College Physics Quiz: Electric Current

20 questions · exam conditions

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Electric CurrentQuestion 1 of 20

In a household circuit, a $100 \text{ W}$ light bulb and a $1500 \text{ W}$ space heater are both designed to operate at $120 \text{ V}$ . When both devices are connected in series (instead of the normal parallel household connection) to a $120 \text{ V}$ source, what is the current in the circuit?

0.42 \text{ A}

0.78 \text{ A}

1.25 \text{ A}

13.3 \text{ A}

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College Physics Quiz

College Physics Quiz: Electric Current

Practice Electric Current in College Physics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Electric Current, giving you a quick way to practice the rules, question types, and explanations that matter most for College Physics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$0.42 \text{ A}$
$0.78 \text{ A}$ (correct answer)
$1.25 \text{ A}$
$13.3 \text{ A}$

Explanation: First find the resistance of each device using

P = \frac{V^2}{R}

. For the light bulb:

R_1 = \frac{(120)^2}{100} = \frac{14400}{100} = 144 \text{ Ω}

. For the space heater:

R_2 = \frac{(120)^2}{1500} = \frac{14400}{1500} = 9.6 \text{ Ω}

. In series, total resistance is

R_{total} = 144 + 9.6 = 153.6 \text{ Ω}

. Current is

I = \frac{V}{R_{total}} = \frac{120}{153.6} = 0.78 \text{ A}

. Choice A uses incorrect power calculations. Choice C assumes parallel connection. Choice D adds the individual currents incorrectly.

Question 2

What is the effect of adding a resistor in series on the total $\vec{R}$ ?

Total $\vec{R}$ increases because the equivalent resistance is the sum (correct answer)
Total $\vec{R}$ decreases because the current splits into more paths
Total $\vec{R}$ becomes the reciprocal sum, $1/\vec{R}_\text{eq}=\sum 1/\vec{R}$
Total $\vec{R}$ becomes equal to the smallest resistor in the circuit

Explanation: This question tests understanding of electric current in DC circuits, specifically series resistance addition. Electric current is defined as the flow of electric charge through a conductor, driven by a voltage difference. Ohm's Law (V=IR) is a key principle, explaining how voltage, current, and resistance are interrelated. Adding in series increases total R as the sum. The correct answer is based on series equivalent resistance. A common distractor might suggest parallel reciprocal, misconfiguring. Teaching strategies include resistance combination worksheets and circuit building activities.

Question 3

How does total $\vec{I}$ in a parallel circuit relate to branch currents?

Total $\vec{I}$ equals the sum of currents in all parallel branches (correct answer)
Total $\vec{I}$ equals the current in the branch with largest $\vec{R}$
Total $\vec{I}$ equals each branch current because $\vec{V}$ is shared
Total $\vec{I}$ is smaller than any branch current due to conservation laws

Explanation: This question tests understanding of electric current in DC circuits, specifically parallel current relations. Electric current is defined as the flow of electric charge through a conductor, driven by a voltage difference. Ohm's Law (V=IR) is a key principle, explaining how voltage, current, and resistance are interrelated. Total I sums branch currents. The correct answer is based on current division and conservation. A common distractor might suggest equality to each, like series. Teaching strategies include parallel network simulations and multi-ammeter experiments.

Question 4

In a series circuit, what happens to $\vec{I}$ when total $\vec{R}$ increases?

$\vec{I}$ decreases for fixed source $\vec{V}$ , consistent with Ohm's law (correct answer)
$\vec{I}$ increases for fixed source $\vec{V}$ , consistent with Ohm's law
$\vec{I}$ remains unchanged because only voltage drops change in series
$\vec{I}$ becomes zero only if $\vec{V}$ increases at the same time

Explanation: This question tests understanding of electric current in DC circuits, specifically current with increasing series R. Electric current is defined as the flow of electric charge through a conductor, driven by a voltage difference. Ohm's Law (V=IR) is a key principle, explaining how voltage, current, and resistance are interrelated. Increasing total R decreases I at fixed V. The correct answer is based on Ohm's inverse relationship. A common distractor might suggest increase, reversing the relation. Teaching strategies include variable resistor demos and series circuit analyses.

Question 5

A wire carries a steady current of $3.0 \text{ A}$ . If the cross-sectional area of the wire is $2.0 \times 10^{-6} \text{ m}^2$ and the number density of free electrons is $8.5 \times 10^{28} \text{ electrons/m}^3$ , what is the drift velocity of the electrons?

$1.1 \times 10^{-4} \text{ m/s}$ (correct answer)
$2.2 \times 10^{-4} \text{ m/s}$
$4.4 \times 10^{-4} \text{ m/s}$
$8.8 \times 10^{-4} \text{ m/s}$
$1.8 \times 10^{-3} \text{ m/s}$

Explanation: When you encounter current and electron motion problems, you need to connect macroscopic current with microscopic electron behavior using the drift velocity equation. Current is related to drift velocity by:

I = nAv_d e

, where

I

is current,

n

is number density of free electrons,

A

is cross-sectional area,

v_d

is drift velocity, and

e

is electron charge (

1.6 \times 10^{-19}

C). Solving for drift velocity:

v_d = \frac{I}{nAe}

Substituting the given values:

v_d = \frac{3.0}{(8.5 \times 10^{28})(2.0 \times 10^{-6})(1.6 \times 10^{-19})}

v_d = \frac{3.0}{2.72 \times 10^{4}} = 1.1 \times 10^{-4} \text{ m/s}

Answer A (

1.1 \times 10^{-4} \text{ m/s}

) is correct. Answer B (

2.2 \times 10^{-4} \text{ m/s}

) results from forgetting to include the electron charge

e

in the denominator. Answer C (

4.4 \times 10^{-4} \text{ m/s}

) comes from using only half the cross-sectional area, perhaps confusing radius with area calculations. Answer D (

8.8 \times 10^{-4} \text{ m/s}

) occurs when both the electron charge is omitted AND the area is halved. Remember that drift velocity is surprisingly slow—electrons move at millimeters per second even though electrical signals propagate near light speed. Always include all four quantities (

I

n

A

e

) in your calculation, and double-check that your result has the characteristically small magnitude typical of drift velocities.

Question 6

A wire has a resistance of $6.0 \, \Omega$ and carries a current of $2.0 \text{ A}$ . If the wire is replaced with one having twice the cross-sectional area but the same length and material, what current will flow when the same voltage is applied?

$1.0 \text{ A}$
$2.0 \text{ A}$
$4.0 \text{ A}$ (correct answer)
$8.0 \text{ A}$
$12.0 \text{ A}$

Explanation: When you encounter resistance problems involving changes to wire dimensions, you need to understand how resistance depends on the physical properties of the conductor. Resistance is given by

R = \rho \frac{L}{A}

, where

\rho

is resistivity,

L

is length, and

A

is cross-sectional area. Let's find the original voltage using Ohm's law:

V = IR = (2.0 \text{ A})(6.0 \, \Omega) = 12 \text{ V}

. When the wire's cross-sectional area doubles while length and material stay the same, the new resistance becomes:

R_{new} = \rho \frac{L}{2A} = \frac{1}{2} \cdot \rho \frac{L}{A} = \frac{R_{original}}{2} = 3.0 \, \Omega

. With the same voltage applied to this lower resistance:

I_{new} = \frac{V}{R_{new}} = \frac{12 \text{ V}}{3.0 \, \Omega} = 4.0 \text{ A}

. Answer A (

1.0 \text{ A}

) incorrectly assumes resistance increases with area, leading to half the original current. Answer B (

2.0 \text{ A}

) represents the original current, ignoring the change in wire dimensions entirely. Answer D (

8.0 \text{ A}

) makes an error in the resistance calculation, possibly confusing the relationship between area and resistance. The correct answer is C (

4.0 \text{ A}

). Study tip: Remember that resistance is inversely proportional to cross-sectional area. When area increases, resistance decreases, allowing more current to flow at the same voltage. Always work through these problems systematically: find the original voltage, determine the new resistance, then calculate the new current.

Question 7

The current in a circuit decreases from $4.0 \text{ A}$ to $1.0 \text{ A}$ over a time interval of $0.5 \text{ s}$ . What is the average rate of change of current?

$-2.0 \text{ A/s}$
$-6.0 \text{ A/s}$ (correct answer)
$-8.0 \text{ A/s}$
$2.5 \text{ A/s}$
$6.0 \text{ A/s}$

Explanation: When you encounter questions about rates of change in physics, you're dealing with how quickly one quantity changes with respect to another. The average rate of change is simply the total change divided by the time interval over which it occurs. To find the average rate of change of current, you need to calculate:

\text{Average rate} = \frac{\text{Final current} - \text{Initial current}}{\text{Time interval}}

Substituting the given values:

\text{Average rate} = \frac{1.0 \text{ A} - 4.0 \text{ A}}{0.5 \text{ s}} = \frac{-3.0 \text{ A}}{0.5 \text{ s}} = -6.0 \text{ A/s}

The negative sign is crucial—it indicates that the current is decreasing over time. Answer A (

-2.0 \text{ A/s}

) likely comes from incorrectly using the average of the initial and final currents (2.5 A) and dividing by something, rather than finding the change. Answer C (

-8.0 \text{ A/s}

) might result from arithmetic errors, perhaps multiplying instead of dividing somewhere in the calculation. Answer D (

2.5 \text{ A/s}

) represents the average current value rather than the rate of change, and it's missing the negative sign that indicates the decreasing trend. Remember that rate of change problems always follow the pattern: change in quantity divided by change in time. Pay close attention to the sign—increasing quantities give positive rates, while decreasing quantities give negative rates. This concept appears frequently in circuits when analyzing how electrical quantities vary over time.

Question 8

Two identical copper wires are connected in parallel between two points. If the current through the first wire is $2.5 \text{ A}$ , what is the current through the second wire?

$1.25 \text{ A}$
$2.5 \text{ A}$ (correct answer)
$5.0 \text{ A}$
$7.5 \text{ A}$
$10.0 \text{ A}$

Explanation: When you encounter parallel circuit problems, remember that components connected in parallel experience the same voltage difference across their terminals. Since these are identical copper wires with the same resistance, they'll behave identically under the same conditions. In a parallel circuit, the voltage across each branch is equal to the source voltage. Using Ohm's law (

V = IR

), if both wires have identical resistance

R

and experience the same voltage

V

, they must carry the same current:

I = V/R

. Since the first wire carries

2.5 \text{ A}

, the second wire must also carry

2.5 \text{ A}

. This makes (B) $2.5 \text{ A}$ correct. (A) $1.25 \text{ A}$ represents a common misconception that current splits equally between branches regardless of resistance. This would only be true if you incorrectly assumed the total current was

2.5 \text{ A}

and divided it between two wires. (C) $5.0 \text{ A}$ might result from confusing this with how currents add in parallel circuits. While the total current from the source would be

5.0 \text{ A}

(

2.5 + 2.5

), this isn't the current through the second wire alone. (D) $7.5 \text{ A}$ has no basis in parallel circuit analysis and likely represents random calculation errors. Study tip: For parallel circuits, remember "same voltage, currents add." Identical resistors in parallel always carry identical currents. The key insight is that parallel components share voltage, not current.

Question 9

A $12 \text{ V}$ battery is connected to a circuit containing two resistors: $4.0 \, \Omega$ and $8.0 \, \Omega$ in series. What is the current flowing through the $8.0 \, \Omega$ resistor?

$0.5 \text{ A}$
$1.0 \text{ A}$ (correct answer)
$1.5 \text{ A}$
$3.0 \text{ A}$
$6.0 \text{ A}$

Explanation: When you encounter resistors in series with a voltage source, remember that the current is the same throughout the entire circuit - this is the fundamental principle that makes these problems straightforward. To find the current, you need the total resistance and apply Ohm's law. In series circuits, resistances add directly:

R_{total} = 4.0 \, \Omega + 8.0 \, \Omega = 12.0 \, \Omega

. Using Ohm's law (

V = IR

), the current is

I = \frac{V}{R_{total}} = \frac{12 \text{ V}}{12.0 \, \Omega} = 1.0 \text{ A}

. Since current is identical through all components in a series circuit, the current through the

8.0 \, \Omega

resistor is

1.0 \text{ A}

, making B correct. Choice A (

0.5 \text{ A}

) likely comes from incorrectly using just the

8.0 \, \Omega

resistor with half the voltage (

6 \text{ V}

), perhaps thinking the voltage splits equally. Choice C (

1.5 \text{ A}

) might result from using only the

8.0 \, \Omega

resistor with the full

12 \text{ V}

, ignoring the

4.0 \, \Omega

resistor entirely. Choice D (

3.0 \text{ A}

) could come from using only the

4.0 \, \Omega

resistor with the full voltage. The key insight for series circuits: current is constant throughout, but you must use the total resistance to find that current. Don't be tempted to analyze individual resistors in isolation - the presence of other resistors affects the overall current flow.

Question 10

If the cross-sectional area of a wire is doubled while keeping the current constant, how does the drift velocity of the charge carriers change?

It increases by a factor of 4
It increases by a factor of 2
It remains the same
It decreases by a factor of 2 (correct answer)
It decreases by a factor of 4

Explanation: When you encounter questions about current and wire properties, think about the fundamental relationship between current, charge carrier density, and drift velocity. This tests your understanding of how charge moves through conductors. Current is defined by the equation

I = nAv_d q

, where

I

is current,

n

is the number density of charge carriers,

A

is the cross-sectional area,

v_d

is drift velocity, and

q

is the charge per carrier. Since the problem states that current remains constant while area doubles, you can solve for how drift velocity changes. If

I

is constant and

A

doubles (while

n

and

q

remain unchanged for the same material), then

v_d

must decrease by a factor of 2 to maintain the same current. Think of it like water flow: if you double the pipe's cross-sectional area but want the same flow rate, the water must move half as fast. Choice A (increases by factor of 4) incorrectly suggests drift velocity increases with area squared. Choice B (increases by factor of 2) mistakenly assumes drift velocity is directly proportional to area, perhaps confusing it with total charge flow capacity. Choice C (remains the same) ignores the inverse relationship between area and drift velocity when current is held constant. Remember this key relationship: for constant current, drift velocity is inversely proportional to cross-sectional area. Larger wires carry the same current with slower-moving charge carriers, which is why thick wires are preferred for high-current applications—they reduce the kinetic energy and heating effects.

Question 11

If the number density of charge carriers in a conductor is increased by $50\%$ while keeping the current and cross-sectional area constant, how does the drift velocity change?

It increases by $50\%$
It increases by $33\%$
It remains constant
It decreases by $33\%$ (correct answer)
It decreases by $50\%$

Explanation: This question tests your understanding of the relationship between current, charge carrier density, and drift velocity in conductors. When analyzing current flow, you need to consider how these three quantities interact through the fundamental equation for current. Current in a conductor is given by

I = nqAv_d

, where

n

is the number density of charge carriers,

q

is the charge per carrier,

A

is the cross-sectional area, and

v_d

is the drift velocity. Since current and cross-sectional area remain constant, and the charge per carrier doesn't change, we have

nv_d = \text{constant}

. If the number density increases by 50%, then

n_{new} = 1.5n_{original}

. For the product

nv_d

to remain constant, the drift velocity must decrease proportionally:

v_{d,new} = \frac{v_{d,original}}{1.5} = \frac{2}{3}v_{d,original}

. This represents a decrease of

\frac{1}{3}

or 33%. Choice A incorrectly assumes drift velocity increases proportionally with carrier density. Choice B makes the same proportional error but with the wrong percentage calculation. Choice C wrongly suggests that adding more charge carriers doesn't affect how fast each one needs to move to maintain the same current. Choice D correctly recognizes the inverse relationship—more carriers means each one moves slower to carry the same total current. Remember this inverse relationship: when charge carrier density increases while current stays constant, drift velocity must decrease proportionally. This appears frequently on physics exams testing current and conductivity concepts.

Question 12

A $24 \text{ V}$ battery is connected to a series combination of a $2.0 \, \Omega$ resistor and a $4.0 \, \Omega$ resistor. If an ammeter is inserted in series with these resistors, what current will it read?

$2.0 \text{ A}$
$4.0 \text{ A}$ (correct answer)
$6.0 \text{ A}$
$8.0 \text{ A}$
$12.0 \text{ A}$

Explanation: When you encounter a circuit problem with resistors in series, remember that the current is the same throughout the entire circuit, and you need to find the total resistance before applying Ohm's law. In this series circuit, the total resistance equals the sum of individual resistances:

R_{total} = 2.0 \, \Omega + 4.0 \, \Omega = 6.0 \, \Omega

. Using Ohm's law (

V = IR

), you can solve for current:

I = \frac{V}{R} = \frac{24 \text{ V}}{6.0 \, \Omega} = 4.0 \text{ A}

. The ammeter will read this same current regardless of where it's placed in the series circuit. Looking at the wrong answers: Choice (A) gives

2.0 \text{ A}

, which you'd get if you mistakenly divided the voltage by only one of the resistors (

24 \text{ V} \div 12 \, \Omega = 2.0 \text{ A}

, but there's no

12 \, \Omega

resistor here). Choice (C) gives

6.0 \text{ A}

, which equals the total resistance value - this suggests confusing resistance with current or forgetting to actually divide voltage by resistance. Choice (D) gives

8.0 \text{ A}

, which you might get from incorrectly using parallel resistance formulas or other calculation errors. Study tip: For series circuits, always remember "add resistances, same current everywhere." First find total resistance by adding individual resistances, then use Ohm's law once. Don't overthink where the ammeter is placed - in series circuits, current is identical at every point.

Question 13

In a conductor, electrons move with an average drift velocity of $2.0 \times 10^{-4} \text{ m/s}$ . If the electron density is $6.0 \times 10^{28} \text{ m}^{-3}$ and the cross-sectional area is $1.5 \times 10^{-6} \text{ m}^2$ , what is the magnitude of the current?

$0.96 \text{ A}$
$1.44 \text{ A}$
$2.88 \text{ A}$ (correct answer)
$4.32 \text{ A}$
$5.76 \text{ A}$

Explanation: When you encounter current calculations involving drift velocity, you're dealing with the microscopic picture of electrical conduction. Current represents the total charge flowing past a point per unit time, which depends on how fast charge carriers move (drift velocity) and how many are available. The relationship between current and drift velocity is:

I = nqv_dA

, where

n

is the charge carrier density,

q

is the charge per carrier,

v_d

is drift velocity, and

A

is cross-sectional area. Substituting the given values:

n = 6.0 \times 10^{28} \text{ m}^{-3}

q = 1.6 \times 10^{-19} \text{ C}

(elementary charge),

v_d = 2.0 \times 10^{-4} \text{ m/s}

, and

A = 1.5 \times 10^{-6} \text{ m}^2

I = (6.0 \times 10^{28})(1.6 \times 10^{-19})(2.0 \times 10^{-4})(1.5 \times 10^{-6})

I = 2.88 \text{ A}

This confirms answer C is correct. Answer A (0.96 A) likely results from forgetting to include the cross-sectional area in the calculation. Answer B (1.44 A) might come from using an incorrect charge value or making an arithmetic error in the powers of 10. Answer D (4.32 A) could result from incorrectly multiplying by an extra factor or mishandling the scientific notation. Remember this formula:

I = nqv_dA

. The key insight is that current depends on both the microscopic motion (drift velocity) and the macroscopic geometry (area and carrier density). Always double-check your powers of 10 in these calculations.

Question 14

A conducting wire has a current density of $2.5 \times 10^6 \text{ A/m}^2$ . If the cross-sectional area of the wire is $3.0 \times 10^{-6} \text{ m}^2$ , what is the total current in the wire?

$7.5 \text{ A}$ (correct answer)
$8.3 \times 10^{11} \text{ A}$
$0.75 \text{ A}$
$2.5 \text{ A}$
$15.0 \text{ A}$

Explanation: When you encounter current density problems, you're working with the relationship between how much current flows through a given cross-sectional area. Current density (J) represents current per unit area, so to find total current, you multiply current density by the cross-sectional area. The calculation is straightforward:

I = J \times A

, where

I

is current,

J

is current density, and

A

is cross-sectional area. Substituting the given values:

I = (2.5 \times 10^6 \text{ A/m}^2) \times (3.0 \times 10^{-6} \text{ m}^2)

I = 2.5 \times 3.0 \times 10^6 \times 10^{-6} = 7.5 \times 10^0 = 7.5 \text{ A}

This confirms answer A is correct. Looking at the wrong answers: Answer B (

8.3 \times 10^{11} \text{ A}

) results from incorrectly dividing current density by area instead of multiplying, or making a significant arithmetic error with the exponents. Answer C (

0.75 \text{ A}

) comes from a decimal placement error—likely forgetting to properly handle the powers of 10 or misplacing a decimal point. Answer D (

2.5 \text{ A}

) suggests you might have ignored the cross-sectional area entirely and just used the current density value. Remember this key relationship: current density tells you current per unit area, so multiply by total area to get total current. The units work out perfectly—A/m² times m² gives you A. Always check that your final answer has reasonable units and magnitude for the physical situation.

Question 15

In a parallel circuit, if the voltage across each branch is $12 \text{ V}$ and the branch currents are $2.0 \text{ A}$ , $3.0 \text{ A}$ , and $1.5 \text{ A}$ , what is the equivalent resistance of the entire circuit?

$1.2 \, \Omega$
$1.8 \, \Omega$ (correct answer)
$2.4 \, \Omega$
$4.0 \, \Omega$
$6.5 \, \Omega$

Explanation: When you encounter parallel circuit problems, remember that all branches share the same voltage, but currents add up. The key is finding the total current first, then using Ohm's law to find equivalent resistance. Since this is a parallel circuit, you know the voltage across each branch is the same (12 V). To find the equivalent resistance, you need the total current flowing from the source. In parallel circuits, the total current equals the sum of all branch currents:

I_{total} = 2.0 + 3.0 + 1.5 = 6.5 \text{ A}

. Now apply Ohm's law to the entire circuit:

R_{eq} = \frac{V}{I_{total}} = \frac{12 \text{ V}}{6.5 \text{ A}} = 1.8 \, \Omega

. This confirms answer B is correct. Let's examine why the other answers are wrong. Answer A (

1.2 \, \Omega

) would result if you incorrectly used

I_{total} = 10 \text{ A}

, perhaps by miscalculating the sum. Answer C (

2.4 \, \Omega

) corresponds to using only

5.0 \text{ A}

, which might happen if you accidentally excluded one branch current. Answer D (

4.0 \, \Omega

) would result from using

3.0 \text{ A}

as your total current, suggesting you only considered the largest branch current instead of summing all branches. Study tip: For parallel circuits, always remember "voltage same, currents add." Write down

I_{total} = I_1 + I_2 + I_3 + ...

first, then use

R_{eq} = \frac{V}{I_{total}}

. Double-check your current addition—it's the most common error source in these problems.

Question 16

A $1.5 \text{ V}$ battery is connected to a resistor, causing a current of $0.3 \text{ A}$ to flow. If a $4.5 \text{ V}$ battery replaces the $1.5 \text{ V}$ battery, what will be the new current?

$0.1 \text{ A}$
$0.3 \text{ A}$
$0.9 \text{ A}$ (correct answer)
$1.35 \text{ A}$
$2.7 \text{ A}$

Explanation: This question tests your understanding of Ohm's Law, which states that voltage, current, and resistance are related by

V = IR

. When you encounter circuit problems involving these three quantities, always remember that resistance remains constant for a given resistor unless temperature changes significantly. First, let's find the resistance of the circuit. Using the initial conditions:

R = \frac{V}{I} = \frac{1.5 \text{ V}}{0.3 \text{ A}} = 5 \text{ Ω}

. This resistance stays the same when we change batteries. Now with the

4.5 \text{ V}

battery, we can find the new current:

I = \frac{V}{R} = \frac{4.5 \text{ V}}{5 \text{ Ω}} = 0.9 \text{ A}

. Notice that tripling the voltage (from

1.5 \text{ V}

4.5 \text{ V}

) triples the current (from

0.3 \text{ A}

0.9 \text{ A}

) because current is directly proportional to voltage when resistance is constant. Choice A (

0.1 \text{ A}

) incorrectly assumes an inverse relationship between voltage and current. Choice B (

0.3 \text{ A}

) suggests current stays constant regardless of voltage, which violates Ohm's Law. Choice D (

1.35 \text{ A}

) might result from incorrectly multiplying the original current by the voltage ratio

\frac{4.5}{1.0}

instead of

\frac{4.5}{1.5}

. For Ohm's Law problems, always identify what stays constant (usually resistance) and use proportional reasoning: if voltage triples, current triples too. This direct relationship is fundamental to understanding electrical circuits.

Question 17

A copper wire has a cross-sectional area of $4.0 \times 10^{-6} \text{ m}^2$ and carries a current of $8.0 \text{ A}$ . If the drift velocity of electrons is $1.5 \times 10^{-4} \text{ m/s}$ , what is the number density of free electrons in the copper?

$4.2 \times 10^{28} \text{ m}^{-3}$
$6.3 \times 10^{28} \text{ m}^{-3}$
$8.3 \times 10^{28} \text{ m}^{-3}$ (correct answer)
$1.2 \times 10^{29} \text{ m}^{-3}$
$2.1 \times 10^{29} \text{ m}^{-3}$

Explanation: When you encounter problems involving current and electron motion in conductors, you're dealing with the microscopic picture of electrical conduction. The key relationship connects macroscopic current to the microscopic motion of charge carriers. The fundamental equation here is

I = nAv_d e

, where

I

is current,

n

is the number density of free electrons,

A

is cross-sectional area,

v_d

is drift velocity, and

e

is the elementary charge (

1.6 \times 10^{-19}

C). Solving for number density:

n = \frac{I}{Av_d e}

. Substituting the given values:

n = \frac{8.0 \text{ A}}{(4.0 \times 10^{-6} \text{ m}^2)(1.5 \times 10^{-4} \text{ m/s})(1.6 \times 10^{-19} \text{ C})}

n = \frac{8.0}{9.6 \times 10^{-29}} = 8.3 \times 10^{28} \text{ m}^{-3}

This confirms answer C is correct. Answer A (

4.2 \times 10^{28}

) results from accidentally doubling the elementary charge or making a similar arithmetic error. Answer B (

6.3 \times 10^{28}

) suggests confusion in the calculation, possibly mixing up the order of operations. Answer D (

1.2 \times 10^{29}

) is too high by about 50%, likely from forgetting to include one of the given parameters or using incorrect units. Remember this formula:

n = \frac{I}{Av_d e}

. Always double-check that you're using the elementary charge (

1.6 \times 10^{-19}

C) and that all units are consistent. The number density of free electrons in metals is typically on the order of

10^{28}

10^{29}

per cubic meter.

Question 18

The current through a resistor varies according to $I(t) = 2.0 + 1.5t$ , where $I$ is in amperes and $t$ is in seconds. What is the total charge that flows through the resistor from $t = 0$ to $t = 4.0 \text{ s}$ ?

$8.0 \text{ C}$
$12.0 \text{ C}$
$16.0 \text{ C}$
$20.0 \text{ C}$ (correct answer)
$32.0 \text{ C}$

Explanation: When you encounter a time-varying current and need to find total charge, remember that current is the rate of charge flow:

I = \frac{dq}{dt}

. To find the total charge that flows over a time interval, you need to integrate the current function. Given

I(t) = 2.0 + 1.5t

, the charge is found by integrating from

t = 0

t = 4.0

q = \int_0^{4.0} I(t) \, dt = \int_0^{4.0} (2.0 + 1.5t) \, dt

q = \left[2.0t + \frac{1.5t^2}{2}\right]_0^{4.0} = \left[2.0t + 0.75t^2\right]_0^{4.0}

q = [2.0(4.0) + 0.75(4.0)^2] - [0] = 8.0 + 0.75(16) = 8.0 + 12.0 = 20.0 \text{ C}

Answer D is correct. Answer A (

8.0 \text{ C}

) represents only the constant current contribution, ignoring the time-dependent term entirely. Answer B (

12.0 \text{ C}

) captures only the variable current contribution, missing the constant term. Answer C (

16.0 \text{ C}

) likely results from an integration error, perhaps incorrectly handling the coefficient in the quadratic term. Study tip: For any time-varying current problem, always integrate

I(t)

over the given time interval to find charge. Don't forget that integration of polynomials follows the power rule:

\int t^n dt = \frac{t^{n+1}}{n+1}

. Double-check by evaluating each term separately.

Question 19

Two resistors of $3.0 \, \Omega$ and $6.0 \, \Omega$ are connected in parallel to a $9.0 \text{ V}$ battery. What is the current through the $3.0 \, \Omega$ resistor?

$1.0 \text{ A}$
$1.5 \text{ A}$
$3.0 \text{ A}$ (correct answer)
$4.5 \text{ A}$
$6.0 \text{ A}$

Explanation: When analyzing parallel circuits, remember that each resistor experiences the same voltage as the battery, but currents through individual resistors depend on their resistance values according to Ohm's law. In this parallel circuit, both the

3.0 \, \Omega

and

6.0 \, \Omega

resistors are directly connected across the

9.0 \text{ V}

battery. This means each resistor has

9.0 \text{ V}

across it, regardless of what other resistors are present. To find the current through the

3.0 \, \Omega

resistor, apply Ohm's law:

I = \frac{V}{R} = \frac{9.0 \text{ V}}{3.0 \, \Omega} = 3.0 \text{ A}

. This confirms answer C is correct. Now let's examine why the other answers are wrong. Answer A (

1.0 \text{ A}

) would result from incorrectly using the equivalent resistance of the parallel combination (

2.0 \, \Omega

) instead of the individual resistor's resistance. Answer B (

1.5 \text{ A}

) represents the current through the

6.0 \, \Omega

resistor, showing confusion about which resistor the question asks about. Answer D (

4.5 \text{ A}

) is the total current from the battery, which you'd find by adding the individual currents or using the equivalent resistance—but this isn't what the question requests. Study tip: In parallel circuits, always remember that voltage is the same across each branch, but current divides inversely with resistance. Higher resistance means lower current. Create a mental checklist: identify the voltage across the specific resistor, then apply Ohm's law directly to that resistor alone.

Question 20

A circuit contains three resistors in series: $5.0 \, \Omega$ , $10.0 \, \Omega$ , and $15.0 \, \Omega$ . If the current through the $10.0 \, \Omega$ resistor is $0.40 \text{ A}$ , what is the current through the $15.0 \, \Omega$ resistor?

$0.27 \text{ A}$
$0.40 \text{ A}$ (correct answer)
$0.60 \text{ A}$
$1.2 \text{ A}$
$1.8 \text{ A}$

Explanation: When you encounter series circuit problems, remember that current behaves very differently from voltage. In a series circuit, there's only one path for current to flow, which creates a fundamental rule: the current must be identical through every component. Think of current like water flowing through a single pipe with different narrow sections (resistors). The same amount of water that enters one section must exit and continue to the next section—it has nowhere else to go. This is exactly how electric current behaves in series circuits. Since the current through the

10.0 \, \Omega

resistor is

0.40 \text{ A}

, the current through every other resistor in this series circuit must also be

0.40 \text{ A}

. This includes the

15.0 \, \Omega

resistor, making (B) $0.40 \text{ A}$ correct. The wrong answers represent common misconceptions: (A) $0.27 \text{ A}$ suggests incorrectly thinking current decreases as it encounters larger resistance values. (C) $0.60 \text{ A}$ and (D) $1.2 \text{ A}$ might come from incorrectly applying ratios of resistance values or confusing series rules with parallel circuit behavior, where currents do split and vary. Key strategy: When you see "series circuit" and "current through," immediately think "same current everywhere." This is one of the most tested concepts in circuit analysis. Don't let the different resistance values fool you—they affect voltage drops across each resistor, but current remains constant throughout the entire series path.