PHY 131 Practice Exam 2 – Step-by-Step Physics Guidance

Q1. Suppose that in the wire on the left the electron drift speed is $v_d$. What is the electron drift speed in the wire on the right side?

Background

Topic: Current Density and Drift Velocity in Conductors

This question tests your understanding of how current, cross-sectional area, and drift velocity are related in wires of different diameters but the same material and length.

Key Terms and Formulas:

• Current density: $J = \frac{I}{A}$

• Drift velocity: $J = nqv_d$

• Area of a cylinder: $A = \pi r^2$

Step-by-Step Guidance

1. Recall that the current $I$ is the same through both wires since they are connected in series.

2. Write the expression for current in terms of drift velocity: $I = nqv_dA$.

3. Since $n$ and $q$ are the same for both wires, relate the drift velocities using the areas: $v_{d,1}A_1 = v_{d,2}A_2$.

4. Calculate the ratio of the areas using the diameters given (1.0 mm and 2.0 mm).

Try solving on your own before revealing the answer!

Final Answer: $v_{d, ext{right}} = v_d/4$

The area increases by a factor of 4, so the drift speed decreases by a factor of 4.

Q2. Suppose that the resistance of the wire on the left is $R$. What is the resistance of the wire on the right?

Background

Topic: Resistance of a Wire

This question tests your understanding of how resistance depends on the dimensions of a wire (length and cross-sectional area).

Key Terms and Formulas:

• Resistance: $R = \rho \frac{L}{A}$

• Area of a cylinder: $A = \pi r^2$

Step-by-Step Guidance

1. Recall that both wires have the same length $L$ and are made of the same material (same $\rho$).

2. Write the resistance for each wire: $R_1 = \rho \frac{L}{A_1}$ and $R_2 = \rho \frac{L}{A_2}$.

3. Express $A_2$ in terms of $A_1$ using the diameters (since $A \propto d^2$).

4. Find the ratio $R_2/R_1$ to see how the resistance changes with area.

Try solving on your own before revealing the answer!

Final Answer: $R_ ext{right} = R/4$

Doubling the diameter increases the area by 4, so resistance decreases by 4.

Q3. What length of a certain metal wire of diameter 0.15 mm is needed for the wire to have a resistance of 15 Ω? The resistivity of this metal is $1.68 \times 10^{-8} \Omega \cdot m$.

Background

Topic: Resistance and Resistivity

This question tests your ability to use the formula for the resistance of a wire to solve for its length.

Key Terms and Formulas:

• Resistance: $R = \rho \frac{L}{A}$

• Area of a wire: $A = \pi r^2$

Step-by-Step Guidance

1. Convert the diameter to radius in meters: $r = 0.15\,\text{mm}/2$.

2. Calculate the cross-sectional area $A$ using $A = \pi r^2$.

3. Rearrange the resistance formula to solve for $L$: $L = \frac{RA}{\rho}$.

4. Plug in the values for $R$, $A$, and $\rho$ (be careful with units).

Try solving on your own before revealing the answer!

Final Answer: $L = 16\,\text{m}$

After plugging in the values and solving, you find the required length is 16 meters.

Q4. An electric device delivers a current of 5.0 A to a device. How many electrons flow through this device in 10 s?

Background

Topic: Electric Current and Charge

This question tests your understanding of the relationship between current, charge, and the number of electrons.

Key Terms and Formulas:

• Current: $I = \frac{Q}{t}$

• Charge of one electron: $e = 1.602 \times 10^{-19}\,\text{C}$

• Number of electrons: $n = \frac{Q}{e}$

Step-by-Step Guidance

1. Calculate the total charge $Q$ that flows in 10 seconds using $Q = It$.

2. Divide the total charge by the elementary charge $e$ to find the number of electrons: $n = Q/e$.

Try solving on your own before revealing the answer!

Final Answer: $n = 3.1 \times 10^{20}$ electrons

This is the number of electrons passing through the device in 10 seconds at 5.0 A.

Q5. How much current will be flowing through a 40.0 m length of cylindrical metal wire with radius 4.00 mm if it is connected to a source supplying 16.0 V? The resistivity of this metal is $1.68 \times 10^{-8} \Omega \cdot m$. Assume ideal conditions.

Background

Topic: Ohm's Law and Resistance of a Wire

This question tests your ability to calculate current using Ohm's Law and the formula for resistance of a wire.

Key Terms and Formulas:

• Ohm's Law: $I = \frac{V}{R}$

• Resistance: $R = \rho \frac{L}{A}$

• Area: $A = \pi r^2$

Step-by-Step Guidance

1. Calculate the cross-sectional area $A$ using the given radius (convert mm to m).

2. Calculate the resistance $R$ using $R = \rho \frac{L}{A}$.

3. Use Ohm's Law to solve for current: $I = \frac{V}{R}$.

Try solving on your own before revealing the answer!

Final Answer: $I = 1200\,\text{A}$

With the large cross-sectional area and short length, the current is very high for the given voltage.

Q6. A light bulb is connected to a 110-V source. What is the resistance of this bulb if it is a 100-W bulb?

Background

Topic: Power in Electric Circuits

This question tests your ability to relate power, voltage, and resistance in an electric circuit.

Key Terms and Formulas:

• Power: $P = \frac{V^2}{R}$

• Rearranged: $R = \frac{V^2}{P}$

Step-by-Step Guidance

1. Write the formula relating power, voltage, and resistance.

2. Rearrange to solve for resistance: $R = \frac{V^2}{P}$.

3. Plug in the values for $V$ and $P$.

Try solving on your own before revealing the answer!

Final Answer: $R = 120\,\Omega$

This is the resistance that allows the bulb to dissipate 100 W at 110 V.

Q7. In the circuit shown in the figure, four identical resistors labeled A to D are connected to a battery as shown. S1 and S2 are switches. Which of the following actions would result in the GREATEST amount of current through resistor A?

Background

Topic: Series and Parallel Circuits

This question tests your understanding of how closing switches affects the current through a specific resistor in a combination circuit.

Key Terms and Formulas:

• Series and parallel resistor combinations

• Current division in circuits

Step-by-Step Guidance

1. Analyze the circuit for each switch configuration (open/closed) to see how the total resistance changes.

2. Recall that the current through A depends on the total resistance in the circuit.

3. Determine which configuration minimizes total resistance, maximizing current through A.

Try solving on your own before revealing the answer!

Final Answer: Closing both switches (option b)

This configuration puts the most resistors in parallel, minimizing total resistance and maximizing current through A.