Question 1

In metric units, the dimensions of current are A) coulombs. B) coulombs·second. C) coulombs/second. D) coulombs·second².

Accepted Answer

In the International System of Units (SI), the unit of electric current is the ampere, which is defined as one coulomb of charge passing through a given point per second. Therefore, the dimension of current is coulombs per second.

Question 2

The direction of the conventional current flow is best described as
A) the direction of a hypothetical positive charge in the circuit.
B) the actual direction of the negatively charged electrons in the circuit.
C) the path the electrons follow from the positive to the negative terminals of the battery.
D) the path of ions from the positive to the negative terminals of the battery.

Accepted Answer

Conventional current flow is defined as the direction a positive charge would move, which is from the positive terminal to the negative terminal of a power source. This convention was established before the actual charge carriers (electrons) were discovered, and electrons, being negatively charged, actually move in the opposite direction.

Question 3

A current of 15 mA is passing through a wire. The time it will take for 15 C of charge to pass through a given point in the wire is
A) 1 ms.
B) 1.0 s.
C) 100 s.
D) 1000 s.

Accepted Answer

D

Question 4

A total of 3.0 $\times$ 10¹⁵ electrons pass through a wire in 48.0 ms. The current is A) 10 $\mu$ A. B) 0.10 mA. C) 15 mA. D) 100 pA.

Accepted Answer

The answer of A total of 3.0 $\times$ 10¹⁵ electrons...

Question 5

A 3.0-V potential is attached to a 1000-$\Omega$ resistor. The number of electrons that pass through the wire in one hour is A) 7.7 $\times$ 10¹⁹. B) 6.7 $\times$ 10¹⁹. C) 5.7 $\times$ 10¹⁹. D) 4.7 $\times$ 10¹⁹.

Accepted Answer

First, calculate the current using Ohm's Law: I = V/R = 3V/1000Ω = 0.003A. Then, find the total charge: Q = I × t = 0.003A × 3600s = 10.8C. Finally, calculate the number of electrons: number of electrons = Q/e = 10.8C / 1.6 × 10^-19C/electron ≈ 6.7 × 10^19 electrons.

Question 6

The resistance of a wire is 1.2  $\Omega$. The potential difference across the wire when 150 C pass through the wire in 15 seconds is
A) 1.2 V.
B) 12 V.
C) 120 V.
D) 120 mV.

Accepted Answer

The resistance of the wire is given as 1.2 ohms. To find the potential difference, we first need to find the current using the charge (Q) and the time (t), with the formula I = Q/t. Here, Q = 150 C and t = 15 s, so I = 150 C / 15 s = 10 A. Then, using Ohm's law (V = IR), where I = 10 A and R = 1.2 ohms, the potential difference V = 10 A * 1.2 ohms = 12 V.

Question 7

A current that is initially1.5 A decreases linearly to zero over 2.0 seconds. The total charge that passes through the wire is
A) 1.5 C.
B) 0.75 C.
C) 3.0 C.
D) 0 C.

Accepted Answer

The answer of A current that is initially1.5 A decreases...

Question 8

The electric field in a uniform wire is related to the potential across the wire, $\Delta$ V, and the length of the wire, $l$ , by A)$\Delta$ V· $l$ B) $\Delta$ V/ $l$ C) $l$ /$\Delta$ V. D) $l$ ²/$\Delta$ V.

Accepted Answer

The electric field in a uniform wire is given by the potential difference (voltage) across the wire divided by the length of the wire, hence $E = \frac{V}{l}$.

Question 9

The potential difference between two ends of a 45-cm length of wire is 1.5 V. The magnitude of the electric field in the wire is
A) 3.3mV/m.
B) 3.3 V/m.
C) 0.67 V/m.
D) 6.7 V/m.

Accepted Answer

The magnitude of the electric field in the wire is given by $E = \frac{V}{d}$, where $V$ is the potential difference and $d$ is the distance. Here, $V = 1.5 \, \text{V}$ and $d = 45 \, \text{cm} = 0.45 \, \text{m}$. Thus, $E = \frac{1.5}{0.45} = 3.33 \, \text{V/m}$, which rounds to $3.3 \, \text{V/m}$.

Question 10

The potential difference between two ends of a 45-cm length of copper wire (1.7 $\times$ 10^-8 $\Omega$·m) is 1.5 V. The magnitude of the current density in the wire is A) 1.0 $\times$ 10⁸ A/m². B) 1.5 $\times$ 10⁸ A/m². C) 2.0 $\times$ 10⁸ A/m². D) 2.5 $\times$ 10⁸ A/m².

Accepted Answer

The current density $ J $ is given by $ J = \frac{V}{\rho \cdot L} $, where $ V = 1.5 \, \text{V} $, $ \rho = 1.7 \times 10^{-8} \, \Omega \cdot \text{m} $, and $ L = 0.45 \, \text{m} $. Calculating gives $ J = \frac{1.5}{1.7 \times 10^{-8} \times 0.45} = 2.0 \times 10^8 \, \text{A/m}^2 $.

Question 11

In metric units the dimensions of resistivity are
A) ohm.
B) ohm·second.
C) ohm·meter.
D) ohm/meter.

Accepted Answer

Resistivity is defined as the resistance of a conductor of unit length and unit cross-sectional area. Its SI unit is ohm-meter (Ω·m), which corresponds to the resistance (in ohms) of a conductor with a length of one meter and a cross-sectional area of one square meter.

Question 12

The resistivity of a conductor depends on the collision time by an equation of the form A) $\rho$ proportional . B)$\rho$ proportional to ^-1. C)$\rho$ proportional to ^-2. D)$\rho$0 proportional to ².

Accepted Answer

The resistivity of a conductor is inversely proportional to the collision time, meaning as the collision time increases, the resistivity decreases, and vice versa. This is represented by the equation form where resistivity is proportional to the inverse of collision time ($\tau^{-1}$).

Question 13

If the average collision time of an electron increases by a factor of 2, the resistivity of the material will
A) remain the same.
B) increase by a factor of 2.
C) decrease by a factor of 2.
D) increase by a factor of 4.

Accepted Answer

The resistivity of a material is inversely proportional to the average collision time of electrons. If the collision time increases by a factor of 2, the resistivity decreases by a factor of 2.

Question 14

A wire whose circular cross section is not uniform carries a current I. Relative to the larger region, the drift velocity in the narrower region is
A) larger.
B) smaller.
C) the same.

Accepted Answer

The drift velocity is inversely proportional to the cross-sectional area of the wire. In the narrower region, the cross-sectional area is smaller, leading to a larger drift velocity compared to the larger region.

Question 15

A 2.0-m-long piece of wire has a diameter of 2.0 mm and a resistance of 0.018  $\Omega$. The resistivity of the material is
A) $2.8 \times 10 ^ { - 8 } \Omega \cdot \mathrm { m } .$
B) $1.3 \times 10 ^ { - 7 } \Omega \cdot \mathrm { m } .$
C) $1.4 \times 10 ^ { - 8 } \Omega \mathrm { m } .$
D) $1.3 \times 10 ^ { 4 } \Omega \cdot \mathrm { m }$

Accepted Answer

The resistivity $\rho$ of a material is given by $\rho = \frac{RA}{L}$, where $R$ is the resistance, $A$ is the cross-sectional area, and $L$ is the length. The area $A$ of the wire can be calculated using the formula for the area of a circle, $A = \pi r^2$, where $r$ is the radius of the wire. Given the diameter is 2.0 mm, the radius is 1.0 mm or $1.0 \times 10^{-3}$ m. Thus, $A = \pi (1.0 \times 10^{-3})^2 = 3.14 \times 10^{-6} \, \text{m}^2$. Substituting the given values into the resistivity formula gives $\rho = \frac{0.018 \, \Omega \cdot 3.14 \times 10^{-6} \, \text{m}^2}{2.0 \, \text{m}} = 2.8 \times 10^{-8} \, \Omega \cdot \text{m}$.

Question 16

A current of 1.5 is produced through a conductor whose resistance is 0.50  $\Omega$. The potential necessary to produce this current is
A) 3.0 V.
B) 0.75 V.
C) 0.33 V.
D) 1.1 V.

Accepted Answer

The answer of A current of 1.5 is produced through...

Question 17

A wire's diameter decreases linearly from d_i to d_f = d_i/2. If the wire has a current of I at the end whose diameter is d_i, then the current at the end whose diameter is d_i/2 A) decreases by a factor of 2. B) increases by a factor of 2. C) decreases by a factor of 4. D) remains the same.

Accepted Answer

The current flowing through a conductor is independent of its cross-sectional area, meaning it remains constant regardless of changes in the wire's diameter.

Question 18

A wire's diameter decreases linearly from d_i to d_f = d_i/2. If the electron drift velocity at the other end of the wire is v_i, then the drift velocity at the end whose diameter is d_i/2 A) decreases by a factor of 2. B) increases by a factor of 2. C) increases by a factor of 4. D) remains the same.

Accepted Answer

The drift velocity of electrons in a conductor is inversely proportional to the cross-sectional area of the conductor. When the diameter of the wire is halved, its cross-sectional area decreases by a factor of 4 (since area ∝ diameter^2), leading to a fourfold increase in the electron drift velocity.

Question 19

In metric units the dimensions of resistance are
A) volt.
B) volt·ampere.
C) ampere·volt.
D) volt/ampere.

Accepted Answer

Resistance is measured in ohms, which can be expressed as volts per ampere (V/A) according to Ohm's law (V = IR).

Question 20

In metric units the dimensions of current density are consistent with A) ampere. B) ampere·meter. C) ampere/meter². D) ampere/meter³.

Accepted Answer

Current density is defined as the amount of current flowing per unit area. The correct unit is amperes per square meter, which is option C (ampere/meter^2).

Quiz 27: Currents and Ohms Law

In metric units, the dimensions of current are

The direction of the conventional current flow is best described as

A current of 15 mA is passing through a wire. The time it will take for 15 C of charge to pass through a given point in the wire is

A total of 3.0 $\times$ 10¹⁵ electrons pass through a wire in 48.0 ms. The current is

A 3.0-V potential is attached to a 1000- $\Omega$ resistor. The number of electrons that pass through the wire in one hour is

The resistance of a wire is 1.2 $\Omega$ . The potential difference across the wire when 150 C pass through the wire in 15 seconds is

A current that is initially1.5 A decreases linearly to zero over 2.0 seconds. The total charge that passes through the wire is

The electric field in a uniform wire is related to the potential across the wire, $\Delta$ V, and the length of the wire, $l$ , by

The potential difference between two ends of a 45-cm length of wire is 1.5 V. The magnitude of the electric field in the wire is

The potential difference between two ends of a 45-cm length of copper wire (1.7 $\times$ 10^-8 $\Omega$ ·m) is 1.5 V. The magnitude of the current density in the wire is

In metric units the dimensions of resistivity are

If the average collision time of an electron increases by a factor of 2, the resistivity of the material will

A wire whose circular cross section is not uniform carries a current I. Relative to the larger region, the drift velocity in the narrower region is

A 2.0-m-long piece of wire has a diameter of 2.0 mm and a resistance of 0.018 $\Omega$ . The resistivity of the material is

A current of 1.5 is produced through a conductor whose resistance is 0.50 $\Omega$ . The potential necessary to produce this current is

A wire's diameter decreases linearly from d_i to d_f = d_i/2. If the wire has a current of I at the end whose diameter is d_i, then the current at the end whose diameter is d_i/2

A wire's diameter decreases linearly from d_i to d_f = d_i/2. If the electron drift velocity at the other end of the wire is v_i, then the drift velocity at the end whose diameter is d_i/2

In metric units the dimensions of resistance are

In metric units the dimensions of current density are consistent with

Quiz 27: Currents and Ohms Law

In metric units, the dimensions of current are

The direction of the conventional current flow is best described as

A current of 15 mA is passing through a wire. The time it will take for 15 C of charge to pass through a given point in the wire is

A total of 3.0 ×\times× 1015 electrons pass through a wire in 48.0 ms. The current is

A 3.0-V potential is attached to a 1000- Ω\OmegaΩ resistor. The number of electrons that pass through the wire in one hour is

The resistance of a wire is 1.2 Ω\OmegaΩ . The potential difference across the wire when 150 C pass through the wire in 15 seconds is

A current that is initially1.5 A decreases linearly to zero over 2.0 seconds. The total charge that passes through the wire is

The electric field in a uniform wire is related to the potential across the wire, Δ\DeltaΔ V, and the length of the wire, lll , by