Q: Power units and metric prefixes Confirm whether the electrical power value 4,200 W is numerically equal to 4.2 kW when expressed using the kilo (10^3) prefix.

Introduction / Context: Working across datasheets and test reports, engineers routinely convert raw watt values using SI prefixes to keep numbers readable and comparable. Misapplying prefixes can lead to 10× or 1000× errors with costly consequences, especially in power budgeting and thermal design. Given Data / Assumptions: Base unit for power: watt (W). kilo (k) denotes multiplication by 10^3. No change in the physical quantity, only unit scaling. Concept / Approach: To convert W to kW, divide by 10^3. The relationship is P(kW) = P(W) / 1000. Conversely, multiply kW by 1000 to return to W. This conversion is independent of AC or DC context, power factor, or frequency; it is purely a unit scaling within the SI system. Step-by-Step Solution: Start with 4,200 W.Apply the kilo conversion: 4,200 W / 1000 = 4.2 kW.Confirm that reversing the conversion yields 4.2 kW × 1000 = 4,200 W.Conclude that 4,200 W and 4.2 kW are numerically consistent. Verification / Alternative check: Express in megawatts: 4,200 W = 0.0042 MW because 1 MW = 10^6 W. Each form is consistent as long as the power-of-ten factor is applied correctly. Why Other Options Are Wrong: Incorrect: contradicts basic SI prefix scaling.Power factor, AC/DC, and frequency do not alter unit conversion; they affect how power is measured or computed, not how watts convert to kilowatts. Common Pitfalls: Confusing k (10^3) with K (kelvin); writing “KW” instead of “kW”; mixing k (10^3) with Ki (kibi, 2^10) which is not used for SI power units. Final Answer: Correct

Q: Metric prefixes in electronics documentation Assess the statement: “The metric prefix giga (symbol G) represents one billion, that is 10^9.”

Introduction / Context: Standardized prefixes keep communication clear across disciplines. In electronics, you will see gigahertz for radio bands, gigabytes for storage marketing (with caveats), and gigohms for very high resistances. Understanding what “giga” means avoids order-of-magnitude mistakes in design and procurement. Given Data / Assumptions: SI metric prefix “giga”, symbol G. We use the International System of Units (SI), not binary prefixes, unless explicitly stated. Quantities may include frequency (Hz), resistance (Ω), power (W), etc. Concept / Approach: By SI definition, giga denotes a multiplication factor of 10^9. This is exact and does not depend on the underlying physical quantity. While computer memory historically used base-2 interpretations (e.g., 1 GB ≈ 2^30 bytes), SI clarifies that binary prefixes are different (gibi, Gi = 2^30). In scientific and engineering notation, G always means 10^9. Step-by-Step Solution: Map prefix to power: giga → 10^9.Apply to examples: 3.2 GHz = 3.2 × 10^9 Hz; 5 GΩ = 5 × 10^9 Ω.Distinguish from binary: 1 GiB = 2^30 bytes ≈ 1.073 × 10^9 bytes, not 10^9 exactly.Conclude the statement is correct within SI usage. Verification / Alternative check: Cross-check with other prefixes: mega (M) = 10^6, kilo (k) = 10^3, tera (T) = 10^12. The sequence increases by powers of 10^3 as expected, placing giga squarely at 10^9. Why Other Options Are Wrong: Incorrect or 10^6: contradicts the SI standard.“Varies with base-2” confuses SI with binary prefixes; in SI, G ≠ 2^30.“Not part of SI” is false; giga is an established SI prefix. Common Pitfalls: Using uppercase/lowercase incorrectly (G for giga vs. g for gram); mixing GB and GiB; assuming unit symbols alter the meaning of the prefix (they do not). Final Answer: Correct

Q: Recognizing proper scientific notation Evaluate the statement: “The number 560,000, as written, is already in scientific notation.”

Introduction / Context: Scientific notation standardizes how we write large and small numbers: a × 10^n with 1 ≤ a < 10. This convention improves readability, prevents zeros from being dropped or added accidentally, and facilitates clear comparison of magnitudes in datasheets and calculations. Given Data / Assumptions: Number given: 560,000. No unit specified; the concept is unit-agnostic. We test compliance with strict scientific notation, not “approximate” styles. Concept / Approach: In scientific notation, the mantissa a must be at least 1 and less than 10. The exponent n is an integer. Writing 560,000 does not show a power of ten or a normalized mantissa. The correct scientific notation is 5.6 × 10^5 (or 5.600 × 10^5 if more significant digits are justified by measurement). Step-by-Step Solution: Express 560,000 with one nonzero digit to the left of the decimal: 5.6.Count decimal shifts: five places to the left from 560,000 to 5.6 → 10^5.Combine: 5.6 × 10^5.Conclude: 560,000 as typed is not scientific notation. Verification / Alternative check: Engineering notation groups powers by 3: 560,000 = 560 × 10^3. This is engineering notation (mantissa between 1 and 999), not scientific notation. Both are valid formats, but the rules differ; the prompt asked about scientific notation specifically. Why Other Options Are Wrong: Correct: ignores the requirement to show the power of ten and normalized mantissa.Unit-based or rounding-based claims do not change the definition of scientific notation.“Engineering but not scientific” is close conceptually, but the original 560,000 is not written as 560 × 10^3 either. Common Pitfalls: Assuming any compact form qualifies; omitting the exponent; confusing engineering notation (steps of 10^3) with scientific notation (1 to 10 mantissa). Final Answer: Incorrect

Question 1

Current addition and unit conversion: Add 21 mA and 8000 µA, and express the result in milliamperes (mA).

Accepted Answer

Introduction / Context: Combining quantities with different SI prefixes requires careful unit conversion. In circuit analysis, currents may be reported in microamperes or milliamperes depending on sensor ranges or measurement equipment. Correct conversion prevents order-of-magnitude errors in power and sizing calculations. Given Data / Assumptions: I1 = 21 mA. I2 = 8000 µA. We must report the total in mA. Concept / Approach: Use micro to milli conversion: 1 mA = 1000 µA. Convert µA to mA, add, and keep consistent units. Always write a short conversion equation to avoid mental slips, especially under exam pressure. Step-by-Step Solution: Convert 8000 µA to mA: 8000 µA / 1000 = 8 mA.Add: 21 mA + 8 mA = 29 mA.Therefore, the sum expressed in mA is 29 mA. Verification / Alternative check: Convert everything to amperes: 21 mA = 0.021 A; 8000 µA = 0.008 A; total = 0.029 A = 29 mA. The alternate route confirms the same result. Why Other Options Are Wrong: 21.8 mA: Treats 8000 µA as 0.8 mA; off by a factor of 10. 218 mA or 290 mA: Off by factors of 10 or 100; incorrect unit scaling. Common Pitfalls: Misplacing decimal points when converting between micro and milli; remember that µ → m divides by 1000, and m → µ multiplies by 1000. Final Answer: 29 mA

Question 2

Fundamental frequency unit in SI: The unit of frequency is:

Accepted Answer

Introduction / Context: Frequency—the number of cycles or events per second—is central in signal processing, power systems, and communications. Correct unit usage is essential for interpreting specifications like sampling rates, clock speeds, and mains frequency. Given Data / Assumptions: Frequency f measures cycles per second. 1 hertz (Hz) = 1 s^−1. We distinguish frequency from electrical current (A), power (W), and time (s). Concept / Approach: The SI derived unit for frequency is the hertz, defined as one event per second. While second appears in the definition, second itself is a base unit of time; watt and ampere are unrelated quantities representing power and electric current respectively. Therefore, hertz uniquely identifies frequency. Step-by-Step Solution: Recall definition: frequency f = number of cycles / time.Unit derivation: f units = s^−1 → named hertz (Hz).Select hertz as the correct unit. Verification / Alternative check: Check common values: mains frequency 50/60 Hz, audio range up to 20 kHz, RF signals in MHz/GHz—all stated in hertz multiples, confirming the standard. Why Other Options Are Wrong: Ampere: Unit of current, not frequency. Watt: Unit of power, not frequency. Second: Base time unit; its inverse defines frequency but is not itself the frequency unit name. Common Pitfalls: Confusing angular frequency (rad/s) with frequency in Hz; remember ω = 2 * π * f. Final Answer: hertz

Question 3

Scientific notation check — evaluate the statement: “Subtracting 18 × 10^2 from 480 × 10^1 gives 30 × 10^2.” Is this arithmetic statement valid?

Accepted Answer

Introduction / Context: Scientific notation is a compact way to handle large and small numbers in electronics, physics, and engineering calculations. This problem asks you to validate a subtraction expressed in scientific-notation style and to recognize appropriate normalization. Given Data / Assumptions: Two quantities: 480 × 10^1 and 18 × 10^2. We are checking the raw arithmetic equivalence, not significant figures or unit conversions. No units are attached; treat both numbers as pure magnitudes. Concept / Approach: The easiest way to verify is to convert both terms into ordinary integers and perform the subtraction, then re-express the result in scientific notation. Alternatively, align powers of ten and subtract the mantissas accordingly. Step-by-Step Solution: Compute each term: 480 × 10^1 = 4800; 18 × 10^2 = 1800.Subtract: 4800 − 1800 = 3000.Express the result: 3000 = 30 × 10^2 (since 30 × 100 = 3000).Therefore, the statement is arithmetically correct. Verification / Alternative check: Normalize further to strict scientific notation if desired: 3000 = 3.0 × 10^3. Both 30 × 10^2 and 3.0 × 10^3 represent the same value; the latter is preferred when the mantissa is between 1 and 10. Why Other Options Are Wrong: Incorrect: The arithmetic does check out; the equality holds. Valid only if units cancel / Cannot be determined: Units are not involved here; it is purely numeric. Common Pitfalls: Forgetting to align exponents or misplacing the decimal when renormalizing. Always convert to ordinary numbers or match exponents before subtracting. Final Answer: Correct — 480 × 10^1 − 18 × 10^2 equals 30 × 10^2.

Question 4

Fundamental units — in SI (International System of Units), which unit is the correct unit of electric charge used throughout electronics and physics?

Accepted Answer

Introduction / Context: Clear understanding of SI units is essential in circuit analysis, electromagnetics, and instrumentation. Electric charge is a primary quantity from which current, voltage relationships, and capacitor behavior are derived. This item checks if you can identify the proper SI unit for charge. Given Data / Assumptions: Use of SI base and derived units. Standard definitions per the SI Brochure. No need to convert between systems; just identify the correct unit. Concept / Approach: The coulomb (symbol C) is the SI unit of electric charge. One coulomb corresponds to the charge transported by a current of one ampere in one second, so 1 C = 1 A * 1 s. Recognizing this definition also connects charge to capacitor equations such as Q = C_elec * V, where Q is charge (C), C_elec is capacitance (F), and V is voltage (V). Step-by-Step Solution: Recall the definition of current: I = dQ/dt.Rearrange to Q: Q = ∫ I dt, giving coulombs when amperes are integrated over seconds.Select the unit “coulomb” as the measure of charge. Verification / Alternative check: Capacitor formula Q = C_elec * V uses coulombs on the left when capacitance is in farads and voltage in volts (since 1 F = 1 C/V). Why Other Options Are Wrong: Ampere: Unit of electric current (charge per second), not charge itself. Volt: Unit of electric potential difference. Ohm: Unit of resistance. Common Pitfalls: Confusing charge with current due to their close relationship; remember current is the rate of flow of charge. Final Answer: coulomb

Question 5

Symbols for physical quantities — evaluate the claim: “Only electrical quantities have symbols associated with them.” Choose the most accurate assessment.

Accepted Answer

Introduction / Context: Symbols are a universal language across science and engineering, allowing concise expression of equations and relationships. The statement claims exclusivity for electrical quantities, which is a misconception. This question tests your awareness of standard scientific notation practices. Given Data / Assumptions: Conventional use of symbols in physics, chemistry, and engineering disciplines. Standard SI symbols and widely accepted variable names. Focus on whether symbolism is unique to electrical topics. Concept / Approach: In mechanics, kinematics uses x for position, v for velocity, a for acceleration, t for time, m for mass, and F for force. Thermodynamics uses T for temperature, p for pressure, V for volume, and n for amount of substance. Chemistry uses c for concentration and K for equilibrium constants. Electricity and magnetism use V for voltage, I for current, R for resistance, C for capacitance, and so on. Clearly, symbolic notation is not limited to electrical quantities. Step-by-Step Solution: Identify the claim: symbolism is exclusive to electrical quantities.Provide counterexamples from mechanics, thermodynamics, and chemistry.Conclude the claim is incorrect. Verification / Alternative check: Review any introductory physics text: the first chapters already employ symbolic variables for non-electrical topics, confirming universal use. Why Other Options Are Wrong: Correct — exclusive to electrical: demonstrably false. Correct only in SI: SI formalizes units and symbols across all fields, not only electricity. Undecidable: Not subjective; convention is well established. Common Pitfalls: Assuming that because one is currently studying circuits, symbols must be unique to that context. They are not. Final Answer: Incorrect — all physical sciences use symbols for quantities (e.g., m, t, v, p).

Question 6

Electrical safety principle — evaluate the statement: “Voltage, not current, determines the severity of an electric shock.” Choose the most accurate assessment based on physiology and circuit theory.

Accepted Answer

Introduction / Context: Understanding shock hazards requires both physiology and basic circuit theory. Safety standards, medical data, and engineering practice all agree that the magnitude and duration of current through the human body is the primary determinant of injury risk, with voltage acting as a driver that sets current according to the body’s impedance. Given Data / Assumptions: Ohm’s law applies: I = V / R_body. Physiological thresholds are current-based (for example, perception, let-go, ventricular fibrillation thresholds). Contact duration and current path (hand-to-hand, hand-to-foot) matter. Concept / Approach: Severity correlates with current and exposure time. Standards often cite approximate thresholds such as 1 mA (perception), 5–10 mA (pain and muscle contraction), 30 mA (respiratory effects, let-go threshold), and 50–100 mA (risk of ventricular fibrillation), depending on conditions. Voltage contributes by driving current through body resistance; dry skin may present high resistance, while wet skin drastically lowers it, increasing current for the same voltage. Step-by-Step Solution: Express relationship: I = V / R_body.Recognize injury scales and thresholds are in amperes or milliamperes, not volts.Conclude that current magnitude and duration primarily determine severity; voltage is an important factor only as it affects current. Verification / Alternative check: Safety codes (e.g., IEC/IEEE guidance) set limits based on allowable touch current and energy, not a fixed “safe voltage” without context, reinforcing current-centric risk. Why Other Options Are Wrong: Voltage alone determines danger: Ignores variability of body resistance and contact conditions. Correct only for 50/60 Hz: While frequency matters for physiological response, the core metric remains current across frequencies. Depends solely on footwear/humidity: These affect R_body, but current and time are still the decisive metrics. Common Pitfalls: Believing a low voltage is always safe or a high voltage always fatal. Context (path, time, skin condition) and resulting current are key. Final Answer: Incorrect — severity depends primarily on current through the body; voltage influences current via resistance.

Question 7

Scientific notation multiplication — evaluate the statement: “(5 × 10^6) × (4 × 10^3) = 20 × 10^18.” Is this mathematically correct?

Accepted Answer

Introduction / Context: Accurate manipulation of powers of ten is crucial in electronics (for example, converting microfarads to farads or kilohertz to hertz) and in data-sheet calculations. This question asks you to validate a multiplication written in scientific notation. Given Data / Assumptions: Numbers: 5 × 10^6 and 4 × 10^3. Standard rules for scientific notation apply. No units are required; evaluation is purely numerical. Concept / Approach: When multiplying numbers in scientific notation, multiply the mantissas and add the exponents: (a × 10^m) × (b × 10^n) = (a*b) × 10^(m+n). Afterward, normalize the mantissa if necessary so that it lies between 1 and 10, adjusting the exponent accordingly. Step-by-Step Solution: Multiply mantissas: 5 × 4 = 20.Add exponents: 10^6 × 10^3 = 10^(6+3) = 10^9.Intermediate product: 20 × 10^9.Normalize mantissa: 20 × 10^9 = 2.0 × 10^10. Verification / Alternative check: Compute in plain numbers: 5,000,000 × 4,000 = 20,000,000,000 = 2.0 × 10^10. The statement “20 × 10^18” is off by nine orders of magnitude in the exponent. Why Other Options Are Wrong: Correct: Incorrect because the exponents must be added (6 + 3 = 9), not multiplied or concatenated to 18. Correct only if significant figures are ignored: Significant figures do not change the exponent arithmetic. Cannot be decided without units: Units are irrelevant to the numeric operation. Common Pitfalls: Confusing exponent addition with multiplication, or forgetting to normalize mantissas. Always multiply coefficients first, then add exponents and normalize. Final Answer: Incorrect — the correct product is 2.0 × 10^10, not 20 × 10^18.

Question 8

Scientific notation and powers of ten in electronics  Verify the statement: the voltage value 0.00015 V can be correctly written in scientific notation as 1.5 × 10^–4 V.

Accepted Answer

Introduction / Context: Scientific notation is essential in electronics because measured quantities often span many orders of magnitude. Representing numbers as a mantissa multiplied by a power of ten reduces confusion, keeps units consistent, and minimizes transcription errors when dealing with very small or very large values such as microvolts, megaohms, and nanofarads. Given Data / Assumptions: Given decimal quantity: 0.00015 V. Target format: scientific notation a × 10^n with 1 ≤ a < 10. Unit remains volts (V); we are not converting units to mV or µV. Concept / Approach: To write a number in scientific notation, move the decimal point so that exactly one nonzero digit remains to its left. Count the places moved; left-to-right movement for small values yields a negative exponent. The numeric value must remain identical after scaling by the proper power of ten, and the unit stays unchanged unless an explicit unit conversion is performed. Step-by-Step Solution: Start with 0.00015 V.Move the decimal 4 places to the right to get 1.5.Compensate with 10^–4 because moving right increases the number; the negative exponent brings it back: 1.5 × 10^–4.Attach the original unit: 1.5 × 10^–4 V. Verification / Alternative check: Compute 1.5 × 10^–4 = 1.5 / 10,000 = 0.00015, confirming exact equality. If desired, convert to microvolts: 0.00015 V = 150 µV (since 1 V = 10^6 µV). Both representations are consistent. Why Other Options Are Wrong: Incorrect: contradicts the correct power-of-ten placement.0.15 × 10^–3 V equals 1.5 × 10^–4 V numerically, but the mantissa is not between 1 and 10, so it is not standard scientific notation.“Only for millivolts” and “cannot be expressed” are false; scientific notation is universal across units. Common Pitfalls: Mixing up sign of the exponent; changing the unit without adjusting the exponent; writing a mantissa outside 1 to 10 range and calling it “scientific notation.” Final Answer: Correct

Question 9

Metrology foundations: distinguishing precision from accuracy  Evaluate the statement: “Precision refers to the difference between a measured value and the accepted (true) value.”

Accepted Answer

Introduction / Context: In electronics labs and field testing, engineers constantly judge the quality of measurements. Two foundational terms are accuracy and precision. Confusing them leads to poor instrument selection, incorrect uncertainty budgets, and misleading conclusions about whether a design meets specification. Given Data / Assumptions: We compare a measurement process to a known accepted (true) value. We consider repeatability (scatter across repeated trials) and closeness to the true value. Random and systematic errors can both be present. Concept / Approach: Accuracy describes closeness to the accepted value. It is about correctness. Precision describes repeatability or spread when measuring the same quantity multiple times under identical conditions. It is about consistency. A system can be precise but inaccurate (tight groupings far from the bullseye), accurate but imprecise (average near truth but with wide scatter), neither, or both. The statement in the prompt defines accuracy, not precision. Step-by-Step Solution: Identify the phrase “difference between measured value and accepted value.”Recognize this is the definition of accuracy (or error magnitude), not precision.Recall that precision concerns variability among repeated measurements (standard deviation, repeatability).Conclude that the given statement mislabels the concept; it is incorrect. Verification / Alternative check: Plot repeated readings as a histogram. Small spread (low standard deviation) indicates high precision. Compare the mean of these readings to the accepted value. A small bias (mean minus accepted) indicates high accuracy. The two qualities are independent axes of measurement quality. Why Other Options Are Wrong: Correct: would propagate the misconception.Conditional/device/unit-based options: accuracy vs. precision definitions do not depend on meter type, systematic error being zero, or the unit system. Common Pitfalls: Using “precise” when you mean “accurate”; trusting a high-resolution readout as proof of accuracy; ignoring calibration and traceability when claiming accuracy; equating a single reading with precision, which requires repetition. Final Answer: Incorrect

Question 10

Power units and metric prefixes  Confirm whether the electrical power value 4,200 W is numerically equal to 4.2 kW when expressed using the kilo (10^3) prefix.

Accepted Answer

Introduction / Context: Working across datasheets and test reports, engineers routinely convert raw watt values using SI prefixes to keep numbers readable and comparable. Misapplying prefixes can lead to 10× or 1000× errors with costly consequences, especially in power budgeting and thermal design. Given Data / Assumptions: Base unit for power: watt (W). kilo (k) denotes multiplication by 10^3. No change in the physical quantity, only unit scaling. Concept / Approach: To convert W to kW, divide by 10^3. The relationship is P(kW) = P(W) / 1000. Conversely, multiply kW by 1000 to return to W. This conversion is independent of AC or DC context, power factor, or frequency; it is purely a unit scaling within the SI system. Step-by-Step Solution: Start with 4,200 W.Apply the kilo conversion: 4,200 W / 1000 = 4.2 kW.Confirm that reversing the conversion yields 4.2 kW × 1000 = 4,200 W.Conclude that 4,200 W and 4.2 kW are numerically consistent. Verification / Alternative check: Express in megawatts: 4,200 W = 0.0042 MW because 1 MW = 10^6 W. Each form is consistent as long as the power-of-ten factor is applied correctly. Why Other Options Are Wrong: Incorrect: contradicts basic SI prefix scaling.Power factor, AC/DC, and frequency do not alter unit conversion; they affect how power is measured or computed, not how watts convert to kilowatts. Common Pitfalls: Confusing k (10^3) with K (kelvin); writing “KW” instead of “kW”; mixing k (10^3) with Ki (kibi, 2^10) which is not used for SI power units. Final Answer: Correct

Question 11

Metric prefixes in electronics documentation  Assess the statement: “The metric prefix giga (symbol G) represents one billion, that is 10^9.”

Accepted Answer

Introduction / Context: Standardized prefixes keep communication clear across disciplines. In electronics, you will see gigahertz for radio bands, gigabytes for storage marketing (with caveats), and gigohms for very high resistances. Understanding what “giga” means avoids order-of-magnitude mistakes in design and procurement. Given Data / Assumptions: SI metric prefix “giga”, symbol G. We use the International System of Units (SI), not binary prefixes, unless explicitly stated. Quantities may include frequency (Hz), resistance (Ω), power (W), etc. Concept / Approach: By SI definition, giga denotes a multiplication factor of 10^9. This is exact and does not depend on the underlying physical quantity. While computer memory historically used base-2 interpretations (e.g., 1 GB ≈ 2^30 bytes), SI clarifies that binary prefixes are different (gibi, Gi = 2^30). In scientific and engineering notation, G always means 10^9. Step-by-Step Solution: Map prefix to power: giga → 10^9.Apply to examples: 3.2 GHz = 3.2 × 10^9 Hz; 5 GΩ = 5 × 10^9 Ω.Distinguish from binary: 1 GiB = 2^30 bytes ≈ 1.073 × 10^9 bytes, not 10^9 exactly.Conclude the statement is correct within SI usage. Verification / Alternative check: Cross-check with other prefixes: mega (M) = 10^6, kilo (k) = 10^3, tera (T) = 10^12. The sequence increases by powers of 10^3 as expected, placing giga squarely at 10^9. Why Other Options Are Wrong: Incorrect or 10^6: contradicts the SI standard.“Varies with base-2” confuses SI with binary prefixes; in SI, G ≠ 2^30.“Not part of SI” is false; giga is an established SI prefix. Common Pitfalls: Using uppercase/lowercase incorrectly (G for giga vs. g for gram); mixing GB and GiB; assuming unit symbols alter the meaning of the prefix (they do not). Final Answer: Correct

Question 12

Recognizing proper scientific notation  Evaluate the statement: “The number 560,000, as written, is already in scientific notation.”

Accepted Answer

Introduction / Context: Scientific notation standardizes how we write large and small numbers: a × 10^n with 1 ≤ a < 10. This convention improves readability, prevents zeros from being dropped or added accidentally, and facilitates clear comparison of magnitudes in datasheets and calculations. Given Data / Assumptions: Number given: 560,000. No unit specified; the concept is unit-agnostic. We test compliance with strict scientific notation, not “approximate” styles. Concept / Approach: In scientific notation, the mantissa a must be at least 1 and less than 10. The exponent n is an integer. Writing 560,000 does not show a power of ten or a normalized mantissa. The correct scientific notation is 5.6 × 10^5 (or 5.600 × 10^5 if more significant digits are justified by measurement). Step-by-Step Solution: Express 560,000 with one nonzero digit to the left of the decimal: 5.6.Count decimal shifts: five places to the left from 560,000 to 5.6 → 10^5.Combine: 5.6 × 10^5.Conclude: 560,000 as typed is not scientific notation. Verification / Alternative check: Engineering notation groups powers by 3: 560,000 = 560 × 10^3. This is engineering notation (mantissa between 1 and 999), not scientific notation. Both are valid formats, but the rules differ; the prompt asked about scientific notation specifically. Why Other Options Are Wrong: Correct: ignores the requirement to show the power of ten and normalized mantissa.Unit-based or rounding-based claims do not change the definition of scientific notation.“Engineering but not scientific” is close conceptually, but the original 560,000 is not written as 560 × 10^3 either. Common Pitfalls: Assuming any compact form qualifies; omitting the exponent; confusing engineering notation (steps of 10^3) with scientific notation (1 to 10 mantissa). Final Answer: Incorrect

Question 13

Significant figures and rounding rules  Decide whether this rounding statement is correct: “The number 24.5, rounded to two significant digits, becomes 25.”

Accepted Answer

Introduction / Context: Significant figures communicate the precision of measured or computed values. Rounding properly preserves the intended precision without overstating certainty. This is crucial in tolerance stacks, component value reporting, and lab notes. Given Data / Assumptions: Original value: 24.5 (three significant digits). Target: two significant digits. Standard rounding-to-nearest with 5 causing a round up of the preceding digit. Concept / Approach: To round to two significant digits, keep the first two nonzero digits and examine the next digit. If the next digit is 5 or greater, increase the last kept digit by one; otherwise, leave it. The position of the decimal point and whether trailing zeros are written can affect how many significant figures are perceived, but the rounded numeric value follows these rules consistently. Step-by-Step Solution: Identify two significant digits to keep: “2” and “4”.Look at the third digit: “5”. Because it is 5, round up the “4”.Compute: 24.5 → 25 (to two significant digits).Interpretation: “25” has two significant digits by default; if one needs to signal two significant digits explicitly, writing “25” is acceptable. Writing “25.” can imply two sig figs as well in some conventions. Verification / Alternative check: Compare with rounding to one significant digit: 24.5 → 20 (since the next digit “4” is less than 5 after the “2”). For three significant digits, 24.5 remains 24.5. The two-digit target clearly yields 25 by standard rules. Why Other Options Are Wrong: Incorrect / 24 / 24.0: each conflicts with the round-up rule when the following digit is exactly 5.“Only if 25.”: the decimal point can clarify sig figs but is not mandatory here; “25” already conveys two significant digits. Common Pitfalls: Confusing significant digits with decimal places; assuming “.0” always increases sig figs; using bank rounding rules unintentionally instead of standard round-to-nearest for technical reporting. Final Answer: Correct

Question 14

Fundamental SI quantities in electronics  Select the physical quantity measured in joules (J), the SI unit commonly encountered in energy storage, work, and heat calculations.

Accepted Answer

Introduction / Context: Units anchor equations to physical meaning. In electronics, the joule appears in capacitor energy (E = 0.5 * C * V^2), inductor energy (E = 0.5 * L * I^2), resistive heating (E = P * t), and battery capacity measurements (watt-hours converted to joules). Knowing what the joule measures prevents dimensional mistakes in design and analysis. Given Data / Assumptions: We seek the physical quantity corresponding to the SI unit joule (J). Other options list common but different SI units. No special conditions; standard SI definitions apply. Concept / Approach: The joule is the SI unit of energy (and work). One joule equals one newton-meter, which also equals one watt-second because power in watts multiplied by time in seconds yields energy in joules. In electrical terms, 1 J = 1 V * 1 C (one volt times one coulomb), reinforcing its connection to stored or transferred energy. Step-by-Step Solution: Match “joule” to “energy/work.”Cross-relate: watt (power) × second (time) = joule (energy).Apply to components: capacitors and inductors store energy expressible in joules.Select “energy” as the correct quantity. Verification / Alternative check: Dimensional analysis: volts × coulombs = joules. Since V = J/C by definition, the product returns joules consistently in circuit energy computations. Why Other Options Are Wrong: time: measured in seconds (s).frequency: measured in hertz (Hz = s^–1).charge: measured in coulombs (C).electric field strength: SI unit volt per meter (V/m), not joule. Common Pitfalls: Mixing energy (J) with power (W); reporting battery capacity only in mAh and forgetting that energy depends on voltage (Wh → J via 1 Wh = 3600 J). Final Answer: energy

Question 15

Frequency unit conversion  Convert the radio frequency 106,100,000 Hz into megahertz (MHz). Choose the numerically correct value.

Accepted Answer

Introduction / Context: Frequency specifications routinely use hertz, kilohertz, megahertz, and gigahertz. Correct unit scaling is crucial when comparing oscillator ratings, radio channels, and clock sources. A simple order-of-magnitude error can derail a design or cause an off-by-1000 mistake in code or documentation. Given Data / Assumptions: Given frequency: 106,100,000 Hz. 1 MHz = 10^6 Hz. We only convert units; the physical frequency remains identical. Concept / Approach: To convert from hertz to megahertz, divide the numeric value by 10^6. This is a direct SI prefix conversion. Ensure commas do not mislead during division, and avoid rounding until the last step if precision matters. Step-by-Step Solution: Write the conversion factor: MHz = Hz / 10^6.Compute: 106,100,000 Hz / 1,000,000 = 106.1 MHz.Check significant figures: the original has four to five meaningful digits; 106.1 preserves appropriate precision.State the result: 106.1 MHz. Verification / Alternative check: Express as scientific notation: 106,100,000 Hz = 1.061 × 10^8 Hz. Dividing by 10^6 gives 1.061 × 10^2 MHz = 106.1 MHz. Both pathways agree. Why Other Options Are Wrong: 1,061 / 10,610 / 106,100: each treats the factor as 10^3 repeatedly or ignores the 10^6 divisor.0.1061: corresponds to dividing by 10^9 (gigahertz conversion), not megahertz. Common Pitfalls: Confusing MHz with kHz or GHz; misplacing decimal points; mixing engineering notation (e.g., 106.1 MHz) with scientific notation and altering the meaning inadvertently. Final Answer: 106.1

Question 16

Multiplying numbers in scientific notation  Compute (12.6 × 10^6) * (10.5 × 10^3) and express the result in proper scientific notation (mantissa between 1 and 10).

Accepted Answer

Introduction / Context: Scientific notation accelerates multiplication and division by separating magnitude (10^n) from the significant digits (mantissa). This is particularly helpful in electronics when combining values across very different scales, such as gains, frequencies, or time constants. Given Data / Assumptions: Numbers: 12.6 × 10^6 and 10.5 × 10^3. We seek normalized scientific notation (1 ≤ mantissa < 10). Arithmetic is exact to at least three significant digits. Concept / Approach: Multiply mantissas and add exponents: (a × 10^m) * (b × 10^n) = (a*b) × 10^(m+n). If the mantissa product is not between 1 and 10, re-normalize by adjusting the exponent accordingly. Keep an appropriate number of significant figures based on inputs (three sig figs here are reasonable: 12.6 and 10.5 each have three). Step-by-Step Solution: Multiply mantissas: 12.6 * 10.5 = 132.3.Add exponents: 10^6 * 10^3 = 10^(6+3) = 10^9.Combine: 132.3 × 10^9.Normalize mantissa: 132.3 × 10^9 = 1.323 × 10^11 (move decimal left two places, add 2 to exponent). Verification / Alternative check: Use a rounded mantissa to three significant digits consistent with inputs: 1.323 × 10^11 ≈ 1.32 × 10^11. A calculator check confirms the numeric value: 12.6e6 * 10.5e3 = 1.323e11, matching the normalized result. Why Other Options Are Wrong: 13.2 × 10^11 and 13.2 × 10^12: mantissas are not normalized (should be between 1 and 10) and/or exponents are off by a power of ten.1.32 × 10^12: exponent too large by 10×.1.323 × 10^10: exponent too small by 10×. Common Pitfalls: Forgetting to renormalize the mantissa; adding exponents incorrectly; rounding too early, which can introduce noticeable errors when cascaded across multiple steps. Final Answer: 1.32 × 10^11

Question 17

Scaling small decimals with powers of ten  Find the power of ten n such that 0.0136 equals 13.6 × 10^n. Choose the correct exponent.

Accepted Answer

Introduction / Context: Expressing values as a mantissa times a power of ten lets you shift the decimal point without changing the quantity. This skill is fundamental when converting between unit prefixes (milli, micro, nano) or reformatting numbers for clarity in reports and calculations. Given Data / Assumptions: Decimal value: 0.0136. Target form: 13.6 × 10^n with mantissa 13.6 (not normalized scientific notation, but an equivalent scaling). We need to find the exponent n that preserves equality. Concept / Approach: Moving the decimal point to the right increases the mantissa; to keep the numeric value unchanged, the power of ten must compensate with a negative exponent. Specifically, shifting the decimal point k places to the right requires multiplying by 10^k and simultaneously multiplying by 10^–k to maintain equality when expressing the factor explicitly. Step-by-Step Solution: Start with 0.0136.To reach 13.6, move the decimal point three places to the right: 0.0136 → 13.6.Each right shift corresponds to multiplying by 10. Three shifts correspond to ×10^3.To keep the value identical, include ×10^–3 as the compensating factor: 0.0136 = 13.6 × 10^–3. Verification / Alternative check: Compute 13.6 × 10^–3 = 13.6 / 1000 = 0.0136, confirming equality. Alternatively, write 0.0136 in scientific notation: 1.36 × 10^–2. Multiplying the mantissa by 10 (to 13.6) requires reducing the exponent by 1 (from –2 to –3), consistent with the result. Why Other Options Are Wrong: 10^–2 or 10^–1: do not compensate for three-place shift; values would be 0.136 or 1.36 respectively.10^–4: overshoots; would produce 0.00136.10^0: leaves 13.6 unchanged, not 0.0136. Common Pitfalls: Confusing the direction of decimal movement with exponent sign; insisting on normalized mantissa unnecessarily when the problem specifies a fixed mantissa of 13.6. Final Answer: 10^–3

Question 18

Engineering unit conversion — capacitance scaling: 1300 × 10^–12 farads is equivalent to 1300 what unit of capacitance?

Accepted Answer

Introduction / Context: Electronic component values are commonly expressed using engineering prefixes to keep numbers readable. Being fluent with pico, nano, micro, and milli simplifies interpreting datasheets, selecting parts, and checking calculations in circuits that use capacitors, resistors, and inductors. Given Data / Assumptions: The given quantity is 1300 × 10^–12 farads. We must express the same value using a standard SI-prefixed capacitance unit. Valid SI prefixes near 10^–12 are: pico (10^–12), nano (10^–9), micro (10^–6), milli (10^–3). Concept / Approach: Match the power of ten to the correct SI prefix, then rewrite the number so that the numerical part is convenient (typically between 1 and 9999 with an engineering prefix step of 10^3). Since 10^–12 corresponds to pico, multiplying any number by 10^–12 farads yields a value in picofarads directly. Step-by-Step Solution: Identify prefix: 10^–12 = pico (p).Rewrite: 1300 × 10^–12 F = 1300 pF.Optionally, compress to other prefixes: 1300 pF = 1.3 nF = 0.0013 µF.Select the answer requested in the prompt format: 1300 pF. Verification / Alternative check: Convert to farads as a decimal: 1300 × 10^–12 F = 1.3 × 10^–9 F = 1.3 nF. This cross-check confirms the pico-to-nano equivalence and validates that 1300 pF is correct. Why Other Options Are Wrong: nF would be 1.3 nF, not 1300 nF. µF would be 0.0013 µF, not 1300 µF. mF and F are much larger and do not match the power of ten given. Common Pitfalls: Confusing p (pico, 10^–12) with n (nano, 10^–9) or µ (micro, 10^–6), and forgetting that shifting prefixes by 10^3 changes the decimal point by three places. Final Answer: pF

Question 19

Current unit conversion — express in milliampere: 7450 × 10^–5 amperes equals 74.5 what unit?

Accepted Answer

Introduction / Context: Converting between base SI units and engineering prefixes is routine in electronics. Expressing a current in milliamperes is often more intuitive for circuit analysis and component ratings, such as the forward current of diodes or the quiescent current of ICs. Given Data / Assumptions: Quantity: 7450 × 10^–5 A. We want an equivalent form stated as 74.5 (numeric) followed by the correct unit symbol. Useful relations: 1 mA = 10^–3 A; 1 µA = 10^–6 A; 1 nA = 10^–9 A. Concept / Approach: Simplify the scientific notation to a decimal in amperes, then change units to a convenient engineering prefix so that the numeric part matches 74.5. Recognize that milliamperes are thousandths of an ampere, so multiplying amperes by 1000 converts to milliamperes. Step-by-Step Solution: Compute in amperes: 7450 × 10^–5 A = 0.0745 A.Convert to mA: 0.0745 A × 1000 mA/A = 74.5 mA.Confirm numeric match: the form required is “74.5 ___”, hence unit = mA.State final unit: mA (milliamperes). Verification / Alternative check: Convert directly: 10^–5 A = 0.01 mA. Then 7450 × 0.01 mA = 74.5 mA. Both paths agree, confirming the result. Why Other Options Are Wrong: A (amperes) would be 0.0745 A, not 74.5 A. µA would be 74,500 µA, not 74.5 µA. nA would be 74,500,000 nA. kA is orders of magnitude too large. Common Pitfalls: Losing track of decimal placement when switching between prefixes, and confusing 10^–5 with 10^–3 (milli) or 10^–6 (micro). Always rewrite to base units first if unsure. Final Answer: mA

Question 20

Scientific notation alignment — express with 10^1: The number 5200 × 10^–2 is equal to what value multiplied by 10^1?

Accepted Answer

Introduction / Context: Proficiency with scientific notation prevents mistakes when scaling numbers or matching units. In electronics, powers of ten appear everywhere: impedance at different frequencies, RC time constants, or converting between prefixes like kilo and milli. This item practices re-expressing a quantity with a different power of ten while keeping the value identical. Given Data / Assumptions: Original expression: 5200 × 10^–2. Target form: X × 10^1 (find X). Basic rule: shifting the exponent by +3 requires shifting the coefficient by ÷1000, etc. Concept / Approach: First simplify the original form to a plain number, then rebuild it using the required base of 10^1. Alternatively, keep the value constant by adjusting the coefficient and exponent inversely: if the exponent increases by +3, the coefficient must be divided by 10^3, and vice versa. Step-by-Step Solution: Compute 5200 × 10^–2 = 5200 / 100 = 52.Express 52 using 10^1: since 10^1 = 10, write 52 as 5.2 × 10.Hence X = 5.2 so that 5.2 × 10^1 = 52.Check: 5.2 × 10 = 52, consistent with the simplified value. Verification / Alternative check: Adjust exponents: 5200 = 5.2 × 10^3. Then 5200 × 10^–2 = 5.2 × 10^(3–2) = 5.2 × 10^1. This directly yields X = 5.2. Why Other Options Are Wrong: 52 would demand 52 × 10^1 = 520, not 52. 0.52 × 10^1 = 5.2, not 52. 520 × 10^1 = 5200, and 0.052 × 10^1 = 0.52; both incorrect. Common Pitfalls: Forgetting that changing the exponent requires compensating the coefficient so the overall value is unchanged, and mixing up decimal shifts in the wrong direction. Final Answer: 5.2

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