Resistance and Conductors Lesson Test (40 Questions)

Answer: D

Correct answer: Ohm. Resistance describes how strongly a material or component opposes current, and one ohm is the resistance that allows 1 A to flow when 1 V is applied, so 1 Ω = 1 V/A. The unit is named after Georg Ohm, who described the relationship

The volt, ampere, and watt measure the other key quantities: voltage, current, and power.

• A: A volt measures electric potential difference, not electrical resistance.
• B: An ampere measures electric current rather than opposition to current.
• C: A watt measures the rate of energy transfer, or power.
• D (correct): The ohm (Ω) is the SI derived unit of electrical resistance.

Answer: A

Correct answer: The opposition a material or component presents to electric current. Resistance is the opposition to current flow, measured in ohms. Voltage is the push and current is the flow; resistance is a property of the part itself.

• A (correct): Resistance is the property of a path that opposes the flow of charge, measured in ohms.
• B: Storing charge is not what resistance does; a resistor converts energy to heat rather than storing it.
• C: That describes current, measured in amperes, which flows through a part.
• D: That describes voltage, which is applied across a part, not a property of the part.

Answer: D

Correct answer: Gold resists corrosion, so the contact surface stays clean. Gold plating is used on connectors because it resists corrosion. Silver is actually the best ordinary metal conductor, and copper is a close second.

• A: Silver, not gold, is the best ordinary metal conductor, with copper close behind.
• B: Gold is a metal conductor; a contact plating must conduct, not insulate.
• C: Gold costs far more than copper; low cost is not the reason it is used.
• D (correct): Gold is chosen mainly because it does not corrode, which keeps contact resistance low over time.

Answer: A

Correct answer: Semiconductor. Pure silicon conducts poorly because its four outer electrons are locked into covalent bonds, yet a small energy boost from heat or light frees some of them to carry current. Adding tiny amounts of impurities, called doping, raises conductivity in a controlled way, which is what makes transistors and solar cells possible. The tempting mistake is calling silicon a conductor because it is used in electronics; metals conduct far better, and silicon's usefulness comes from controllable, not high, conductivity.

• A (correct): Silicon's resistivity lies between that of conductors and insulators, and doping can control its conductivity.
• B: Insulators such as dry porcelain or PVC have very high resistivity and cannot be made to conduct by doping.
• C: Good conductors are metals such as silver, copper, and aluminum with many free electrons.
• D: Ionic conductors are electrolytes such as salt water, where ions carry the charge.

Answer: B

Correct answer: Ohmmeter. An ohmmeter works by sending a small known current from its own internal battery through the component and measuring the resulting voltage, then reporting resistance in ohms. Because it supplies its own current, it must be used on a circuit that is powered off, or the reading will be wrong and the meter may be damaged. Most technicians use the resistance setting of a digital multimeter for this job.

• A: A thermometer measures temperature, which can affect but does not directly equal resistance.
• B (correct): An ohmmeter measures electrical resistance in ohms; use it only under safe conditions.
• C: An ammeter measures electric current, in amperes.
• D: A voltmeter measures potential difference and is not the named resistance instrument.

Answer: B

Correct answer: Sodium and chloride ions drifting toward opposite electrodes. In an electrolyte the moving charges are ions. Salt in water separates into positive sodium ions and negative chloride ions that drift toward opposite electrodes.

• A: Doping and holes belong to semiconductors such as silicon, not to electrolytes.
• B (correct): Dissolved salt separates into charged ions, and those ions are the moving charges in an electrolyte.
• C: Neutral molecules carry no net charge, so their motion cannot be an electric current.
• D: Free electrons in a lattice describe conduction in metals, not in a salt solution.

Answer: D

Correct answer: Ω. The ohm is represented by the Greek capital letter omega, Ω, chosen partly because a capital O could be mistaken for the number zero. Each main electrical unit has its own symbol: V for volts, A for amperes, C for coulombs, and W for watts. Resistor values are often written with prefixes, such as 4.7 kΩ for 4,700 ohms or 1 MΩ for one million ohms.

• A: V is the volt symbol for potential difference.
• B: A is the ampere symbol for current, not resistance.
• C: C is the coulomb symbol for electric charge.
• D (correct): Ω is the standard symbol for the resistance unit ohm.

Answer: C

Correct answer: 3,300 Ω. One kilohm is 1,000 ohms, so 3.3 k$\Omega$ =

$\Omega$.

• A: This multiplies by only 10 instead of by 1,000.
• B: This treats the k as mega, multiplying by one million instead of one thousand.
• C (correct): Kilo means one thousand, so 3.3 times 1,000 gives 3,300 ohms.
• D: This multiplies by 100 instead of by 1,000.

Answer: C

Correct answer: Effectively infinite, so no current flows through it. A break in the path, such as a blown fuse, is an open circuit. Its resistance is effectively infinite, so no current flows, even though the source voltage may appear across the gap.

• A: Nearly zero resistance describes a short circuit, which is too easy a path, not a break.
• B: The gap has no connection to the load's value; a break blocks current entirely.
• C (correct): A blown fuse is an open circuit: there is no path, so the resistance is effectively infinite.
• D: Once the fuse element melts, the path is broken, so its resistance is no longer its small normal value.

Answer: A

Correct answer: 16 Ω. With material, area, and temperature fixed, R is proportional to length. Twice the length gives

$\Omega$.

• A (correct): Resistance is directly proportional to length, so doubling the length doubles 8 ohms to 16 ohms.
• B: Resistance is not unchanged; a longer path gives more opposition.
• C: This multiplies by 4, squaring the length ratio, which applies only to a diameter change.
• D: This halves the resistance, treating length as if it worked like area.

Answer: D

Correct answer: 15 Ω. Use

• A: This adds voltage and current, which has no physical meaning.
• B: This divides current by voltage, which gives the reciprocal of the resistance.
• C: This multiplies voltage by current, which gives power in watts, not resistance.
• D (correct): R = V/I = 120 V divided by 8 A, which equals 15 ohms.

Answer: B

Correct answer: 1.5 Ω. Area depends on diameter squared, so doubling the diameter gives

times the area. Resistance is inversely proportional to area:

$\Omega$.

• A: This doubles the resistance, but a thicker wire has less resistance, not more.
• B (correct): Doubling the diameter makes the area 4 times larger, so the resistance falls to 6/4 = 1.5 ohms.
• C: This multiplies by 4; the area factor is right, but a larger area lowers resistance.
• D: This divides by 2, treating the diameter ratio as if it were the area ratio without squaring it.

Answer: C

Correct answer: 44 Ω. The temperature change is

• A: This uses the final temperature of 45 degrees instead of the 25 degree change.
• B: This adds only the 0.1 factor itself instead of multiplying it by the 40 ohm starting value.
• C (correct): The rise is 25 degrees, so the factor is 1 + 0.004 x 25 = 1.1, and 40 x 1.1 = 44 ohms.
• D: This subtracts the change, but copper has a positive coefficient, so warming raises its resistance.

Answer: B

Correct answer: 5 Ω. Ohm's law,

rearranges to

when you need resistance. Dividing 15 V by 3 A gives 5 Ω, meaning 5 volts are needed for each ampere of current. The common trap is multiplying voltage by current, which gives power in watts rather than resistance. A useful memory aid is the triangle with V on top and I and R below; cover the unknown to see the formula.

• A: $\frac{3}{15}$ reverses the ratio and gives conductance in siemens.
• B (correct): R = $\frac{V}{I}$ = $\frac{15}{3}$ = 5 Ω under these specified conditions.
• C: Multiplying V by I gives watts, not the requested R in ohms.
• D: Adding volt and ampere numerically is not an electrical resistance formula.

Answer: C

Correct answer: 4.5 W. With current and resistance known, use

• A: This squares 0.3 incorrectly as 0.9 instead of 0.09.
• B: This multiplies current by resistance without squaring the current; 0.3 x 50 is the voltage, 15 V.
• C (correct): P = I squared times R = 0.09 x 50 = 4.5 watts.
• D: This squares the current but forgets to multiply by the 50 ohm resistance.

Answer: B

Correct answer: Ω·m. Resistivity describes the material itself, independent of the object's shape. Rearranging R = ρL/A gives ρ = RA/L, so the units are ohms × square meters ÷ meters, which simplifies to ohm-meters. The common trap is answering plain ohms: that unit describes a particular wire or component, while ohm-meters lets you compare copper with rubber regardless of length or thickness. Copper is about 1.7 × 10⁻⁸ Ω·m; good insulators are many orders of magnitude higher.

• A: Ampere measures current and cannot identify material resistivity.
• B (correct): The quantity ρ in R = ρ$\frac{L}{A}$ has units of ohm-meters.
• C: Volts per second measures a change in voltage over time.
• D: Ohm alone is the unit of object resistance, not bulk resistivity.

Answer: C

Correct answer: 10 W. Find the power with

Twice that is 6 W, so the 10 W rating is the only choice that meets the guideline.

• A: A 1 W part is below the 3 W expected dissipation and would overheat.
• B: A 3 W part would run right at its limit with no safety margin, not at about twice the expected power.
• C (correct): The resistor dissipates 3 W, so the rating should be at least about 6 W; 10 W meets that guideline.
• D: A 1/4 W part is far below the 3 W it must shed and would overheat.

Answer: A

Correct answer: They have the same resistivity, but the long, thin wire has more resistance. Resistivity is a property of the material, so both copper wires share it. Resistance belongs to the object and rises with length and falls with area, so the long, thin wire has more resistance.

• A (correct): Resistivity belongs to the material, while resistance also depends on length and area.
• B: Being the same material gives the same resistivity, not the same resistance; size and shape still matter.
• C: Resistivity does not change with an object's size or shape, only resistance does.
• D: More copper in the cross section gives more paths for current, which lowers resistance.

Answer: C

Correct answer: The meter supplies its own test current, so outside voltage gives false readings and can damage it. An ohmmeter sends a small known current from its internal battery through the part and calculates resistance from the resulting voltage. Voltage from a live circuit adds to that and produces false readings, and it can damage the meter.

• A: An ohmmeter connects across the isolated part; series connection describes how an ammeter is used.
• B: A resistor still has resistance while operating; the problem is interference with the meter, not a zero value.
• C (correct): An ohmmeter computes resistance from its own small internal current, so any circuit voltage spoils the result and may harm the meter.
• D: Switching off does not change the resistor's rated value; it protects the meter and the reading.

Answer: A

Correct answer: 10 AWG. AWG numbers run backward: a smaller number is a thicker wire with a larger cross-sectional area. Larger area means lower resistance, so 10 AWG has the least resistance per foot.

• A (correct): In the AWG system a smaller gauge number means a thicker wire, and the thickest wire has the least resistance.
• B: 14 AWG is thinner than 12 AWG and 10 AWG, so it has more resistance than both.
• C: The largest gauge number is actually the thinnest wire, so it has the most resistance.
• D: 12 AWG is thicker than 14 AWG but still thinner than 10 AWG.

Answer: C

Correct answer: It decreases. NTC stands for negative temperature coefficient. An NTC thermistor's resistance falls strongly as it warms, which is what lets it sense temperature.

• A: Near-zero temperature change in resistance describes alloys such as manganin, not a thermistor.
• B: Zero resistance occurs only in a superconductor below its critical temperature, not in a warm thermistor.
• C (correct): NTC means negative temperature coefficient, so the resistance falls as temperature rises.
• D: Rising resistance with heat describes a PTC thermistor or an ordinary metal, not an NTC type.

Answer: A

Correct answer: The meter can show an equivalent resistance lower than its value. An ohmmeter sends a small test current and measures whatever path that current can take. With another resistor still connected in parallel, the test current splits between both, and parallel resistance is always less than the smallest branch. Turning the power off removes the hazard but not the extra path. To read a component's true value, disconnect at least one of its leads from the circuit first.

• A (correct): A parallel path can reduce the effective resistance seen by the ohmmeter; proper safe isolation matters.
• B: Removing power does not open the parallel path, so the meter still sees a finite combined resistance.
• C: An ohmmeter reports resistance, not voltage, so a 0 V reading does not answer the question.
• D: An in-circuit meter also measures the parallel path, so the reading can differ from the label.

Answer: D

Correct answer: 9.6 Ω. Rearrange

to

• A: This divides voltage by power, which does not give a resistance.
• B: This inverts the formula, dividing power by voltage squared.
• C: Dividing power by voltage gives the current, 12.5 A, not the resistance.
• D (correct): R = V squared divided by P = 14,400/1,500 = 9.6 ohms.

Answer: D

Correct answer: 0.85 Ω. Convert the area: 1 mm${}^2$ = $10^{-6}$ m${}^2$, so A = $2.0\times 10^{-6}$ m${}^2$.

• A: This is ten times too large, a power-of-ten slip in the exponents.
• B: This is ten times too small, a power-of-ten slip in the exponents.
• C: This converts square millimeters as if 1 mm squared were 10^-3 square meters instead of 10^-6.
• D (correct): With A = 2.0 x 10^-6 square meters, R = (1.7 x 10^-8 x 100)/(2.0 x 10^-6) = 0.85 ohms.

Answer: B

Correct answer: A short circuit: very low resistance, so current can become very large. A short circuit is an unintended path with very low resistance that bypasses the load. Because R is so small, current becomes very large until a fuse or breaker opens. Open means no path; short means too easy a path.

• A: This mislabels the fault and wrongly says the bypassed appliance gets more current.
• B (correct): An unintended low-resistance path around the load is a short circuit, and the small resistance allows a large, dangerous current.
• C: An open circuit is a break in the path; here the conductors touch, creating too easy a path instead.
• D: A short has very low resistance, not high, and the current rises rather than falls.

Answer: C

Correct answer: A surface leakage path forms, so it insulates less effectively. Porcelain is a good insulator when clean and dry. Moisture with dissolved salts contains ions that conduct, so a contaminated surface creates a leakage path that reduces how well the insulator works.

• A: Damp, salty grime contains ions that conduct, so it weakens insulation rather than adding to it.
• B: The ceramic material does not become a metal; the problem is a conducting film on its surface.
• C (correct): Moisture and dissolved salts on the surface can carry current, creating leakage around the insulator.
• D: No insulator blocks current under every condition; contamination and moisture reduce its effectiveness.

Answer: D

Correct answer: Its resistance is 5 Ω. Ohm's law, V = IR, links the three quantities, so rearranging gives

volts on top, amperes underneath. A quick check is to plug the answer back in: 2 A × 5 Ω = 10 V, which matches the applied voltage. The usual slip is multiplying 10 by 2, which gives 20, but V × I is power in watts, not resistance. Dividing the other way gives conductance instead.

• A: 0.2 is $\frac{I}{V}$, the conductance in siemens, not the resistance.
• B: 20 is the numerical product VI, giving 20 W, not ohms.
• C: Adding voltage and current numerically does not compute resistance.
• D (correct): The resistance is R = $\frac{V}{I}$ = $\frac{10}{2}$ = 5 Ω.

Answer: C

Correct answer: 40 Ω. On an I-versus-V graph the slope is

• A: This is the slope itself, which equals 1/R in siemens, not R.
• B: This multiplies voltage by current instead of dividing voltage by current.
• C (correct): The slope is 0.25/10 = 0.025 siemens, and R is its reciprocal, 40 ohms.
• D: This makes a decimal-place slip when dividing 10 by 0.25.

Answer: A

Correct answer: It usually has high resistivity, but conditions and breakdown matter. Insulators such as rubber, glass and plastic hold their electrons tightly, so very little charge can move. That resistance is not unlimited: enough voltage can tear electrons loose and cause breakdown, as when lightning crosses air, and heat, moisture or damage can let current leak. The tempting idea that insulators contain no electrons is wrong; every material has electrons, and what differs is how freely they move.

• A (correct): A good insulator restricts charge transport but is not universally perfect.
• B: Insulators contain electrons; the key distinction is their limited mobility.
• C: Real insulating materials can leak current or break down under adverse conditions.
• D: Insulation alone does not guarantee safety around hazardous energized equipment.

Answer: D

Correct answer: 45 W. All of the heater current passes through the contact, so use

• A: This multiplies current by resistance without squaring the current, which gives the voltage drop of 3 V.
• B: This divides current by resistance, which is not a power formula.
• C: This divides current squared by resistance, mixing up $I^{2}R$ and $V^2/R$.
• D (correct): P = I squared times R = 225 x 0.2 = 45 watts, concentrated at the small connection.

Answer: B

Correct answer: 4.5 Ω. Multiply the old resistance by a factor for each change, with the area ratio inverted. Doubling the diameter makes the area

times larger.

• A: This handles length and area correctly but leaves out the higher resistivity of the new metal.
• B (correct): Factors: 1.5 for resistivity, 2 for length, and 1/4 for area, since doubling diameter quadruples area: 6 x 1.5 x 2 / 4 = 4.5 ohms.
• C: This divides by 2 for the diameter change, forgetting that area depends on diameter squared.
• D: This multiplies by the area factor instead of dividing; a larger area lowers resistance.

Answer: B

Correct answer: The copper coil's resistance rises, while the carbon resistor's falls. The sign of the temperature coefficient depends on the material. Copper, a pure metal, has a positive coefficient of about 0.004 per degree Celsius, so its resistance rises. Carbon has a negative coefficient, so its resistance drops as it warms.

• A: Heat raises resistance in most pure metals, but carbon and semiconductors have negative coefficients.
• B (correct): Copper has a positive temperature coefficient, while carbon has a negative one over normal ranges.
• C: This reverses the signs: copper's coefficient is positive and carbon's is negative.
• D: Warming helps carbon and semiconductors conduct, but copper's resistance rises with temperature.

Answer: C

Correct answer: 8.3 A. First find the hot resistance:

The cold filament is about one-tenth of that, 14.4 $\Omega$.

• A: This multiplies the hot resistance by 10 instead of dividing, giving a cold value that is far too large.
• B: This divides the hot resistance by 100 instead of by 10.
• C (correct): The hot resistance is 120^2/100 = 144 ohms; the cold value is about 14.4 ohms, so the surge is 120/14.4, about 8.3 A.
• D: This is the normal operating current through the hot 144 ohm filament, not the turn-on surge.

Answer: A

Correct answer: The 16 Ω resistor dissipates 64 W, four times as much as the other. At a fixed current use

The larger resistance dissipates four times as much.

• A (correct): At the same current, P = I squared R, giving 64 W for 16 ohms and 16 W for 4 ohms.
• B: Equal current does not mean equal power; power also depends on resistance through $I^{2}R$.
• C: This uses I times R without squaring the current; 32 is the voltage across it in volts, not the power.
• D: That reasoning applies at a fixed voltage; in series the current is the same, so the larger resistance heats more.

Answer: A

Correct answer: 8 milliohms. For steady DC, resistance depends on the cross-sectional area of metal. Area scales with diameter squared, so the tube's metal area is

of the rod's.

• A (correct): The tube's cross section is (4 - 1)/4 = 3/4 of the rod's, so R = 6 / 0.75 = 8 milliohms.
• B: This multiplies by the 3/4 area ratio instead of dividing; less metal means more resistance.
• C: This assumes the outer size or surface sets resistance, but the cross-sectional area of metal is what matters.
• D: This uses diameters linearly, treating the area as halved, instead of squaring them.

Answer: B

Correct answer: 2.0 × 10⁻⁸ Ω·m. First find the area:

${}^2$. Then rearrange

• A: This uses pi times d squared for the area, forgetting to divide by 4.
• B (correct): With A = pi(0.001)^2/4, about 7.85 x 10^-7 square meters, rho = RA/L = 0.50 x 7.85 x 10^-7 / 20, about 2.0 x 10^-8.
• C: This uses the diameter squared as the area, leaving out the factor of pi/4.
• D: This inverts the formula, computing RL/A instead of RA/L.

Answer: D

Correct answer: About 5.1 mm. To keep R the same with 1.6 times the resistivity and equal length, the area must be 1.6 times larger. Area scales with diameter squared, so the diameter scales by

• A: This multiplies the diameter by 1.6, but it is the area, not the diameter, that must grow by 1.6.
• B: Same size would give the aluminum 1.6 times the resistance because of its higher resistivity.
• C: This makes the aluminum wire thinner, which would raise its resistance even further.
• D (correct): The area must be 1.6 times larger, so the diameter grows by the square root of 1.6, about 1.26: 4.0 x 1.26 is about 5.1 mm.

Answer: D

Correct answer: Verify zero voltage across the capacitor, then lift one resistor lead and measure. Unplugging does not guarantee a de-energized circuit, because capacitors can store charge after power is removed. Confirm zero voltage with a meter, then disconnect one lead of the resistor so other components cannot form a parallel path, and only then measure resistance.

• A: Unplugged is not the same as de-energized; a large capacitor can hold a charge after power is removed.
• B: An ohmmeter must never be used on a live circuit; it gives false readings and can be damaged.
• C: A charged capacitor can damage the meter and spoil the reading, and other parts in the circuit can form parallel paths.
• D (correct): This confirms the circuit is truly de-energized and isolates the resistor from parallel paths before measuring.

Answer: A

Correct answer: Half as large. The slope is resistance per meter. Y's slope is

times larger, so Y's area is 1/4 of X's. Because area scales with diameter squared, Y's diameter is

of X's.

• A (correct): Y has 4 times the resistance per meter, so 1/4 the area; since area goes with diameter squared, Y's diameter is half of X's.
• B: A steeper line means more resistance per meter, which comes from a thinner wire, not a thicker one.
• C: This reverses the direction and also skips the square root linking area to diameter.
• D: The slope ratio of 4 gives the area ratio, but the diameter ratio is its square root, 1/2, not 1/4.

Answer: B

Correct answer: 12 Ω. Half the length gives half the resistance: 10 $\Omega$ at 20 degrees Celsius. The rise is 50 degrees, so

• A: This halves the resistance for the shorter length but ignores the temperature rise of 50 degrees.
• B (correct): Halving the length gives 10 ohms, and the 50 degree rise multiplies that by 1.2, giving 12 ohms.
• C: This applies the 20 percent change in the wrong direction; copper's resistance rises when heated.
• D: This applies the temperature increase to the full 20 ohms but forgets that the wire was cut in half.

Answer: D

Correct answer: 6 Ω. For the same material and temperature, R is proportional to L/A. Multiply by the length factor and divide by the area factor.

• A: This multiplies by 3 and divides by 2, flipping the roles of length and area.
• B: This multiplies by both factors; a larger area lowers resistance, so it must divide.
• C: This divides by both factors; a longer wire raises resistance, so length must multiply.
• D (correct): Length doubles R and tripled area divides it by 3: 9 times 2 divided by 3 is 6 Ω.

Answer: C

Correct answer: Silver. Silver conducts best of the ordinary metals, with copper a close second. Copper is used for most wiring because it costs far less, aluminum is used where low weight matters, and gold plating is chosen for corrosion resistance.

• A: Copper is a very close second and is used for wiring because it is far cheaper, but it is not the best.
• B: Gold is used to plate connectors because it resists corrosion, not because it conducts best.
• C (correct): Silver has the lowest resistivity of the ordinary metals.
• D: Aluminum is chosen for overhead lines because it is light; its resistivity is about 1.6 times copper's.

Answer: A

Correct answer: Twice as much. Halving the length of the same wire halves its resistance. At a fixed voltage, use

with V unchanged and R halved, the power doubles. The shorter element draws twice the current at the same voltage, so it heats more, not less.

• A (correct): Half the length gives half the resistance, and at fixed voltage P = V squared over R, so power doubles.
• B: This assumes less wire means less heat, but at a fixed voltage a smaller resistance draws more current and more power.
• C: Power would stay the same only if the resistance stayed the same, but halving the length halves R.
• D: Power scales as 1/R at fixed voltage, not 1/R squared; halving R doubles the power.

Answer: B

Correct answer: 70 degrees Celsius. The fractional increase is

so

The winding warmed by 50 degrees, from 20 to about 70 degrees Celsius.

• A: 50 degrees is the temperature rise; it must be added to the starting 20 degrees.
• B (correct): The resistance rose by 6/30 = 0.2, and 0.2 divided by 0.004 is a 50 degree rise, so 20 + 50 = 70 degrees.
• C: This divides the 6 Ω increase by 0.004 without first dividing by the 30 Ω starting value.
• D: This adds the 6 Ω increase directly to the temperature, mixing ohms with degrees.

Answer: D

Correct answer: 250 Ω. Convert first: 12 mA = 0.012 A. Then use

• A: This divides 3 by 12 without converting 12 mA to 0.012 A.
• B: This divides 12 by 3, putting current over voltage and also skipping the conversion.
• C: This divides current by voltage, 0.012 by 3, which gives conductance-type units, not ohms.
• D (correct): Converting 12 mA to 0.012 A and dividing 3 V by 0.012 A gives 250 Ω.

Answer: B

Correct answer: Switch to a higher resistance range, such as 2 MΩ, and measure again. OL on a resistance setting means the value is larger than the range selected. A 220 kΩ resistor cannot be shown on a 2 kΩ range, so she should step up to a range above 220 kΩ, keeping the circuit de-energized and the part isolated.

• A: OL only means the value is above the selected range; the resistor may be perfectly good.
• B (correct): 220 kΩ is far above a 2 kΩ range, so a higher range is needed for a reading.
• C: An ohmmeter supplies its own test current; power must stay off for a safe, valid reading.
• D: Reversing leads does not matter for a plain resistor and does not fix an over-range reading.

Answer: B

Correct answer: Resistivity. Resistivity, symbol $\rho$, describes how strongly a material opposes current regardless of size or shape. Resistance then depends on that resistivity together with length and area, through

• A: Resistance belongs to a particular object and changes with its length and cross-sectional area.
• B (correct): Resistivity is a material property in ohm-meters; every copper wire at one temperature has the same value.
• C: Current depends on the applied voltage and the circuit's resistance, not just on the material.
• D: A wire's voltage drop depends on its resistance and the current it carries.

Answer: C

Correct answer: The kettle, at 12 Ω compared with 240 Ω for the lamp. At a fixed voltage,

With

The high-power kettle has the lower resistance because it must draw more current from the same voltage.

• A: This swaps the two results; the higher-power device at a fixed voltage has the lower resistance.
• B: The values are reversed: 120 squared over 1,200 is 12 Ω, not 240 Ω.
• C (correct): At fixed voltage R = V squared over P, giving 14,400/1,200 = 12 Ω and 14,400/60 = 240 Ω.
• D: These come from dividing power by voltage, which gives current in amperes, not resistance.

Answer: C

Correct answer: 12 Ω. With the same volume, doubling the length means the cross-sectional area must be cut in half. Both changes raise resistance.

• A: This accounts for the doubled length but forgets that stretching also makes the wire thinner.
• B: Length and area do change by the same factor, but in opposite directions, so their effects multiply rather than cancel.
• C (correct): Doubling the length at constant volume halves the area, so R is multiplied by 2 and by 2 again: 12 Ω.
• D: This treats the area as falling to one fourth, as if the diameter were halved, but at constant volume the area only halves.

Answer: A

In a controlled experiment, change only the factor under study. Resistance also depends on material, cross-sectional area and temperature, so those must stay the same while only the length is varied. A small test current helps keep the temperature steady.

• A (correct): Holding every other factor constant lets any change in resistance be credited to length alone.
• B: Changing the material also changes resistivity, so the effect of length cannot be separated out.
• C: Changing both length and diameter at once mixes two effects, so neither can be identified.
• D: A large current heats the wire and changes its resistance during the test; a small current is better.