Q: Basic measurement instruments: Which device is most commonly used to measure electrical potential difference (voltage) across two points in a circuit?

Introduction / Context: Electric circuits are characterized by voltage, current, resistance, and power. Accurate measurement of these quantities is essential for troubleshooting, design validation, and safety. Identifying the right instrument for each quantity is foundational knowledge. Given Data / Assumptions: We want the instrument that measures potential difference (voltage). Measurement is across two nodes in a circuit. Standard bench instruments are considered. Concept / Approach: A voltmeter measures voltage; it is connected in parallel with the component or nodes of interest and should have a high internal resistance to minimize circuit loading. An ammeter measures current (series connection, low internal resistance). An ohmmeter measures resistance (usually with an internal source). A wattmeter measures power (both voltage and current, often with two elements). Step-by-Step Solution: Identify the target quantity: voltage.Select instrument designed for parallel measurement with high input impedance.Answer: Voltmeter. Verification / Alternative check: Digital multimeters (DMMs) combine these functions; the “V” mode corresponds to voltmeter operation with high input impedance (e.g., 10 MΩ). Why Other Options Are Wrong: Ohmmeter: measures resistance, not live circuit voltage.Ammeter: measures current in series.Wattmeter: measures power, a product of voltage and current. Common Pitfalls: Connecting a meter in the wrong configuration (e.g., placing an ammeter in parallel can blow a fuse). Always verify meter mode and leads before measuring. Final Answer: Voltmeter

Q: Waveforms and time–frequency relationship: which statement best describes how the period of a waveform relates to its frequency in basic AC circuit theory?

Introduction / Context: In alternating-current (AC) theory and signal analysis, time-domain and frequency-domain descriptions are linked by a simple inverse relationship. Recognizing how period and frequency relate is foundational for analyzing power systems, audio signals, and communication waveforms. Given Data / Assumptions: Frequency f is measured in hertz (cycles per second). Period T is measured in seconds per cycle. Sine wave behavior is representative for illustrating the f–T relationship. Concept / Approach: The fundamental relationship is f = 1 / T. As the time for one cycle (T) increases, the number of cycles per second (f) decreases proportionally, and vice versa. Amplitude is independent of frequency; and 1.41 (square root of 2) relates RMS and peak of a sine wave’s amplitude, not its frequency. There is no default “assumed” frequency unless specified by a standard within a given context (e.g., regional mains frequency). Step-by-Step Solution: Start with identity: f = 1 / T. Infer monotonic relation: if T goes up, f goes down. Conclude: the correct qualitative statement is “longer period → lower frequency.” Verification / Alternative check: Example: if T = 0.02 s, f = 1 / 0.02 = 50 Hz; if T doubles to 0.04 s, f halves to 25 Hz. This numerical check confirms the inverse relation. Why Other Options Are Wrong: 1.41 factor concerns amplitude (peak vs. RMS), not frequency. Assumed 60 Hz is context-dependent; not a universal default. Amplitude does not set frequency; they are independent attributes. None: incorrect because the period–frequency inverse is correctly stated in option D. Common Pitfalls: Confusing amplitude–RMS relationships with frequency, or assuming regional mains frequency applies universally. Always rely on f = 1 / T to reason about time and frequency. Final Answer: A longer period corresponds to a lower frequency

Q: Capacitor design fundamentals: which single design change will increase the capacitance value of a parallel-plate capacitor, all else being equal?

Introduction / Context: Capacitance measures a component’s ability to store electric charge per unit voltage. For the widely taught parallel-plate model, geometric and material parameters determine the capacitance. Knowing which adjustments raise the capacitance is crucial for design, tuning, and selection of capacitors in circuits. Given Data / Assumptions: Parallel-plate capacitor approximation applies. Plate separation and dielectric permittivity remain unchanged unless stated. We compare the effect of changing a single parameter. Concept / Approach: The ideal formula for a parallel-plate capacitor is C = epsilon * A / d, where epsilon is the absolute permittivity (epsilon = epsilon_r * epsilon_0), A is plate area, and d is plate separation. From this, increasing A increases C linearly, decreasing d increases C, and using a higher-permittivity dielectric increases C. Changing frequency or applied voltage does not change the intrinsic capacitance of a linear capacitor; those affect reactance Xc = 1 / (2 * pi * f * C) and stored energy E = 0.5 * C * V^2, respectively, but not C itself. Step-by-Step Solution: Recall C = epsilon * A / d for a parallel-plate capacitor. Hold epsilon and d constant as implied, vary A. Increasing A → proportionally larger C. Verification / Alternative check: Doubling plate area doubles capacitance in the ideal model. Practical capacitors use stacked or rolled plates to maximize effective area for higher C values, confirming the principle in real-world designs. Why Other Options Are Wrong: Decreasing area reduces C. Increasing frequency changes reactance, not C. Increasing voltage changes stored energy, not C for linear dielectrics. None: invalid because increasing area does increase C. Common Pitfalls: Confusing frequency-dependent reactance with capacitance, and assuming voltage level alters C. Only geometry and dielectric properties set C for linear capacitors. Final Answer: Increasing the area of the plates

Q: Power in resistors: which equation directly computes power dissipation in a resistor when the voltage across it (E) and its resistance (R) are known?

Introduction / Context: Resistor power dissipation can be computed using several equivalent formulas derived from Ohm’s law and the definition of power. Selecting the right form depends on which quantities are known. When voltage and resistance are given, a specific formula avoids intermediate steps and gives power directly. Given Data / Assumptions: Known: voltage across the resistor E, resistance R. Ohm’s law applies: I = E / R. Power is defined as P = I * E for resistive elements. Concept / Approach: Start from P = I * E and substitute Ohm’s law (I = E / R) to eliminate current. This produces P = (E / R) * E = E^2 / R. Alternate forms P = I^2 * R or P = I * E are correct in general but require knowing I explicitly. The question asks for the formula that uses E and R directly. Step-by-Step Solution: Power definition: P = I * E. Ohm’s law: I = E / R. Substitute: P = (E / R) * E = E^2 / R. Verification / Alternative check: Example: E = 10 V, R = 5 ohm → P = E^2 / R = 100 / 5 = 20 W. Using P = I^2 * R with I = 10/5 = 2 A gives 2^2 * 5 = 20 W, confirming equivalence. Why Other Options Are Wrong: P = I^2 * R: correct but needs I, not given directly. P = I * E: correct but again requires I. I = P / E: rearranged relation for current, not power from E and R. None: invalid because P = E^2 / R is correct. Common Pitfalls: Using the wrong form for the known variables or mixing units (ensure volts and ohms yield watts). Always confirm by dimensional analysis: V^2 / ohm yields watts. Final Answer: P = E^2 / R

Q: Source stiffness in circuit design: a 12 V source has internal resistance r_s = 0.5 Ω. For which load resistance values does this source behave “stiff” (i.e., minimal voltage sag under load)?

Introduction / Context: A “stiff” voltage source maintains nearly constant terminal voltage when the load varies. Practically, this occurs when the load resistance is much larger than the source’s internal resistance, making load current small and IR drop inside the source negligible. This question checks whether you can apply the rule-of-thumb comparing R_load and r_s. Given Data / Assumptions: Ideal source voltage: 12 V. Internal resistance r_s = 0.5 Ω. Goal: choose R_load values that cause minimal terminal voltage drop. Concept / Approach: Terminal voltage V_out = V_s * (R_load / (R_load + r_s)). If R_load >> r_s, then R_load / (R_load + r_s) ≈ 1 and V_out ≈ V_s. A common engineering heuristic is R_load ≥ 10 * r_s for “stiff” performance; higher ratios are even better. With r_s = 0.5 Ω, 10 * r_s = 5 Ω, which is the bare minimum threshold; substantially larger values like 50 Ω or 100 Ω produce very small sag and thus clearly “stiff” behavior. Step-by-Step Solution: Compute ratio for 50 Ω: 50 / (50 + 0.5) ≈ 0.990 → about 1% drop. Compute ratio for 100 Ω: 100 / (100 + 0.5) ≈ 0.995 → about 0.5% drop. Compute ratio for 5–6 Ω: 5 / 5.5 ≈ 0.909 (≈9% drop); 6 / 6.5 ≈ 0.923 (≈7.7% drop) → less stiff. Verification / Alternative check: As a rule, R_load ≥ 10 * r_s is borderline; better stiffness with 100 * r_s. Both 50 Ω and 100 Ω satisfy >> 0.5 Ω; the inclusive choice “50 Ω or more” correctly captures the stiffness threshold among the options. Why Other Options Are Wrong: 5 Ω and 6 Ω: load is not much larger than 0.5 Ω; noticeable sag occurs. None: invalid because “50 Ω or more” is appropriate. Common Pitfalls: Forgetting internal resistance or assuming an ideal source. Always compare R_load to r_s to estimate voltage regulation quality. Final Answer: 50 Ω or more

Question 1

A transformer has a primary impedance of 60 Ω and a secondary impedance of 120 Ω. What is the impedance ratio (primary : secondary)?

Accepted Answer

Introduction / Context: Impedance ratio is a straightforward comparison often used when referencing impedances between transformer windings. It relates to the square of the turns ratio but can also be computed directly if both effective impedances are given. Given Data / Assumptions: Z_primary = 60 Ω. Z_secondary = 120 Ω. We seek Z_primary : Z_secondary as a numeric ratio. Concept / Approach: The impedance ratio (primary : secondary) is simply Zp/Zs when both values are expressed in the same units. If needed, turns ratio a satisfies Zp/Zs = (Np/Ns)^2, but here a direct division suffices. Step-by-Step Solution: Compute ratio: Zp/Zs = 60 / 120.Simplify: 60 / 120 = 0.5.Therefore the impedance ratio is 0.5 (primary lower than secondary). Verification / Alternative check: Turns ratio check: If Zp/Zs = 0.5, then Np/Ns = sqrt(0.5) ≈ 0.707, consistent with secondary having higher impedance. Why Other Options Are Wrong: 0.25 and 4: correspond to a 1:2 impedance ratio in the opposite sense.180: unrelated magnitude.None: incorrect because 0.5 is correct. Common Pitfalls: Inverting the ratio (secondary : primary instead of primary : secondary) or mixing turns and impedance ratios. Final Answer: 0.5

Question 2

Sources of alternating current (AC): From which type of device is most AC electrical power generated for homes and industry?

Accepted Answer

Introduction / Context: AC power in grids worldwide is produced at large scale and then transmitted and distributed. Understanding the physical source of this power clarifies why frequency is stable (50 or 60 Hz) and why mechanical energy is central to generation. Given Data / Assumptions: We refer to bulk electrical power used by utilities and consumers. Standard grid frequencies are 50 Hz or 60 Hz. We compare practical devices that could supply sustained AC power. Concept / Approach: Electric generators (alternators) convert mechanical rotational energy into electrical energy via electromagnetic induction. Primary movers include steam turbines, water turbines, wind, or gas turbines. Batteries provide DC and require inversion; oscillators generate signals but not grid-scale power; lightning is transient and uncontrollable. Step-by-Step Solution: Identify required characteristics: continuous, controllable AC at fixed frequency and voltage.Match to device: synchronous/alternator generators naturally produce AC.Eliminate alternatives: batteries are DC; oscillators are signal-level; lightning is sporadic. Verification / Alternative check: Power plants—from hydroelectric to coal, nuclear, wind—use rotating generators synchronized to the grid. Inverters may interface renewables, but the bulk AC standard originates from generator operation. Why Other Options Are Wrong: Batteries: DC sources; need inverters for AC and are not primary grid generators.Oscillators: produce low-power signals; not utility-scale sources.Lightning: non-continuous, destructive transient, not usable as a controlled supply. Common Pitfalls: Confusing small inverters feeding AC loads with the original source of grid AC; assuming any AC waveform generator equals power generation. Final Answer: generators

Question 3

Direction conventions: What is the direction of electron current flow in a conductor under an applied potential?

Accepted Answer

Introduction / Context: There are two current flow conventions: conventional current (positive charge flow) and electron flow (actual charge carriers in metals). Many circuit texts use conventional current from positive to negative; however, electrons in metallic conductors move in the opposite direction. Given Data / Assumptions: Medium: a metal conductor where electrons are majority carriers. Applied potential creates an electric field along the conductor. We are asked specifically about electron current flow. Concept / Approach: Electrons carry negative charge (−q). Under an electric field pointing from positive to negative terminal, electrons drift toward higher potential? Actually, because they are negatively charged, the electric force is opposite the field direction: electrons drift toward the positive terminal. Therefore, electron flow is from the negative terminal to the positive terminal. Step-by-Step Solution: Establish polarity: electric field points from + to −.Apply charge sign: force on negative charges is opposite the field.Deduce electron drift: toward + terminal, i.e., from negative to positive. Verification / Alternative check: In semiconductor P–N junctions, electron diffusion/drift directions are consistent with this rule under applied bias. Conventional current arrows on schematics point opposite to electron flow in metals. Why Other Options Are Wrong: Positive to negative: that is conventional current, not electron flow.Right to left / neutral to positive: arbitrary or physically incorrect in this context. Common Pitfalls: Mixing conventional current with electron flow; forgetting the opposite directions due to charge sign. Final Answer: from negative to positive

Question 4

Electromagnet fundamentals: Which principle best explains why a coil with current behaves like a magnet?

Accepted Answer

Introduction / Context: An electromagnet is created by driving electric current through a coil, producing a magnetic field that can attract ferromagnetic materials or interact with other magnetic fields. This principle underpins motors, generators, relays, and magnetic actuators. Given Data / Assumptions: We are asked for the basic physical principle. We consider steady or slowly varying currents in a wire or coil. No mechanical motion is required to form the magnetic field in the conductor. Concept / Approach: Ampère’s circuital law and the Biot–Savart relationship state that an electric current produces a magnetic field encircling the current path. Winding wire into a coil concentrates and shapes that field; adding an iron core further increases flux via higher permeability. Step-by-Step Solution: Identify the cause: electric current in a conductor.State the effect: creation of a magnetic field around the conductor.Apply to coils: multiple turns superimpose fields to form a strong electromagnet. Verification / Alternative check: Right-hand rule around a straight conductor or solenoid predicts field direction and confirms the magnetic behavior observed in labs and devices. Why Other Options Are Wrong: Induced current in a moving conductor: describes generator action (Faraday’s law), not why a coil with current is magnetic.Pole attraction/repulsion statements: true facts about magnets, but they do not explain the origin of the electromagnet’s field.None: incorrect because the core principle is explicitly stated in option (c). Common Pitfalls: Confusing electromagnetic induction (motion-based voltage generation) with magnetization due to current; mixing cause (current) and observed effects (force between poles). Final Answer: Current flowing through a conductor generates a magnetic field

Question 5

Design guideline for a “stiff” current source: If a current source has an internal resistance of 100 kΩ (in parallel with the source), what is the largest load resistance you would choose so the source still appears stiff?

Accepted Answer

Introduction / Context: A practical current source is modeled as an ideal current source in parallel with a large internal resistance R_int. To appear “stiff,” the load current should be close to the ideal source value and not vary much with changes in load. Choosing an appropriate load resistance is key. Given Data / Assumptions: Internal (parallel) resistance R_int = 100 kΩ. Stiffness rule of thumb: R_int should be at least about 10 times greater than R_load. We seek the largest R_load still meeting that guidance. Concept / Approach: With an ideal current source I_S in parallel with R_int feeding R_load, current division gives I_load = I_S * (R_int / (R_int + R_load)). To keep I_load ≈ I_S, require R_int ≫ R_load. A common heuristic is R_int ≥ 10 * R_load (error <≈ 10%). Step-by-Step Solution: Apply heuristic: R_load ≤ R_int / 10.Compute: R_int / 10 = 100 kΩ / 10 = 10 kΩ.Therefore, the largest advisable load is about 10 kΩ. Verification / Alternative check: Exact division: I_load/I_S = 100k / (100k + 10k) ≈ 0.909, about 9% error—typical stiffness target for preliminary design. Why Other Options Are Wrong: 20 kΩ: ratio only 5:1; larger error (~17%).2 kΩ or 1 kΩ: these are acceptable (even stiffer), but the question asks for the largest value that still qualifies; 10 kΩ is larger.None: incorrect since 10 kΩ fits the design rule. Common Pitfalls: Using series-source intuition (for voltage sources) instead of parallel model; forgetting that “stiff” for current sources means small load compared with internal resistance. Final Answer: 10 k Ohm

Question 6

Basic measurement instruments: Which device is most commonly used to measure electrical potential difference (voltage) across two points in a circuit?

Accepted Answer

Introduction / Context: Electric circuits are characterized by voltage, current, resistance, and power. Accurate measurement of these quantities is essential for troubleshooting, design validation, and safety. Identifying the right instrument for each quantity is foundational knowledge. Given Data / Assumptions: We want the instrument that measures potential difference (voltage). Measurement is across two nodes in a circuit. Standard bench instruments are considered. Concept / Approach: A voltmeter measures voltage; it is connected in parallel with the component or nodes of interest and should have a high internal resistance to minimize circuit loading. An ammeter measures current (series connection, low internal resistance). An ohmmeter measures resistance (usually with an internal source). A wattmeter measures power (both voltage and current, often with two elements). Step-by-Step Solution: Identify the target quantity: voltage.Select instrument designed for parallel measurement with high input impedance.Answer: Voltmeter. Verification / Alternative check: Digital multimeters (DMMs) combine these functions; the “V” mode corresponds to voltmeter operation with high input impedance (e.g., 10 MΩ). Why Other Options Are Wrong: Ohmmeter: measures resistance, not live circuit voltage.Ammeter: measures current in series.Wattmeter: measures power, a product of voltage and current. Common Pitfalls: Connecting a meter in the wrong configuration (e.g., placing an ammeter in parallel can blow a fuse). Always verify meter mode and leads before measuring. Final Answer: Voltmeter

Question 7

In solid-state electronics, silicon atoms arrange themselves in an orderly, repeating three-dimensional pattern. What is this long-range atomic arrangement called in materials science and semiconductor fabrication?

Accepted Answer

Introduction / Context: Understanding how atoms are arranged inside a material is fundamental to electronics. Silicon, the workhorse of integrated circuits, does not exist as isolated atoms inside a chip; instead, it forms a repeating three-dimensional network. The question asks for the correct term used in materials science and semiconductor fabrication for this long-range, ordered structure. Given Data / Assumptions: The material under discussion is elemental silicon used in wafers and devices. Atoms are arranged in a regular, repeating pattern through the bulk of the solid. We are looking for the correct term for the macroscopic orderly arrangement. Concept / Approach: When atoms occupy periodic positions in space, the solid is said to be crystalline. Silicon specifically forms a diamond-cubic crystal structure held together by covalent bonds. While “covalent bond” names the type of bonding between neighboring atoms, and “semiconductor” names the electrical classification, the question asks for the structural arrangement itself, which is a crystal. “Valence orbit” refers to electron shells, not lattice arrangement. Step-by-Step Solution: Identify the phenomenon: long-range periodic atomic order. Match the phenomenon to the correct materials term: crystalline solid. Apply to silicon used in chips: silicon wafers are single crystals grown as ingots, sliced, and polished. Therefore, the correct answer describing the arrangement is “crystal.” Verification / Alternative check: X-ray diffraction patterns from silicon show sharp Bragg peaks, a signature of crystalline order. Amorphous or polycrystalline materials would produce broadened peaks or multiple grain orientations, not the single-crystal pattern typical of wafer-grade silicon. Why Other Options Are Wrong: Covalent bond: bonding type, not the overall arrangement. Semiconductor: electrical property class, not a structural term. Valence orbit: electron shell concept, unrelated to lattice periodicity. None of the above: invalid because “Crystal” is correct. Common Pitfalls: Confusing bonding (covalent) with structure (crystal), or material class (semiconductor) with atomic arrangement. Always separate bonding type, electrical behavior, and lattice structure when classifying materials. Final Answer: Crystal

Question 8

In an electric motor, what is the name of the rotating coil (or set of windings) that turns within the magnetic field to convert electrical energy into mechanical motion?

Accepted Answer

Introduction / Context: Electric motors convert electrical energy into mechanical rotation by placing current-carrying conductors in a magnetic field. The key rotating component carrying these conductors has a specific name that distinguishes it from supporting parts. Correct identification is essential for troubleshooting, design, and maintenance. Given Data / Assumptions: The question refers to the rotating coil or windings inside a motor. Traditional brushed DC motors include an armature, commutator, and brushes. The term sought is the standard textbook name for the rotating winding assembly. Concept / Approach: The rotating assembly with windings is the armature (in many DC machines also called the rotor when contrasted with a stationary stator). Brushes are stationary contacts supplying current to the rotating armature through the commutator. The commutator is a segmented mechanical rectifier on the shaft, not the windings themselves. “Loop” is an informal term and lacks the precision required for component naming. Step-by-Step Solution: Identify the rotating conductor pack: the winding assembly on the rotating core. Recall standard parts: armature (windings), commutator (segments), brushes (contacts). Select the component name that specifically corresponds to the rotating windings: armature. Verification / Alternative check: Motor datasheets and repair manuals refer to armature rewinding, armature resistance tests, and armature balancing, confirming that the windings on the rotating part are collectively the armature. Why Other Options Are Wrong: Brush: stationary current-transfer element. Commutator: segmented copper assembly on the shaft, not the winding. Loop: vague descriptor, not a formal component name. None of the above: incorrect because “Armature” is correct. Common Pitfalls: Mixing “armature” with “rotor” terminology. While many use “rotor” generically, the question asks specifically about the rotating coil windings, which are called the armature in classical DC motor terminology. Final Answer: Armature

Question 9

Electromagnetism application: if electron current flows straight into the page (electrons moving away from you), what is the direction of the induced magnetic field lines encircling the conductor?

Accepted Answer

Introduction / Context: Right-hand and left-hand rules help visualize the magnetic field around a current-carrying conductor. A subtlety arises when the problem states “electron current,” because electron flow is opposite to conventional current. This question tests whether you can correctly translate electron flow into an equivalent conventional current direction and then apply the curling rule for magnetic fields. Given Data / Assumptions: Electron current is specified as moving into the page (away from the observer). Conventional current flows opposite to electron motion. Magnetic field lines around a straight conductor form concentric circles. Concept / Approach: First, convert electron flow to conventional current direction: electrons into the page imply conventional current out of the page. Next, apply the right-hand rule for straight conductors: point the right-hand thumb in the direction of conventional current; the curled fingers show the circular magnetic field direction. For current out of the page, fingers curl counterclockwise as viewed by the observer. Step-by-Step Solution: Electron flow into page → conventional current out of page. Use right-hand rule: thumb out of page, fingers curl around the wire. Observed field direction on the page: counterclockwise circles. Verification / Alternative check: Draw a dot (out of page) to mark conventional current and sketch a few circular arrows around the conductor; standard references show counterclockwise arrows for a current emerging toward the viewer, confirming the result. Why Other Options Are Wrong: Clockwise: would correspond to conventional current into the page. Out/Into this page: those describe field direction normal to the page, not the tangential circular direction. None of the above: incorrect because counterclockwise is correct. Common Pitfalls: Forgetting that electron flow is opposite conventional current or using the left hand by mistake. Always convert to conventional current first, then apply the right-hand rule. Final Answer: Counterclockwise

Question 10

Waveforms and time–frequency relationship: which statement best describes how the period of a waveform relates to its frequency in basic AC circuit theory?

Accepted Answer

Introduction / Context: In alternating-current (AC) theory and signal analysis, time-domain and frequency-domain descriptions are linked by a simple inverse relationship. Recognizing how period and frequency relate is foundational for analyzing power systems, audio signals, and communication waveforms. Given Data / Assumptions: Frequency f is measured in hertz (cycles per second). Period T is measured in seconds per cycle. Sine wave behavior is representative for illustrating the f–T relationship. Concept / Approach: The fundamental relationship is f = 1 / T. As the time for one cycle (T) increases, the number of cycles per second (f) decreases proportionally, and vice versa. Amplitude is independent of frequency; and 1.41 (square root of 2) relates RMS and peak of a sine wave’s amplitude, not its frequency. There is no default “assumed” frequency unless specified by a standard within a given context (e.g., regional mains frequency). Step-by-Step Solution: Start with identity: f = 1 / T. Infer monotonic relation: if T goes up, f goes down. Conclude: the correct qualitative statement is “longer period → lower frequency.” Verification / Alternative check: Example: if T = 0.02 s, f = 1 / 0.02 = 50 Hz; if T doubles to 0.04 s, f halves to 25 Hz. This numerical check confirms the inverse relation. Why Other Options Are Wrong: 1.41 factor concerns amplitude (peak vs. RMS), not frequency. Assumed 60 Hz is context-dependent; not a universal default. Amplitude does not set frequency; they are independent attributes. None: incorrect because the period–frequency inverse is correctly stated in option D. Common Pitfalls: Confusing amplitude–RMS relationships with frequency, or assuming regional mains frequency applies universally. Always rely on f = 1 / T to reason about time and frequency. Final Answer: A longer period corresponds to a lower frequency

Question 11

Basic electric circuits: what is the smallest number of distinct terminals (polarities) required to establish a current flow through a passive load?

Accepted Answer

Introduction / Context: Current requires a driving force and a closed path. In lumped circuit theory, this driving force is a potential difference between two points. The question tests recognition that at least two distinct terminals are needed to define such a difference and complete a circuit for current to flow through a load. Given Data / Assumptions: A passive load (e.g., resistor) needs a voltage applied across it. Voltage is defined as the difference in electric potential between two nodes. A closed loop is formed from a source terminal through the load back to the other source terminal. Concept / Approach: One point alone cannot define a voltage; we need two points to establish a potential difference. Once a source provides a positive and a negative terminal (or two nodes at different potentials), and a conductive path connects them through the load, charges can move and current flows according to Ohm’s law I = E / R. Extra terminals like “neutral” may exist in power systems, but they are not the fundamental minimum requirement for current in a simple DC circuit. Step-by-Step Solution: Recognize voltage requires two nodes: Vab = Va − Vb. Current flows when a loop connects the two terminals through a load. Therefore, the minimum is two terminals of differing potential: positive and negative. Verification / Alternative check: Consider a single battery and a lamp: one wire from battery positive to the lamp, and one wire from lamp back to battery negative completes the circuit. Removing either path or terminal stops the current completely. Why Other Options Are Wrong: One terminal: cannot define voltage or complete a loop. Three or four terminals: not required for basic current flow through a single load. None: invalid because two terminals are both necessary and sufficient. Common Pitfalls: Thinking a single “ground” reference suffices; it still implies a second node relative to ground. Also confusing system wiring conventions with fundamental circuit requirements. Final Answer: Two — a negative terminal and a positive terminal

Question 12

Capacitor design fundamentals: which single design change will increase the capacitance value of a parallel-plate capacitor, all else being equal?

Accepted Answer

Introduction / Context: Capacitance measures a component’s ability to store electric charge per unit voltage. For the widely taught parallel-plate model, geometric and material parameters determine the capacitance. Knowing which adjustments raise the capacitance is crucial for design, tuning, and selection of capacitors in circuits. Given Data / Assumptions: Parallel-plate capacitor approximation applies. Plate separation and dielectric permittivity remain unchanged unless stated. We compare the effect of changing a single parameter. Concept / Approach: The ideal formula for a parallel-plate capacitor is C = epsilon * A / d, where epsilon is the absolute permittivity (epsilon = epsilon_r * epsilon_0), A is plate area, and d is plate separation. From this, increasing A increases C linearly, decreasing d increases C, and using a higher-permittivity dielectric increases C. Changing frequency or applied voltage does not change the intrinsic capacitance of a linear capacitor; those affect reactance Xc = 1 / (2 * pi * f * C) and stored energy E = 0.5 * C * V^2, respectively, but not C itself. Step-by-Step Solution: Recall C = epsilon * A / d for a parallel-plate capacitor. Hold epsilon and d constant as implied, vary A. Increasing A → proportionally larger C. Verification / Alternative check: Doubling plate area doubles capacitance in the ideal model. Practical capacitors use stacked or rolled plates to maximize effective area for higher C values, confirming the principle in real-world designs. Why Other Options Are Wrong: Decreasing area reduces C. Increasing frequency changes reactance, not C. Increasing voltage changes stored energy, not C for linear dielectrics. None: invalid because increasing area does increase C. Common Pitfalls: Confusing frequency-dependent reactance with capacitance, and assuming voltage level alters C. Only geometry and dielectric properties set C for linear capacitors. Final Answer: Increasing the area of the plates

Question 13

Power in resistors: which equation directly computes power dissipation in a resistor when the voltage across it (E) and its resistance (R) are known?

Accepted Answer

Introduction / Context: Resistor power dissipation can be computed using several equivalent formulas derived from Ohm’s law and the definition of power. Selecting the right form depends on which quantities are known. When voltage and resistance are given, a specific formula avoids intermediate steps and gives power directly. Given Data / Assumptions: Known: voltage across the resistor E, resistance R. Ohm’s law applies: I = E / R. Power is defined as P = I * E for resistive elements. Concept / Approach: Start from P = I * E and substitute Ohm’s law (I = E / R) to eliminate current. This produces P = (E / R) * E = E^2 / R. Alternate forms P = I^2 * R or P = I * E are correct in general but require knowing I explicitly. The question asks for the formula that uses E and R directly. Step-by-Step Solution: Power definition: P = I * E. Ohm’s law: I = E / R. Substitute: P = (E / R) * E = E^2 / R. Verification / Alternative check: Example: E = 10 V, R = 5 ohm → P = E^2 / R = 100 / 5 = 20 W. Using P = I^2 * R with I = 10/5 = 2 A gives 2^2 * 5 = 20 W, confirming equivalence. Why Other Options Are Wrong: P = I^2 * R: correct but needs I, not given directly. P = I * E: correct but again requires I. I = P / E: rearranged relation for current, not power from E and R. None: invalid because P = E^2 / R is correct. Common Pitfalls: Using the wrong form for the known variables or mixing units (ensure volts and ohms yield watts). Always confirm by dimensional analysis: V^2 / ohm yields watts. Final Answer: P = E^2 / R

Question 14

Source stiffness in circuit design: a 12 V source has internal resistance r_s = 0.5 Ω. For which load resistance values does this source behave “stiff” (i.e., minimal voltage sag under load)?

Accepted Answer

Introduction / Context: A “stiff” voltage source maintains nearly constant terminal voltage when the load varies. Practically, this occurs when the load resistance is much larger than the source’s internal resistance, making load current small and IR drop inside the source negligible. This question checks whether you can apply the rule-of-thumb comparing R_load and r_s. Given Data / Assumptions: Ideal source voltage: 12 V. Internal resistance r_s = 0.5 Ω. Goal: choose R_load values that cause minimal terminal voltage drop. Concept / Approach: Terminal voltage V_out = V_s * (R_load / (R_load + r_s)). If R_load >> r_s, then R_load / (R_load + r_s) ≈ 1 and V_out ≈ V_s. A common engineering heuristic is R_load ≥ 10 * r_s for “stiff” performance; higher ratios are even better. With r_s = 0.5 Ω, 10 * r_s = 5 Ω, which is the bare minimum threshold; substantially larger values like 50 Ω or 100 Ω produce very small sag and thus clearly “stiff” behavior. Step-by-Step Solution: Compute ratio for 50 Ω: 50 / (50 + 0.5) ≈ 0.990 → about 1% drop. Compute ratio for 100 Ω: 100 / (100 + 0.5) ≈ 0.995 → about 0.5% drop. Compute ratio for 5–6 Ω: 5 / 5.5 ≈ 0.909 (≈9% drop); 6 / 6.5 ≈ 0.923 (≈7.7% drop) → less stiff. Verification / Alternative check: As a rule, R_load ≥ 10 * r_s is borderline; better stiffness with 100 * r_s. Both 50 Ω and 100 Ω satisfy >> 0.5 Ω; the inclusive choice “50 Ω or more” correctly captures the stiffness threshold among the options. Why Other Options Are Wrong: 5 Ω and 6 Ω: load is not much larger than 0.5 Ω; noticeable sag occurs. None: invalid because “50 Ω or more” is appropriate. Common Pitfalls: Forgetting internal resistance or assuming an ideal source. Always compare R_load to r_s to estimate voltage regulation quality. Final Answer: 50 Ω or more

Question 15

Applying Ohm’s law: which formula directly yields the current through a resistor when the voltage across it (E) and its resistance (R) are known?

Accepted Answer

Introduction / Context: Ohm’s law is the cornerstone of circuit analysis, relating voltage (E), current (I), and resistance (R). When you know the voltage across a resistor and its resistance value, a specific rearrangement of Ohm’s law gives current directly. Correctly choosing this form speeds analysis and avoids unnecessary steps. Given Data / Assumptions: Resistive element obeys Ohm’s law. Known quantities: E and R. Desired unknown: I. Concept / Approach: Starting from the canonical relation E = I * R, solve for current by dividing both sides by R to obtain I = E / R. This direct form is used constantly in electronics to compute current from known applied voltage across a component and its resistance value in ohms. Step-by-Step Solution: Write Ohm’s law: E = I * R. Isolate I: divide both sides by R. Get I = E / R as the direct formula. Verification / Alternative check: Unit analysis: volts / ohms = amperes. Example: E = 10 V, R = 5 Ω → I = 2 A, which is consistent with common circuit intuition and power calculations. Why Other Options Are Wrong: E = P * R: relates voltage to power and resistance, not the target form. E = I^2 * R: power relation rearranged incorrectly; actual is P = I^2 * R. I = E * R: incorrect; multiplying would increase units to V·Ω, not amperes. None: invalid because I = E / R is correct. Common Pitfalls: Confusing P = I * E and E = I * R; ensure you are solving specifically for I with known E and R. Final Answer: I = E / R

Question 16

Foundations of electrical quantities: which statement best defines voltage (also called electric potential difference) in basic circuit theory?

Accepted Answer

Introduction / Context: Voltage, current, and resistance are the three primary electrical quantities used throughout electronics. Confusing these terms leads to incorrect reasoning and design mistakes. This question asks for the definition of voltage in clear, practical terms relevant to basic circuits. Given Data / Assumptions: Voltage is measured in volts (V). It is defined between two points (potential difference). It motivates charges to move when a conductive path is available. Concept / Approach: Voltage is analogous to pressure in fluid systems: it represents the potential to do electrical work by moving charges. When a circuit is closed, voltage across a component drives current through it according to Ohm’s law I = E / R. Current, by contrast, is the actual flow of charge (measured in amperes). Statements about particles or electron counts are unrelated to the definition of voltage. Step-by-Step Solution: Clarify roles: voltage is cause (potential difference), current is effect (charge flow). Identify the definition that mentions potential difference across two points and driving effect. Select the option describing voltage as the force (potential difference) that causes electrons to flow. Verification / Alternative check: Examples: a 9 V battery has a 9-volt potential difference between its terminals. Connected across a resistor, it produces current I = 9 / R. Without a closed path, potential exists but no current flows, consistent with the “cause” role of voltage. Why Other Options Are Wrong: Flow of charged particles: that is current, not voltage. Smallest charged particle: describes an electron, not voltage. Number of free electrons: relates to material properties, not the definition of voltage. None: invalid because a correct definition is provided. Common Pitfalls: Using “voltage” and “current” interchangeably. Always remember: voltage is a potential difference; current is movement of charge due to that potential when a path exists. Final Answer: The force that causes electrons to flow (potential difference between two points)

Question 17

Magnetism basics — identifying a magnet’s north pole: Which statement correctly describes the behavior of the north-seeking pole of a bar magnet with respect to the Earth and other magnets?

Accepted Answer

Introduction / Context: This conceptual question checks your understanding of how a magnet’s labeled north pole (the north-seeking pole) interacts with the Earth's magnetic field and with other magnets. It also touches on a common misconception: the region near the Earth's geographic North actually behaves as a magnetic south pole, which is why a compass’s north pole points there. Given Data / Assumptions: We are dealing with a standard bar magnet with clearly labeled north and south poles. “North pole” refers to the end that seeks the Earth’s geographic North when freely suspended (the compass-like behavior). Earth's geographic North corresponds to a magnetic south polarity region. Like poles repel; unlike poles attract between magnets. Concept / Approach: Magnetic poles interact according to two rules: like repels, unlike attracts; and the Earth's magnetic "north" indicator is a bit counterintuitive—compass north is drawn to a magnetic south in the high northern latitudes. Therefore, the north pole of a magnet is attracted toward the Earth's magnetic south (near the geographic North). Between separate magnets, the north pole is attracted to another magnet’s south pole and repelled by its north pole. Step-by-Step Solution: Identify the magnet’s north pole as the “north-seeking” end.Recall that Earth’s geographic North hosts a magnetic south polarity region.Therefore, the magnet’s north pole is attracted toward the Earth’s magnetic south near the geographic North.Cross-check with magnet-to-magnet interactions: north attracts south, repels north. Verification / Alternative check: Hold a freely pivoted bar magnet: the north-seeking end points toward geographic North. This confirms that the region near geographic North must be magnetic south, consistent with electromagnetic conventions. Why Other Options Are Wrong: (a) Random magnetic domains would not produce a strong pole; magnets have aligned domains. (b) The Earth’s “north magnetic pole” phrasing is misleading—near geographic North, the polarity is effectively south. (c) A north pole is attracted to a south pole, not repelled. (e) Incorrect because one option is correct. Common Pitfalls: Confusing geographic labels with magnetic polarity; assuming the Earth’s “north magnetic pole” is a magnetic north in the strict physics sense; forgetting like repels and unlike attracts. Final Answer: It would be attracted toward the Earth's south magnetic pole (near the geographic North).

Question 18

Resistor tolerance application — 100 Ω ± 10%: For a resistor rated at 100 Ω with a tolerance of ±10%, determine the minimum and maximum permissible resistance values.

Accepted Answer

Introduction / Context: Component tolerances define acceptable manufacturing variation. Being able to compute the allowable resistance range from a rated value and a percentage tolerance is a basic skill in circuit design, parts selection, and worst-case analysis. Given Data / Assumptions: Nominal resistance R_nom = 100 Ω. Tolerance = ±10% of the nominal value. We assume standard linear tolerance interpretation (no temperature coefficient or aging effects included). Concept / Approach: The allowable range is R_nom * (1 − tol) to R_nom * (1 + tol), where tol is expressed as a decimal. Here tol = 0.10. Compute both bounds to find the minimum and maximum permissible resistances that still meet specification. Step-by-Step Solution: Compute the decrease: 10% of 100 Ω is 10 Ω.Minimum value: 100 Ω − 10 Ω = 90 Ω.Maximum value: 100 Ω + 10 Ω = 110 Ω.Therefore, the acceptable range is 90 Ω to 110 Ω. Verification / Alternative check: Use ratio form: Lower bound = 0.9 * 100 Ω = 90 Ω; Upper bound = 1.1 * 100 Ω = 110 Ω. The results match the direct arithmetic. Why Other Options Are Wrong: (a) Upper bound 101 Ω is inconsistent with ±10%. (c) and (d) are far too tight (more like 1–2% tolerance). (e) Invalid because a correct range is available. Common Pitfalls: Confusing absolute ohms with percentage points; forgetting to apply the tolerance to both sides; mixing up ±10 Ω with ±10% (coincidentally equal here because the nominal is 100 Ω). Final Answer: 90 Ω to 110 Ω.

Question 19

Atomic structure and charge — definition of a negative ion: Which description correctly defines a negative ion (anion) in terms of its subatomic particle counts?

Accepted Answer

Introduction / Context: Ions are atoms with a net electric charge due to a difference between the number of protons (positive charges) and electrons (negative charges). Understanding the sign of that charge is foundational for chemistry, solid-state physics, and semiconductor device theory. Given Data / Assumptions: Protons carry +1 elementary charge each; electrons carry −1 elementary charge each; neutrons are neutral. “Negative ion” (anion) implies total negative net charge. We compare particle counts to infer the sign of net charge. Concept / Approach: Net charge Q is proportional to (number of protons − number of electrons). If an atom has more electrons than protons, the net charge is negative, and the atom is a negative ion. Neutrons do not affect the sign of charge, only the mass and isotopic identity. Step-by-Step Solution: Identify the key comparison: electrons vs protons.If electrons > protons → net charge negative → negative ion (anion).Therefore, the correct statement is “more electrons than protons.” Verification / Alternative check: Consider neutral atoms: electrons = protons. Adding one electron (without changing protons) yields a net −1 charge, creating a negative ion. This matches the definition. Why Other Options Are Wrong: (a) and (b) compare with neutrons, which do not determine charge. (d) Describes a positive ion (cation). (e) Invalid because a correct option exists. Common Pitfalls: Confusing particle roles (neutrons vs electrons); thinking mass differences affect charge; forgetting that charge is about proton–electron balance. Final Answer: An atom that contains more electrons than protons.

Question 20

Semiconductor materials in electronics manufacturing: Identify the most widely used semiconductor material for integrated circuits and electronic devices.

Accepted Answer

Introduction / Context: Modern electronics are built on semiconductor materials. The choice of substrate affects cost, yield, device performance, and the available processing ecosystem. Silicon dominates mainstream integrated circuit fabrication from microcontrollers to advanced processors. Given Data / Assumptions: We compare candidate materials: copper (a conductor), silicon and germanium (semiconductors). “Most widely used” refers to volume of commercial device fabrication and manufacturing infrastructure. Focus on integrated circuits and general-purpose electronics. Concept / Approach: Silicon offers a native oxide (SiO2) that enables robust, manufacturable MOSFETs; it is abundant, relatively inexpensive, and supported by mature fabrication processes and design tools. Germanium was historically important but is much less common today for standalone devices, though it appears in SiGe technologies. Copper is not a semiconductor; it is often used for interconnects, not as the semiconductor substrate. Step-by-Step Solution: Eliminate non-semiconductors: copper is a metal.Compare semiconductors: silicon vs germanium.Silicon wins on oxide quality, thermal stability, and ecosystem maturity.Therefore, select “Silicon.” Verification / Alternative check: The ubiquity of silicon foundries, process nodes, and the CMOS process standard confirms this choice. Discrete power, analog ICs, and digital logic are predominantly silicon-based. Why Other Options Are Wrong: (a) Copper is not a semiconductor. (c) Germanium is niche today. (d) “All of the above” cannot be correct because copper is not a semiconductor. Common Pitfalls: Confusing interconnect materials with semiconductors; assuming germanium’s historical use implies present dominance. Final Answer: Silicon.

Electronic Principles Questions