Question 1

Using Ohm’s law, how many ohms of resistance are required to allow a current of 720 µA to flow when an applied voltage of 3.6 kV is present (assume linear, dc conditions)?

Accepted Answer

Introduction / Context: Calculating resistance from a specified current and voltage is a direct application of Ohm’s law. Accurate unit conversion (kV to V, µA to A) is crucial to avoid orders-of-magnitude errors that could lead to unsafe component choices. Given Data / Assumptions: Voltage V = 3.6 kV = 3600 V. Current I = 720 µA = 720 * 10^-6 A = 0.00072 A. Ideal resistive behavior; temperature effects neglected. Concept / Approach: Ohm’s law: R = V / I. Convert quantities to SI units first, then divide. Always cross-check reasonableness by computing the implied power to ensure the resistor wattage is practical. Step-by-Step Solution: Convert units: V = 3600 V; I = 0.00072 A.Compute resistance: R = V / I = 3600 / 0.00072.Perform division: 3600 / 0.00072 = 5,000,000 Ω.Express neatly: R = 5 MΩ.Optional power check: P = V * I = 3600 * 0.00072 = 2.592 W → select an adequate wattage rating and high-voltage resistor construction. Verification / Alternative check: Recompute using scientific notation: 3.6 * 10^3 / 7.2 * 10^-4 = (3.6 / 7.2) * 10^(3 + 4) = 0.5 * 10^7 = 5 * 10^6 Ω. Why Other Options Are Wrong: 200 nΩ and 5 kΩ are far too low; they would allow enormous currents. 200 kΩ is too low by a factor of 25. 50 MΩ is ten times too high, limiting current to 72 µA, not 720 µA. Common Pitfalls: Forgetting to convert microamps to amps or kilovolts to volts, producing errors of 10^3–10^6. Final Answer: 5 MΩ

Question 2

With Ohm’s law in mind, if the applied voltage increases while the resistance remains unchanged, what happens to the circuit current (qualitatively)?

Accepted Answer

Introduction / Context: Predicting how current responds to changes in voltage or resistance is foundational for circuit design. This understanding helps in component derating, fuse sizing, and thermal management. Given Data / Assumptions: Resistance is constant (linear resistor). Applied voltage increases. Ambient and temperature effects ignored. Concept / Approach: Ohm's law states I = V / R. With R fixed, any increase in V proportionally increases I. Power also changes: P = V * I = V^2 / R, implying a quadratic rise in power with voltage increases. Step-by-Step Solution: Start from I = V / R.Increase V; R constant implies I must scale up linearly with V.Implication: P rises as V^2 / R, stressing components and possibly requiring higher wattage ratings. Verification / Alternative check: Example: If V rises from 5 V to 10 V across 1 kΩ, current doubles from 5 mA to 10 mA, confirming the direct proportionality. Why Other Options Are Wrong: “Current remains the same” contradicts I = V / R. “Power decreases” is false; power increases with V at constant R. “Resistance decreases” changes a different variable; R is constant by premise. “Current decreases” is opposite of the correct relationship. Common Pitfalls: Neglecting power dissipation when increasing voltage can cause resistor overheating or device failure. Final Answer: current increases

Question 3

From a foundational physics perspective, which list most accurately represents six basic forms of energy commonly introduced in electrical and mechanical engineering contexts?

Accepted Answer

Introduction / Context: Energy classification provides a conceptual map for understanding how systems convert and utilize energy. In entry-level electrical and mechanical engineering, a common enumeration includes electrical, mechanical, radiant (light), thermal (heat), magnetic, and chemical energy. Given Data / Assumptions: Educational context favors broadly used forms seen in labs and industry. Focus on categories relevant to electric machines, electronics, and common physical processes. “Sun” is not a form of energy; it is a source. “Potential” is a mechanical subcategory rather than a distinct everyday list item in this framing. Concept / Approach: Systems convert energy among forms: batteries convert chemical to electrical; motors convert electrical to mechanical; heaters convert electrical to thermal; lamps/LEDs convert electrical to light; inductors and transformers store/transmit energy magnetically. These categories align with the listed six in the correct option. Step-by-Step Solution: Identify forms: electrical, mechanical, light (radiant), heat (thermal), magnetic, chemical.Relate to devices: battery (chemical→electrical), motor (electrical→mechanical), resistor (electrical→heat), lamp/LED (electrical→light), inductor/transformer (electromagnetic/magnetic energy).Conclude the option that matches these categories. Verification / Alternative check: Introductory physics and engineering texts consistently treat these as standard categories in practical engineering examples. Why Other Options Are Wrong: Options with “sun” misuse a source instead of a form. “Potential” is a subset of mechanical energy (gravitational/elastic), not a separate everyday category here. Other lists mix sources and specialized forms (e.g., tidal/wind) rather than the general forms taught alongside electric circuits. Common Pitfalls: Confusing sources (solar) with forms (radiant/light) and mixing system-level phenomena with energy categories. Final Answer: electrical, mechanical, light, heat, magnetic, and chemical

Question 4

Considering standard component families, how many basic types of resistors are commonly recognized at the top level of classification (grouping variants underneath)?

Accepted Answer

Introduction / Context: Resistors are fundamental components used to oppose current flow and set voltages and currents. For selection and teaching purposes, it is helpful to group resistors into broad, top-level categories before exploring specialized variants. Given Data / Assumptions: Top-level classification is desired, not an exhaustive list of subtypes. Common educational convention separates resistors by whether their value is intended to remain fixed or be user-adjustable. Special devices (thermistors, photoresistors, varistors) are considered variants within broader categories. Concept / Approach: The two most common top-level resistor categories are fixed resistors and variable resistors. Fixed resistors include carbon film, metal film, metal oxide, and wirewound. Variable resistors include potentiometers, rheostats, trimmers, and digitally controlled potentiometers. Many sensing resistors (thermistors, LDRs) are often treated as variable/special-purpose resistors due to their resistance dependence on external stimuli. Step-by-Step Solution: Identify the broadest meaningful split: fixed vs variable.Map typical parts: fixed (metal film, wirewound), variable (potentiometer, trimmer).Conclude that two basic types suffice at the top level. Verification / Alternative check: Textbook chapters and component catalogs commonly structure resistor families under fixed and variable headings, with specialized devices listed under variable/sensor categories. Why Other Options Are Wrong: One type ignores clear usage differences. Three or more types typically enumerate subcategories rather than top-level classes. Common Pitfalls: Confusing packaging (through-hole vs SMD) with functional type, and counting special sensors as separate top-level families rather than variable variants. Final Answer: 2

Question 5

In practical electronics, how many terminals (connections) does a potentiometer have for use as a three-terminal adjustable voltage divider?

Accepted Answer

Introduction / Context: A potentiometer (pot) is a variable resistor commonly used to adjust signal levels, set reference voltages, and create user controls (e.g., volume knobs). Understanding its terminals is crucial for correct wiring as a rheostat (2-terminal use) or as a voltage divider (3-terminal use). Given Data / Assumptions: Standard rotary or slider potentiometer. Used as a voltage divider unless otherwise stated. Mechanical wiper provides a variable contact along the resistive track. Concept / Approach: A potentiometer has two end terminals connected to the ends of a resistive element and a third terminal connected to a movable wiper. As the wiper moves, it partitions the total resistance into two parts, producing an adjustable output voltage proportional to the wiper position when the end terminals are tied to a fixed potential difference. Step-by-Step Solution: Identify terminals: two ends of the resistive track plus one wiper.Wire as divider: connect ends to reference voltages (e.g., Vcc and ground); take output from wiper to obtain a variable fraction of Vcc.Alternative 2-terminal use: short one end to the wiper to act as a variable series resistor (rheostat). Verification / Alternative check: Datasheets show three pins labeled A (end), W (wiper), and B (end). Schematics depict the wiper arrow contacting the resistive element. Why Other Options Are Wrong: One or two terminals cannot realize a proper voltage divider. Four or more terminals are found only in specialized multi-gang or dual pots, not a single standard potentiometer section. Common Pitfalls: Leaving the wiper floating or miswiring ends can cause noise or full-scale jumps; always reference the wiper correctly. Final Answer: 3

Question 6

In circuit terminology, which components are explicitly designed to oppose the flow of electric current and thereby control voltage and current levels in a network?

Accepted Answer

Introduction / Context: Managing current and voltage in electronic circuits requires components that introduce controlled opposition to current flow. The primary passive element used for this purpose is the resistor, which obeys Ohm’s law and dissipates power as heat. Given Data / Assumptions: Standard low-voltage electronic circuits are considered. We differentiate functional roles of materials and components. Linear, temperature-stable behavior is assumed for ideal resistors. Concept / Approach: Resistors present a specified resistance R in ohms, producing a voltage drop V = I * R proportional to current I. They are deliberately selected to set bias currents, create voltage dividers, sense current (via small-value shunts), and limit inrush currents. While insulators prevent current altogether and conductors allow it readily, resistors provide controlled, predictable opposition. Step-by-Step Solution: State role: oppose current in a controlled, quantifiable way.Apply Ohm’s law: V = I * R across a resistor.Recognize uses: biasing, limiting, sensing, dividing. Verification / Alternative check: Circuit simulations and measurements consistently show that adding resistance reduces current for a given applied voltage, in line with Ohm’s law. Why Other Options Are Wrong: Insulators ideally block current (infinite resistance) and are not circuit elements for controlled opposition. Conductors facilitate current flow; transformers transfer energy via magnetic coupling; heat exchangers are mechanical thermal devices. Common Pitfalls: Confusing high-resistance insulators with designed resistors; resistors have precise values and tolerances for circuit control. Final Answer: resistors

Question 7

For a 470 Ω resistor carrying 1 mA of current, determine a suitable power (wattage) rating from the given choices, based on calculated dissipation and standard derating practice.

Accepted Answer

Introduction / Context: Selecting an appropriate resistor wattage rating prevents overheating and improves reliability. The rating should exceed the expected power dissipation with ample margin, considering ambient temperature and enclosure conditions. Given Data / Assumptions: Resistance R = 470 Ω. Current I = 1 mA = 0.001 A. Continuous operation; standard ambient conditions; typical derating margin applied. Concept / Approach: Resistor power dissipation is P = I^2 * R or P = V^2 / R. Compute actual dissipation, then choose a commercial wattage rating comfortably above that value. Common preferred ratings are 1/8 W (0.125 W), 1/4 W (0.25 W), 1/2 W (0.5 W), etc. Step-by-Step Solution: Compute power: P = I^2 * R = (0.001)^2 * 470 = 1e-6 * 470 = 0.00047 W.Convert to milliwatts: 0.00047 W = 0.47 mW.Compare with ratings: even 1/8 W = 125 mW greatly exceeds 0.47 mW.Choose a standard rating: 1/4 W is a very common minimum stocked value for through-hole parts, providing abundant margin. Verification / Alternative check: Voltage check: V = I * R = 0.001 * 470 = 0.47 V; P = V * I = 0.47 * 0.001 = 0.00047 W, matching the earlier calculation. Why Other Options Are Wrong: Higher ratings (1/2 W, 1 W, 2 W) are unnecessary here and larger/costlier without benefit. While 1/8 W would also be sufficient electrically, 1/4 W is the more common stocking option in many labs and offers extra robustness; among the provided typical selections, 1/4 W is the best fit. Common Pitfalls: Choosing a rating too close to the calculated power can cause overheating under elevated ambient temperatures or tolerances; include adequate margin. Final Answer: 1/4 watt

Question 8

In variable resistors (potentiometers used as controls), if the resistance does not change uniformly with shaft rotation, how is the device classified?

Accepted Answer

Introduction / Context: Rotary variable resistors are specified by how resistance changes versus shaft angle. Two common laws are linear (uniform change) and tapered (nonuniform change, such as logarithmic or audio taper). Recognizing these behaviors is essential when selecting controls for volume, brightness, and sensor interfaces. Given Data / Assumptions: Shaft rotation is the input variable. Observed resistance change with rotation is nonuniform. Device is a variable resistor/potentiometer used as a user control. Concept / Approach: A linear pot produces a resistance proportional to angle (straight-line R vs. theta). A tapered pot intentionally produces a curved response, commonly logarithmic (audio taper) or antilog (reverse audio), to match human perception or circuit requirements. Nonuniform change directly implies a taper, not a defect, provided the response matches the specified law. Step-by-Step Solution: Identify the R vs. angle behavior from measurement or datasheet.If R increases linearly with angle → linear law.If R increases slowly at first and then faster (or vice versa) → tapered law (log/antilog).Nonuniform change therefore classifies the device as tapered. Verification / Alternative check: Plot resistance at equal angle increments. A straight line indicates linear; a curve indicates taper. Datasheets label these as “A” or “B” laws depending on region/manufacturer conventions. Why Other Options Are Wrong: Linear: Contradicts the observed nonuniformity. Defective: Nonuniformity alone does not indicate failure if the part is specified as tapered. Not wirewound: Wirewound construction does not determine response law; both linear and tapered versions exist. Common Pitfalls: Assuming all pots are linear; many control applications require taper to suit human perception. Misinterpreting logarithmic behavior as a fault during quick bench tests. Final Answer: tapered

Question 9

How is a resistor’s tolerance typically conveyed if it is not printed numerically on the body?

Accepted Answer

Introduction / Context: The tolerance of a resistor indicates the allowed deviation from its nominal value (for example, ±1%, ±5%). Correctly reading tolerance helps ensure precision where needed and allows cost savings where wide tolerance is acceptable. Given Data / Assumptions: Through-hole, color-banded resistors (common in labs and education). No printed numeric tolerance on the body. Standard color-code conventions are in use. Concept / Approach: In the 4-band/5-band color code, one band denotes tolerance: gold (±5%), silver (±10%), brown (±1%), red (±2%), green (±0.5%), blue (±0.25%), violet (±0.1%), gray (±0.05%). If tolerance is not printed, it is recognized by this color band on the resistor's body. Step-by-Step Solution: Identify the tolerance band (often the band spaced apart or last band).Map color to tolerance using standard chart (e.g., gold → ±5%).Record nominal value, multiplier, and tolerance for specification or replacement. Verification / Alternative check: Measure with an ohmmeter and compare to nominal ± tolerance window. This confirms that the color-code-indicated tolerance is consistent with the actual part within measurement accuracy. Why Other Options Are Wrong: Keyed containers/size: Packaging and size do not encode tolerance. Ohmmeter reading: Shows present value, not rated tolerance category. Common Pitfalls: Reading the resistor from the wrong end and misidentifying the tolerance band. Confusing tolerance with temperature coefficient; the latter is not encoded by the same color in basic 4/5-band systems. Final Answer: color code

Question 10

In wire sizing, the cross-sectional area of a round conductor measured in circular mils (cmils) equals the square of its diameter in mils. What is the area for a conductor 1 inch in diameter (1000 mils)?

Accepted Answer

Introduction / Context: Electrical conductors are often specified using circular mils (cmils), a convenient unit for circular cross-sections. It simplifies area calculations because the area in cmils is the square of the diameter in mils, avoiding use of pi for quick estimates and tables. Given Data / Assumptions: Diameter d = 1 inch. 1 inch = 1000 mils (since 1 mil = 0.001 inch). Area (cmils) = d^2 (with d in mils). Concept / Approach: Use the circular mil definition: A_cmils = d_mils^2. This is a unit definition used widely in NEC tables and cable datasheets. It provides a direct proportional relationship for comparing conductor sizes without converting to square inches each time. Step-by-Step Solution: Convert diameter: d = 1 inch = 1000 mils.Apply definition: A = d^2 = (1000)^2 cmils.Compute: A = 1,000,000 cmils. Verification / Alternative check: Relate to square inches: 1 circular mil = (pi/4) * (0.001 in)^2 scaled factor; however, by definition, the shortcut d^2 in mils gives cmils directly, confirming 1,000,000 cmils for 1 inch diameter. Why Other Options Are Wrong: 500 or 100,000 cmils: Far too small for a 1 inch diameter. 500,000,000 cmils: Off by a factor of 500 from the correct square relationship. Common Pitfalls: Forgetting that diameter (not radius) is squared in the cmils definition. Mixing mils and inches without converting; always convert inches to mils first. Final Answer: 1,000,000 cmils

Question 11

Resistor categories with tolerance ratings of ±5% or greater are generally referred to as which class?

Accepted Answer

Introduction / Context: Resistors are grouped by tolerance classes that indicate expected deviation from nominal value. Designers choose precision parts for tight control and general-purpose parts where wider variation is acceptable at lower cost. Given Data / Assumptions: Tolerance threshold of about ±5% or larger. Standard commodity resistors used in many noncritical applications. Concept / Approach: Precision resistors typically have tolerances tighter than ±1% (e.g., ±0.1% to ±1%). General-purpose resistors commonly have ±5% or ±10% tolerances, adequate for pull-ups, biasing, and general signal routing. Tolerance class impacts error budgets and matching performance. Step-by-Step Solution: Identify tolerance: ≥ ±5% falls outside the precision category.Map to class: commodity carbon film or thick-film parts are generally classified as general-purpose.Therefore, the correct classification is general-purpose. Verification / Alternative check: Distributor catalogs categorize ±1% and tighter as precision; ±5% and ±10% are listed as general-purpose, confirming the terminology used in practice. Why Other Options Are Wrong: Precision: Reserved for ±1% or better (sometimes ±2%). SIP: Refers to a package style, not tolerance class. Wirewound: A construction type that can be either precision or power-rated; tolerance is not implied solely by being wirewound. Common Pitfalls: Assuming construction type determines tolerance; process control and trimming dominate tolerance. Confusing package style with tolerance classification. Final Answer: general-purpose

Question 12

Using the relation P = V^2 / R for a resistive load at fixed voltage, what happens to power when the resistance decreases?

Accepted Answer

Introduction / Context: Power relationships are central to sizing components and predicting thermal behavior. For purely resistive loads driven by a constant-voltage source, power depends on both the resistance and the current drawn. Given Data / Assumptions: Resistive load, no reactance. Applied voltage is fixed. Formula of interest: P = V^2 / R. Concept / Approach: The formula P = V^2 / R shows an inverse relationship between power and resistance at constant voltage. Decrease R → denominator decreases → P increases. This also aligns with P = I^2 * R and I = V / R; as R drops, I rises, and the net effect in P = V^2 / R is increased power dissipation. Step-by-Step Solution: Start from P = V^2 / R (V fixed).Let R decrease to R_new < R_old.Then P_new = V^2 / R_new > V^2 / R_old = P_old.Therefore, power increases as resistance decreases. Verification / Alternative check: Pick numbers: V = 10 V. If R = 10 Ω, P = 10^2 / 10 = 10 W. If R halves to 5 Ω, P = 10^2 / 5 = 20 W. Power doubled as resistance halved, confirming the inverse proportionality. Why Other Options Are Wrong: A decrease in power: Opposite of the formulaic result. An increase in ohms: Resistance is a parameter, not an outcome here. A decrease in current: As R drops, I = V / R increases, not decreases. Common Pitfalls: Mixing formulas and forgetting which variable is held constant (voltage vs current). Confusing instantaneous changes with steady-state relations. Final Answer: an increase in power

Question 13

Compared with carbon composition resistors, what is one key advantage of carbon film resistors in low-noise or precision-biased circuits?

Accepted Answer

Introduction / Context: Choosing resistor technology impacts noise, stability, and tolerance. Carbon composition (bulk carbon) and carbon film (deposited carbon on ceramic) are two legacy technologies with distinct characteristics important in audio, sensor biasing, and analog front-ends. Given Data / Assumptions: Comparison is between carbon composition and carbon film. Noise and tolerance are primary concerns. Concept / Approach: Carbon composition resistors exhibit higher excess noise and wider tolerance due to granular conductive paths. Carbon film parts, produced via deposited film and laser trimming, generally have lower excess noise and better tolerance for the same value and size, making them preferable for low-level analog signals. Step-by-Step Solution: Identify performance metric: noise in microvolt-level circuits.Compare technologies: carbon film typically shows lower excess noise index.Conclude: carbon film offers less circuit noise than carbon composition. Verification / Alternative check: Manufacturer datasheets and application notes report improved noise and tighter tolerance (e.g., ±1%–±5%) for carbon film versus typical ±5%–±20% for carbon composition parts. Why Other Options Are Wrong: Smaller size / higher wattage: Size and wattage vary by series; not an inherent carbon film advantage over composition. Poor tolerance: Carbon film usually has better tolerance than composition. Common Pitfalls: Selecting composition parts for sensitive analog stages due to availability, then chasing unexplained hiss or drift. Overlooking environmental stability; film types typically age better. Final Answer: less circuit noise

Question 14

Which formula explicitly shows a direct proportionality between electrical power and voltage (holding the other variable constant)?

Accepted Answer

Introduction / Context: Power relations in circuits are derived from Ohm’s law and definitions of work/energy. Recognizing which expression portrays direct proportionality with voltage is useful when reasoning about how power scales as supply voltage changes. Given Data / Assumptions: Standard definitions: P = V * I, V = I * R. We consider proportionality while holding one variable constant. Concept / Approach: The definition of electric power in any circuit is P = V * I. If current I is held constant (for example, an ideal current source), then P is directly proportional to V. Alternative forms (P = V^2 / R or P = I^2 * R) change the variable being held constant and do not show simple linear proportionality with V alone unless R or I is constant. Step-by-Step Solution: Start from P = V * I.If I is fixed, P ∝ V (a straight-line relationship).Hence, among the options, P = VI directly shows power increasing with voltage when current is constant. Verification / Alternative check: Holding R constant gives P = V^2 / R, a quadratic dependence on V, not a direct linear proportionality. Thus P = VI is the cleanest expression of direct proportionality under constant current conditions. Why Other Options Are Wrong: V = IR and I = V/R: These are Ohm’s law relations, not power formulas. P = IR: Dimensional mismatch; correct form is P = I^2 * R or P = V^2 / R. Common Pitfalls: Confusing “direct proportionality” (linear) with “depends on.” Forgetting to specify which variable is held constant when discussing proportionality. Final Answer: P = VI

Question 15

How many terminals (leads) does a standard potentiometer provide for connection in circuits such as voltage dividers and rheostats?

Accepted Answer

Introduction / Context: A potentiometer is a three-terminal variable resistor used either as a voltage divider (three terminals used) or as a rheostat (two terminals used). Understanding its terminal count and usage modes is foundational in analog interfaces and control circuits. Given Data / Assumptions: Standard rotary or slide potentiometer. Two end terminals connect to the ends of the resistive element; the third is the movable wiper. Concept / Approach: With three terminals, a potentiometer can divide an applied voltage to produce an adjustable fraction at the wiper. If used as a rheostat, only the wiper and one end terminal are connected externally, but the device still physically provides three leads. Step-by-Step Solution: Identify terminals: End A, Wiper, End B.Voltage divider mode: use all three (A–W–B).Rheostat mode: use two (W + A or W + B) while the third remains unused or tied to the wiper for noise reduction. Verification / Alternative check: Datasheets label pins 1–2–3 or A–W–B and show both usage modes. The physical count is consistently three for standard potentiometers. Why Other Options Are Wrong: 1 or 2: Too few for full divider functionality; two are used only in rheostat configuration. 4: Specialized multi-gang or dual pots have more terminals, but a single potentiometer section has three. Common Pitfalls: Confusing the number of connected leads in a given application with the number of available terminals on the device. Miswiring the wiper and end terminals, which reverses the control direction. Final Answer: 3

Question 16

A calculator shows the value 0.00263. Express this as a percentage, rounded to two decimal places.

Accepted Answer

Introduction / Context: Converting decimal numbers to percentages is a basic numerical skill used throughout electronics, statistics, and tolerance calculations. Many specifications and datasheets present errors and tolerances in percent format for easy comparison. Given Data / Assumptions: Decimal value: 0.00263. Requested output: percent form, rounded to two decimal places. Concept / Approach: The conversion rule is simple: percent = decimal * 100. Rounding to two decimals means we present the percentage with two digits after the decimal point, consistent with common reporting conventions. Step-by-Step Solution: Multiply by 100: 0.00263 * 100 = 0.263 %.Round to two decimals: 0.263 % rounds to 0.26 % (since the third decimal digit is 3).Therefore, the reported percentage is 0.26 %. Verification / Alternative check: Reverse the operation: 0.26 % / 100 = 0.0026, which is close to the original 0.00263 with a rounding difference of 0.00003, consistent with two-decimal rounding requirements. Why Other Options Are Wrong: 0.026%: Off by a factor of 10 (equivalent to 0.00026 as a decimal). 2.63% and 26.3%: Off by factors of 10^1 and 10^2; these misplace the decimal point in conversion. Common Pitfalls: Forgetting to multiply by 100, or moving the decimal two places in the wrong direction. Rounding to the wrong number of decimal places or prematurely rounding intermediate steps. Final Answer: 0.26%

Question 17

A 33 kΩ resistor with ±20% tolerance is measured with an ohmmeter. Which reading is within the acceptable tolerance band?

Accepted Answer

Introduction / Context: Tolerance defines the acceptable spread around a resistor’s nominal value. Checking whether a measured resistance falls inside the permissible band prevents unnecessary replacements and ensures circuits perform to design targets. Given Data / Assumptions: Nominal resistance R_nom = 33 kΩ = 33,000 Ω. Tolerance = ±20%. Measured candidates: 26,400 Ω; 24,183 Ω; 6,600 Ω; 39,970 Ω. Concept / Approach: The acceptable range is computed as R_nom * (1 ± tolerance). For ±20%, limits are 0.8 * R_nom and 1.2 * R_nom. Any measurement within [lower, upper] is acceptable. Step-by-Step Solution: Compute lower limit: R_low = 0.8 * 33,000 = 26,400 Ω.Compute upper limit: R_high = 1.2 * 33,000 = 39,600 Ω.Compare readings: 26,400 Ω (on the lower limit) → acceptable; 24,183 Ω (below lower limit) → unacceptable; 6,600 Ω (far below) → unacceptable; 39,970 Ω (above upper limit) → unacceptable. Verification / Alternative check: Express deviations in percent relative to 33,000 Ω: 26,400 Ω is −20%; 39,970 Ω is +21.1%, exceeding the +20% cap. The band is inclusive of the endpoints, so 26,400 Ω qualifies. Why Other Options Are Wrong: 24,183 Ω: More than −20% deviation; out of tolerance. 6,600 Ω: Extremely out of tolerance (−80%). 39,970 Ω: Greater than +20% deviation; out of tolerance. Common Pitfalls: Mistaking kΩ and Ω during calculation; always convert to the same unit. Rounding the limits incorrectly; compute limits first, then compare. Final Answer: 26400 ohms

Question 18

Conductor sizing and heating effect: For a given current level in basic electrical wiring, what happens when the wire has a smaller cross-sectional area (thinner conductor)?

Accepted Answer

Introduction / Context: When selecting conductors for power or signal wiring, cross-sectional area directly impacts resistance and heating. A thinner wire has higher resistance per unit length, which increases power loss and temperature rise for the same current. This question tests understanding of how geometry affects resistive heating and safety margins in practical circuits. Given Data / Assumptions: The same material (for example, copper) and the same wire length are assumed. The current through the conductor is held constant for comparison. Steady-state conditions, ignoring skin effect and frequency-dependent phenomena. Concept / Approach: Resistance of a uniform conductor is R = rho * L / A, where rho is resistivity, L is length, and A is cross-sectional area. Joule heating in the wire equals P_loss = I^2 * R for a given current I. Therefore, decreasing A increases R and thus increases I^2 * R heating, all else equal. Step-by-Step Solution: Start from R = rho * L / A.Decrease in A → increase in R (inversely proportional).Power loss: P_loss = I^2 * R; with I fixed, higher R gives higher P_loss.More loss manifests as more heat, raising conductor temperature unless mitigated by better cooling or lower duty cycle. Verification / Alternative check: Compare AWG tables: smaller gauge numbers (thicker wires) have lower resistance and lower temperature rise for a given current. Thermal ratings and ampacity charts in wiring standards corroborate that thinner wires must carry lower current to avoid overheating. Why Other Options Are Wrong: Less heat / more conductance / less resistance: opposite of the R versus A relationship. No change: contradicts R = rho * L / A and P_loss = I^2 * R. Common Pitfalls: Holding voltage constant instead of current can change current as R changes; however, the core trend that smaller A increases R remains true. Ignoring length; doubling length also doubles R, compounding heating. Final Answer: more heat

Question 19

Ohm’s law proportionality check: If resistance remains constant in a circuit, how are current (I) and voltage (V) related?

Accepted Answer

Introduction / Context: Ohm’s law is the cornerstone of DC and low-frequency AC circuit analysis. Understanding proportionality with a fixed resistance explains how sensors, voltage dividers, and loads behave as supply voltage varies. This question reinforces the linear relationship between voltage and current when resistance does not change. Given Data / Assumptions: Resistance R is constant (temperature and nonlinearity ignored). Simple two-terminal resistive element. Quasi-static conditions where reactive effects are negligible. Concept / Approach: Ohm’s law states V = I * R. With R fixed, I = V / R, so I is directly proportional to V. Doubling V doubles I; halving V halves I. This is the basis of linear resistor behavior and enables predictable control of current by setting voltage or resistance. Step-by-Step Solution: Start with V = I * R.Solve for I: I = V / R.Since R is a constant, I ∝ V (direct proportionality).Graphically, I-V is a straight line through the origin with slope 1/R. Verification / Alternative check: Measure current at two voltages keeping the same resistor. The ratio I2/I1 equals V2/V1. This linearity is observed in standard resistors within rated power and temperature ranges. Why Other Options Are Wrong: Inversely proportional: would require I ∝ 1/V, which contradicts Ohm’s law for fixed R. “The same” or “unable to produce energy”: not a relationship statement and not relevant. “Unrelated unless frequency is specified”: frequency matters for reactive elements, not ideal resistors. Common Pitfalls: Confusing constant-resistance behavior with constant-current sources (where current is fixed and voltage varies). Ignoring temperature coefficients that can slightly change R in real components. Final Answer: directly proportional

Question 20

Mechanical variable resistors in audio paths: In practical audio circuits, which audible symptom often reveals a worn or dirty potentiometer (mechanically variable resistor)?

Accepted Answer

Introduction / Context: Potentiometers (pots) are widely used as volume and tone controls. Mechanical wear, dust, and oxidation can degrade contact quality between the wiper and resistive track, injecting noise into the audio signal. Recognizing the symptom helps target maintenance like cleaning or replacement. Given Data / Assumptions: A standard carbon or cermet track potentiometer used in an analog audio path. Signal present while the knob is being adjusted. Room-temperature operation; no exotic failures assumed. Concept / Approach: A noisy, intermittent wiper contact modulates the signal and produces a characteristic crackle or “scratchy” noise, especially while the control is being rotated or slid. The noise is due to contact resistance fluctuations and microscopic arcing across contaminated or oxidized surfaces. Step-by-Step Solution: Wiper moves across the resistive track to set attenuation.Contaminants or wear create random resistance jumps (contact noise).These abrupt changes appear as wideband noise mixed with the program audio.Audible result: scratchy/crackling sound while moving the control, often reduced when stationary. Verification / Alternative check: Cleaning with contact cleaner usually improves or eliminates the noise. Measuring resistance while moving the shaft often shows erratic readings. Replacing the potentiometer resolves persistent issues due to track wear. Why Other Options Are Wrong: Lack of bass or excessive treble: typically due to EQ or coupling capacitor issues, not pot scratchiness. Variable volume: expected behavior of a functioning potentiometer, not necessarily a fault. Loss of stereo separation: usually a balance/pan or channel wiring issue. Common Pitfalls: Confusing dirty switch contacts with pot noise; both can crackle, but pot noise correlates with knob motion. Using inappropriate lubricants that attract dust and worsen the problem. Final Answer: scratchy noise

Resistance and Power Questions