Question 1

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 2

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 3

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?

Accepted Answer

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 4

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

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