Q: Foundations — electrical power is defined as the rate of energy transfer (work per unit time), not as “how fast current is flowing.” Evaluate the statement.

Introduction / Context: Power is one of the three pillars of Ohm’s law problems, alongside voltage and current. Misstatements such as “power measures how fast current flows” confuse learners and lead to dimensional errors. This item clarifies the physical meaning of electrical power and its correct formulas. Given Data / Assumptions: Standard circuit quantities: voltage V, current I, resistance R. Time-average power in passive components; instantaneous definitions extend naturally. Ohmic behavior unless otherwise stated. Concept / Approach: Electrical power P is energy per unit time delivered to or drawn from a component. In lumped circuits, instantaneous power is p(t) = v(t) * i(t). For resistors under DC, P = V * I = I^2 * R = V^2 / R. None of these expressions measure “how fast current flows”; that idea is captured by current I itself (charge per unit time). Power tells you the rate at which electrical energy converts to heat, mechanical work, or other forms. Step-by-Step Solution: State correct definition: P = energy/time.Relate to circuit variables: P = V * I for any element.For resistors, use P = I^2 * R or P = V^2 / R for practical sizing and thermal calculations.Conclude that calling power “how fast current flows” is incorrect. Verification / Alternative check: Dimensional analysis: V (J/C) times I (C/s) yields J/s (watts). No “speed” quantity appears; current already equals charge per time, while power equals energy per time. Why Other Options Are Wrong: Correct/only DC/depends solely on R/needs frequency: These distract from the fundamental definition P = V * I applicable across DC and AC. Common Pitfalls: Confusing current magnitude with power; forgetting that a high current at near-zero voltage can still imply low power (and vice versa). Final Answer: Incorrect — power is the rate of energy transfer, not the “speed” of current.

Q: Energy equivalence — verify the statement: consuming 4 kWh of electrical energy is equivalent to drawing 2000 W of power for 2 hours (energy = power * time).

Introduction / Context: Utility bills quote energy in kilowatt-hours (kWh). Converting between energy and power-time products is essential both for household consumption and engineering estimates. The statement equates 4 kWh with 2000 W sustained for 2 hours; this is a straightforward application of E = P * t. Given Data / Assumptions: Energy E = 4 kWh. Proposed power-time pair: P = 2000 W, t = 2 h. We consider real power; reactive aspects average to zero for energy billing. Concept / Approach: Convert units carefully: 1 kW = 1000 W. Energy in kWh is the time integral of kilowatts. If power is constant, E = P * t. Therefore, 2000 W * 2 h = (2 kW) * 2 h = 4 kWh. The equality holds irrespective of DC or AC, provided P denotes real power over the interval. Step-by-Step Solution: Convert 2000 W to kW: 2000 W = 2 kW.Multiply by time: 2 kW * 2 h = 4 kWh.Compare to the stated energy: both are 4 kWh.Conclude that the statement is correct. Verification / Alternative check: Another equivalent: 1000 W for 4 h or 4000 W for 1 h also yield 4 kWh. Any P, t pair with P(kW) * t(h) = 4 satisfies the same energy. Why Other Options Are Wrong: Incorrect: contradicts the direct unit calculation. “Only DC” or “only at unity PF”: Energy calculation E = ∫ P_real dt remains valid in AC; utility meters measure real energy. “Need voltage”: Voltage is unnecessary if power is already given. Common Pitfalls: Mixing up kWh (energy) and kW (power); forgetting to convert W to kW before multiplying by hours. Final Answer: Correct — 2000 W for 2 h equals 4 kWh of energy.

Q: Unit scaling — choose the most concise engineering notation for 7,500,000 W (watts) from the following options.

Introduction / Context: Expressing large or small quantities in appropriate SI prefixes makes technical communication clearer. Power is commonly stated in W, kW, MW, or GW depending on scale. The goal is to choose the compact, correct prefix form for 7,500,000 W. Given Data / Assumptions: 1 kW = 10^3 W. 1 MW = 10^6 W. 1 GW = 10^9 W. Concept / Approach: Convert watts to the nearest convenient SI-prefixed unit by dividing by powers of ten. Since 7,500,000 W = 7.5 × 10^6 W, the megawatt scale is appropriate. Reporting the value as 7.5 MW is concise and standard in power systems and specifications. Step-by-Step Solution: Write 7,500,000 W as 7.5 × 10^6 W.Recognize that 1 MW = 10^6 W.Therefore, 7,500,000 W = 7.5 MW.Select 7.5 MW as the best expression. Verification / Alternative check: Back-convert: 7.5 MW × 10^6 W/MW = 7,500,000 W, confirming equivalence. Why Other Options Are Wrong: 7.5 kW: Off by a factor of 1000; 7.5 kW = 7,500 W. 7,500 MW: Off by a factor of 1000; equals 7.5 × 10^9 W. 7,500 GW: Off by a factor of 10^6; equals 7.5 × 10^12 W. 0.0075 GW: Numerically equal to 7.5 MW but far less standard than 7.5 MW for this magnitude; the best concise form is MW. Common Pitfalls: Confusing k (10^3), M (10^6), and G (10^9); writing overly large numbers without prefixes, reducing readability. Final Answer: 7.5 MW

Q: Ohm’s law calculation — a resistor has 12 V across it and carries 500 µA of current. Compute the resistance value and choose the correct option.

Introduction / Context: Direct application of Ohm’s law V = I * R allows quick back-calculation of a resistor value from measured voltage and current. Careful attention to micro- units prevents decimal mistakes that are common in troubleshooting and exams. Given Data / Assumptions: Voltage across the resistor, V = 12 V. Current through the resistor, I = 500 µA = 500 × 10^−6 A = 0.0005 A. Ohmic behavior (constant R). Concept / Approach: Rearrange Ohm’s law to R = V / I. Convert microamperes to amperes before dividing to avoid a 10^6 error. Finally, express the answer in a convenient SI multiple (kΩ) for readability. Step-by-Step Solution: Convert current: 500 µA = 0.0005 A.Compute resistance: R = V / I = 12 / 0.0005 = 24,000 Ω.Express in kΩ: 24,000 Ω = 24 kΩ.Select the option that matches 24 kΩ. Verification / Alternative check: Cross-check using I = V / R with R = 24 kΩ: I = 12 / 24,000 = 0.0005 A = 500 µA, confirming consistency. Why Other Options Are Wrong: 600 Ω, 2.4 kΩ, 6 kΩ, 60 kΩ: Each corresponds to using an incorrect unit conversion or arithmetic slip when dividing 12 by 0.0005. Common Pitfalls: Forgetting that 1 µA = 10^−6 A; mixing up milli and micro; failing to convert before dividing. Final Answer: 24 kΩ

Q: Direct Ohm’s law implication — for a fixed-value resistor, if the voltage applied across it increases, how does the current through it change (assuming temperature effects are negligible)?

Introduction / Context: Ohm’s law gives an immediate relationship between the voltage across a resistor and the current through it. Understanding this proportionality is fundamental to predicting how changes in supply voltage affect branch currents in simple and complex circuits alike. Given Data / Assumptions: Resistor value R is constant and positive. Temperature and self-heating effects are negligible for the change considered. Polarity is fixed; we are discussing magnitude changes. Concept / Approach: Ohm’s law states V = I * R. Holding R constant implies I = V / R. Therefore, if V increases, I must increase proportionally. This is the definition of a linear, ohmic element. The exact numeric change requires R, but the qualitative direction (increase) does not. Step-by-Step Solution: Start from I = V / R for a fixed R.Consider a higher applied V: numerator increases while denominator is unchanged.Therefore I increases linearly with V.Conclude that current increases when voltage increases across a fixed resistor. Verification / Alternative check: Plot I vs. V: the line I = (1/R) * V has positive slope 1/R. Any movement to the right along the V-axis moves up along the I-axis. Why Other Options Are Wrong: Decreases/Stays the same: Contradicts I = V / R for constant R. Cannot be determined: While the exact value needs R, the trend is determined. Reverses direction: Direction is set by polarity; magnitude increase does not flip direction. Common Pitfalls: Confusing constant-current sources (which enforce I) with resistors (which follow Ohm’s law); ignoring that significant heating can change R slightly, but the ideal law remains linear. Final Answer: Increases

Q: Ohm’s law relationship — resistance vs current: In a simple conductor at constant temperature supplied by a fixed voltage, resistance (R) and current (I) are _________.

Introduction / Context: Ohm’s law, expressed as V = I * R, links voltage (V), current (I), and resistance (R). Understanding how any two of these quantities co-vary when the third is held constant is essential for quick circuit intuition and troubleshooting. Given Data / Assumptions: Voltage V is held constant (e.g., a regulated supply). Temperature and material properties are constant so R does not change with I. Linear (ohmic) device behavior is assumed. Concept / Approach: From V = I * R with V constant, solve for I: I = V / R. This shows that current varies inversely with resistance. Doubling R halves I; halving R doubles I. The proportionality is direct only between V and I when R is constant, not between I and R at constant V. Step-by-Step Solution: Start with V = I * R.Hold V constant → I = V / R.Therefore, as R increases, I decreases (inverse relationship).Conversely, as R decreases, I increases. Verification / Alternative check: Example: With V = 10 V, R = 1 kΩ → I = 10 mA; if R = 2 kΩ → I = 5 mA (halved), confirming inverse proportionality. Why Other Options Are Wrong: Directly proportional: true for V and I when R is constant, not for I vs R at constant V.Not related / similar to voltage: contradicts Ohm’s law. Common Pitfalls: Mixing “direct” and “inverse” proportionality when switching which variable is held constant; forgetting temperature effects that can change R in real parts. Final Answer: inversely proportional

Q: Energy–power linkage — unit equivalence: The number of joules used in 1 second is always equal to the number of _______. (Hint: energy per unit time.)

Introduction / Context: Power quantifies the rate at which energy is transferred or converted. In SI, one watt is defined as one joule per second. Recognizing this equivalence helps in quickly converting between energy and power in specifications and calculations. Given Data / Assumptions: Energy is measured in joules (J). Time interval is 1 second (s). We want the unit that equals J/s. Concept / Approach: Power P is defined as P = energy / time. Therefore, when 1 joule is used each second, the rate is 1 watt. The other listed units measure different quantities: siemens (conductance), coulombs (charge), amperes (current). Step-by-Step Solution: Write P = E / t.Insert E = joules and t = seconds → P has units J/s.Recognize J/s is named “watt”.Therefore choose “watts”. Verification / Alternative check: Check dimensional consistency: 1 W = 1 N·m/s = 1 kg·m^2/s^3, aligning with joule-per-second identity. Why Other Options Are Wrong: Siemens: unit of conductance.Coulombs: unit of electric charge.Amperes: unit of current (coulombs per second). Common Pitfalls: Confusing power with energy (watt-hours vs watts); mixing up ampere (A) with watt (W). Final Answer: watts

Q: Power calculation from energy and time: What is the power if 200 J of energy are used in 10 s?

Introduction / Context: Many practical problems present energy consumed over a time interval, from which you must compute average power. This is a direct application of the definition of power in the time domain. Given Data / Assumptions: Energy E = 200 J. Time t = 10 s. Average power P = E / t. Concept / Approach: Use the basic relation: power equals energy divided by time. Ensure units are SI to avoid conversion errors. The arithmetic should be straightforward as the numbers are chosen to divide cleanly. Step-by-Step Solution: P = E / t.Substitute: P = 200 J / 10 s.Compute: P = 20 J/s.Recognize J/s = W → P = 20 W. Verification / Alternative check: If the same 200 J were used in 20 s, power would halve to 10 W; if used in 5 s, it would double to 40 W. This proportionality confirms the calculation. Why Other Options Are Wrong: 2 W and 5 W: too low by factors of 10 and 4 respectively.50 W: too high by a factor of 2.5. Common Pitfalls: Forgetting that “per second” means divide by time; confusing power (W) with energy (J). Final Answer: 20 W

Question 1

Ohm’s law proportional change — repaired stem: In a fixed-resistance DC circuit, the current is initially 8 mA. If the current is increased to 24 mA with the same resistance, the source voltage must have tripled. Evaluate this claim.

Accepted Answer

Introduction / Context: This item originally referenced a “given circuit” without data. Applying the Recovery-First Policy, we minimally repaired the stem by stating the initial current explicitly and assuming a fixed-resistance DC circuit. The concept being tested is the direct proportionality between current and voltage for a constant resistance, as stated by Ohm’s law. Given Data / Assumptions: Initial current I1 = 8 mA. Final current I2 = 24 mA. Resistance R is constant (temperature and tolerance effects neglected). DC source; wires and contacts ideal for concept. Concept / Approach: Ohm’s law gives I = V / R or equivalently V = I * R. With R fixed, voltage and current are directly proportional. Therefore, any multiplicative change in current requires the same multiplicative change in voltage. Tripling the current requires tripling the voltage at constant resistance. Step-by-Step Solution: Write V1 = I1 * R and V2 = I2 * R.Form the ratio V2 / V1 = (I2 * R) / (I1 * R) = I2 / I1.Compute I2 / I1 = 24 mA / 8 mA = 3.Conclude V2 / V1 = 3, i.e., the source voltage must have tripled. Verification / Alternative check: Numerical example: If R = 1 kΩ, then V1 = 8 mA * 1 kΩ = 8 V and V2 = 24 mA * 1 kΩ = 24 V, confirming a 3x increase. The specific R cancels in the ratio, so the conclusion does not depend on its value. Why Other Options Are Wrong: Incorrect: Conflicts with direct proportionality at constant resistance. Only correct for AC sources: The proportionality holds for DC and RMS AC in resistive circuits. Only correct if resistance decreases: Resistance is specified constant; no change is required or implied. Common Pitfalls: Forgetting to specify that resistance is constant; mixing peak and RMS when dealing with AC; assuming the conclusion depends on a particular resistance value when the ratio makes it independent. Final Answer: Correct.

Question 2

General trend — resistance vs. current: In a simple circuit obeying Ohm’s law, increasing the total resistance (with source voltage fixed) will generally decrease the current. Assess this statement.

Accepted Answer

Introduction / Context: Understanding qualitative trends from Ohm’s law is as important as calculating exact numbers. Designers often need to know how changing a component will affect the rest of the circuit. This question checks whether you can correctly predict the direction of change in current when resistance increases while the source voltage remains the same. Given Data / Assumptions: Ohmic circuit with V_source held constant. Total effective resistance increases (e.g., adding series resistance). Temperature effects and non-linearities neglected for the concept check. Concept / Approach: From I = V / R, for a fixed voltage V, current I is inversely proportional to resistance R. Therefore, increasing R reduces I, and decreasing R increases I. This is a general principle across DC resistive circuits and the RMS behavior of AC circuits composed solely of resistors. Step-by-Step Solution: Hold V constant and consider two resistance values R1 and R2.If R2 > R1, then I2 = V / R2 < V / R1 = I1.Therefore the current must decrease when resistance increases.This inverse relationship is independent of the particular numeric values as long as V is unchanged. Verification / Alternative check: Series networks demonstrate the trend clearly: adding a series resistor raises total R and lowers current drawn from the source. Power dissipation also shifts; for a given V, reducing current reduces total input power P = V * I. Why Other Options Are Wrong: Incorrect: Opposes the inverse relationship from Ohm’s law. Correct only if resistance halves / only at low voltage: The relationship is not conditional on specific magnitudes, only on V being held constant. Common Pitfalls: Confusing effects of changing voltage with changing resistance; overlooking that parallel additions reduce resistance and thus increase current for a given V (the opposite trend when branches are added in parallel). Final Answer: Correct.

Question 3

Ohm’s law basics — the “voltage drop” across a resistor is simply the potential difference measured between its two terminals. In circuit analysis, is this definition appropriate?

Accepted Answer

Introduction / Context: The phrase “voltage drop” appears constantly in electronics. Students sometimes overcomplicate it, thinking it requires current flow or special conditions. In standard circuit theory, the voltage drop across any element is the electric potential difference between its two terminals, regardless of whether the element is a source, a resistor, or a reactive component. Given Data / Assumptions: Lumped, ideal circuit elements are used for analysis. Voltage is defined as potential at one node minus potential at another node. Measurement is taken directly across the two terminals of the resistor. Concept / Approach: Voltage is a state variable associated with nodes. For any two points A and B, V_AB = V(A) - V(B). When the two points are the terminals of a resistor, this difference is colloquially called the “voltage drop across the resistor.” Ohm’s law then relates current through the resistor and the drop across it: V = I * R. The concept does not require that current be nonzero at the instant of measurement; with I = 0, the drop could be zero, but the definition still stands. Step-by-Step Solution: Identify the two resistor terminals (call them Node 1 and Node 2).Measure or compute V(Node 1) and V(Node 2) with respect to a common reference.Compute the voltage drop as V_drop = V(Node 1) - V(Node 2).Interpret the sign convention as needed for passive sign convention and power calculations. Verification / Alternative check: A high-impedance voltmeter placed across the resistor reads this potential difference directly. SPICE node-voltage results confirm the same definition numerically for DC, AC, and transient simulations. Why Other Options Are Wrong: Incorrect: Conflicts with standard definitions of potential difference. Applies only in DC: The definition holds for AC and time-varying signals. Applies only when current is nonzero: Voltage can be defined even with I = 0. Depends on power rating: Power rating is unrelated to the definition of voltage. Common Pitfalls: Confusing the definition of voltage with conditions for power dissipation; assuming “drop” implies energy loss only (sources can also have terminal voltages). Final Answer: Correct — the voltage drop across a resistor is the potential difference between its terminals.

Question 4

Foundations — electrical power is defined as the rate of energy transfer (work per unit time), not as “how fast current is flowing.” Evaluate the statement.

Accepted Answer

Introduction / Context: Power is one of the three pillars of Ohm’s law problems, alongside voltage and current. Misstatements such as “power measures how fast current flows” confuse learners and lead to dimensional errors. This item clarifies the physical meaning of electrical power and its correct formulas. Given Data / Assumptions: Standard circuit quantities: voltage V, current I, resistance R. Time-average power in passive components; instantaneous definitions extend naturally. Ohmic behavior unless otherwise stated. Concept / Approach: Electrical power P is energy per unit time delivered to or drawn from a component. In lumped circuits, instantaneous power is p(t) = v(t) * i(t). For resistors under DC, P = V * I = I^2 * R = V^2 / R. None of these expressions measure “how fast current flows”; that idea is captured by current I itself (charge per unit time). Power tells you the rate at which electrical energy converts to heat, mechanical work, or other forms. Step-by-Step Solution: State correct definition: P = energy/time.Relate to circuit variables: P = V * I for any element.For resistors, use P = I^2 * R or P = V^2 / R for practical sizing and thermal calculations.Conclude that calling power “how fast current flows” is incorrect. Verification / Alternative check: Dimensional analysis: V (J/C) times I (C/s) yields J/s (watts). No “speed” quantity appears; current already equals charge per time, while power equals energy per time. Why Other Options Are Wrong: Correct/only DC/depends solely on R/needs frequency: These distract from the fundamental definition P = V * I applicable across DC and AC. Common Pitfalls: Confusing current magnitude with power; forgetting that a high current at near-zero voltage can still imply low power (and vice versa). Final Answer: Incorrect — power is the rate of energy transfer, not the “speed” of current.

Question 5

Ohm’s devices — for an ideal ohmic resistor, the current–voltage (I–V) relationship is linear and passes through the origin. Does the claim “current and voltage have a nonlinear relationship” hold for ohmic conductors?

Accepted Answer

Introduction / Context: Ohm’s law states that for an ohmic conductor the voltage across it is directly proportional to the current through it at a fixed temperature and physical state. This implies a straight-line I–V characteristic through the origin. The given claim asserts nonlinearity, which is not true for ideal resistors used in introductory circuit analysis. Given Data / Assumptions: Component is an ideal ohmic resistor. Temperature and physical conditions are constant. Small-signal and large-signal operation remain within the linear region. Concept / Approach: The defining relation of an ohmic resistor is V = I * R with constant R. If R is constant, doubling I doubles V. The graph of V vs. I is a straight line with slope R. Many real devices (diodes, lamps, thermistors) are indeed nonlinear, but the ohmic resistor model is linear by definition and underpins the superposition and Thevenin/Norton techniques learned early in circuits. Step-by-Step Solution: Start with Ohm’s law: V = I * R.Hold R constant; vary I: V changes proportionally.Plot V vs. I: a straight line through the origin with slope R.Conclude that for ohmic resistors the relationship is linear, not nonlinear. Verification / Alternative check: Bench measurements with a variable supply and ammeter show linear scaling of V and I for carbon film or metal film resistors operated within ratings. SPICE resistor elements are ideal linear by default. Why Other Options Are Wrong: “All elements are nonlinear”: diodes/transistors are, but ohmic resistors are modeled as linear. “Only at high temperature” and “depends on tolerance”: temperature and tolerance shift R but do not make the I–V relation mathematically nonlinear over the operating range. “Because power depends on I^2”: that does not change the linear V–I law. Common Pitfalls: Generalizing nonlinearity from semiconductor examples to all components; confusing power formulas with I–V linearity. Final Answer: Incorrect — ohmic resistors have a linear I–V relationship.

Question 6

Resistor value vs. wattage — is a resistor’s numeric resistance (in ohms) directly proportional to its power handling capability (watt rating), or are these independent specifications?

Accepted Answer

Introduction / Context: Resistor datasheets list at least two fundamentally different specifications: resistance value (ohms) and power rating (watts). Confusing these leads to poor part selection and overheating in products. The question probes whether the two are directly proportional, which they are not. Given Data / Assumptions: Commercial resistors with specified ohmic values and watt ratings. Ambient conditions and mounting affect derating but do not couple ohms to watts mathematically. Thermal design governs power rating via size and materials. Concept / Approach: Resistance is set by material resistivity, geometry, and construction; power rating is set by how much heat the body can dissipate while staying within safe temperature rise. You can buy a 10 Ω resistor rated at 1/4 W or 10 W; the ohmic value does not force a specific watt rating. Conversely, two resistors of the same wattage can have wildly different resistances. Step-by-Step Solution: Identify what resistance means: opposition to current (V = I * R).Identify what power rating means: sustained thermal limit (P_max) before damage.Note that P dissipated = I^2 * R = V^2 / R for a given operating point, but P_max is a thermal/mechanical spec, not a function of nominal R.Conclude the lack of direct proportionality between ohms and watts. Verification / Alternative check: Compare catalog parts: the same 1 kΩ value is offered in 1/8 W, 1/4 W, 1/2 W, 1 W, etc., illustrating independence of specs. Why Other Options Are Wrong: “Higher ohms → higher watt”: demonstrably false from catalogs. “Only for wirewound” or “because P = I^2 * R”: the equation gives operating power, not rated power; construction sets rating. “Indeterminate without tolerance/TC”: those are separate parameters. Common Pitfalls: Assuming higher resistance dissipates more heat; in fact, dissipation depends on the applied V or I and the environment. Final Answer: Incorrect — resistance and watt rating are independent specifications.

Question 7

Energy equivalence — verify the statement: consuming 4 kWh of electrical energy is equivalent to drawing 2000 W of power for 2 hours (energy = power * time).

Accepted Answer

Introduction / Context: Utility bills quote energy in kilowatt-hours (kWh). Converting between energy and power-time products is essential both for household consumption and engineering estimates. The statement equates 4 kWh with 2000 W sustained for 2 hours; this is a straightforward application of E = P * t. Given Data / Assumptions: Energy E = 4 kWh. Proposed power-time pair: P = 2000 W, t = 2 h. We consider real power; reactive aspects average to zero for energy billing. Concept / Approach: Convert units carefully: 1 kW = 1000 W. Energy in kWh is the time integral of kilowatts. If power is constant, E = P * t. Therefore, 2000 W * 2 h = (2 kW) * 2 h = 4 kWh. The equality holds irrespective of DC or AC, provided P denotes real power over the interval. Step-by-Step Solution: Convert 2000 W to kW: 2000 W = 2 kW.Multiply by time: 2 kW * 2 h = 4 kWh.Compare to the stated energy: both are 4 kWh.Conclude that the statement is correct. Verification / Alternative check: Another equivalent: 1000 W for 4 h or 4000 W for 1 h also yield 4 kWh. Any P, t pair with P(kW) * t(h) = 4 satisfies the same energy. Why Other Options Are Wrong: Incorrect: contradicts the direct unit calculation. “Only DC” or “only at unity PF”: Energy calculation E = ∫ P_real dt remains valid in AC; utility meters measure real energy. “Need voltage”: Voltage is unnecessary if power is already given. Common Pitfalls: Mixing up kWh (energy) and kW (power); forgetting to convert W to kW before multiplying by hours. Final Answer: Correct — 2000 W for 2 h equals 4 kWh of energy.

Question 8

Unit scaling — choose the most concise engineering notation for 7,500,000 W (watts) from the following options.

Accepted Answer

Introduction / Context: Expressing large or small quantities in appropriate SI prefixes makes technical communication clearer. Power is commonly stated in W, kW, MW, or GW depending on scale. The goal is to choose the compact, correct prefix form for 7,500,000 W. Given Data / Assumptions: 1 kW = 10^3 W. 1 MW = 10^6 W. 1 GW = 10^9 W. Concept / Approach: Convert watts to the nearest convenient SI-prefixed unit by dividing by powers of ten. Since 7,500,000 W = 7.5 × 10^6 W, the megawatt scale is appropriate. Reporting the value as 7.5 MW is concise and standard in power systems and specifications. Step-by-Step Solution: Write 7,500,000 W as 7.5 × 10^6 W.Recognize that 1 MW = 10^6 W.Therefore, 7,500,000 W = 7.5 MW.Select 7.5 MW as the best expression. Verification / Alternative check: Back-convert: 7.5 MW × 10^6 W/MW = 7,500,000 W, confirming equivalence. Why Other Options Are Wrong: 7.5 kW: Off by a factor of 1000; 7.5 kW = 7,500 W. 7,500 MW: Off by a factor of 1000; equals 7.5 × 10^9 W. 7,500 GW: Off by a factor of 10^6; equals 7.5 × 10^12 W. 0.0075 GW: Numerically equal to 7.5 MW but far less standard than 7.5 MW for this magnitude; the best concise form is MW. Common Pitfalls: Confusing k (10^3), M (10^6), and G (10^9); writing overly large numbers without prefixes, reducing readability. Final Answer: 7.5 MW

Question 9

Ohm’s law calculation — a resistor has 12 V across it and carries 500 µA of current. Compute the resistance value and choose the correct option.

Accepted Answer

Introduction / Context: Direct application of Ohm’s law V = I * R allows quick back-calculation of a resistor value from measured voltage and current. Careful attention to micro- units prevents decimal mistakes that are common in troubleshooting and exams. Given Data / Assumptions: Voltage across the resistor, V = 12 V. Current through the resistor, I = 500 µA = 500 × 10^−6 A = 0.0005 A. Ohmic behavior (constant R). Concept / Approach: Rearrange Ohm’s law to R = V / I. Convert microamperes to amperes before dividing to avoid a 10^6 error. Finally, express the answer in a convenient SI multiple (kΩ) for readability. Step-by-Step Solution: Convert current: 500 µA = 0.0005 A.Compute resistance: R = V / I = 12 / 0.0005 = 24,000 Ω.Express in kΩ: 24,000 Ω = 24 kΩ.Select the option that matches 24 kΩ. Verification / Alternative check: Cross-check using I = V / R with R = 24 kΩ: I = 12 / 24,000 = 0.0005 A = 500 µA, confirming consistency. Why Other Options Are Wrong: 600 Ω, 2.4 kΩ, 6 kΩ, 60 kΩ: Each corresponds to using an incorrect unit conversion or arithmetic slip when dividing 12 by 0.0005. Common Pitfalls: Forgetting that 1 µA = 10^−6 A; mixing up milli and micro; failing to convert before dividing. Final Answer: 24 kΩ

Question 10

Thermal design — which resistor specification tends to increase as the physical surface area (size and heat-dissipation area) of the resistor increases?

Accepted Answer

Introduction / Context: Choosing a resistor is not only about ohms; the wattage (power rating) determines how much heat the part can safely dissipate. Physical size and surface area play a major role in thermal performance. This question links form factor to a key datasheet limit used in reliable design. Given Data / Assumptions: Ambient convection and board copper area influence cooling. Larger body = more surface area for heat transfer (all else equal). Comparing mainstream through-hole and SMD resistor families. Concept / Approach: Power rating is specified as the maximum continuous power the resistor can dissipate without exceeding its allowable temperature rise. Heat leaves primarily via convection, conduction to leads/PCB, and radiation; increasing surface area improves all three paths, so the allowable wattage typically increases with size. Resistance value is independent of physical area; voltage rating depends on geometry and film thickness but does not scale simply with area; tolerance is a trimming/manufacturing attribute. Step-by-Step Solution: Identify the thermal-limited parameter: power rating (watts).Relate heat transfer to surface area: more area → better cooling → higher allowable P.Note independence from ohmic value: the same value can be purchased in different wattages.Conclude that increasing surface area increases the resistor’s power rating. Verification / Alternative check: Compare datasheets: 0603 resistors are typically 0.1 W, 0805 ~0.125–0.25 W, 1206 ~0.25–0.5 W, larger packages higher, demonstrating the area–wattage trend. Why Other Options Are Wrong: Resistance: Set by material and geometry but not monotonic with body area across series. Current rating: Not a standard standalone resistor spec; current limit derives from power and temperature rise. Voltage rating: Tied to creepage, film thickness, and construction, not purely to surface area. Tolerance: Manufacturing/trim-related, independent of area. Common Pitfalls: Assuming “bigger resistor means higher resistance”; size mostly signals higher wattage, not ohmic value. Final Answer: Power rating

Question 11

Direct Ohm’s law implication — for a fixed-value resistor, if the voltage applied across it increases, how does the current through it change (assuming temperature effects are negligible)?

Accepted Answer

Introduction / Context: Ohm’s law gives an immediate relationship between the voltage across a resistor and the current through it. Understanding this proportionality is fundamental to predicting how changes in supply voltage affect branch currents in simple and complex circuits alike. Given Data / Assumptions: Resistor value R is constant and positive. Temperature and self-heating effects are negligible for the change considered. Polarity is fixed; we are discussing magnitude changes. Concept / Approach: Ohm’s law states V = I * R. Holding R constant implies I = V / R. Therefore, if V increases, I must increase proportionally. This is the definition of a linear, ohmic element. The exact numeric change requires R, but the qualitative direction (increase) does not. Step-by-Step Solution: Start from I = V / R for a fixed R.Consider a higher applied V: numerator increases while denominator is unchanged.Therefore I increases linearly with V.Conclude that current increases when voltage increases across a fixed resistor. Verification / Alternative check: Plot I vs. V: the line I = (1/R) * V has positive slope 1/R. Any movement to the right along the V-axis moves up along the I-axis. Why Other Options Are Wrong: Decreases/Stays the same: Contradicts I = V / R for constant R. Cannot be determined: While the exact value needs R, the trend is determined. Reverses direction: Direction is set by polarity; magnitude increase does not flip direction. Common Pitfalls: Confusing constant-current sources (which enforce I) with resistors (which follow Ohm’s law); ignoring that significant heating can change R slightly, but the ideal law remains linear. Final Answer: Increases

Question 12

Resistor sizing (repaired) — “The power rating for the resistor in the given circuit should be at least ___.” Without the circuit diagram, voltage, current, or resistance values, can we determine the minimum wattage?

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Introduction / Context: Choosing a resistor wattage requires estimating how much power the part will dissipate in its actual operating conditions. The original prompt references a “given circuit,” but no values or diagram are supplied. This repaired question tests whether you recognize that minimum wattage cannot be selected without knowing voltage, current, or resistance, and ideally a safety margin strategy. Given Data / Assumptions: No schematic or numeric values are provided. We seek the minimum continuous power rating that avoids overheating. Standard practice includes a safety margin above calculated dissipation. Concept / Approach: The dissipated power in a resistor is P = V^2 / R = I^2 * R = V * I (all equivalent when quantities refer to the same element). To select a wattage, compute the worst-case power from the known operating conditions and multiply by a margin (commonly 1.5× to 2× or per organizational standards) to account for tolerances, ambient temperature, and transient stress. Without V, I, or R, the required P cannot be calculated. Step-by-Step Solution: Identify the available data: none of V, I, or R are given; diagram absent.Recall the formulas: P = V^2 / R = I^2 * R = V * I.Note the need for worst-case conditions and derating (e.g., choose 2× the computed P).Conclude that a numeric wattage cannot be selected from the information provided. Verification / Alternative check: If a circuit later specifies R = 1 kΩ with 10 V across it, P = 10^2 / 1000 = 0.1 W; a common choice would be ≥ 0.25 W to include margin. Different values would change the required wattage, proving the dependence on missing data. Why Other Options Are Wrong: “Always 1/4 W” or “always 1 W”: Rules of thumb can fail badly in power circuits or in very low-power designs. “Equal to V^2/R with no margin” or “I^2*R/2”: Both ignore safety margins or invent unjustified factors. Common Pitfalls: Picking wattage by habit without calculation; forgetting ambient temperature, airflow, pulse loads, or board copper area that affect thermal rise. Final Answer: Cannot be determined from the information provided — resistor wattage selection requires operating values and a margin.

Question 13

Resistor power rating selection (repaired for solvability): A DC bias resistor in a small-signal amplifier is measured to dissipate about 0.18 W under worst-case conditions. According to good design practice (choose the next higher standard wattage …

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Introduction / Context: Choosing an appropriate resistor power rating is part of reliable circuit design. The rating must exceed the actual worst-case power dissipation so the part does not overheat, drift, or fail prematurely. A common rule is to select the next higher standard rating above the calculated dissipation, typically leaving a margin (for example, running at 50–70% of the rated wattage). Given Data / Assumptions: Measured or calculated worst-case resistor dissipation P_diss ≈ 0.18 W (continuous). Ambient conditions are typical; no forced cooling. Use standard through-hole ratings such as 1/8 W, 1/4 W, 1/2 W, 1 W, 2 W, etc. Concept / Approach: First determine the smallest standard rating that is greater than the worst-case dissipation. Then verify that the resulting operating point keeps the resistor comfortably below its limit. Designers often target no more than about 50–70% of the rated wattage in normal service to accommodate tolerances and temperature rise. Step-by-Step Solution: Compute or note P_diss ≈ 0.18 W.List the nearby standard ratings: 0.125 W (1/8 W), 0.25 W (1/4 W), 0.5 W (1/2 W).Select the next higher available rating above 0.18 W → 0.25 W (1/4 W).Check margin: 0.18 / 0.25 = 0.72 → about 72% of rating (acceptable for many low-cost builds; 1/2 W would give extra thermal headroom if the environment is hot). Verification / Alternative check: Use P = I^2 * R or P = V^2 / R to recompute the worst case and compare to the data sheet’s derating curves. If the resistor will be in a high-ambient enclosure, consider stepping up to 1/2 W for additional safety margin. Why Other Options Are Wrong: 1/2 W, 1 W, 2 W: These exceed the minimum required; while safer thermally, they are not the “at least” value needed by the question.1/8 W (not listed): would be below the dissipation and unsafe. Common Pitfalls: Choosing a rating equal to the calculated dissipation with no margin; forgetting that supply tolerance, component tolerance, and ambient temperature can increase real dissipation; ignoring enclosure derating curves. Final Answer: 1/4 W

Question 14

Ohm’s law relationship — resistance vs current: In a simple conductor at constant temperature supplied by a fixed voltage, resistance (R) and current (I) are _________.

Accepted Answer

Introduction / Context: Ohm’s law, expressed as V = I * R, links voltage (V), current (I), and resistance (R). Understanding how any two of these quantities co-vary when the third is held constant is essential for quick circuit intuition and troubleshooting. Given Data / Assumptions: Voltage V is held constant (e.g., a regulated supply). Temperature and material properties are constant so R does not change with I. Linear (ohmic) device behavior is assumed. Concept / Approach: From V = I * R with V constant, solve for I: I = V / R. This shows that current varies inversely with resistance. Doubling R halves I; halving R doubles I. The proportionality is direct only between V and I when R is constant, not between I and R at constant V. Step-by-Step Solution: Start with V = I * R.Hold V constant → I = V / R.Therefore, as R increases, I decreases (inverse relationship).Conversely, as R decreases, I increases. Verification / Alternative check: Example: With V = 10 V, R = 1 kΩ → I = 10 mA; if R = 2 kΩ → I = 5 mA (halved), confirming inverse proportionality. Why Other Options Are Wrong: Directly proportional: true for V and I when R is constant, not for I vs R at constant V.Not related / similar to voltage: contradicts Ohm’s law. Common Pitfalls: Mixing “direct” and “inverse” proportionality when switching which variable is held constant; forgetting temperature effects that can change R in real parts. Final Answer: inversely proportional

Question 15

Energy–power linkage — unit equivalence: The number of joules used in 1 second is always equal to the number of _______. (Hint: energy per unit time.)

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Introduction / Context: Power quantifies the rate at which energy is transferred or converted. In SI, one watt is defined as one joule per second. Recognizing this equivalence helps in quickly converting between energy and power in specifications and calculations. Given Data / Assumptions: Energy is measured in joules (J). Time interval is 1 second (s). We want the unit that equals J/s. Concept / Approach: Power P is defined as P = energy / time. Therefore, when 1 joule is used each second, the rate is 1 watt. The other listed units measure different quantities: siemens (conductance), coulombs (charge), amperes (current). Step-by-Step Solution: Write P = E / t.Insert E = joules and t = seconds → P has units J/s.Recognize J/s is named “watt”.Therefore choose “watts”. Verification / Alternative check: Check dimensional consistency: 1 W = 1 N·m/s = 1 kg·m^2/s^3, aligning with joule-per-second identity. Why Other Options Are Wrong: Siemens: unit of conductance.Coulombs: unit of electric charge.Amperes: unit of current (coulombs per second). Common Pitfalls: Confusing power with energy (watt-hours vs watts); mixing up ampere (A) with watt (W). Final Answer: watts

Question 16

Power calculation from energy and time: What is the power if 200 J of energy are used in 10 s?

Accepted Answer

Introduction / Context: Many practical problems present energy consumed over a time interval, from which you must compute average power. This is a direct application of the definition of power in the time domain. Given Data / Assumptions: Energy E = 200 J. Time t = 10 s. Average power P = E / t. Concept / Approach: Use the basic relation: power equals energy divided by time. Ensure units are SI to avoid conversion errors. The arithmetic should be straightforward as the numbers are chosen to divide cleanly. Step-by-Step Solution: P = E / t.Substitute: P = 200 J / 10 s.Compute: P = 20 J/s.Recognize J/s = W → P = 20 W. Verification / Alternative check: If the same 200 J were used in 20 s, power would halve to 10 W; if used in 5 s, it would double to 40 W. This proportionality confirms the calculation. Why Other Options Are Wrong: 2 W and 5 W: too low by factors of 10 and 4 respectively.50 W: too high by a factor of 2.5. Common Pitfalls: Forgetting that “per second” means divide by time; confusing power (W) with energy (J). Final Answer: 20 W

Question 17

Ohm’s law (repaired for solvability): A DC source drives a 2.5 kΩ load and the measured current is 10 mA. What is the supply voltage in the circuit?

Accepted Answer

Introduction / Context: Ohm’s law is the foundational relationship among voltage, current, and resistance in linear circuits. Given any two of the three quantities, the third is determined uniquely. This problem asks you to compute the supply voltage that produces a given current through a known resistance. Given Data / Assumptions: Resistance R = 2.5 kΩ (i.e., 2500 Ω). Current I = 10 mA (i.e., 0.010 A). Steady DC conditions; wiring drops are negligible. Concept / Approach: Use V = I * R. Be careful with unit prefixes: kilo (10^3) for ohms and milli (10^-3) for amperes. Multiply the base units to avoid mistakes, then convert back to convenient engineering units (volts). Step-by-Step Solution: Write V = I * R.Insert I = 0.010 A and R = 2500 Ω.Compute V = 0.010 * 2500 = 25 V.Therefore, the supply voltage is 25 V. Verification / Alternative check: Cross-check with current: I = V / R = 25 / 2500 = 0.010 A = 10 mA, confirming consistency. Why Other Options Are Wrong: 2.5 V: would give 1 mA, not 10 mA.250 V and 2.5 kV: would produce 100 mA and 1 A respectively, far above the stated 10 mA. Common Pitfalls: Mixing kilo and milli prefixes; forgetting to convert mA to A before multiplying. Final Answer: 25 V

Question 18

Voltage across a resistor — apply Ohm’s law with unit prefixes: What is the voltage across a 2.2 kΩ resistor if 3 mA of current flows through it?

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Introduction / Context: This is a direct substitution problem using Ohm’s law. Correct handling of kilo-ohms and milliamperes is critical to avoid order-of-magnitude errors in the final voltage. Given Data / Assumptions: R = 2.2 kΩ = 2200 Ω. I = 3 mA = 0.003 A. Linear resistor, steady state. Concept / Approach: Use V = I * R, ensuring that units are in base SI prior to multiplication. Then express the answer in volts with sensible significant figures reflecting the inputs. Step-by-Step Solution: Convert: R = 2200 Ω, I = 0.003 A.Compute V = I * R = 0.003 * 2200.Multiply: 0.003 * 2200 = 6.6 V.Answer: 6.6 V across the resistor. Verification / Alternative check: Back-check current: I = V / R = 6.6 / 2200 ≈ 0.003 A = 3 mA, confirming the result. Why Other Options Are Wrong: 7.33 V: not supported by 2.2 kΩ at 3 mA.1.4 V and 0.8 V: too small by factors of about 4.7 and 8.25 respectively. Common Pitfalls: Leaving R in kΩ and I in mA without adjusting; misplacing decimal points when using prefixes. Final Answer: 6.6 V

Question 19

Current from voltage across a resistor — apply Ohm’s law: The voltage across a 5 kΩ resistor is 25 V. The current through the resistor is ______.

Accepted Answer

Introduction / Context: Given the voltage across and resistance of a component, Ohm’s law provides the current. Careful unit handling (kΩ) avoids common errors in the milliampere range. Given Data / Assumptions: V = 25 V. R = 5 kΩ = 5000 Ω. Linear resistor, steady state. Concept / Approach: Use I = V / R. Convert kilo-ohms to ohms before dividing to get amperes, then express the result in milliamperes if convenient for readability. Step-by-Step Solution: I = V / R = 25 / 5000 A.Compute: 25 / 5000 = 0.005 A.Convert: 0.005 A = 5 mA.Therefore, current is 5 mA. Verification / Alternative check: Check voltage drop: V = I * R = 0.005 * 5000 = 25 V, consistent. Why Other Options Are Wrong: 200 mA / 125 mA / 75 mA: far too large for 25 V across 5 kΩ; would require much lower resistance values. Common Pitfalls: Forgetting to convert kΩ to Ω, leading to answers 1000× too large or small. Final Answer: 5 mA

Ohm's Law Questions