Wire A has length L and cross-sectional area A, with resistance R. Wire B is made of the same material but has length 2L and cross-sectional area 2A. What is Wire B's resistance?
AR_B = 2R — doubling the length doubles the resistance
BR_B = R/2 — doubling the area halves the resistance
CR_B = R — doubling both length and area leaves resistance unchanged
DR_B = 4R — both factors double the resistance independently
R = ρL/A. For Wire B: R_B = ρ(2L)/(2A) = ρL/A = R. Doubling the length increases resistance (longer path, more electron-lattice collisions) and doubling the area decreases resistance (more parallel paths for current to share) — the two effects cancel exactly. Options A and B each account for only one of the two geometric changes. Option D wrongly treats both as multiplicative increases. This tests whether students understand L and A as having opposite effects on resistance, not just the formula mechanically.
Question 2 Multiple Choice
An engineer designing a long-distance power transmission line must choose between transmitting 1 MW at high voltage (low current) or low voltage (high current). Which minimizes energy loss in the wires, and why?
ALow voltage, high current — current carries the energy, so more current means more efficient delivery
BHigh voltage, low current — since power loss is P = I²R, reducing current reduces losses quadratically while transmitting the same power
CIt doesn't matter — the total power transmitted is the same either way, so losses are identical
DHigh voltage, high current — both parameters must be maximized to overcome wire resistance
Power loss in a wire is P_loss = I²R. The wire's resistance R is fixed by its material and geometry. Since power transmitted is P = IV, delivering the same power at higher voltage requires proportionally less current. Halving the current reduces I²R losses by a factor of four — a quadratic benefit. This is why long-distance transmission uses voltages in the hundreds of kilovolts, stepped down by transformers at the destination. Options A and C misunderstand the I² scaling: current is not neutral in its effect on losses.
Question 3 True / False
Both metals and semiconductors show increasing electrical resistivity as temperature rises.
TTrue
FFalse
Answer: False
Metals and semiconductors behave oppositely. In metals, rising temperature increases lattice vibrations, which scatter conduction electrons more frequently and increase resistivity (positive temperature coefficient α). In semiconductors, rising temperature thermally excites more charge carriers into the conduction band, increasing conductivity and decreasing resistivity (negative temperature coefficient). This opposite behavior is fundamental to semiconductor device operation — thermistors are designed to exploit this temperature sensitivity, and the sign of α is one of the key distinctions between metallic and semiconducting materials.
Question 4 True / False
Two wires made of different materials can have the same electrical resistance even if their resistivities differ by orders of magnitude.
TTrue
FFalse
Answer: True
Resistance R = ρL/A depends on the combination of material and geometry. A high-resistivity material can produce the same R as a low-resistivity material if compensating geometry is used — shorter length, larger cross-section, or both. For instance, nichrome (high ρ) can be made into a short, thick wire with the same resistance as a long, thin copper wire (low ρ). Resistivity is a material property; resistance is the result of material and shape together. This is why different resistor types can have identical resistance values despite using very different materials.
Question 5 Short Answer
Why do electrical engineers design long-distance power transmission lines to carry power at very high voltage and low current, rather than low voltage and high current?
Think about your answer, then reveal below.
Model answer: Power loss in transmission lines is P = I²R, where R is the wire's resistance (fixed by material and geometry) and I is the current. The loss scales with the square of current — doubling the current quadruples the loss, while halving the current reduces it by three-quarters. Since transmitted power P = IV, the same power can be delivered at high voltage with proportionally low current. Stepping up voltage before transmission dramatically reduces I, and therefore reduces I²R losses quadratically. The wire's resistance cannot easily be changed over long distances, so controlling current through high-voltage transmission is the practical solution.
This is the central motivation for the AC power grid's transformer infrastructure. Transformers efficiently convert between voltage levels, allowing generation at practical voltages, transmission at extremely high voltages (100–765 kV), and distribution at safe consumer voltages. The entire system architecture is a direct consequence of the I² dependence in Joule heating.