Physics 2 Quiz: Current Voltage Resistance And Power

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Current Voltage Resistance And PowerQuestion 1 of 9

Two wires, Wire 1 and Wire 2, are made of the same material and carry the same current $I$ . Wire 1 has length $L$ and radius $r$ . Wire 2 has length $2L$ and radius $2r$ . Which of the following correctly compares the power dissipated in each wire and provides the correct reasoning?

Wire 2 dissipates more power than Wire 1, because it is longer and therefore has more resistance, and power is proportional to resistance for a fixed current.

Wire 1 dissipates more power than Wire 2, because although Wire 2 is longer, its larger cross-sectional area reduces its resistance by a greater factor than the length increases it, resulting in lower resistance and thus lower power at fixed current.

Both wires dissipate equal power, because the increase in length and the increase in radius cancel exactly when computing

P = I^2 R

Wire 2 dissipates more power than Wire 1, because its greater volume means more charge carriers are accelerated by the field, increasing the total energy transferred per unit time regardless of cross-section effects on resistance.

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Physics 2 Quiz

Physics 2 Quiz: Current Voltage Resistance And Power

Practice Current Voltage Resistance And Power in Physics 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Current Voltage Resistance And Power, giving you a quick way to practice the rules, question types, and explanations that matter most for Physics 2.

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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.

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Question 1

Wire 2 dissipates more power than Wire 1, because it is longer and therefore has more resistance, and power is proportional to resistance for a fixed current.
Wire 1 dissipates more power than Wire 2, because although Wire 2 is longer, its larger cross-sectional area reduces its resistance by a greater factor than the length increases it, resulting in lower resistance and thus lower power at fixed current. (correct answer)
Both wires dissipate equal power, because the increase in length and the increase in radius cancel exactly when computing $P = I^2 R$ .
Wire 2 dissipates more power than Wire 1, because its greater volume means more charge carriers are accelerated by the field, increasing the total energy transferred per unit time regardless of cross-section effects on resistance.

Explanation: When a question compares resistance and power across wires of different geometry, your first move should always be to write out the resistance formula:

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

, where

A = \pi r^2

. This tells you that resistance scales linearly with length but inversely with the square of the radius — and that asymmetry is the key to this problem. For Wire 1:

R_1 = \frac{\rho L}{\pi r^2}

. For Wire 2:

R_2 = \frac{\rho (2L)}{\pi (2r)^2} = \frac{2\rho L}{4\pi r^2} = \frac{1}{2} \cdot \frac{\rho L}{\pi r^2} = \frac{R_1}{2}

. Doubling the length doubles resistance, but doubling the radius quadruples the cross-sectional area, cutting resistance by four — net result is Wire 2 has half the resistance of Wire 1. Since both carry the same current

I

, and

P = I^2 R

, Wire 1 dissipates twice the power of Wire 2. That confirms B is correct. A is wrong because it only considers the effect of length on resistance, completely ignoring the radius. Doubling the radius has a larger effect than doubling the length, so resistance actually decreases. C is wrong because the length and radius changes do not cancel — the radius change wins by a factor of two. D is a seductive but physically incorrect argument; power dissipation is determined by

P = I^2R

, not by the volume of charge carriers present. Your takeaway: whenever a wire's dimensions change, always compute

R = \frac{\rho L}{\pi r^2}

explicitly — radius changes are easy to underestimate because they enter squared.

Question 2

A resistor is connected to a battery with terminal voltage $V_T$ . When a second identical resistor is connected in parallel with the first, the battery's internal resistance $r$ causes the terminal voltage to drop to $V_T'$ . The EMF of the battery is $\mathcal{E}$ and the internal resistance is $r$ .

After the second resistor is added in parallel, the power dissipated in the internal resistance of the battery changes by what factor compared to before the second resistor was added? Assume each external resistor has resistance $R$ .

The power in $r$ increases by a factor of 4, because the total current drawn from the battery doubles when the external resistance is halved, and power scales as $I^2$ .
The power in $r$ increases by a factor of 2, because halving the external resistance doubles the current and power scales linearly with current in the internal resistor.
The power in $r$ increases by a factor of $\left(\dfrac{2(R+r)}{R+2r}\right)^2$ , because the current does not exactly double when $r \neq 0$ , and the factor depends on both $R$ and $r$ in a way that does not reduce to a simple integer. (correct answer)
The power in $r$ increases by a factor of 4 when $r \ll R$ , and by a factor less than 4 for any finite $r$ , but the exact expression cannot be written as a closed-form ratio of $R$ and $r$ .

Explanation: Whenever you see a question about power dissipation in a battery's internal resistance, your instinct should be to track how the total current changes — and to remember that internal resistance prevents current from simply doubling when external resistance halves. Before the second resistor is added, the total current is

I_1 = \dfrac{\mathcal{E}}{R + r}

. After adding an identical resistor in parallel, the external resistance becomes

\dfrac{R}{2}

, so the new current is

I_2 = \dfrac{\mathcal{E}}{\frac{R}{2} + r} = \dfrac{2\mathcal{E}}{R + 2r}

. The power dissipated in

r

scales as

I^2 r

, so the ratio of new to old power is:

\frac{P_{\text{new}}}{P_{\text{old}}} = \left(\frac{I_2}{I_1}\right)^2 = \left(\frac{2\mathcal{E}}{R+2r} \cdot \frac{R+r}{\mathcal{E}}\right)^2 = \left(\frac{2(R+r)}{R+2r}\right)^2

This is exactly what C states. The factor depends on both

R

and

r

and does not simplify to a clean integer unless

r = 0

. A is tempting but wrong — it assumes current exactly doubles, which would only be true if

r = 0

. With finite

r

, the denominator increases by less than a factor of 2, so the current increase is less than double, making the power factor less than 4. B commits two errors: it assumes current doubles and incorrectly treats power as linear in current rather than quadratic (

P = I^2 r

, not

Ir

). D is wrong because the exact expression can be written in closed form — it's precisely the ratio in C. Study tip: Whenever internal resistance is mentioned, resist the urge to treat the battery as ideal. Always recompute the current using the full loop:

I = \dfrac{\mathcal{E}}{R_{\text{ext}} + r}

, then apply

P = I^2 r

Question 3

A conducting wire of resistivity $\rho$ , length $L$ , and uniform circular cross-section of radius $r$ is bent into a closed loop and connected to nothing (it is a standalone loop with no external source). A student argues that since the wire forms a complete circuit, a steady current must flow around the loop.

Which of the following is the most physically rigorous explanation for why no steady current flows in this isolated loop, and what does this imply about the relationship between voltage and current in this context?

No steady current flows because the wire's resistance $R = \rho L/(\pi r^2)$ is nonzero; by Ohm's law $I = V/R$ , any nonzero resistance prevents current flow unless an external voltage is applied, confirming that resistance is the barrier to current.
No steady current flows because the electrons in the wire are in thermal equilibrium and their random thermal velocities average to zero net flow; this is a statistical effect unrelated to the concept of voltage or resistance.
No steady current flows because there is no potential difference (voltage) around the loop to drive charge carriers; without an EMF source, there is no sustained electric field along the wire to maintain drift, so $V = 0$ implies $I = 0$ regardless of the value of $R$ . (correct answer)
No steady current flows because the closed loop has no net charge, and current requires a separation of charge between two terminals; without terminals, the concepts of voltage, resistance, and current do not apply to the loop at all.

Explanation: Whenever you see a question about current flow in a circuit, your first instinct should be to ask: what is driving the current? Current doesn't flow simply because a path exists — it flows because something pushes charge carriers along that path. In a conducting loop with no external source, there is no EMF — no battery, no changing magnetic flux, no chemical potential difference — meaning there is no sustained electric field along the wire. Without a potential difference to do work on electrons, there is no net force driving drift, and therefore no steady current. This is the essence of C:

V = 0

implies

I = 0

through Ohm's law, and the value of

R

is simply irrelevant when there's nothing to drive the current in the first place. A gets the logic exactly backwards. Resistance doesn't prevent current — it limits current when a voltage is present. A superconductor (

R = 0

) with no EMF also carries no steady current. Resistance is not the barrier here; the absence of a driving voltage is. B contains a true statement — thermal velocities do average to zero — but this is a consequence of equilibrium, not the root cause rooted in circuit physics. The question asks for the most rigorous circuit-theory explanation, and thermal statistics alone misses the role of EMF and potential difference entirely. D is wrong because charge separation between terminals is not required for voltage, resistance, and current to be meaningful concepts. These quantities apply to loops and distributed circuits quite generally. Study tip: Always identify the EMF source before analyzing any circuit. No EMF → no voltage → no current, regardless of how the path is shaped or what its resistance is.

Question 4

A cylindrical conductor of length $L$ and cross-sectional area $A$ carries a steady current $I$ . The free electron number density is $n$ and each electron carries charge $-e$ .

Which of the following correctly expresses the drift velocity $v_d$ of the conduction electrons in terms of the given quantities, and what is the direction of $v_d$ relative to the conventional current direction $\hat{I}$ ?

$v_d = \frac{I}{nAe}$ , directed opposite to $\hat{I}$ , because electrons are negatively charged and move against the conventional current direction. (correct answer)
$v_d = \frac{I}{nAe}$ , directed parallel to $\hat{I}$ , because the magnitude formula is the same regardless of charge sign and current defines the electron flow direction.
$v_d = \frac{I}{nA(-e)}$ , which is negative, indicating the drift speed is undefined without knowing the sign convention for the area vector of the conductor.
$v_d = \frac{nAe}{I}$ , directed opposite to $\hat{I}$ , because resistance increases with electron density, inverting the relationship between current and drift velocity.

Explanation: When you see a question about drift velocity and current direction, you need to hold two ideas simultaneously: the magnitude formula derived from charge flow, and the direction correction due to electron charge sign. Start from the fundamental current definition. Current

I

represents the net positive charge flowing past a cross-section per unit time. In a conductor, electrons (charge

-e

) drift with speed

v_d

through

n

electrons per unit volume across area

A

. The magnitude of current is

I = nAev_d

, which rearranges to

v_d = \frac{I}{nAe}

. Notice that

n

A

e

, and

I

are all treated as positive magnitudes here — the formula gives you the speed of drift. Because conventional current is defined as the direction positive charges would flow, and electrons carry negative charge, electrons must physically move opposite to the conventional current direction

\hat{I}

. Answer A captures both facts correctly. Answer B fails on direction: it wrongly claims electrons move parallel to conventional current. This confuses the motion of positive charge carriers with actual electron motion — a classic trap. Answer C inserts

-e

directly into the denominator, producing a negative "drift velocity," and then invents a nonsensical conclusion about undefined speed. The sign of charge is already accounted for in the directional reasoning, not by making the formula negative. Answer D inverts the relationship entirely, placing

I

in the denominator. Higher current means faster drift, not slower — so this inverted formula has no physical basis. Study tip: Always derive drift velocity using magnitudes (

v_d = \frac{I}{nAe}

), then separately determine direction by remembering that electrons move opposite to

\hat{I}

. Keeping these two steps distinct prevents sign errors.

Question 5

A current $I$ flows through a resistor of resistance $R$ , and the same current flows through a second resistor of resistance $4R$ . A student wishes to express the ratio of the voltage across the second resistor to the power dissipated in the first resistor. Which of the following is the correct expression for this ratio, and what are its SI units?

The ratio is $\dfrac{4R}{I}$ , with SI units of $\Omega\cdot\text{A}^{-1}$ , because voltage equals $4IR$ and power equals $IR$ (one factor of current multiplied by resistance), so one factor of $I$ cancels.
The ratio is $\dfrac{4}{I}$ , with SI units of $\text{A}^{-1}$ , because the resistance factors cancel when dividing voltage by power, leaving only the inverse of the current. (correct answer)
The ratio is $\dfrac{4R}{I}$ , with SI units of $\Omega$ (ohms), because $\text{V}/\text{W} = \text{V}/(\text{V}\cdot\text{A}) = \text{A}^{-1}$ , which equals one ohm when resistance is held constant.
The ratio is $\dfrac{4}{R}$ , with SI units of $\Omega^{-1} = \text{S}$ (siemens), because dividing voltage (proportional to $IR$ ) by power (proportional to $I^2 R$ ) cancels one factor of $IR$ , leaving $1/R$ .

Explanation: When a question asks you to form a ratio of two electrical quantities, your first job is to write out each quantity symbolically, simplify the algebra, then track the units separately as a check. The voltage across the second resistor (resistance

4R

) is

V_2 = I(4R) = 4IR

. The power dissipated in the first resistor (resistance

R

) is

P_1 = I^2 R

. Dividing them gives:

\frac{V_2}{P_1} = \frac{4IR}{I^2R} = \frac{4}{I}

One factor of

I

and one factor of

R

cancel, leaving

\dfrac{4}{I}

. For units:

\dfrac{\text{V}}{\text{W}} = \dfrac{\text{V}}{\text{V} \cdot \text{A}} = \text{A}^{-1}

, confirming the SI units are inverse amperes. That makes B correct. Choice A gets the numerator right (

4IR

) but incorrectly writes power as

IR

instead of

I^2R

, leaving an extra factor of

R

that shouldn't be there — and then misidentifies the units as

\Omega \cdot \text{A}^{-1}

rather than

\text{A}^{-1}

. Choice C arrives at the right expression

\dfrac{4R}{I}

(same error as A) and then contradicts itself by correctly stating

\text{V}/\text{W} = \text{A}^{-1}

, yet claims those units equal ohms — a false equivalence. Choice D flips the cancellation: it correctly cancels one power of

IR

but loses track of the remaining

I

, producing

1/R

instead of

4/I

. Strategy tip: Always write the full symbolic expression for every quantity before canceling — sloppy cancellation is the trap in nearly every ratio problem on this exam.

Question 6

A student claims: 'Voltage is the energy per unit charge delivered to a circuit element, so a 9 V battery always delivers 9 J of energy to every coulomb of charge that flows through any external resistor connected to it, regardless of the internal resistance of the battery.' Which of the following most precisely identifies the flaw in this claim?

The claim is flawed because voltage is defined as energy per unit charge only for capacitors, not for resistors; for resistors, energy per unit charge is defined by the current and the resistance separately.
The claim is flawed because the 9 V refers to the battery's EMF, which represents the total energy per unit charge supplied by the chemical reaction; the energy per unit charge delivered to the external resistor is the terminal voltage $V_T = \mathcal{E} - Ir$ , which is less than 9 V whenever current flows through the nonzero internal resistance $r$ . (correct answer)
The claim is flawed because energy per unit charge depends on the resistance of the external element; a higher-resistance external resistor receives more energy per coulomb than a lower-resistance one, so 9 J/C is only correct for one specific resistance value.
The claim is flawed because voltage is not defined as energy per unit charge; rather, it is defined as the work done per unit time by the electric field, so the student is using the wrong definition of voltage from the outset.

Explanation: When analyzing battery circuits, you must carefully distinguish between two related but different quantities: EMF and terminal voltage. The EMF (

\mathcal{E}

) represents the total energy per unit charge that the battery's chemical reaction supplies — this is the 9 V stamped on the battery. However, any real battery also has an internal resistance

r

, and whenever current

I

flows, some energy is dissipated inside the battery itself. The voltage actually available at the battery's terminals — and therefore delivered to the external resistor — is the terminal voltage:

V_T = \mathcal{E} - Ir

. Only when

r = 0

(an ideal battery) or

I = 0

does the terminal voltage equal the full EMF. This is exactly the flaw in the student's claim, making B correct. Choice A is wrong because "energy per unit charge" (i.e., voltage) is a universal definition that applies to any circuit element — resistors, capacitors, and inductors alike. It's not restricted to capacitors. Choice C is subtly tempting but incorrect. The terminal voltage depends on the current drawn (which depends on external resistance), but for a given circuit, every coulomb passing through the external resistor receives the same terminal voltage — not a value that varies by resistance type. Choice D is flatly wrong. Voltage is defined as energy per unit charge (

V = W/q

), not power (energy per unit time). That definition belongs to power:

P = W/t

. Study tip: Whenever a problem mentions a battery with internal resistance, immediately write

V_T = \mathcal{E} - Ir

— the EMF and terminal voltage are never interchangeable once current flows.

Question 7

An engineer doubles the operating voltage across a fixed resistor while simultaneously replacing it with a resistor of half the original resistance. By what factor does the power dissipated change, and which formula most directly yields this result?

Power increases by a factor of 4, most directly from $P = V^2/R$ : doubling $V$ multiplies the numerator by $2^2 = 4$ , while halving $R$ has no additional effect on the ratio because the change in resistance is already absorbed into the new operating conditions.
Power increases by a factor of 4, most directly from $P = IV$ : the current doubles because voltage doubles while resistance is unchanged, and with voltage also doubled, $P' = (2I)(2V) = 4P$ .
Power increases by a factor of 8 from $P = V^2/R$ , but only a factor of 4 from $P = I^2R$ , because the two formulas are not equivalent when both voltage and resistance change simultaneously.
Power increases by a factor of 8, most directly from $P = V^2/R$ : doubling $V$ contributes a factor of $2^2 = 4$ in the numerator, and halving $R$ contributes an additional factor of $1/(1/2) = 2$ in the denominator, giving an overall factor of $4 \times 2 = 8$ . (correct answer)

Explanation: When both voltage and resistance change simultaneously, your most reliable strategy is to apply the power formula that directly contains both variables:

P = V^2/R

. This lets you track each change independently without introducing intermediate steps that can cause errors. Starting with

P = V^2/R

, doubling the voltage replaces

V

with

2V

, so the numerator becomes

(2V)^2 = 4V^2

— a factor of 4. Halving the resistance replaces

R

with

R/2

in the denominator, which means dividing by a smaller number:

\frac{4V^2}{R/2} = \frac{8V^2}{R} = 8P

. Both changes multiply together, giving an overall factor of 8. Answer D captures this correctly. Answer A correctly identifies

P = V^2/R

as the right formula but then claims halving

R

has "no additional effect" — a critical error. Halving the denominator always doubles the result, and you cannot ignore it. Answer B makes a subtler mistake: it uses

P = IV

and doubles the current, but current actually quadruples here (

I = V/R

: voltage doubles and resistance halves, so

I' = 2V/(R/2) = 4I

). Plugging in correctly gives

P' = (4I)(2V) = 8P

, not 4P. Answer C introduces a false claim that the power formulas

P = V^2/R

and

P = I^2R

give different answers — they are always equivalent and will both yield a factor of 8 if applied correctly. Your takeaway: when multiple quantities change, list each substitution explicitly and never skip a step. The formulas never contradict each other — if they seem to, you've made a substitution error.

Question 8

A copper wire and a nichrome wire are connected end-to-end (in series) and carry the same current $I$ . Copper has resistivity $\rho_{Cu} \approx 1.7 \times 10^{-8}\,\Omega\cdot\text{m}$ and nichrome has resistivity $\rho_{Ni} \approx 1.1 \times 10^{-6}\,\Omega\cdot\text{m}$ . Both wires have the same length and the same circular cross-sectional area.

A student measures the voltage across each wire separately and finds that the voltage across the nichrome wire is approximately 65 times the voltage across the copper wire. The student then claims: 'Since the nichrome wire has 65 times the voltage drop, it must also have 65 times the current.' Which of the following best evaluates this claim and correctly identifies the relevant circuit principle?

The student's claim is correct: Ohm's law states $V = IR$ , so if $V$ is 65 times larger across nichrome and both wires have the same resistance, the current must be 65 times larger through the nichrome wire to produce the larger drop.
The student's claim is partially correct: while the current is not exactly 65 times larger, it is somewhat larger through the nichrome wire because the higher resistance causes charge to accumulate at the junction, creating a slight current imbalance between the two wires.
The student's claim is incorrect: the larger voltage across nichrome means it has lower resistance by $V = IR$ , so nichrome actually draws less current than copper when connected to the same source, contradicting the student's reasoning.
The student's claim is incorrect: the two wires are in series, so they must carry the same current $I$ by Kirchhoff's current law (charge conservation). The 65× voltage ratio directly reflects the 65× resistance ratio ( $\rho_{Ni}/\rho_{Cu}$ ), not a current difference. (correct answer)

Explanation: Whenever you see resistors connected in series, the single most important fact to anchor on is that series elements share the same current — always. This question tests whether you can separate the concepts of current and voltage in a series circuit. Since the copper and nichrome wires are connected end-to-end with no branching paths, every coulomb of charge that flows through one wire must flow through the other. By Kirchhoff's current law (charge conservation), the current

I

is identical in both wires. The voltage drop across each wire is given by

V = IR

, and since resistance scales with resistivity via

R = \rho L / A

, the ratio of voltages equals the ratio of resistivities:

V_{Ni}/V_{Cu} = \rho_{Ni}/\rho_{Cu} \approx 1.1\times10^{-6}/1.7\times10^{-8} \approx 65

. The 65× voltage difference reflects the 65× resistance difference — not any current difference. D is correct. Choice A commits the student's exact error: it assumes that a larger voltage implies a larger current, ignoring that resistance is also 65× larger in the nichrome wire. Choice B introduces a fictional "charge accumulation" effect at the junction that causes a current imbalance — this doesn't happen in a simple series circuit; junctions between series elements do develop surface charges, but these enforce equal current flow, not unequal flow. Choice C inverts the reasoning entirely, incorrectly concluding that higher voltage means lower resistance. Study tip: In series circuits, current is the quantity that stays the same; voltage divides. In parallel circuits, it's the reverse — voltage stays the same, current divides. Keeping this distinction sharp will help you avoid the trap in choices A and B on any circuit question.

Question 9

A physicist measures the potential difference across a circuit element and the current through it as a function of temperature. At low temperatures, the element obeys Ohm's law with resistance $R_0$ . As temperature rises, the resistance increases linearly: $R(T) = R_0(1 + \alpha T)$ , where $\alpha > 0$ and $T$ is the temperature above a reference.

The element is connected to an ideal voltage source of fixed EMF $\mathcal{E}$ . As temperature increases from $T_1$ to $T_2 > T_1$ , which of the following correctly describes both the change in current and the change in power dissipated in the element?

Current decreases and power dissipated decreases, because with fixed voltage, $I = \mathcal{E}/R(T)$ falls as $R$ rises, and $P = \mathcal{E}^2/R(T)$ also falls as $R$ increases. (correct answer)
Current decreases and power dissipated increases, because although fewer electrons flow per second, each electron dissipates more energy due to increased scattering at higher temperatures.
Current decreases and power dissipated remains constant, because the decrease in current is exactly offset by the increase in resistance such that $P = I^2 R$ is conserved at fixed voltage.
Current increases and power dissipated decreases, because higher temperature increases the thermal energy of electrons, augmenting the drift velocity while simultaneously increasing collisions that waste energy.

Explanation: When a resistive element is connected to a fixed voltage source, two formulas govern everything:

I = \mathcal{E}/R(T)

and

P = \mathcal{E}^2/R(T)

. Notice that both depend on

R(T)

in the denominator — this is the key insight the question is testing. Since

R(T) = R_0(1 + \alpha T)

with

\alpha > 0

, resistance increases as temperature rises from

T_1

T_2

. Plugging into both formulas: a larger denominator means

I

decreases and

P

decreases simultaneously. That's exactly what A describes, making it the correct answer. B is tempting because it correctly identifies that current decreases, but the claim that power increases is wrong. With fixed

\mathcal{E}

, power is

\mathcal{E}^2/R

, not determined by "energy per electron" reasoning. More resistance means less power delivered, full stop. C makes a subtle algebra error. While

P = I^2 R

, both

I

and

R

are not independent —

I

itself depends on

R

. Substituting

I = \mathcal{E}/R

gives

P = \mathcal{E}^2/R

, which clearly does change with temperature. The offset described in C doesn't happen. D reverses the direction of current change entirely. Higher temperature increases resistance through more frequent collisions, which reduces drift velocity and therefore reduces current — thermal energy does not augment electron flow. Study tip: For fixed-voltage circuits, always rewrite current and power in terms of

\mathcal{E}

and

R

only — this eliminates confusion about whether changes "cancel out."