Resistance, Resistivity, and Ohm's Law - AP Physics C Electricity and Magnetism

A potential difference $V$ is applied across an ohmic resistor, resulting in a current $I$ . If the potential difference is doubled to $2V$ while the temperature of the resistor remains constant, what is the new current?
Answer: $2I$ For an ohmic resistor, the resistance $R$ is constant. According to Ohm's Law, $I = V/R$ . The new current is $I_{\text{new}} = V_{\text{new}}/R = (2V)/R = 2(V/R) = 2I$ . The current is directly proportional to the potential difference.
A potential difference of $1.0\,\text{V}$ is applied across a cylindrical wire of length $2.0\,\text{m}$ and radius $0.50\,\text{mm}$ . A current of $4.0\,\text{A}$ is measured. What is the resistivity of the wire's material?
Answer: $9.8 \times 10^{-8}\,\Omega \cdot \text{m}$ First, calculate the resistance using Ohm's Law: $R = V/I = 1.0\,\text{V} / 4.0\,\text{A} = 0.25\,\Omega$ . Next, calculate the cross-sectional area: $A = \pi r^2 = \pi (0.50 \times 10^{-3}\,\text{m})^2 \approx 7.85 \times 10^{-7}\,\text{m}^2$ . Finally, use the resistance formula to find resistivity: $\rho = RA/L = (0.25\,\Omega)(7.85 \times 10^{-7}\,\text{m}^2) / (2.0\,\text{m}) \approx 9.8 \times 10^{-8}\,\Omega \cdot \text{m}$ .
A cylindrical conductor has length $L$ , radius $r$ , and resistance $R$ . A second conductor made of the same material has length $L/2$ and radius $2r$ . What is the resistance of the second conductor?
Answer: $R/8$ The resistance $R$ is given by $R = \rho L/A = \rho L/(\pi r^2)$ . The new resistance $R_{\text{new}}$ is for a wire with length $L_{\text{new}} = L/2$ and radius $r_{\text{new}} = 2r$ . So, $R_{\text{new}} = \rho (L/2) / (\pi (2r)^2) = \rho (L/2) / (4\pi r^2) = (1/8) (\rho L / (\pi r^2)) = R/8$ .
Considering the information provided, a lightbulb in a household circuit obeys $V=IR$ ; resistance $R$ (Ω) controls current, and resistivity $\rho$ (in $\Omega\cdot\text{m}$ ) is intrinsic to the filament material. What happens to current if resistance increases and voltage remains constant according to Ohm's Law ( $V=IR$ )?
Answer: Current decreases as resistance increases This question tests understanding of resistance, resistivity, and Ohm's Law in AP Physics C: Electricity and Magnetism. Ohm's Law V = IR shows that for constant voltage, current and resistance have an inverse relationship, expressed as I = V/R. In the lightbulb circuit example, when the filament's resistance increases while the applied voltage remains constant, the current through the bulb must decrease to satisfy Ohm's Law. Choice B is correct because it accurately states that current decreases as resistance increases when voltage is constant, following directly from the mathematical relationship I = V/R. Choice A is incorrect as it suggests a direct relationship between current and resistance, which would violate Ohm's Law and conservation principles. Students should understand that increased resistance means more opposition to current flow, like a dimmer switch increasing resistance to reduce bulb brightness. Practice with real circuits helps: a 60W bulb has lower resistance than a 40W bulb, so it draws more current at the same voltage.
The filament of an incandescent light bulb is made of tungsten. As current passes through it, its temperature increases significantly. How does this temperature increase affect the filament's resistance, and why?
Answer: The resistance increases because the increased thermal vibrations of the tungsten atoms impede the flow of electrons. Tungsten is a metallic conductor. For such materials, resistivity increases with temperature. The increased temperature causes the metal's lattice ions to vibrate with greater amplitude, which increases the frequency of collisions with conduction electrons, thus impeding their flow and increasing resistance.
Two cylindrical wires, X and Y, are made of different materials and have different dimensions. The resistance of Wire X is greater than the resistance of Wire Y. Which of the following statements provides a sufficient condition for the resistivity of material X to be greater than that of material Y?
Answer: The ratio of length to area ( $L/A$ ) for wire X is less than or equal to that of wire Y. We are given $R_X > R_Y$ , which means $(\rho_X L_X / A_X) > (\rho_Y L_Y / A_Y)$ . We want to find a condition that guarantees $\rho_X > \rho_Y$ . If we know that $(L_X / A_X) \le (L_Y / A_Y)$ , then for the inequality $R_X > R_Y$ to hold, it must be true that $\rho_X > \rho_Y$ . The other options do not provide sufficient information.
Two wires of identical length and diameter are connected to identical batteries. Wire 1 is made of copper ( $\rho_C \approx 1.7 \times 10^{-8}\,\Omega \cdot \text{m}$ ) and Wire 2 is made of nichrome ( $\rho_N \approx 1.1 \times 10^{-6}\,\Omega \cdot \text{m}$ ). How does the current $I_C$ in the copper wire compare to the current $I_N$ in the nichrome wire?
Answer: $I_C$ is much greater than $I_N$ . Since the wires have identical dimensions, their resistances are directly proportional to their resistivities ( $R = \rho L/A$ ). Copper has a much lower resistivity than nichrome, so the copper wire has a much lower resistance. With identical batteries (same voltage $V$ ), the current ( $I=V/R$ ) will be much larger in the wire with lower resistance, so $I_C$ is much greater than $I_N$ .
A material is classified as non-ohmic. What does this classification imply about the material's resistance?
Answer: The resistance varies as the potential difference across it or the current through it changes. A non-ohmic material is one that does not follow Ohm's Law, which states that the ratio of voltage to current (the resistance) is constant. Therefore, for a non-ohmic material, the resistance changes depending on the operating conditions like voltage or current.
The resistance $R$ of a uniform cylindrical conductor is given by $R = \rho L/A$ . What are the SI units for resistivity, $\rho$ ?
Answer: $\text{ohm} \cdot \text{meter}$ Rearranging the formula for resistivity gives $\rho = RA/L$ . Substituting the SI units for each quantity: $\rho$ has units of $(\Omega \cdot \text{m}^2) / \text{m} = \Omega \cdot \text{m}$ .
A potential difference of $12\,\text{V}$ is maintained across an ohmic resistor with a resistance of $150\,\Omega$ . What is the current flowing through the resistor?
Answer: $0.080\,\text{A}$ According to Ohm's Law, the current $I$ is given by $I = V/R$ . Substituting the given values: $I = 12\,\text{V} / 150\,\Omega = 0.080\,\text{A}$ .
Considering the information provided, a lightbulb in a household circuit follows Ohm's Law $V=IR$ , where $R$ (in $\Omega$ ) limits $I$ (in A). Resistivity $\rho$ (in $\Omega\cdot\text{m}$ ) is a material property affecting $R$ . What happens to current if resistance increases and voltage remains constant according to $V=IR$ ?
Answer: Current decreases because $I = V/R$ This question tests understanding of resistance, resistivity, and Ohm's Law in AP Physics C: Electricity and Magnetism. Ohm's Law V = IR rearranges to I = V/R, establishing that current is inversely proportional to resistance when voltage remains constant in a circuit. In the lightbulb example, if the filament's resistance increases while the household circuit maintains constant voltage, the current through the bulb must decrease according to the relationship I = V/R. Choice B is correct because it states that current decreases and provides the correct mathematical relationship I = V/R that governs this inverse proportionality. Choice D is incorrect because it misinterprets Ohm's Law - current does not become equal to resistance; rather, current equals voltage divided by resistance, and they have different units (amperes vs ohms). Students should practice dimensional analysis with Ohm's Law to verify relationships and avoid unit confusion. Emphasize that in household circuits, voltage is typically fixed by the power company, making resistance the primary variable for controlling current in devices.
The current $I$ through a circuit component is measured as a function of the potential difference $V$ across it. A plot of $I$ as a function of $V$ is a straight line passing through the origin with a slope of $0.05\,\text{A/V}$ . What is the resistance of the component?
Answer: $20\,\Omega$ The graph shows $I$ (y-axis) versus $V$ (x-axis). For an ohmic resistor, the slope of this graph is $m = \Delta I / \Delta V$ . According to Ohm's Law, $R = V/I$ , so the resistance is the reciprocal of the slope: $R = 1/m = 1/(0.05\,\text{A/V}) = 20\,\Omega$ .
A uniform potential difference is applied across the ends of a cylindrical ohmic conductor. If the potential difference is tripled, what is the effect on the average drift velocity of the charge carriers within the conductor, assuming its properties remain unchanged?
Answer: It increases by a factor of three. Tripling the potential difference $V$ across an ohmic conductor triples the current $I$ ( $I=V/R$ ). The current is related to drift velocity $v_d$ by $I = nqAv_d$ , where $n, q,$ and $A$ are constants for the wire. Therefore, if $I$ is tripled, $v_d$ must also be tripled.
Ohm's Law is an empirical relationship often written as $V=IR$ . It is a very useful model in circuit analysis. Which statement best describes the applicability of Ohm's Law?
Answer: It accurately describes materials for which the ratio of potential difference to current is constant over a wide range of conditions. Ohm's Law is not a fundamental law but an empirical model that holds true for certain materials, called ohmic materials. For these materials, the resistance $R = V/I$ is constant over a range of voltages and temperatures. Many materials, such as semiconductors and insulators, do not follow this linear relationship.
Considering the information provided, a household device follows $V=IR$ ; resistance $R$ (Ω) is the circuit-level opposition to current, while resistivity $\rho$ (in $\Omega\cdot\text{m}$ ) is intrinsic to the material. What happens to current if resistance increases and voltage remains constant according to Ohm's Law ( $V=IR$ )?
Answer: Current decreases because $I=V/R$ This question tests understanding of resistance, resistivity, and Ohm's Law in AP Physics C: Electricity and Magnetism. Ohm's Law V = IR can be rearranged to I = V/R, clearly showing that current equals voltage divided by resistance. In the household device example, when resistance increases while voltage stays constant, the current must decrease according to this inverse relationship shown in the equation I = V/R. Choice B is correct because it both states the correct outcome (current decreases) and provides the correct mathematical relationship (I = V/R) that explains why this happens. Choice A is incorrect because it presents a mathematically impossible equation I = VR, which would have incorrect units and suggest current increases with resistance. Students should always check their equation rearrangements using dimensional analysis: I[A] = V[V]/R[Ω] gives correct units, while I = VR would give units of V·Ω, not amperes. Understanding the mathematical form reinforces the conceptual inverse relationship.
A rod of length $L$ and uniform cross-sectional area $A$ is made of a material whose resistivity varies along its length according to the function $\rho(x) = \rho_0(1 + x^2/L^2)$ , where $x$ is the distance from one end. What is the total resistance of the rod?
Answer: $\frac{4\rho_0 L}{3A}$ The resistance $dR$ of an infinitesimal slice of length $dx$ is $dR = \rho(x) \frac{dx}{A}$ . To find the total resistance, we integrate from $x=0$ to $x=L$ : $R = \int_0^L \frac{\rho_0}{A}(1 + \frac{x^2}{L^2}) dx = \frac{\rho_0}{A} \left[x + \frac{x^3}{3L^2}\right]_0^L = \frac{\rho_0}{A} \left(L + \frac{L^3}{3L^2}\right) = \frac{\rho_0}{A} \left(L + \frac{L}{3}\right) = \frac{4\rho_0 L}{3A}$ .
The potential difference $V$ across a circuit component is measured as a function of the current $I$ through it. A plot of $V$ as a function of $I$ results in a straight line passing through the origin with a slope of $200\,\text{V/A}$ . What is the resistance of the component?
Answer: $200\,\Omega$ From Ohm's Law, $V=IR$ . A graph of $V$ (y-axis) versus $I$ (x-axis) for an ohmic resistor is a straight line with slope $m = \Delta V / \Delta I$ . This slope is equal to the resistance $R$ . Therefore, the resistance of the component is $200\,\Omega$ .
Based on the provided data, which of the following conclusions is most justified?
Answer: The element is ohmic with a resistance of $4.0\,\Omega$ . To determine if the element is ohmic, we calculate the ratio $R = V/I$ for each data point: $2.0/0.50 = 4.0\,\Omega$ , $4.0/1.00 = 4.0\,\Omega$ , $6.0/1.50 = 4.0\,\Omega$ , and $8.0/2.00 = 4.0\,\Omega$ . Since the ratio is constant, the element is ohmic with a resistance of $4.0\,\Omega$ .
A resistor is shaped like a truncated cone of length $L$ and has uniform resistivity $\rho$ . The radius varies linearly with position $x$ from $r_1$ at $x=0$ to $r_2$ at $x=L$ . The current is directed along the x-axis. Which of the following expressions represents the resistance of a thin disk of the material with thickness $dx$ at position $x$ ?
Answer: $\rho \frac{dx}{\pi [r(x)]^2}$ The resistance of a small element is given by $dR = \rho \frac{d(\text{length})}{A}$ . For a thin disk of thickness $dx$ oriented perpendicular to the current flow, the length is $dx$ and the cross-sectional area is $A(x) = \pi [r(x)]^2$ . Therefore, the resistance of the thin disk is $dR = \rho \frac{dx}{\pi [r(x)]^2}$ .
Wire A has length $L$ and radius $r$ . Wire B, made from the same uniform material, has length $2L$ and radius $r/2$ . What is the ratio of the resistance of Wire B to the resistance of Wire A, $R_B/R_A$ ?
Answer: $8$ Resistance is given by $R = \rho L/A$ , where $A = \pi r^2$ . For Wire A, $R_A = \rho L/(\pi r^2)$ . For Wire B, $L_B=2L$ and $r_B=r/2$ , so $A_B = \pi(r/2)^2 = \pi r^2/4$ . Thus, $R_B = \rho(2L)/(\pi r^2/4) = 8(\rho L/(\pi r^2)) = 8R_A$ . The ratio $R_B/R_A$ is 8.

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