A circuit element has 8V across it and 3A flowing into its positive terminal. Using the passive sign convention, P = +24W. What does this mean?
AThe element supplies 24W to the rest of the circuit — it is acting as a source
BThe element absorbs 24W — it is a load converting electrical energy to another form
CThe element stores 24W — it is a reactive element like a capacitor or inductor
DThe sign is ambiguous without knowing whether the element is a resistor or a source
Under the passive sign convention, current entering the positive terminal with positive voltage gives P = +VI > 0, meaning the element *absorbs* power (it is a load: resistor, motor, etc.). P < 0 would indicate a source supplying power. Many students invert this, thinking positive power means the element is 'doing work on' the circuit — but positive power means work is being done *on* the element. A battery or generator would show negative P under this convention.
Question 2 Multiple Choice
A transmission line has resistance 2 Ω and must deliver 10 kW of power. Compare the resistive power loss when transmitting at 100V versus at 10,000V.
AThe loss is the same at both voltages — only total power and resistance determine it
BAt 100V the current is 100A giving I²R = 20,000W loss; at 10,000V the current is 1A giving I²R = 2W loss
CAt higher voltage, I²R loss increases because both voltage and current contribute
DThe loss depends only on the voltage across the line, not the current through it
P = IV, so for fixed power P, current I = P/V. At 100V: I = 100A → loss = I²R = 10,000 × 2 = 20,000W (exceeding the intended delivery). At 10,000V: I = 1A → loss = 1 × 2 = 2W — negligible. Resistive losses scale as I², so the ratio of losses is (100/1)² = 10,000-fold. This is why the power grid transmits at high voltage: high voltage forces low current, and I²R loss is exquisitely sensitive to current.
Question 3 True / False
Kirchhoff's voltage law (the sum of voltages around any closed loop equals zero) is mathematically equivalent to energy conservation in that loop.
TTrue
FFalse
Answer: True
KVL states ΣV = 0 around any closed loop. Multiplying each term by the common loop current I gives Σ(VI) = ΣP = 0 — total power in the loop is zero. This means sources supply exactly as much power as loads absorb. KVL and energy conservation are two expressions of the same underlying constraint, which is why both are always satisfied simultaneously in a valid circuit.
Question 4 True / False
A resistor can supply power to a circuit if a sufficiently large current is forced through it by an external source.
TTrue
FFalse
Answer: False
Resistors can only dissipate (absorb) power — they can never supply it. The power dissipated is P = I²R = V²/R, which is always non-negative regardless of the direction or magnitude of current. The I² term ensures the result is always positive. Physically, a resistor converts electrical energy to heat; there is no mechanism by which it can return energy to the circuit. Only sources (batteries, generators, dependent sources) supply power.
Question 5 Short Answer
Why do power transmission lines operate at high voltage and low current rather than low voltage and high current? Use P = I²R in your explanation.
Think about your answer, then reveal below.
Model answer: For a fixed amount of power to be delivered (P = IV is constant), choosing high voltage means the current must be low (I = P/V). The power dissipated as heat in the transmission line's resistance is P_loss = I²R. Because the loss scales with the *square* of current, even a modest reduction in current dramatically reduces losses: doubling the voltage halves the current and cuts resistive losses by a factor of four. High-voltage transmission therefore minimizes wasted energy over long distances.
This is why the power grid steps up voltage to hundreds of kilovolts for long-distance transmission and then steps it back down near homes and businesses. The I²R formula makes the design principle clear: line resistance is fixed, so minimizing I is the only way to minimize losses. The quadratic dependence makes high-voltage transmission especially efficient at scale.