An infinite two-dimensional square grid is constructed with identical resistors, each of resistance $R$, connected at every junction node. A battery supplies a net current $I$ that enters the grid at node $A$ and exits at an adjacent node $B$. Which of the following claims correctly gives the equivalent resistance $R_{eq}$ between nodes $A$ and $B$, along with the correct physical justification?(a) $R_{eq} = \dfrac{R}{2}$, because injecting current $I$ into node $A$ sends $\dfrac{I}{4}$ through the resistor between $A$ and $B$, and extracting current $I$ from node $B$ draws $\dfrac{I}{4}$ through the same resistor, giving a net current of $\dfrac{I}{2}$ through that resistor. , (b) $R_{eq} = \dfrac{R}{4}$, because four identical paths branch symmetrically from node $A$, so the total current $I$ splits equally into four parallel paths, each having an equivalent resistance of $R$. , (c) $R_{eq} = \dfrac{R}{2}$, because the infinite network surrounding the edge between $A$ and $B$ acts as a single equivalent resistor of resistance $R$ connected in parallel with the resistor directly between $A$ and $B$. , (d) $R_{eq} = R$, because the current takes the path of least resistance through the single resistor connecting nodes $A$ and $B$, while all surrounding indirect paths through the infinite grid present infinite total resistance. :

An infinite two-dimensional square grid is constructed with identical resistors, each of resistance $R$, connected at every junction node. A battery supplies a net current $I$ that enters the grid at node $A$ and exits at an adjacent node $B$. Which of the following claims correctly gives the equivalent resistance $R_{eq}$ between nodes $A$ and $B$, along with the correct physical justification? • Nerd Notes

AP Physics

Unit 1 - Vectors and Kinematics

MCQ

Mathematical

Advanced

Step	Reasoning
Identify the target quantity and select the superposition method.	The equivalent resistance is defined by $R_{eq} = \dfrac{V_{AB}}{I}$. Because direct series-parallel reduction is impossible for an infinite grid, we decompose the total current flow into two independent, symmetric states using superposition.
Analyze state 1: inject current $I$ into node $A$ and let it flow out to infinity. \[ I_1 = \dfrac{I}{4} \]	By the four-fold rotational symmetry of the square grid, the current entering node $A$ must divide equally among the four resistors attached to node $A$.
Analyze state 2: extract current $I$ from node $B$ drawn from infinity. \[ I_2 = \dfrac{I}{4} \]	By the same four-fold rotational symmetry, the current pulled out of node $B$ draws equal current through each of the four resistors attached to node $B$.
Superimpose state 1 and state 2 to find the total current through resistor $AB$. \[ I_{AB} = I_1 + I_2 = \dfrac{I}{4} + \dfrac{I}{4} = \dfrac{I}{2} \]	Superimposing both flows gives a net current $I$ entering node $A$ and exiting node $B$. In both sub-problems, the current flows in the same direction along the direct segment from $A$ to $B$.
Apply Ohm’s law across the resistor between $A$ and $B$ to solve for $R_{eq}$. \[ V_{AB} = I_{AB} R = \left(\dfrac{I}{2}\right) R \implies R_{eq} = \dfrac{V_{AB}}{I} = \dfrac{R}{2} \]	The potential difference $V_{AB}$ across the direct resistor of resistance $R$ depends only on the net current $I_{AB}$ passing through it.

Why each choice is correct or incorrect:

(A) This is the correct answer.

(B) Incorrectly assumes the four branches meeting at node A act as simple parallel branches each with equivalent resistance R, yielding $R_{eq} = R/4$.

(C) Reaches the correct numerical value $R/2$ through unsound reasoning, asserting without justification that the remaining infinite network simplifies to a single resistor of value R.

(D) Falsely assumes that adding parallel paths increases total resistance to infinity, incorrectly ignoring current flow through the rest of the grid.

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Intermediate

Mathematical

MCQ

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Advanced

Mathematical

MCQ

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Beginner

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An ideal circuit consists of an inductor of inductance $L$ connected in series with a capacitor of capacitance $C$. At time $t = 0$, the capacitor carries an initial charge $Q_0$ and the current in the circuit is zero. Which of the following graphs best represents the charge $q(t)$ on the capacitor as a function of time $t$?

Intermediate

Mathematical

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A long, straight cylindrical wire of radius $R$ carries a steady total current $I$ distributed uniformly across its circular cross section. The permeability of free space is $\mu_0$. Which of the following expressions represents the total magnetic energy stored per unit length inside the volume of the wire?

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A circular region of radius $R$ in the $xy$-plane contains a time-dependent, non-uniform magnetic field directed perpendicular to the plane. Inside the region ($r \le R$), the magnitude of the magnetic field is given by $B(r,t) = C t \left(\dfrac{r}{R}\right)^2$, where $C$ is a positive constant and $r$ is the radial distance from the central axis. Outside the region ($r > R$), the magnetic field is zero. Which of the following expressions gives the magnitude of the induced electric field $E(r)$ as a function of distance $r$ from the central axis inside the region ($r < R$)?

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In the circuit shown, an ideal battery with potential difference $V_0$ is connected to three resistors ($R_1 = R$, $R_2 = 2R$, $R_3 = R$), an ideal inductor $L$, and a switch $S$. The switch $S$ has been closed for a long time. At time $t = 0$, switch $S$ is opened. Which of the following expressions represents the magnitude of the current $I(t)$ through resistor $R_2$ as a function of time $t$ for $t \ge 0$?

Step	Reasoning
Identify the target quantity and select the superposition method.	The equivalent resistance is defined by \(R_{eq} = \dfrac{V_{AB}}{I}\). Because direct series-parallel reduction is impossible for an infinite grid, we decompose the total current flow into two independent, symmetric states using superposition.
Analyze state 1: inject current \(I\) into node \(A\) and let it flow out to infinity. \[ I_1 = \dfrac{I}{4} \]	By the four-fold rotational symmetry of the square grid, the current entering node \(A\) must divide equally among the four resistors attached to node \(A\).
Analyze state 2: extract current \(I\) from node \(B\) drawn from infinity. \[ I_2 = \dfrac{I}{4} \]	By the same four-fold rotational symmetry, the current pulled out of node \(B\) draws equal current through each of the four resistors attached to node \(B\).
Superimpose state 1 and state 2 to find the total current through resistor \(AB\). \[ I_{AB} = I_1 + I_2 = \dfrac{I}{4} + \dfrac{I}{4} = \dfrac{I}{2} \]	Superimposing both flows gives a net current \(I\) entering node \(A\) and exiting node \(B\). In both sub-problems, the current flows in the same direction along the direct segment from \(A\) to \(B\).
Apply Ohm’s law across the resistor between \(A\) and \(B\) to solve for \(R_{eq}\). \[ V_{AB} = I_{AB} R = \left(\dfrac{I}{2}\right) R \implies R_{eq} = \dfrac{V_{AB}}{I} = \dfrac{R}{2} \]	The potential difference \(V_{AB}\) across the direct resistor of resistance \(R\) depends only on the net current \(I_{AB}\) passing through it.

Kinematics	Forces
\(\Delta x = v_i t + \frac{1}{2} at^2\)	\(F = ma\)
\(v = v_i + at\)	\(F_g = \frac{G m_1 m_2}{r^2}\)
\(v^2 = v_i^2 + 2a \Delta x\)	\(f = \mu N\)
\(\Delta x = \frac{v_i + v}{2} t\)	\(F_s =-kx\)
\(v^2 = v_f^2 \,-\, 2a \Delta x\)

Circular Motion	Energy
\(F_c = \frac{mv^2}{r}\)	\(KE = \frac{1}{2} mv^2\)
\(a_c = \frac{v^2}{r}\)	\(PE = mgh\)
\(T = 2\pi \sqrt{\frac{r}{g}}\)	\(KE_i + PE_i = KE_f + PE_f\)
	\(W = Fd \cos\theta\)

Momentum	Torque and Rotations
\(p = mv\)	\(\tau = r \cdot F \cdot \sin(\theta)\)
\(J = \Delta p\)	\(I = \sum mr^2\)
\(p_i = p_f\)	\(L = I \cdot \omega\)

Simple Harmonic Motion	Fluids
\(F = -kx\)	\(P = \frac{F}{A}\)
\(T = 2\pi \sqrt{\frac{l}{g}}\)	\(P_{\text{total}} = P_{\text{atm}} + \rho gh\)
\(T = 2\pi \sqrt{\frac{m}{k}}\)	\(Q = Av\)
\(x(t) = A \cos(\omega t + \phi)\)	\(F_b = \rho V g\)
\(a = -\omega^2 x\)	\(A_1v_1 = A_2v_2\)

Constant	Description
[katex]g[/katex]	Acceleration due to gravity, typically [katex]9.8 , \text{m/s}^2[/katex] on Earth’s surface
[katex]G[/katex]	Universal Gravitational Constant, [katex]6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2[/katex]
[katex]\mu_k[/katex] and [katex]\mu_s[/katex]	Coefficients of kinetic ([katex]\mu_k[/katex]) and static ([katex]\mu_s[/katex]) friction, dimensionless. Static friction ([katex]\mu_s[/katex]) is usually greater than kinetic friction ([katex]\mu_k[/katex]) as it resists the start of motion.
[katex]k[/katex]	Spring constant, in [katex]\text{N/m}[/katex]
[katex] M_E = 5.972 \times 10^{24} , \text{kg} [/katex]	Mass of the Earth
[katex] M_M = 7.348 \times 10^{22} , \text{kg} [/katex]	Mass of the Moon
[katex] M_M = 1.989 \times 10^{30} , \text{kg} [/katex]	Mass of the Sun

Variable	SI Unit
[katex]s[/katex] (Displacement)	[katex]\text{meters (m)}[/katex]
[katex]v[/katex] (Velocity)	[katex]\text{meters per second (m/s)}[/katex]
[katex]a[/katex] (Acceleration)	[katex]\text{meters per second squared (m/s}^2\text{)}[/katex]
[katex]t[/katex] (Time)	[katex]\text{seconds (s)}[/katex]
[katex]m[/katex] (Mass)	[katex]\text{kilograms (kg)}[/katex]

Variable	Derived SI Unit
[katex]F[/katex] (Force)	[katex]\text{newtons (N)}[/katex]
[katex]E[/katex], [katex]PE[/katex], [katex]KE[/katex] (Energy, Potential Energy, Kinetic Energy)	[katex]\text{joules (J)}[/katex]
[katex]P[/katex] (Power)	[katex]\text{watts (W)}[/katex]
[katex]p[/katex] (Momentum)	[katex]\text{kilogram meters per second (kgm/s)}[/katex]
[katex]\omega[/katex] (Angular Velocity)	[katex]\text{radians per second (rad/s)}[/katex]
[katex]\tau[/katex] (Torque)	[katex]\text{newton meters (Nm)}[/katex]
[katex]I[/katex] (Moment of Inertia)	[katex]\text{kilogram meter squared (kgm}^2\text{)}[/katex]
[katex]f[/katex] (Frequency)	[katex]\text{hertz (Hz)}[/katex]

Prefix	Symbol	Power of Ten
Pico-	p	[katex]10^{-12}[/katex]
Nano-	n	[katex]10^{-9}[/katex]
Micro-	µ	[katex]10^{-6}[/katex]
Milli-	m	[katex]10^{-3}[/katex]
Centi-	c	[katex]10^{-2}[/katex]
Deci-	d	[katex]10^{-1}[/katex]
(Base unit)	–	[katex]10^{0}[/katex]
Deca- or Deka-	da	[katex]10^{1}[/katex]
Hecto-	h	[katex]10^{2}[/katex]
Kilo-	k	[katex]10^{3}[/katex]
Mega-	M	[katex]10^{6}[/katex]
Giga-	G	[katex]10^{9}[/katex]
Tera-	T	[katex]10^{12}[/katex]

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