Properties of Electrical Materials practice questions

Answer: B) Phosphorus is a donor; its fifth valence electron is weakly bound and easily freed to conduct

Phosphorus (group V) has one more valence electron than silicon needs for bonding. That loosely bound electron is easily freed into the conduction band, so phosphorus is a donor and the material is n-type.

• AIncorrect. This describes a group III dopant such as boron, which makes p-type silicon. Phosphorus is in group V and has five valence electrons.
• BCorrect. Four of phosphorus's five valence electrons form covalent bonds with silicon. The fifth needs very little energy to enter the conduction band, so electrons become the majority carriers.
• CIncorrect. Donors give up electrons, not holes. Adding holes would make the material p-type.
• DIncorrect. An acceptor that captures electrons creates holes, which makes the material p-type. Phosphorus does not act as an acceptor in silicon.

Answer: D) Much wider band gap

• AIncorrect. That describes metals.
• BIncorrect. Metals have no gap.
• CIncorrect. Insulators have higher resistivity.
• DCorrect. Few electrons can reach the conduction band.

Answer: A) Its DC resistance is zero, and it expels magnetic flux from its interior

Below its critical temperature (and within its critical field and current), a superconductor has zero DC resistance and expels magnetic flux (the Meissner effect). Zero resistance alone would describe only a hypothetical perfect conductor.

• ACorrect. Below Tc (and below the critical field and current), a superconductor carries DC current with zero resistance. It also shows the Meissner effect, which pushes magnetic flux out of its bulk.
• BIncorrect. A residual resistance that levels off is how an ordinary metal behaves at low temperature. A superconductor's resistance drops abruptly to zero at Tc.
• CIncorrect. That would describe a hypothetical 'perfect conductor'. A superconductor also expels magnetic flux (the Meissner effect), so the field does not pass freely through its bulk.
• DIncorrect. Superconductivity is a sharp transition at the critical temperature Tc, which is above absolute zero. The resistance does not fall gradually toward zero at 0 K.

Answer: C) Increases

• AIncorrect. It changes with temperature.
• BIncorrect. That's typical of semiconductors.
• CCorrect. Metals have a positive temperature coefficient.
• DIncorrect. That's superconductivity at very low temperatures.

Answer: A) A larger diameter and a lower resistance

AWG numbers run inversely to conductor size, so 10 AWG is thicker than 14 AWG. With R = ρL/A, the thicker conductor has lower resistance for the same length and material.

• ACorrect. In AWG, a smaller gauge number means a thicker wire. 10 AWG therefore has more cross-sectional area than 14 AWG. Since R = ρL/A, a larger area at the same length and resistivity gives a lower resistance.
• BIncorrect. This reverses the AWG scale. A higher gauge number (14) is the thinner wire, not the lower one (10).
• CIncorrect. 10 AWG is not the thinner wire. Also, for the same material and length, a smaller cross-section would raise resistance, not lower it.
• DIncorrect. 10 AWG is indeed larger, but resistance varies inversely with cross-sectional area. The larger conductor has lower resistance.

Answer: B) 1.1 eV

• AIncorrect. That's much larger.
• BCorrect. Germanium is about 0.67 eV.
• CIncorrect. Metals have no band gap.
• DIncorrect. That's a wide-bandgap material.

Answer: A) Energy spent reorienting magnetic domains as the core traverses its B–H loop

Hysteresis loss is the energy dissipated each cycle as magnetic domains are realigned, and it is proportional to the area of the B–H loop. Eddy-current loss and copper loss are separate loss mechanisms.

• ACorrect. Each AC cycle drives the core around its B–H loop. The area enclosed by the loop is energy per unit volume per cycle that is lost as heat while magnetic domains are realigned. Soft magnetic materials with narrow loops reduce this loss.
• BIncorrect. This describes eddy-current loss, the other part of core loss. It is reduced by laminating the core or raising its resistivity, not by narrowing the B–H loop.
• CIncorrect. This is copper (I²R) loss in the windings. It depends on load current and is not a core-material loss.
• DIncorrect. Leakage flux shows up as leakage reactance, which causes voltage drop. It is not the energy dissipated by cycling the core's magnetization.

Answer: D) Maximum electric field before breakdown

• AIncorrect. It's the breakdown field.
• BIncorrect. That's the dielectric constant.
• CIncorrect. It's the breakdown field.
• DCorrect. Often in kV/mm.

Answer: A) Conductivity rises because absorbed photons create extra electron-hole pairs

In a photoconductor, photons with energy above the band gap generate extra free carriers (electron-hole pairs). This raises the material's conductivity, so its resistance drops under illumination.

• ACorrect. Photons with energy above the band gap lift electrons from the valence band to the conduction band. This adds free electrons and holes, which raises conductivity and lowers resistance.
• BIncorrect. Lattice heating can change mobility slightly, but the main effect in a photoconductor is extra carrier generation. That increases conductivity rather than decreasing it.
• CIncorrect. A photon needs at least the band gap energy to create an electron-hole pair across the gap. Light below that energy is mostly not absorbed and does not cause the intrinsic photoconductive effect.
• DIncorrect. This mixes up photoconductivity with the photovoltaic effect in a p-n junction. A photoconductor is a resistor whose conductance changes with light. It needs an external bias and does not generate its own voltage.

Answer: A) 120 Ω

• ACorrect. \(R = R_0(1 + \alpha\Delta T) = 100(1.2)\).
• BIncorrect. Multiply α by ΔT.
• CIncorrect. Recheck the calculation.
• DIncorrect. A positive α raises resistance.