate — 2 W” (option a) is below required 2.5 W; the resistor will overheat. “Adequate — 1.5 W” (option c) miscalculates power. “Not adequate — 5 W dissipation” (option d) overstates the load; required dissipation is 2.5 W, not 5 W. Common Pitfalls: Ignoring derating for temperature, using nameplate ratings at 25 °C without considering enclosure heating, and rounding current/voltage before squaring in P = I^2 * R. Final Answer: Not adequate — actual dissipation is 2.5 W; select at least a 3 W resistor (with margin).

Question 19

Conductor design principle: Does increasing a conductor’s cross-sectional area reduce its resistance and therefore affect how much it opposes current?

Accepted Answer

Introduction / Context: Wire sizing impacts voltage drop, heating, and efficiency. Understanding how geometry influences resistance helps in selecting conductors for power distribution and signal integrity. Given Data / Assumptions: Uniform conductor of length L and cross-sectional area A. Material has resistivity ρ at the operating temperature. Direct current conditions (but the geometric relation holds for AC resistance ignoring skin effect). Concept / Approach: The basic relationship for resistance is R = ρ * L / A. For fixed material (ρ constant) and length L, R is inversely proportional to area. Doubling area halves resistance, which reduces voltage drop and I^2R heating for a given current. Step-by-Step Solution: State the model: R = ρ * L / A.Hold ρ and L constant; vary A.Observe inverse relationship: as A increases, R decreases proportionally. Verification / Alternative check: Compare AWG tables: lower AWG numbers (thicker wires, larger area) exhibit lower resistance per unit length, confirming the formula. Why Other Options Are Wrong: “Only length matters” (option b) contradicts R ∝ L / A. “Only resistivity matters” (option c) ignores geometry. “Increasing area increases resistance” (option d) reverses the proportionality. Common Pitfalls: Overlooking temperature dependence of ρ and ignoring skin effect at high frequencies, which effectively reduces the conductive area. Final Answer: Applies — larger cross-section reduces resistance (R ∝ 1/A).

Question 20

Ohm’s law behavior in a purely resistive circuit: If the applied voltage increases while resistance is constant, does the current increase as well?

Accepted Answer

Introduction / Context: Ohm’s law is foundational for DC and low-frequency AC analysis in resistive circuits. This question checks proportional relationships among voltage, current, and resistance. Given Data / Assumptions: Purely resistive element with constant resistance R (no reactive components). Voltage V is varied; R is held constant. Temperature changes are negligible so R does not drift. Concept / Approach: Ohm’s law states V = I * R. For a fixed R, current I = V / R, so I is directly proportional to V. Doubling V doubles I, and halving V halves I. This proportionality holds for both DC and AC at frequencies where the resistor’s behavior is ideal. Step-by-Step Solution: Start with V = I * R.Solve for current: I = V / R with R constant.Conclude: as V increases, I increases proportionally. Verification / Alternative check: Power relation P = V * I = V^2 / R shows power grows with the square of voltage at fixed R, consistent with I rising linearly with V. Why Other Options Are Wrong: Current unchanged (option b) contradicts I = V / R. Current decreases (option c) reverses the dependence. “Depends on power factor” (option d) is irrelevant in a purely resistive load where power factor is 1. Common Pitfalls: Ignoring that real resistors heat and change resistance slightly; within normal ranges, the ideal model is accurate for analysis. Final Answer: Yes — current increases with voltage when R is constant.

Resistance and Power Questions