Answer Key

Chapter 24: Review Questions Solutions

Teacher Resource
AP Physics C: E&M

1. Instantaneous Current

(a) Current is the time-derivative of charge: \(I(t) = \frac{dQ}{dt}\).

\(I(t) = \frac{d}{dt}(3t^2 - 4t + 5) = 6t - 4 \text{ A}\)

(b) Set \(I(t) = 0\): \(6t - 4 = 0 \Rightarrow 6t = 4 \Rightarrow t = \frac{2}{3} \text{ s} \approx 0.67 \text{ s}\).

2. The Drude Model

Increased temperature causes the lattice ions to vibrate with greater amplitude. This increases the probability of collisions between free electrons and the lattice, thereby decreasing the mean free time (\(\tau\)). According to the Drude model:

\[ \rho = \frac{m}{n e^2 \tau} \]

Since resistivity \(\rho\) is inversely proportional to \(\tau\), a decrease in \(\tau\) results in an increase in \(\rho\).

3. Non-Uniform Current Density

Total current is the flux of the current density vector through the cross-section:

\[ I = \int \vec{J} \cdot d\vec{A} = \int_0^R J_0 \left( 1 - \frac{r}{R} \right) (2\pi r dr) \] \[ I = 2\pi J_0 \int_0^R \left( r - \frac{r^2}{R} \right) dr = 2\pi J_0 \left[ \frac{r^2}{2} - \frac{r^3}{3R} \right]_0^R \] \[ I = 2\pi J_0 \left( \frac{R^2}{2} - \frac{R^2}{3} \right) = 2\pi J_0 \left( \frac{R^2}{6} \right) = \frac{\pi J_0 R^2}{3} \]

4. Empirical Nature

Ohm's Law is empirical because it is a material property observed under specific conditions (like constant temperature), not a universal law derived from fundamental symmetries or conservation principles.

Example non-Ohmic components: Diodes, transistors, or vacuum tubes.

5. Variable Geometry (Frustum)

Radius as a function of distance \(x\): \(r(x) = r_1 + \frac{r_2 - r_1}{L}x\). Area \(A(x) = \pi [r(x)]^2\).

\[ R = \int_0^L \frac{\rho dx}{\pi [r_1 + \frac{r_2-r_1}{L}x]^2} \] Let \(u = r_1 + \frac{r_2-r_1}{L}x\), then \(du = \frac{r_2-r_1}{L}dx\). \[ R = \frac{\rho L}{\pi(r_2-r_1)} \int_{r_1}^{r_2} u^{-2} du = \frac{\rho L}{\pi(r_2-r_1)} \left[ -\frac{1}{u} \right]_{r_1}^{r_2} = \frac{\rho L}{\pi(r_2-r_1)} \left( \frac{1}{r_1} - \frac{1}{r_2} \right) = \frac{\rho L}{\pi r_1 r_2} \]

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6. Power Dissipation

Total power dissipation for a constant voltage source is \(P = \frac{V^2}{R_{eq}}\).

Series: \(R_{eq} = R_1 + R_2\) (Resistance increases)
Parallel: \(R_{eq} = \frac{R_1 R_2}{R_1 + R_2}\) (Resistance decreases)

Since the parallel configuration has a lower equivalent resistance, it will result in greater total power dissipation.

7. The Real Battery

(a) Current: Using the terminal voltage formula \(V = \mathcal{E} - Ir\):

\(10.5 = 12.0 - I(0.5) \Rightarrow 0.5I = 1.5 \Rightarrow I = 3.0 \text{ A}\)

(b) Load Resistance: Using Ohm's Law across the load \(V = IR\):

\(10.5 = (3.0)R \Rightarrow R = 3.5 \text{ \(\Omega\)}\)

8. Network Analysis

To solve for 3 unknown branch currents, we need 3 independent equations. This is typically achieved using 1 Junction Rule equation and 2 Loop Rule equations.

The Loop Rule (\(\sum \Delta V = 0\)) is a statement of the Conservation of Energy, as it implies the net work done on a charge moving around a closed loop is zero.

9. Loading Effects

(a) Ammeter in Parallel: Ammeters have near-zero resistance. Connecting one in parallel creates a short circuit. The extremely high current flow can blow the meter's fuse or destroy its internal circuitry.

(b) Voltmeter in Series: Voltmeters have near-infinite resistance. If placed in series, the total circuit resistance becomes massive, causing the current to drop to nearly zero and preventing the load from receiving functional power.

10. Advanced Symmetry

In a cube of identical resistors, current enters at one corner and splits. Because all paths from that corner look identical, the current must divide equally (\(I/3\)) into the three adjacent edges.

The three nodes at the ends of those first edges are symmetric relative to the source and sink. Therefore, there is no physical reason for their potentials to differ. These equipotential points can be "shorted" together in a theoretical model without changing the circuit's behavior, greatly simplifying the calculation of equivalent resistance.

End of Solutions — Circuit Mastery Unit

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Review Questions Key

Review Questions Key

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