College Physics Quiz: Resistance Resistivity And Ohms Law

20 questions · exam conditions

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Resistance Resistivity And Ohms LawQuestion 1 of 20

Two wires are made of the same material but have different dimensions. Wire A has length $L$ and cross-sectional area $A$ . Wire B has length $2L$ and cross-sectional area $A/2$ . What is the ratio of the resistance of wire B to the resistance of wire A?

1:1

2:1

4:1

1:2

1:4

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College Physics Quiz

College Physics Quiz: Resistance Resistivity And Ohms Law

Practice Resistance Resistivity And Ohms Law in College Physics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Resistance Resistivity And Ohms Law, giving you a quick way to practice the rules, question types, and explanations that matter most for College Physics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$1:1$
$2:1$
$4:1$ (correct answer)
$1:2$
$1:4$

Explanation: When you encounter resistance problems involving different wire dimensions, always start with the fundamental resistance formula:

R = \rho \frac{L}{A}

, where

\rho

is resistivity (a material property),

L

is length, and

A

is cross-sectional area. For Wire A:

R_A = \rho \frac{L}{A}

For Wire B:

R_B = \rho \frac{2L}{A/2} = \rho \frac{2L \cdot 2}{A} = \rho \frac{4L}{A}

The ratio of resistances is:

\frac{R_B}{R_A} = \frac{\rho \frac{4L}{A}}{\rho \frac{L}{A}} = \frac{4L}{A} \cdot \frac{A}{L} = 4

This gives us a 4:1 ratio, confirming answer C. Let's examine why the other options are incorrect: A) 1:1 would only be true if both length and area changes exactly canceled out, but doubling length while halving area creates a net 4× increase in resistance. B) 2:1 incorrectly assumes only the length change matters (2L vs L) while ignoring the area reduction, or vice versa. D) 1:2 gets the ratio backwards and significantly underestimates the effect of the combined dimensional changes. Study tip: Remember that resistance increases with length and decreases with area. When both changes work in the same direction (longer length AND smaller area both increase resistance), multiply their individual effects. Here, 2× length increase and 2× area decrease combine to give 2 × 2 = 4× total resistance increase.

Question 2

A $12 \, V$ battery is connected across a resistor, and a current of $0.75 \, A$ flows through the circuit. If the battery is replaced with a $9 \, V$ battery and the same resistor is used, what current will flow through the circuit?

$0.56 \, A$ (correct answer)
$0.60 \, A$
$0.67 \, A$
$0.75 \, A$
$1.00 \, A$

Explanation: This question tests your understanding of Ohm's Law, which states that voltage, current, and resistance are related by

V = IR

. When you see a circuit problem involving a battery replacement, think about how changing one variable affects the others while keeping resistance constant. First, let's find the resistance using the initial conditions. With

V = 12 \, V

and

I = 0.75 \, A

, we can solve for resistance:

R = \frac{V}{I} = \frac{12}{0.75} = 16 \, \Omega

. This resistance remains the same when we swap batteries. Now with the

9 \, V

battery and the same

16 \, \Omega

resistor, we can find the new current:

I = \frac{V}{R} = \frac{9}{16} = 0.5625 \, A

, which rounds to

0.56 \, A

. Looking at the wrong answers: Choice B (

0.60 \, A

) might result from incorrectly assuming the resistance is

15 \, \Omega

instead of

16 \, \Omega

. Choice C (

0.67 \, A

) could come from using

\frac{9}{13.5}

, perhaps from a calculation error in finding resistance. Choice D (

0.75 \, A

) represents the trap of thinking current stays constant regardless of voltage—this ignores Ohm's Law entirely. The correct answer is A)

0.56 \, A

. Study tip: In Ohm's Law problems, always identify what stays constant. Here, resistance is fixed while voltage changes, so current must change proportionally. Practice calculating each variable systematically: find the unknown quantity first, then apply it to the new conditions.

Question 3

When a voltage of $6.0 \, V$ is applied across a certain resistor, the power dissipated is $18 \, W$ . What is the resistance of this resistor?

$0.33 \, \Omega$
$2.0 \, \Omega$ (correct answer)
$3.0 \, \Omega$
$12 \, \Omega$
$108 \, \Omega$

Explanation: When you encounter a problem involving voltage, power, and resistance, you're working with the fundamental relationships in electrical circuits. The key is knowing which power formula to use based on what information you're given. Since you have voltage (

V = 6.0 \, V

) and power (

P = 18 \, W

), the most direct approach is to use the power formula

P = \frac{V^2}{R}

. Rearranging to solve for resistance:

R = \frac{V^2}{P}

. Substituting the values:

R = \frac{(6.0 \, V)^2}{18 \, W} = \frac{36 \, V^2}{18 \, W} = 2.0 \, \Omega

This confirms answer choice B is correct. Let's examine why the other options are wrong. Choice A (

0.33 \, \Omega

) would result if you incorrectly calculated

\frac{P}{V^2} = \frac{18}{36}

, essentially using the reciprocal of the correct formula. Choice C (

3.0 \, \Omega

) comes from the error

\frac{V \times P}{V} = \frac{6.0 \times 18}{6.0} = 18

, which has no physical basis. Choice D (

12 \, \Omega

) results from incorrectly using

R = \frac{V \times P}{V} = \frac{6.0 \times 18}{6.0} \times 2 = 36/3

, another computational mistake. Remember the three power formulas:

P = IV

P = I^2R

, and

P = \frac{V^2}{R}

. Choose the one that directly uses your given variables to avoid unnecessary steps and potential errors from calculating intermediate values like current.

Question 4

Three identical resistors, each with resistance $R$ , are connected in different configurations. Configuration A has all three in series, Configuration B has all three in parallel, and Configuration C has two in parallel connected in series with the third. What is the ratio of the total resistance of Configuration A to Configuration C?

$6:5$
$3:2$
$2:1$ (correct answer)
$3:1$
$9:2$

Explanation: When analyzing resistor networks, you need to apply the fundamental rules for series and parallel combinations systematically. For resistors in series, total resistance equals the sum of individual resistances. For parallel resistors, use

\frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} + ...

Let's find each configuration's total resistance. Configuration A (all series):

R_A = R + R + R = 3R

Configuration C requires two steps since it's a combination circuit. First, find the resistance of the two parallel resistors:

\frac{1}{R_{parallel}} = \frac{1}{R} + \frac{1}{R} = \frac{2}{R}

, so

R_{parallel} = \frac{R}{2}

. Then add the third resistor in series:

R_C = \frac{R}{2} + R = \frac{3R}{2}

The ratio of Configuration A to C is:

\frac{R_A}{R_C} = \frac{3R}{\frac{3R}{2}} = \frac{3R \times 2}{3R} = 2

, giving us

2:1

. Answer A (

6:5

) likely comes from incorrectly calculating the parallel combination or making arithmetic errors. Answer B (

3:2

) might result from forgetting to add the series resistor in Configuration C, comparing

3R

to just

R

. Answer D (

3:1

) could come from mistakenly comparing Configuration A to Configuration B (all parallel), where

R_B = \frac{R}{3}

. Remember: always work combination circuits step-by-step, solving parallel sections first, then incorporating them into the series calculation. Draw the circuit if needed to visualize the current path clearly.

Question 5

A heating element is designed to operate at $120 \, V$ and draw $10 \, A$ of current. If the voltage is reduced to $100 \, V$ and the resistance remains constant, what is the new power consumption?

$694 \, W$
$833 \, W$ (correct answer)
$1000 \, W$
$1200 \, W$
$1440 \, W$

Explanation: This question tests your understanding of electrical power relationships and how changes in voltage affect power consumption when resistance remains constant. First, you need to find the resistance of the heating element using the original conditions. With Ohm's law,

R = \frac{V}{I} = \frac{120 \, V}{10 \, A} = 12 \, \Omega

. Since the problem states resistance remains constant, this value stays the same when voltage changes. When voltage drops to

100 \, V

, you can find the new power using

P = \frac{V^2}{R}

. Substituting:

P = \frac{(100 \, V)^2}{12 \, \Omega} = \frac{10000}{12} = 833 \, W

. This confirms answer (B). Let's examine why the other answers are incorrect. (A) 694 W might result from calculation errors or using an incorrect resistance value. (C) 1000 W represents a common trap—students might assume power scales linearly with voltage, calculating

1200 \times \frac{100}{120} = 1000

, but power actually depends on voltage squared. (D) 1200 W is the original power consumption (

P = VI = 120 \times 10 = 1200 \, W

), which some students might choose if they forget that reducing voltage decreases power. Study tip: Remember that power has multiple formulas:

P = VI

P = I^2R

, and

P = \frac{V^2}{R}

. Choose the most convenient one based on what remains constant. When resistance is constant and voltage changes,

P = \frac{V^2}{R}

is your best friend because it directly shows the quadratic relationship between voltage and power.

Question 6

Two copper wires have the same length but different diameters. Wire X has diameter $d$ and wire Y has diameter $2d$ . If wire X has a resistance of $16 \, \Omega$ , what is the resistance of wire Y?

$2 \, \Omega$
$4 \, \Omega$ (correct answer)
$8 \, \Omega$
$32 \, \Omega$
$64 \, \Omega$

Explanation: When you encounter wire resistance problems, remember that resistance depends on the material's resistivity, length, and cross-sectional area according to

R = \rho \frac{L}{A}

, where

\rho

is resistivity,

L

is length, and

A

is cross-sectional area. Since both wires are copper with the same length, only the cross-sectional area differs. The area of a circular wire is

A = \pi r^2 = \pi (d/2)^2 = \frac{\pi d^2}{4}

. Wire X has area

A_X = \frac{\pi d^2}{4}

, while wire Y has area

A_Y = \frac{\pi (2d)^2}{4} = \frac{\pi \cdot 4d^2}{4} = \pi d^2

. Therefore, wire Y has four times the cross-sectional area of wire X. Since resistance is inversely proportional to area, wire Y has one-fourth the resistance of wire X:

R_Y = \frac{R_X}{4} = \frac{16\,\Omega}{4} = 4\,\Omega

. This confirms answer (B) is correct. Looking at the wrong answers: (A)

2\,\Omega

suggests the resistance decreases by a factor of 8, which would require the diameter to increase by

2\sqrt{2}

, not 2. (C)

8\,\Omega

implies the resistance only halves, corresponding to doubling the area rather than quadrupling it. (D)

32\,\Omega

incorrectly suggests resistance increases with diameter, showing a fundamental misunderstanding of the inverse relationship. Study tip: Remember that doubling the diameter quadruples the area (since area depends on

d^2

), which quarters the resistance. Watch for this

d^2

relationship in wire problems.

Question 7

A variable resistor is adjusted so that when $15 \, V$ is applied across it, the current is $0.75 \, A$ . If the resistor is then adjusted to triple its resistance value, what current will flow when the same $15 \, V$ is applied?

$0.25 \, A$ (correct answer)
$0.50 \, A$
$0.75 \, A$
$1.5 \, A$
$2.25 \, A$

Explanation: When you encounter problems involving variable resistors and changing resistance values, you're working with Ohm's Law:

V = IR

, where voltage, current, and resistance are directly related. First, let's find the initial resistance. With

V = 15 \, V

and

I = 0.75 \, A

, we can solve for resistance:

R = \frac{V}{I} = \frac{15}{0.75} = 20 \, \Omega

. When the resistor is adjusted to triple its resistance, the new resistance becomes

R_{new} = 3 \times 20 = 60 \, \Omega

. Since the same voltage (

15 \, V

) is applied, we can find the new current:

I_{new} = \frac{V}{R_{new}} = \frac{15}{60} = 0.25 \, A

. Looking at the answer choices: Choice A (

0.25 \, A

) is correct—this is exactly what we calculated. Choice B (

0.50 \, A

) represents doubling the resistance instead of tripling it, a common calculation error. Choice C (

0.75 \, A

) is the original current, which would only be correct if resistance remained unchanged. Choice D (

1.5 \, A

) incorrectly assumes that tripling resistance triples current, showing a fundamental misunderstanding of Ohm's Law. Remember the key relationship: current and resistance are inversely proportional when voltage is constant. If resistance increases by a factor of 3, current decreases by the same factor. This inverse relationship is crucial for solving circuit problems efficiently—when one goes up, the other must go down proportionally.

Question 8

A $9.0 \, V$ battery is connected to a circuit containing a $12 \, \Omega$ resistor and an ideal ammeter in series. The ammeter reads $0.60 \, A$ . What is the internal resistance of the battery?

$0 \, \Omega$
$3.0 \, \Omega$ (correct answer)
$6.0 \, \Omega$
$12 \, \Omega$
$15 \, \Omega$

Explanation: When you encounter a circuit problem involving a battery with internal resistance, remember that real batteries don't provide their full emf to the external circuit. The battery's internal resistance acts like an additional resistor in series with the external components. To solve this, apply Ohm's law to the complete circuit. The battery's emf (

9.0 \, V

) must equal the voltage drop across the total resistance. The total resistance includes both the external

12 \, \Omega

resistor and the unknown internal resistance

r

. Using Ohm's law:

V_{emf} = I \times R_{total}

9.0 \, V = 0.60 \, A \times (12 \, \Omega + r)

Solving for the total resistance:

R_{total} = \frac{9.0 \, V}{0.60 \, A} = 15 \, \Omega

Since

R_{total} = R_{external} + r

15 \, \Omega = 12 \, \Omega + r

Therefore:

r = 3.0 \, \Omega

Choice A (

0 \, \Omega

) assumes an ideal battery with no internal resistance, which would give a current of

\frac{9.0}{12} = 0.75 \, A

. Choice C (

6.0 \, \Omega

) would result from incorrectly calculating the total resistance as

18 \, \Omega

. Choice D (

12 \, \Omega

) might come from confusing the internal resistance with the external resistance. Study tip: Always remember that in real battery problems, the emf equals the current times the sum of ALL resistances in the circuit, including internal resistance. This is a fundamental concept that appears frequently in circuit analysis.

Question 9

Two identical light bulbs are connected in series to a $12 \, V$ battery. Each bulb is rated at $6 \, V$ , $3 \, W$ . What is the actual power dissipated by each bulb in this series configuration?

$1.5 \, W$
$3.0 \, W$ (correct answer)
$4.5 \, W$
$6.0 \, W$
$12 \, W$

Explanation: When analyzing circuits with bulbs in series, you need to understand how voltage divides and how this affects power consumption compared to the bulbs' ratings. First, find each bulb's resistance using its rating:

R = \frac{V^2}{P} = \frac{6^2}{3} = 12 \, \Omega

. In the series circuit, the total resistance is

24 \, \Omega

, so the current is

I = \frac{12 \, V}{24 \, \Omega} = 0.5 \, A

. Since the bulbs are identical and in series, they split the

12 \, V

equally—each gets

6 \, V

. The actual power dissipated by each bulb is

P = I^2R = (0.5)^2 \times 12 = 3 \, W

, which matches their rating because they're receiving their rated voltage. Choice A (

1.5 \, W

) would be incorrect if you mistakenly thought the bulbs get less than their rated voltage and calculated power as half the rating. Choice C (

4.5 \, W

) might result from incorrectly assuming the bulbs somehow get more power than rated. Choice D (

6 \, W

) would be wrong if you thought each bulb receives the full battery voltage, leading to

P = \frac{V^2}{R} = \frac{12^2}{12} = 12 \, W

per bulb—but this ignores that voltage divides in series. The correct answer is B:

3.0 \, W

. Study tip: In series circuits, always check if identical components receive their rated voltage after accounting for voltage division. When they do, they operate at their rated power. This is why two

6 \, V

bulbs work perfectly with a

12 \, V

source in series.

Question 10

A resistor dissipates $36 \, W$ when connected to a $12 \, V$ source. If the same resistor is connected to a $6 \, V$ source, what power does it dissipate?

$4.5 \, W$
$9.0 \, W$ (correct answer)
$18 \, W$
$24 \, W$
$72 \, W$

Explanation: When you encounter power dissipation problems with resistors, remember that power depends on both voltage and resistance, and there are multiple equivalent formulas to use depending on what information you have. First, let's find the resistance of this resistor. Using

P = \frac{V^2}{R}

, we can solve for

R

36 = \frac{12^2}{R} = \frac{144}{R}

R = \frac{144}{36} = 4 \, \Omega

Now that we know the resistance is

4 \, \Omega

, we can find the power dissipated when connected to a

6 \, V

source:

P = \frac{V^2}{R} = \frac{6^2}{4} = \frac{36}{4} = 9 \, W

Looking at the wrong answers: Choice A (

4.5 \, W

) comes from incorrectly assuming power scales linearly with voltage—dividing the original power by 2 since voltage was halved. Choice C (

18 \, W

) results from mistakenly thinking power is directly proportional to voltage rather than voltage squared. Choice D (

24 \, W

) likely comes from calculation errors or misapplying formulas. The key insight is that power is proportional to voltage squared (

P \propto V^2

) when resistance is constant. Since voltage was halved from

12 \, V

6 \, V

, the power becomes

(1/2)^2 = 1/4

of the original:

36 \times \frac{1}{4} = 9 \, W

. Study tip: Remember the three power formulas—

P = VI

P = I^2R

, and

P = \frac{V^2}{R}

—and always identify which variables remain constant in the problem to choose the most efficient approach.

Question 11

A carbon resistor has a resistance of $1000 \, \Omega$ at $20°C$ . If the temperature coefficient of resistance for carbon is $-5.0 \times 10^{-4} \, /°C$ , what is the resistance at $-30°C$ ?

$975 \, \Omega$
$1000 \, \Omega$
$1025 \, \Omega$ (correct answer)
$1050 \, \Omega$
$1075 \, \Omega$

Explanation: When you encounter temperature coefficient problems, you're dealing with how material properties change with temperature. The key relationship is:

R = R_0[1 + \alpha(T - T_0)]

, where

R_0

is the initial resistance,

\alpha

is the temperature coefficient, and

(T - T_0)

is the temperature change. Here, you start with

R_0 = 1000 \, \Omega

T_0 = 20°C

, and need to find the resistance at

T = -30°C

. The temperature change is

\Delta T = -30°C - 20°C = -50°C

. Notice that carbon has a negative temperature coefficient (

\alpha = -5.0 \times 10^{-4} \, /°C

), meaning its resistance increases as temperature decreases. Substituting into the formula:

R = 1000[1 + (-5.0 \times 10^{-4})(-50)] = 1000[1 + 0.025] = 1000 \times 1.025 = 1025 \, \Omega

Looking at the wrong answers: Choice A (

975 \, \Omega

) represents what you'd get if you forgot that carbon's temperature coefficient is negative—treating it as positive would decrease resistance with decreasing temperature. Choice B (

1000 \, \Omega

) ignores the temperature effect entirely. Choice D (

1050 \, \Omega

) likely comes from calculation errors, possibly doubling the temperature coefficient effect. Remember that negative temperature coefficients (common in semiconductors and carbon) mean resistance increases as temperature decreases. Always pay attention to the sign of both the temperature coefficient and the temperature change—getting either wrong will lead you to an incorrect answer.

Question 12

A wire has a resistance of $R$ when its length is $L$ and cross-sectional area is $A$ . If the wire is stretched uniformly until its length becomes $3L$ while maintaining the same volume, what is the new resistance?

$R/9$
$R/3$
$3R$
$9R$ (correct answer)
$27R$

Explanation: This question tests your understanding of how electrical resistance changes when a conductor's dimensions are altered while preserving volume. The key insight is recognizing that stretching affects both length and cross-sectional area, and resistance depends on both factors. Resistance follows the formula

R = \rho \frac{L}{A}

, where

\rho

is resistivity (a material property that stays constant),

L

is length, and

A

is cross-sectional area. When the wire stretches from length

L

3L

, you must determine how the cross-sectional area changes. Since volume remains constant:

V = L \times A = 3L \times A_{new}

. Solving for the new area:

A_{new} = \frac{A}{3}

. The wire becomes three times longer but one-third as thick. Now calculate the new resistance:

R_{new} = \rho \frac{3L}{A/3} = \rho \frac{3L \times 3}{A} = 9 \times \rho \frac{L}{A} = 9R

. The answer is D)

9R

. Looking at the wrong answers: A)

R/9

incorrectly assumes resistance decreases, ignoring that length increases more significantly than area. B)

R/3

might result from only considering the area decrease while forgetting the length increase. C)

3R

represents a common error where students only account for the length tripling but forget that area also changes. Remember this pattern: when a wire stretches while maintaining constant volume, length increases by factor

n

and area decreases by factor

n

, so resistance increases by

n^2

. Always consider both geometric changes simultaneously.

Question 13

A cylindrical wire made of aluminum has a diameter of $2.0 \, mm$ and a length of $5.0 \, m$ . If the resistivity of aluminum is $2.8 \times 10^{-8} \, \Omega \cdot m$ , what is the resistance of the wire?

$0.045 \, \Omega$ (correct answer)
$0.089 \, \Omega$
$0.18 \, \Omega$
$0.35 \, \Omega$
$0.71 \, \Omega$

Explanation: This question tests your understanding of electrical resistance in conductors, specifically how the physical properties of a wire determine its resistance. When you encounter resistance problems, always think about the fundamental relationship: resistance depends on the material's resistivity and the wire's geometry. To find the resistance, you'll use the formula

R = \rho \frac{L}{A}

, where

\rho

is resistivity,

L

is length, and

A

is cross-sectional area. First, calculate the cross-sectional area of the cylindrical wire. With a diameter of

2.0 \, mm

, the radius is

1.0 \, mm = 1.0 \times 10^{-3} \, m

. The area is

A = \pi r^2 = \pi (1.0 \times 10^{-3})^2 = 3.14 \times 10^{-6} \, m^2

. Now substitute into the resistance formula:

R = (2.8 \times 10^{-8}) \frac{5.0}{3.14 \times 10^{-6}} = 0.045 \, \Omega

. This confirms answer A is correct. Answer B (

0.089 \, \Omega

) likely results from using the diameter instead of radius when calculating area, giving an area four times too small. Answer C (

0.18 \, \Omega

) might come from using the circumference rather than area in the denominator. Answer D (

0.35 \, \Omega

) could result from multiple unit conversion errors or incorrectly applying the resistance formula. Remember: resistance problems always require careful attention to geometry. Convert all units to standard SI units first, and double-check that you're using radius (not diameter) for circular cross-sections. The area calculation is usually where errors occur.

Question 14

A wire carrying a current of $2.5 \, A$ has a voltage drop of $7.5 \, V$ across a $1.5 \, m$ length. If the wire has a circular cross-section with radius $0.5 \, mm$ , what is the resistivity of the wire material?

$1.6 \times 10^{-6} \, \Omega \cdot m$ (correct answer)
$2.4 \times 10^{-6} \, \Omega \cdot m$
$3.1 \times 10^{-6} \, \Omega \cdot m$
$4.7 \times 10^{-6} \, \Omega \cdot m$
$6.3 \times 10^{-6} \, \Omega \cdot m$

Explanation: When you encounter a problem involving current, voltage, and wire dimensions, you're dealing with electrical resistance and resistivity. Resistivity is a material property that tells you how strongly a substance opposes electric current flow, regardless of the object's shape or size. To find resistivity, you need to combine Ohm's law with the resistance formula for a wire. First, use Ohm's law (

V = IR

) to find the resistance:

R = V/I = 7.5 \, V / 2.5 \, A = 3.0 \, \Omega

. Next, apply the wire resistance formula:

R = \rho L/A

, where

\rho

is resistivity,

L

is length, and

A

is cross-sectional area. For a circular wire,

A = \pi r^2 = \pi (0.5 \times 10^{-3})^2 = 7.85 \times 10^{-7} \, m^2

. Solving for resistivity:

\rho = RA/L = (3.0)(7.85 \times 10^{-7})/1.5 = 1.57 \times 10^{-6} \, \Omega \cdot m

, which rounds to

1.6 \times 10^{-6} \, \Omega \cdot m

. Choice A is correct. Choice B (

2.4 \times 10^{-6}

) likely results from using diameter instead of radius when calculating area. Choice C (

3.1 \times 10^{-6}

) suggests forgetting to divide by length or making an area calculation error. Choice D (

4.7 \times 10^{-6}

) combines multiple errors, possibly using diameter and incorrect unit conversions. Remember: always convert units to meters before calculating, and use radius (not diameter) for circular cross-sections. The two-step approach—find resistance first, then resistivity—prevents calculation errors.

Question 15

A certain material has a resistivity that increases linearly with temperature according to $\rho(T) = \rho_0(1 + \alpha T)$ , where $\rho_0 = 1.5 \times 10^{-8} \, \Omega \cdot m$ at $T = 0°C$ and $\alpha = 4.0 \times 10^{-3} \, /°C$ . What is the resistivity at $50°C$ ?

$1.2 \times 10^{-8} \, \Omega \cdot m$
$1.5 \times 10^{-8} \, \Omega \cdot m$
$1.8 \times 10^{-8} \, \Omega \cdot m$ (correct answer)
$2.1 \times 10^{-8} \, \Omega \cdot m$
$4.5 \times 10^{-8} \, \Omega \cdot m$

Explanation: When you encounter a resistivity temperature problem, you're dealing with the linear relationship between a material's electrical resistance properties and temperature. The key is recognizing that most conductors have resistivity that increases with temperature due to increased atomic vibrations that impede electron flow. Given the formula

\rho(T) = \rho_0(1 + \alpha T)

, you can directly substitute the values. At

T = 50°C

\rho(50) = 1.5 \times 10^{-8}(1 + 4.0 \times 10^{-3} \times 50)

\rho(50) = 1.5 \times 10^{-8}(1 + 0.2)

\rho(50) = 1.5 \times 10^{-8} \times 1.2 = 1.8 \times 10^{-8} \, \Omega \cdot m

This confirms answer C is correct. Looking at the wrong answers: A)

1.2 \times 10^{-8}

represents a decrease in resistivity, which would occur if you mistakenly subtracted the temperature term instead of adding it. B)

1.5 \times 10^{-8}

is simply

\rho_0

, suggesting you forgot to account for the temperature increase entirely. D)

2.1 \times 10^{-8}

is too high, likely from calculation errors like using

\alpha T = 0.4

instead of

0.2

. Remember: resistivity problems with linear temperature dependence are straightforward substitution exercises. Always check that your final answer shows an increase for typical conductors when temperature rises, and double-check your arithmetic with the temperature coefficient calculation.

Question 16

A cylindrical wire has resistance $R$ when its length is $L$ and cross-sectional area is $A$ . If the wire is stretched to twice its original length while keeping the total volume constant, what is the new resistance in terms of $R$ ?

$2R$
$4R$ (correct answer)
$R/2$
$R/4$

Explanation: When the wire is stretched to twice its length while keeping volume constant, the new length becomes

2L

and the new cross-sectional area becomes

A/2

(since volume

= L \cdot A

must remain constant). Using

R = \rho L/A

, the new resistance is

R_{new} = \rho(2L)/(A/2) = \rho \cdot 2L \cdot 2/A = 4\rho L/A = 4R

. Choice A incorrectly accounts for only the length change. Choice C assumes the area increases rather than decreases. Choice D incorrectly inverts the relationship.

Question 17

Two identical resistors are connected in parallel across a battery. If one of the resistors suddenly develops an internal break (becomes an open circuit), what happens to the current through the remaining resistor?

The current doubles because the total resistance increases
The current doubles because all current must now flow through one path (correct answer)
The current remains the same because the voltage across each resistor is unchanged
The current is halved because only one resistor remains in the circuit

Explanation: Initially, each resistor carries current

I = V/R

where

V

is the battery voltage. When one resistor opens, the total current that was previously split between two parallel paths (

2V/R

) must now flow entirely through the remaining resistor, so the current through it becomes

2V/R = 2I

, doubling. Choice A has the wrong reasoning about resistance. Choice C ignores that the current distribution changes. Choice D incorrectly applies series circuit reasoning to a parallel situation.

Question 18

A variable resistor is adjusted so that its resistance varies according to $R(t) = R_0(1 + \alpha t)$ where $\alpha > 0$ is a constant. If this resistor is connected to a constant voltage source $V$ , how does the power dissipated in the resistor change with time?

Power decreases as $P(t) = \frac{V^2}{R_0(1 + \alpha t)}$ due to inverse relationship with resistance (correct answer)
Power increases linearly with time since resistance increases linearly with time
Power remains constant since the voltage source maintains constant energy delivery
Power increases exponentially since both voltage and resistance effects compound over time

Explanation: When analyzing circuits with time-varying components, you need to apply basic circuit laws while carefully tracking how each quantity changes with time. Since power dissipated in a resistor is given by

P = \frac{V^2}{R}

, and the voltage source is constant while resistance varies as

R(t) = R_0(1 + \alpha t)

, you can substitute directly:

P(t) = \frac{V^2}{R_0(1 + \alpha t)}

. As time increases and

\alpha > 0

, the denominator grows larger, making the power smaller. This inverse relationship means power decreases over time, confirming answer A. Option B incorrectly assumes a direct relationship between resistance and power. While resistance does increase linearly, power depends on the inverse of resistance, so it actually decreases. Option C falls into the trap of confusing constant voltage with constant power. A constant voltage source maintains fixed voltage across its terminals, but the power it delivers depends on the circuit's resistance—as resistance changes, so does power. Option D incorrectly suggests exponential behavior and misunderstands how voltage and resistance interact. The voltage remains constant, and there's no compounding effect that would create exponential growth. Study tip: Remember that

P = \frac{V^2}{R}

shows power is inversely proportional to resistance when voltage is constant. When you see time-varying circuit elements, substitute the time-dependent expressions into the basic power formulas and analyze the mathematical relationship—don't rely on intuition about what "should" happen to power.

Question 19

A student measures the voltage across a resistor as 12.0 V and the current through it as 3.00 A. The student then connects a second identical resistor in parallel with the first. If the voltage source maintains constant voltage, what will be the new current reading through the original resistor?

1.50 A because the current is split equally between the parallel resistors
12.0 A because the resistance is halved while voltage remains constant
6.00 A because the total circuit current doubles and flows through the original path
3.00 A because the voltage across the original resistor remains unchanged (correct answer)

Explanation: When analyzing parallel resistor circuits, the key insight is that parallel branches maintain the same voltage across each component, regardless of how many branches you add. Let's work through this systematically. Initially, you have one resistor with 12.0 V across it and 3.00 A through it. Using Ohm's law, this resistor has a resistance of

R = V/I = 12.0\text{ V}/3.00\text{ A} = 4.00\text{ Ω}

. When you add an identical resistor in parallel, each resistor still experiences the full 12.0 V from the voltage source. Since the original resistor's resistance (4.00 Ω) and the voltage across it (12.0 V) both remain unchanged, the current through it must also remain 3.00 A according to Ohm's law. Now let's examine why the other answers miss the mark. Choice A incorrectly assumes the original current gets divided between the resistors, but this confuses series behavior with parallel behavior. Choice B correctly notes that total circuit resistance halves, but incorrectly applies this to find current through one resistor rather than total current from the source. Choice C recognizes that total circuit current doubles (which it does, from 3.00 A to 6.00 A), but wrongly suggests this affects the current through the original resistor. Study tip: Remember that in parallel circuits, voltage is the same across each branch, while currents add up. Each parallel branch operates independently - adding more branches doesn't change the current through existing branches, only the total current drawn from the source.

Question 20

A heating element is designed to operate at 1200 W when connected to 120 V. Due to thermal expansion, its resistance increases by 15% when hot. What is the actual power consumption when the element reaches its operating temperature?

1035 W because power decreases proportionally with increased resistance
1380 W because both resistance and temperature effects increase the power consumption
1200 W because the heating element is designed to maintain constant power output
1044 W because the resistance increase reduces current flow through the element (correct answer)

Explanation: When analyzing heating elements and resistance changes, you need to understand how power, voltage, current, and resistance interact through Ohm's law and the power equation. Start by finding the cold resistance using

P = \frac{V^2}{R}

. With 1200 W at 120 V:

R_{cold} = \frac{V^2}{P} = \frac{(120)^2}{1200} = 12 \text{ ohms}

When the element heats up, its resistance increases by 15%:

R_{hot} = 12 \times 1.15 = 13.8 \text{ ohms}

The actual power at operating temperature becomes:

P_{actual} = \frac{V^2}{R_{hot}} = \frac{(120)^2}{13.8} = 1044 \text{ W}

Option A incorrectly suggests a proportional decrease (15% resistance increase = 15% power decrease), but the relationship between resistance and power is inverse, not linear. Option B makes the fundamental error of thinking resistance increases boost power consumption—this ignores that higher resistance actually restricts current flow. Option C assumes the heating element somehow maintains constant power output, but real heating elements are passive resistors that follow Ohm's law; they don't actively regulate their power consumption. Option D correctly recognizes that increased resistance reduces current flow (

I = \frac{V}{R}

), which in turn reduces power consumption since

P = VI

. Study tip: For heating element problems, always remember that resistance typically increases with temperature for most materials, and higher resistance means lower power consumption at constant voltage. Calculate the cold resistance first, then apply the temperature effect.