AP Physics C: Electricity and Magnetism Practice Quiz: Resistance, Resistivity, and Ohm's Law

Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026

Test your understanding with short quizzes. This quiz has 14 questions to check your progress.

Question 1 of 14

A cylindrical copper wire has a resistance R. If the wire is replaced by another copper wire with twice the length and twice the diameter, what is the new resistance?

A)R/2B)RC)2RD)4R

All Questions (14)

A cylindrical copper wire has a resistance R. If the wire is replaced by another copper wire with twice the length and twice the diameter, what is the new resistance?

A) R/2

B) R

C) 2R

D) 4R

Correct Answer: A

Resistance is given by the formula R = ρl/A, where A is the cross-sectional area. The area of a cylinder is A = πr² = π(d/2)² = (πd²)/4. Thus, resistance is proportional to l/d². The new length is l' = 2l and the new diameter is d' = 2d. The new resistance R' is proportional to l'/(d')² = (2l)/(2d)² = 2l/4d² = (1/2)(l/d²). Therefore, the new resistance is R/2.

Two wires, X and Y, are made of different materials and have the same length and cross-sectional area. The resistance of wire X is greater than the resistance of wire Y. Which of the following statements about the resistivity of the materials is correct?

A) The resistivity of material X is greater than the resistivity of material Y.

B) The resistivity of material Y is greater than the resistivity of material X.

C) The resistivities are equal.

D) The relationship cannot be determined without knowing the potential difference across the wires.

Correct Answer: A

The formula for resistance is R = ρl/A. Since the length (l) and cross-sectional area (A) are the same for both wires, the resistance (R) is directly proportional to the resistivity (ρ). Because wire X has a greater resistance than wire Y, the material of wire X must have a greater resistivity.

A potential difference of 12 V is applied across a resistor with a resistance of 4.0 Ω. What is the current flowing through the resistor?

A) 0.33 A

B) 3.0 A

C) 16 A

D) 48 A

Correct Answer: B

According to Ohm's law, current (I) is related to potential difference (ΔV) and resistance (R) by the equation I = ΔV/R. Substituting the given values: I = 12 V / 4.0 Ω = 3.0 A.

A conductive element in a circuit obeys Ohm's law. If the potential difference across the element is doubled, what happens to the current flowing through it?

A) The current is quartered.

B) The current is halved.

C) The current remains the same.

D) The current is doubled.

Correct Answer: D

Ohm's law states that I = ΔV/R. For an element that obeys Ohm's law, the resistance R is constant. Therefore, the current I is directly proportional to the potential difference ΔV. If the potential difference is doubled, the current will also be doubled.

A wire of length L and cross-sectional area A has a resistance R. This wire is connected to a battery with a constant potential difference ΔV, resulting in a current I. The wire is then stretched to a new length of 2L while its volume remains constant. What is the new current flowing through the wire?

A) I/4

B) I/2

C) I

D) 4I

Correct Answer: A

The volume of the wire is V = A * L. If the volume is constant and the length is doubled (L' = 2L), the new area must be halved (A' = A/2). The new resistance is R' = ρL'/A' = ρ(2L)/(A/2) = 4(ρL/A) = 4R. According to Ohm's law, the new current is I' = ΔV/R' = ΔV/(4R) = (1/4)I.

A resistor is to be made from a wire of a specific material. Which of the following combinations of length (l) and cross-sectional area (A) will result in the greatest resistance?

A) l = L, A = A₀

B) l = 2L, A = 2A₀

C) l = L/2, A = A₀/2

D) l = 2L, A = A₀/2

Correct Answer: D

Resistance is calculated using the formula R = ρl/A. To maximize R, we need to maximize the length (l) and minimize the cross-sectional area (A), which means maximizing the ratio l/A. For option D, the ratio is 2L / (A₀/2) = 4L/A₀, which is the largest ratio among the choices.

The electrical resistance of a metallic conductor is primarily dependent on which of its physical properties?

A) Its resistivity, length, and cross-sectional area.

B) Its mass and density.

C) The potential difference applied across it.

D) The current flowing through it.

Correct Answer: A

The provided content states that the resistance of a resistor is proportional to its resistivity (ρ) and length (l) and is inversely proportional to its cross-sectional area (A), as given by the equation R = ρl/A. These are the intrinsic physical properties that determine resistance.

A cylindrical resistor of length L and uniform cross-sectional area A is made of a material whose resistivity ρ(l) increases linearly with the distance l from one end, such that ρ(l) = kl where k is a constant. Which expression correctly represents the total resistance of this resistor?

A) R = kL/A

B) R = kL²/A

C) R = kL²/(2A)

D) R = k/A

Correct Answer: C

For a material with varying resistivity, the total resistance is found by integrating along its length: R = ∫(ρ(l)dl)/A. In this case, ρ(l) = kl. We integrate from l=0 to l=L: R = ∫[0 to L] (kl/A) dl = (k/A) ∫[0 to L] l dl = (k/A) [l²/2] from 0 to L = (k/A) (L²/2 - 0) = kL²/(2A).

The graph shows the current (I) as a function of potential difference (ΔV) for a circuit element. The line passes through the origin and the point (10 V, 2 A). What is the resistance of the element?

A) 0.2 Ω

B) 2.0 Ω

C) 5.0 Ω

D) 20 Ω

Correct Answer: C

Ohm's law is I = ΔV/R, which can be rearranged to R = ΔV/I. The resistance is the inverse of the slope of an I vs. ΔV graph. Using the point from the graph (ΔV = 10 V, I = 2 A), we can calculate the resistance: R = 10 V / 2 A = 5.0 Ω.

Resistor 1 is made of a material with resistivity ρ₁. It has a length L₁ and a radius r₁. Resistor 2 is made of a material with resistivity ρ₂ = 2ρ₁. It has a length L₂ = 2L₁ and a radius r₂ = 2r₁. What is the ratio of the resistance of Resistor 2 to Resistor 1 (R₂/R₁)?

A) 1/2

B) 1

C) 2

D) 4

Correct Answer: B

The resistance is given by R = ρl/A = ρl/(πr²). For Resistor 1: R₁ = ρ₁L₁/(πr₁²). For Resistor 2: R₂ = ρ₂L₂/(πr₂²) = (2ρ₁)(2L₁)/(π(2r₁)²) = 4ρ₁L₁/(4πr₁²) = ρ₁L₁/(πr₁²). Therefore, R₂ = R₁, and the ratio R₂/R₁ is 1.

The equation R = ∫(ρ(l)dl)/A is used to calculate the resistance of a non-uniform resistor. What does the term (ρ(l)dl)/A represent in this integral?

A) The total resistance of the object.

B) The resistance of an infinitesimally thin slice of the object at position l.

C) The average resistivity of the object.

D) The current density at position l.

Correct Answer: B

The integral sums up the contributions to the total resistance from all parts of the resistor. The term inside the integral, dR = (ρ(l)dl)/A, represents the differential resistance of an infinitesimally thin slice of the resistor of length dl at a position l, where the resistivity is ρ(l) and the cross-sectional area is A.

Two resistors, A and B, are connected in separate circuits to identical batteries. The current flowing through resistor A is twice the current flowing through resistor B. What can be concluded about their resistances?

A) The resistance of A is twice the resistance of B (R_A = 2R_B).

B) The resistance of B is twice the resistance of A (R_B = 2R_A).

C) The resistances are equal (R_A = R_B).

D) The resistance of A is four times the resistance of B (R_A = 4R_B).

Correct Answer: B

According to Ohm's law, R = ΔV/I. Since the batteries are identical, the potential difference ΔV is the same for both circuits. We are given that I_A = 2I_B. Therefore, R_A = ΔV/I_A and R_B = ΔV/I_B. Substituting I_A = 2I_B into the equation for R_A gives R_A = ΔV/(2I_B) = (1/2)(ΔV/I_B) = (1/2)R_B. This means the resistance of B is twice the resistance of A (R_B = 2R_A).

Two rods, 1 and 2, have the same length L and cross-sectional area A. Rod 1 has a uniform resistivity ρ₀. Rod 2 has a resistivity that varies linearly from 0 at one end (l=0) to 2ρ₀ at the other end (l=L), so ρ(l) = (2ρ₀/L)l. How does the resistance of Rod 2 (R₂) compare to the resistance of Rod 1 (R₁)?

A) R₂ < R₁

B) R₂ = R₁

C) R₂ > R₁

D) The relationship depends on the value of L and A.

Correct Answer: B

First, find the resistance of Rod 1: R₁ = ρ₀L/A. Next, find the resistance of Rod 2 using the integral formula: R₂ = ∫[0 to L] (ρ(l)dl)/A = ∫[0 to L] ((2ρ₀/L)l)/A dl = (2ρ₀/AL) ∫[0 to L] l dl = (2ρ₀/AL) [l²/2] from 0 to L = (2ρ₀/AL) (L²/2) = ρ₀L/A. Therefore, R₂ = R₁.

A rectangular block of a conductive material has dimensions L, 2L, and 3L. A potential difference ΔV is applied across the block. To achieve the maximum possible current, between which two faces should the potential difference be applied?

A) The faces with area L x 2L.

B) The faces with area L x 3L.

C) The faces with area 2L x 3L.

D) The current is the same regardless of the faces chosen.

Correct Answer: C

To maximize current (I = ΔV/R), resistance (R) must be minimized. Resistance is R = ρl/A. To minimize R, we must minimize the length (l) the current travels and maximize the cross-sectional area (A). Applying the potential difference across the faces with area 2L x 3L means the current travels the shortest distance, l = L, through the largest area, A = 6L². This combination gives the smallest resistance (R = ρL/6L² = ρ/6L) and therefore the maximum current.