Question 1

Energy billing calculation: If a device uses 600 W of electric power continuously for 60 hours, how much electrical energy is consumed in kilowatt-hours (kWh)?

Accepted Answer

Introduction / Context:Utilities bill energy in kilowatt-hours. Converting power usage over time to kWh is an essential practical skill for estimating electricity costs and comparing the efficiency of appliances and power supplies. Given Data / Assumptions: Power P = 600 W. Operating time t = 60 h. We assume constant power over the interval. Concept / Approach:Energy in kWh is computed as (power in kW) * (time in hours). Convert watts to kilowatts by dividing by 1,000, then multiply by hours of operation. Step-by-Step Solution: Convert power: 600 W = 0.6 kW.Multiply by time: 0.6 kW * 60 h = 36 kWh.Thus, the total energy consumed is 36 kWh. Verification / Alternative check:Compute in watt-hours: 600 W * 60 h = 36,000 Wh. Convert to kWh by dividing by 1,000 → 36 kWh. Both methods agree. Why Other Options Are Wrong: 3.6 kWh: Off by a factor of 10 (would match 600 W for 6 h). 10 kWh and 1 kWh: Result from incorrect time or power conversion assumptions. Common Pitfalls: Forgetting to convert W to kW or misreading hours vs minutes. Final Answer:36 kWh

Question 2

Practical energy usage: A device rated at 400 W runs for 30 hours. Calculate the energy consumed in kilowatt-hours (kWh) for billing purposes.

Accepted Answer

Introduction / Context:Translating device power ratings and operating time into energy (kWh) helps estimate electricity costs and compare efficiency across appliances. This is a straightforward application of energy = power × time with proper unit conversion to kilowatts and hours. Given Data / Assumptions: Power P = 400 W. Time t = 30 h. Power is assumed constant during operation. Concept / Approach:Energy (kWh) = (power in kW) * (time in hours). Convert 400 W to 0.4 kW, then multiply by 30 h to find the energy used. Step-by-Step Solution: Convert power: 400 W = 0.4 kW.Compute energy: 0.4 kW * 30 h = 12 kWh.Therefore, the device consumes 12 kWh over 30 hours. Verification / Alternative check:As watt-hours: 400 W * 30 h = 12,000 Wh. Divide by 1,000 → 12 kWh. Both paths give the same result. Why Other Options Are Wrong: 13.3 kWh and 1.3 kWh: Result from arithmetic or decimal placement errors. 1.2 kWh: Off by a factor of 10; would be correct for 400 W over 3 hours, not 30 hours. Common Pitfalls: Mistaking W for kW or mixing hours with minutes, leading to tenfold errors. Final Answer:12 kWh

Question 3

Power–voltage–resistance calculation: A 220 Ω resistor dissipates 3 W of power in steady state. What is the voltage across the resistor?

Accepted Answer

Introduction / Context:Relating power, voltage, and resistance using Ohm's law and power formulas is routine in circuit design. This problem reinforces the use of P = V^2 / R or equivalently V = √(P * R), which is common when choosing component power ratings and verifying safe operating points. Given Data / Assumptions: Resistance R = 220 Ω. Power P = 3 W. Resistor assumed ideal; DC or RMS values are equivalent for computation. Concept / Approach:Use the power–voltage relation for a resistor: P = V^2 / R. Solving for V gives V = sqrt(P * R). Perform the multiplication and take the square root to obtain the voltage magnitude. Step-by-Step Solution: Start: V = sqrt(P * R).Compute P * R = 3 * 220 = 660.Take square root: V = sqrt(660) ≈ 25.69 V.Rounded to one decimal place: ≈ 25.7 V. Verification / Alternative check:Check using I = V / R: I ≈ 25.69 / 220 ≈ 0.1168 A. Power P = I^2 * R ≈ (0.1168)^2 * 220 ≈ 3.0 W. The numbers are consistent, confirming the solution. Why Other Options Are Wrong: 73.3 V and 257 V: Would imply much higher power dissipation than 3 W for a 220 Ω resistor. 2.5 V: Far too low; would yield P ≈ 0.028 W. Common Pitfalls: Using P = V * I without substituting I = V / R correctly can lead to algebra mistakes. Forgetting to take the square root when solving V^2 = P * R. Final Answer:25.7 V

Question 4

Efficiency of a power supply: If a supply delivers 0.6 W of output power while drawing 0.7 W of input power, what is its efficiency expressed as a percentage?

Accepted Answer

Introduction / Context:Efficiency quantifies how effectively a device converts input power to useful output power. In power electronics, higher efficiency means less waste heat, smaller heatsinks, and improved energy savings—critical in portable devices and large-scale systems alike. Given Data / Assumptions: Output power P_out = 0.6 W. Input power P_in = 0.7 W. Efficiency η is defined as output/input in percentage. Concept / Approach:The efficiency formula is η (%) = (P_out / P_in) * 100. Substitute the given numbers to compute the ratio and convert to percent. Ensure consistent units (both in watts). Step-by-Step Solution: η = (P_out / P_in) * 100.η = (0.6 / 0.7) * 100.η = 0.857142... * 100 = 85.7142... % ≈ 85.7%. Verification / Alternative check:If the input is 0.7 W and 85.7% is delivered as output, the rest is loss: 0.7 W * (1 − 0.857) ≈ 0.1 W dissipated as heat, which is consistent with 0.6 W output. The arithmetic closes properly. Why Other Options Are Wrong: 8.57% and 4.28%: Off by a factor of 10 or more; would imply extremely poor conversion. 42.85%: Approximately half the correct value; likely from using (P_in − P_out)/P_in incorrectly. Common Pitfalls: Mixing input and output in the denominator. Always divide by input power for efficiency. Rounding too early; keep sufficient precision until the final percentage is stated. Final Answer:85.7%

Question 5

Power–voltage relationship: A 75 Ω load is consuming 2 W of power from a DC/AC source. Approximately what is the output voltage of the supply, assuming ideal conditions?

Accepted Answer

Introduction / Context:Many practical problems in basic electrical engineering require moving between power, voltage, and resistance. Recognizing the correct formula and applying it cleanly allows you to size power supplies and verify whether a load is being driven correctly. This question targets the relationship among power (P), voltage (V), and resistance (R). Given Data / Assumptions: Load resistance R = 75 Ω. Load power P = 2 W (steady state). Assume ideal source and purely resistive load (no reactive elements). We seek the approximate source/output voltage across the load. Concept / Approach: For a resistive load, the power–voltage relation is P = V^2 / R. Rearranging gives V = sqrt(P * R). This formula is valid for DC and for RMS values in AC when the load is purely resistive. Step-by-Step Solution: Start with P = V^2 / R.Solve for voltage: V = sqrt(P * R).Substitute numbers: V = sqrt(2 W * 75 Ω) = sqrt(150) ≈ 12.247 V.Round to a sensible figure: approximately 12 V. Verification / Alternative check: Compute power back from 12 V: P = V^2 / R = 144 / 75 = 1.92 W, which is near 2 W. Using 12.25 V gives about 2.0 W exactly, so 12 V is a good approximate choice from the options. Why Other Options Are Wrong: 120 V would yield P = 120^2 / 75 ≈ 192 W (far too high). 1.2 V gives ≈ 0.019 W (too low). 6 V gives 0.48 W (too low). 9 V gives 1.08 W (still too low). Common Pitfalls: Mixing up P = V * I with P = V^2 / R; forgetting to use RMS values for AC; mis-squaring or mis-taking the square root. Final Answer: 12 V

Question 6

Unit conversion of power: Convert 0.025 W into an equivalent expression using standard metric submultiples.

Accepted Answer

Introduction / Context:Engineering problems frequently require converting between watts and submultiples such as milliwatts and microwatts. Mastering metric prefixes avoids scale errors and enables quick interpretation of specifications for sensors, amplifiers, and power supplies. Given Data / Assumptions: Given power P = 0.025 W. Metric prefixes: 1 mW = 10^-3 W and 1 µW = 10^-6 W. We want an exactly equivalent, properly scaled expression. Concept / Approach: Use unit equivalences and multiply/divide by powers of ten. Converting watts to milliwatts requires multiplying by 10^3. Converting watts to microwatts requires multiplying by 10^6. Step-by-Step Solution: Convert to milliwatts: 0.025 W * 10^3 = 25 mW.Check as microwatts: 0.025 W * 10^6 = 25,000 µW.Compare to options: 25 mW matches exactly; 2,500 µW equals 0.0025 W (too small).Therefore, the correct equivalent is 25 mW. Verification / Alternative check: Reverse conversion: 25 mW / 1000 = 0.025 W. This confirms the accuracy. Why Other Options Are Wrong: 2,500 µW = 2.5 mW = 0.0025 W (ten times smaller). 0.00025 mW = 2.5 * 10^-7 W (far smaller). 25 kW is 25,000 W (enormously larger). 250 µW = 0.00025 W (much smaller). Common Pitfalls: Misplacing decimal points, confusing mW with µW, and forgetting that k (kilo) is 10^3 while m (milli) is 10^-3. Final Answer: 25 mW

Question 7

Energy calculation: How many watt-hours correspond to using a 65 W load continuously for 18 h (assume constant power)?

Accepted Answer

Introduction / Context:Utility billing and battery sizing both rely on energy rather than instantaneous power. Energy in watt-hours is power (in watts) multiplied by time (in hours). This problem reinforces that core relationship for steady loads. Given Data / Assumptions: Power P = 65 W (constant). Time t = 18 h. We ignore power factor and assume purely resistive or already RMS/real power. Concept / Approach: Energy E (Wh) = P (W) * t (h). This linear scaling directly gives energy consumption over time for a steady load. Step-by-Step Solution: Write the formula: E = P * t.Substitute values: E = 65 * 18 Wh.Compute: 65 * 18 = 1,170.Therefore, E = 1,170 Wh. Verification / Alternative check: Convert to kWh: 1,170 Wh / 1000 = 1.17 kWh. For a 65 W device, 1.17 kWh over 18 h is reasonable (about the energy of a small appliance run for part of a day). Why Other Options Are Wrong: 11.7 Wh is off by a factor of 100. 11,700 Wh is 10× too large. 117,000 Wh is 100× too large. 117 Wh is 10× too small. Common Pitfalls: Mixing Wh and kWh, or converting hours to seconds unnecessarily and introducing errors. Always keep consistent units. Final Answer: 1,170 Wh

Question 8

Energy from voltage and resistance: A 15 V source is applied across a 12 Ω resistor. How much electrical energy is used in three minutes (expressed in watt-hours)?

Accepted Answer

Introduction / Context:Translating circuit parameters to energy consumed over a time interval is vital for estimating battery life and energy costs. With voltage and resistance known, you can compute power and then energy by accounting for time. Given Data / Assumptions: Voltage V = 15 V. Resistance R = 12 Ω. Time t = 3 minutes = 0.05 h = 180 s. Purely resistive load; steady operation. Concept / Approach: Compute power first: P = V^2 / R. Then compute energy: E = P * t. Keep units consistent; since answers are in Wh, express time in hours. Step-by-Step Solution: Power: P = V^2 / R = 15^2 / 12 = 225 / 12 = 18.75 W.Convert time to hours: t = 3 min = 3 / 60 h = 0.05 h.Energy: E = P * t = 18.75 * 0.05 = 0.9375 Wh.Round reasonably: ≈ 0.938 Wh. Verification / Alternative check: In joules: E = P * t = 18.75 W * 180 s = 3,375 J. Convert to Wh: 3,375 J / 3600 ≈ 0.9375 Wh. Both paths agree. Why Other Options Are Wrong: 56.25 Wh would require one hour at 56.25 W, not 0.05 h at 18.75 W. 5.6 Wh and 938 Wh are off by large factors. 0.0938 Wh is 10× too small. Common Pitfalls: Forgetting to convert minutes to hours when answers are in Wh; using P = V * I without computing I correctly; arithmetic slips with squares. Final Answer: 0.938 Wh

Question 9

Resistor power rating: A 120 Ω resistor must safely carry a maximum current of 25 mA. What minimum power rating should be specified (choose the next sensible standard rating above the calculated dissipation)?

Accepted Answer

Introduction / Context:Choosing an appropriate resistor power rating prevents overheating and premature failure. The rating must exceed the expected worst-case dissipation, ideally with safety margin and considering ambient conditions. Given Data / Assumptions: Resistance R = 120 Ω. Maximum current I_max = 25 mA = 0.025 A. Assume continuous operation and pick at least the next higher standard rating. Concept / Approach: Resistor power dissipation is P = I^2 * R. After computing the required wattage, select a commercially available rating above that value (e.g., 1/8 W, 1/4 W, 1/2 W, etc.). Step-by-Step Solution: Compute P = I^2 * R = (0.025)^2 * 120 = 0.000625 * 120 = 0.075 W.Required minimum is 0.075 W (75 mW).Choose the next sensible standard rating above 75 mW → 150 mW (0.15 W) or 1/4 W (250 mW). Among provided options, 150 mW is the smallest that exceeds the need.Therefore, select 150 mW. Verification / Alternative check: Temperature rise and tolerances suggest margin. 150 mW provides 2× nominal headroom over 75 mW dissipation, a reasonable conservative choice. Why Other Options Are Wrong: 15 mW is below requirement; would overheat. 4.8 W and 480 mW are overkill compared to the actual need; while safe, they are not the 'minimum' practical choice here. 75 mW equals the exact dissipation with zero margin, not advisable as a rating. Common Pitfalls: Choosing a rating equal to calculated power with no margin; mis-converting milliamps to amps; forgetting squared dependence on current. Final Answer: 150 mW

Question 10

Power in a resistor from a battery: A 68 Ω resistor is connected directly across a 3 V battery. What is the resistor’s power dissipation?

Accepted Answer

Introduction / Context:Computing power dissipation is essential for part selection and thermal design. Given voltage and resistance, you can directly compute power without finding current separately by using P = V^2 / R. Given Data / Assumptions: Voltage V = 3 V (DC source). Resistance R = 68 Ω. Purely resistive load, steady state. Concept / Approach: Use the relation P = V^2 / R, which derives from combining Ohm’s law (V = I * R) with P = V * I. This is often the fastest way when V and R are directly given. Step-by-Step Solution: Compute numerator: V^2 = 3^2 = 9.Compute power: P = 9 / 68 ≈ 0.13235 W.Convert to milliwatts: 0.13235 W ≈ 132 mW.Thus, the resistor dissipates about 132 mW. Verification / Alternative check: Find current first: I = V / R = 3 / 68 ≈ 0.0441 A. Then power P = I^2 * R ≈ (0.0441)^2 * 68 ≈ 0.132 W. Both methods agree. Why Other Options Are Wrong: 13.2 mW is 10× too small. 22.6 mW does not match any correct derivation. 226 mW is too large. 6.8 mW is far too small. Common Pitfalls: Rounding too early; confusing mW and W; forgetting to square the voltage in P = V^2 / R. Final Answer: 132 mW

Question 11

Back-solve current from power: A 3.3 kΩ resistor is dissipating 0.25 W under steady DC conditions. What current is flowing through the resistor?

Accepted Answer

Introduction / Context:Given any two of power, resistance, and current for a resistor, you can find the third. This is particularly useful for verifying whether a resistor is within safe operating limits or diagnosing abnormal heating. Given Data / Assumptions: Power P = 0.25 W. Resistance R = 3.3 kΩ = 3300 Ω. DC or RMS AC steady operation. Concept / Approach: Use the formula P = I^2 * R. Solving for current gives I = sqrt(P / R). Convert units carefully to avoid errors. Step-by-Step Solution: Compute ratio: P / R = 0.25 / 3300 ≈ 7.5758 * 10^-5.Take square root: I = sqrt(7.5758 * 10^-5) ≈ 0.0087 A.Convert to milliamps: 0.0087 A = 8.7 mA.Therefore, the current is approximately 8.7 mA. Verification / Alternative check: Check via P = V * I and V = I * R: If I ≈ 8.7 mA, then V ≈ 0.0087 * 3300 ≈ 28.7 V. Power P ≈ 28.7 V * 0.0087 A ≈ 0.25 W. Consistent. Why Other Options Are Wrong: 87 mA would give P ≈ (0.087)^2 * 3300 ≈ 25 W (far too high). 8.7 µA is 1000× too small. 8.7 A is unrealistic (thousands of watts). 0.87 mA yields negligible power relative to 0.25 W. Common Pitfalls: Forgetting the square root; mis-converting kΩ to Ω; mixing mA and A. Final Answer: 8.7 mA

Question 12

Battery capacity interpretation: A power supply can deliver 6 A continuously for 3.5 h. What is its ampere-hour (Ah) rating under these conditions?

Accepted Answer

Introduction / Context:Ampere-hour (Ah) rating expresses how much charge a source can deliver over time. It is widely used for batteries and backup supplies. Translating current and runtime into Ah is straightforward but must be done carefully. Given Data / Assumptions: Current capability I = 6 A (assumed constant). Duration t = 3.5 h. Assume ideal conditions with negligible rate effects or temperature derating. Concept / Approach: Ampere-hours = current (A) * time (h). This is effectively the total charge delivered, since 1 Ah = 3600 coulombs. While real batteries vary with discharge rate, the arithmetic remains the basic rating method. Step-by-Step Solution: Compute capacity: Ah = 6 A * 3.5 h = 21 Ah.No unit conversion beyond hours is needed.Therefore, the device's capacity is 21 Ah under stated conditions.This directly reflects charge = I * t. Verification / Alternative check: In coulombs: Q = I * t = 6 * 3.5 * 3600 ≈ 75,600 C. Convert back: 75,600 C / 3600 ≈ 21 Ah, confirming the result. Why Other Options Are Wrong: 58 Ah or 35 Ah do not match the simple product. 2.1 Ah and 0.58 Ah are off by factors of 10 or more. Common Pitfalls: Confusing Ah (capacity) with Wh (energy), which would also require voltage. Misreading hours (e.g., minutes) can lead to large errors. Final Answer: 21 Ah

Question 13

Average daily energy use: Over a 14-day period your meter shows 18 kWh consumed. What is the average daily energy usage expressed in kWh per day?

Accepted Answer

Introduction / Context:Electric utility data are typically given in kilowatt-hours over billing periods. Estimating average daily usage helps compare different periods and identify energy-saving opportunities. Here we compute the daily average from a total over two weeks. Given Data / Assumptions: Total energy E_total = 18 kWh. Time window = 14 days. Assume uniform average use over the period for calculation purposes. Concept / Approach: Average daily energy = total energy / number of days. Because the question requests kWh per day, no time conversion is needed beyond dividing by days. Step-by-Step Solution: Compute average: E_avg_per_day = 18 kWh / 14 days.Perform division: 18 / 14 ≈ 1.2857.Round suitably: ≈ 1.286 kWh per day.Therefore, the average daily energy use is about 1.286 kWh. Verification / Alternative check: Cross-check: At 1.286 kWh/day for 14 days, total ≈ 18.0 kWh, matching the meter reading. Why Other Options Are Wrong: 12.85 kWh/day would imply 179.9 kWh over 14 days, not 18 kWh. 535 kWh and 252 kWh are grossly inconsistent. 0.1286 kWh/day is 10× too small. Common Pitfalls: Confusing kW (power) with kWh (energy). Forgetting to divide by the exact number of days in the period. Final Answer: 1.286 kWh

Question 14

Instantaneous power from energy and time: If 300 J of electrical energy are consumed uniformly in 15 s, what is the average power during that interval?

Accepted Answer

Introduction / Context:Power quantifies the rate at which energy is used. When total energy over a time interval is known, average power is computed by dividing energy by time. This is fundamental in both electrical and mechanical systems. Given Data / Assumptions: Total energy E = 300 J. Time interval t = 15 s. Assume uniform consumption to interpret as average power over the period. Concept / Approach: Use the definition P = E / t. Joule per second equals watt (1 J/s = 1 W). Ensure consistent SI units before dividing. Step-by-Step Solution: Write formula: P = E / t.Substitute values: P = 300 J / 15 s.Compute: P = 20 J/s = 20 W.Therefore, the power is 20 W. Verification / Alternative check: Check units: joules divided by seconds is watts. 20 W over 15 s gives 300 J, confirming the calculation. Why Other Options Are Wrong: 2 W and 0.2 W are too small by factors of 10 or 100. 200 W and 2,000 W are too large by factors of 10 or 100. Common Pitfalls: Converting to Wh unnecessarily (introduces rounding); mixing seconds and minutes; misplacing the decimal when dividing. Final Answer: 20 W

Question 15

Power equivalence with mixed units: For a circuit operating at 12 V with a current of 40 mA, what is the power expressed in different unit forms?

Accepted Answer

Introduction / Context:Being able to compute electrical power and express it in equivalent units is a core skill in basic circuit analysis and electronics troubleshooting. This question evaluates your fluency with P = V * I and with converting between watts (W), milliwatts (mW), and microwatts (µW). Given Data / Assumptions: Voltage V = 12 V (direct current or RMS equivalent). Current I = 40 mA = 0.04 A. Power relation P = V * I applies to resistive loads (or RMS quantities). Concept / Approach:Use the fundamental power equation P = V * I, then convert the numerical result into different unit scales. Remember that 1 W = 1,000 mW and 1 W = 1,000,000 µW. Consistent unit conversion is crucial for component selection and power budgeting. Step-by-Step Solution: Compute power in watts: P = 12 * 0.04 = 0.48 W.Convert to milliwatts: 0.48 W = 0.48 * 1,000 = 480 mW.Convert to microwatts: 0.48 W = 0.48 * 1,000,000 = 480,000 µW. Verification / Alternative check:Cross-check using P = I^2 * R or P = V^2 / R if resistance is known; here, dimensional checks confirm the unit conversions. The three numerical forms represent the same physical power. Why Other Options Are Wrong: Each individual value is correct; therefore, the best choice is the inclusive option stating all are correct. Common Pitfalls: Forgetting to convert milliamperes to amperes before multiplying by volts. Confusing mW and µW scaling factors (10^3 vs 10^6). Final Answer:all of the above

Question 16

Analog ohmmeter interpretation: When the pointer on an analog ohmmeter deflects close to zero ohms, what does this indicate about the resistor under test?

Accepted Answer

Introduction / Context:Reading an analog ohmmeter correctly is essential for diagnosing component health and wiring faults. The meter's scale is typically non-linear, with zero ohms at full-scale deflection, indicating very low resistance or a short condition. Given Data / Assumptions: An analog ohmmeter is used on a de-energized component. The pointer moves near the zero-ohms end of the scale. Test leads and internal battery are assumed good and zeroed. Concept / Approach:Ohmmeters apply a small internal voltage and measure resulting current to infer resistance. A near-zero reading indicates very high current for a given internal voltage, meaning very low resistance—consistent with a short or near-short path. Step-by-Step Reasoning: Observe pointer position: near zero on ohms scale → very low resistance.Interpretation: low resistance through the device under test implies a shorted component or closed path.Conclusion: the resistor is effectively shorted (resistance ≈ 0 Ω). Verification / Alternative check:Swap test leads or measure a known-good resistor to confirm meter function. If another meter confirms ≈ 0 Ω, the diagnosis stands. Why Other Options Are Wrong: Overheated: A temperature state, not directly indicated by resistance reading. Open: Would show infinite or very high resistance (pointer toward ∞/left). Reversed: Polarity is not relevant for passive resistors on an ohmmeter. Common Pitfalls: Forgetting to zero the analog ohmmeter before measurement. Measuring in-circuit and reading parallel paths as low resistance; isolate the component if uncertain. Final Answer:shorted

Question 17

Power from energy–time data: If 12,000 J of energy are used in 400 ms, what is the power delivered?

Accepted Answer

Introduction / Context:Converting between energy and power is fundamental in electrical engineering and physics. Short-duration events often involve large instantaneous power despite modest energy totals. This question tests correct unit handling and the formula P = E / t. Given Data / Assumptions: Energy E = 12,000 J. Time t = 400 ms = 0.4 s. Power is average over the stated interval. Concept / Approach:Use the relation Power = Energy / Time. Ensure time is in seconds so that watts (joules per second) are obtained directly. Convert milliseconds to seconds prior to division. Step-by-Step Solution: Convert time: 400 ms = 0.4 s.Compute power: P = 12,000 / 0.4 = 30,000 W.Express in kilowatts: 30,000 W = 30 kW. Verification / Alternative check:Dimensional analysis: J / s = W. Quick estimate: if 12 kJ were used in 1 s → 12 kW; in 0.4 s, the power is higher by factor 1/0.4 = 2.5 → 30 kW, confirming the result. Why Other Options Are Wrong: 30 W and 3 W: Off by factors of 1,000 or 10,000 due to unit conversion errors. 300 kW: Too large by a factor of 10; likely from misplacing a decimal when converting time. Common Pitfalls: Treating milliseconds as minutes or forgetting to convert to seconds. Confusing energy (J) with power (W) and mixing formulas. Final Answer:30 kW

Question 18

Power dissipation vs. resistor ratings: A 33 Ω 0.5 W resistor and a 330 Ω 0.5 W resistor are each connected across a 12 V source (parallel branches). Which resistor(s) will overheat based on power dissipation?

Accepted Answer

Introduction / Context:Selecting safe resistor power ratings prevents failures and fire hazards. This problem reinforces the calculation of power from a fixed voltage across resistors and comparison with nameplate ratings. Given Data / Assumptions: Two separate resistors are each placed directly across 12 V. Ratings: 0.5 W (half-watt) each. Resistances: 33 Ω and 330 Ω. Concept / Approach:With a known voltage across a resistor, use P = V^2 / R to compute dissipation. If the computed power exceeds the rated wattage, the part is likely to overheat in continuous service. Step-by-Step Solution: For 33 Ω: P = 12^2 / 33 = 144 / 33 ≈ 4.36 W >> 0.5 W → overheats.For 330 Ω: P = 144 / 330 ≈ 0.436 W < 0.5 W → within rating.Conclusion: Only the 33 Ω resistor overheats. Verification / Alternative check:Check current for 33 Ω: I = V / R = 12 / 33 ≈ 0.364 A; P = V * I ≈ 12 * 0.364 ≈ 4.37 W. Consistent with the V^2 / R result. Why Other Options Are Wrong: 330 Ω: Calculated power is below 0.5 W. Both resistors / neither resistor: Contradicted by the computed values. Common Pitfalls: Forgetting that resistors in separate branches each experience the source voltage. Using P = I^2 * R without first finding I correctly. Final Answer:33 Ω

Question 19

Choosing a safe resistor rating: For a carbon-composition resistor expected to dissipate up to 1.2 W continuously, which nominal power rating should be selected?

Accepted Answer

Introduction / Context:Proper power derating is critical for reliability. Resistors should be chosen with a power rating above the expected dissipation to handle tolerances, ambient temperature, and intermittent surges without overheating. Given Data / Assumptions: Maximum expected dissipation ≈ 1.2 W. Standard resistor power ratings are considered (e.g., 0.5 W, 1 W, 2 W, 5 W). Continuous operation and typical ambient conditions. Concept / Approach:Select the next higher standard rating above the calculated worst-case power. This provides margin for tolerance, temperature rise, and component aging. A common practice is to operate resistors at 50–60% of their rated power for longevity. Step-by-Step Reasoning: Computed need: 1.2 W continuous.Closest higher standard rating: 2 W (1 W is below requirement; 0.5 W far below).Therefore select a 2 W resistor; 5 W would be over-conservative and larger/costlier. Verification / Alternative check:Applying a derating guideline (e.g., use ≤ 60% of rating): 1.2 W / 0.6 = 2.0 W minimum rating. This corroborates the 2 W choice. Why Other Options Are Wrong: 1 W and 0.5 W: Below the required dissipation, likely to overheat. 5 W: Electrically safe but unnecessarily large and costly for a 1.2 W need. Common Pitfalls: Selecting a rating equal to calculated power with no margin. Ignoring ambient temperature and ventilation, which further justify using a higher rating. Final Answer:2 W

Question 20

Compute resistor power: When 8 mA flows through a 12 kΩ resistor, what is the resulting power dissipation?

Accepted Answer

Introduction / Context:Translating current through a known resistance into power helps ensure components operate within safe limits. This is commonly needed when verifying resistor wattage in bias networks and load resistors. Given Data / Assumptions: I = 8 mA = 0.008 A. R = 12 kΩ = 12,000 Ω. DC or RMS values implied; use standard power formulas. Concept / Approach:Use P = I^2 * R. Alternatively, compute voltage first via V = I * R and then apply P = V * I. Both yield the same result when done consistently in SI units. Step-by-Step Solution: Compute P = I^2 * R = (0.008)^2 * 12,000.(0.008)^2 = 0.000064; multiply: 0.000064 * 12,000 = 0.768 W.Convert to milliwatts: 0.768 W = 768 mW. Verification / Alternative check:Find V: V = I * R = 0.008 * 12,000 = 96 V. Then P = V * I = 96 * 0.008 = 0.768 W. Both methods agree. Why Other Options Are Wrong: 7.68 mW: Off by a factor of 100; likely from moving the decimal incorrectly. 7.68 W and 76.8 W: Too large; would require far higher current or resistance. Common Pitfalls: Forgetting to convert kΩ to Ω and mA to A before substituting. Dropping zeros when squaring the current. Final Answer:768 mW

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