Electric Current - College Physics

1. What is the effect of adding a resistor in series on total $\vec{R}$?
   Answer: Total $\vec{R}$ increases because series resistances add algebraically This question tests understanding of electric current in DC circuits, specifically the effect on total resistance when adding components in series. 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. In a series circuit, adding a resistor increases the total resistance because resistances add algebraically, reducing overall current for a fixed voltage. The correct answer is based on the rule for series resistances, R_total = R1 + R2. A common distractor might incorrectly suggest that total R decreases, confusing series with parallel configurations. Teaching strategies include using circuit simulations to visualize resistance addition and reinforcing the differences between series and parallel circuits through hands-on experiments.
2. Two identical batteries, each with voltage $V$, are connected in series to a resistor $R$. If one battery is then removed and the circuit is reconnected with just one battery, how does the current change?
   Answer: The current decreases by a factor of 2 When analyzing circuits with batteries and resistors, you need to apply Ohm's law ($I = V/R$) and understand how voltages combine in series connections. Initially, two identical batteries with voltage $V$ each are connected in series to resistor $R$. In a series circuit, voltages add together, so the total voltage across the resistor is $V + V = 2V$. Using Ohm's law, the initial current is $I_1 = \frac{2V}{R}$. When one battery is removed, only one battery with voltage $V$ remains connected to the same resistor $R$. The new current becomes $I_2 = \frac{V}{R}$. Comparing the two currents: $\frac{I_2}{I_1} = \frac{V/R}{2V/R} = \frac{1}{2}$. This means the current decreases by a factor of 2. Choice A suggests the current increases by a factor of 4, which would require the voltage to quadruple rather than halve. Choice B claims the current increases by a factor of 2, incorrectly assuming that removing a battery somehow increases voltage. Choice C states the current remains the same, which would only be true if voltage stayed constant—but removing a series battery definitely reduces total voltage. The key insight is recognizing that batteries in series add their voltages together. When you remove one battery from a series connection, you're halving the driving voltage, which directly halves the current through any given resistance. Always remember: in series circuits, voltages add, and current is the same throughout the circuit.
3. 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?
   Answer: $1.1 \times 10^{-4} \text{ m/s}$ 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.
4. How does the voltage across each component in parallel compare to total supplied $\vec{V}$?
   Answer: Each component has the same $\vec{V}$ as the supply across the parallel network This question tests understanding of electric current in DC circuits, specifically parallel voltage comparison. 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. Each component voltage matches supplied V. The correct answer is based on parallel voltage sharing. A common distractor might suggest zero, thinking short. Teaching strategies include voltmeter placements and parallel vs. series contrasts.
5. 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?
   Answer: $1.8 \, \Omega$ 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.
6. In a basic battery–resistor–switch loop, what occurs when the switch is open?
   Answer: $\vec{I}$ is essentially zero because the circuit is not a closed conducting path This question tests understanding of electric current in DC circuits, specifically the behavior of open circuits. 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. When a switch is open in a loop, the circuit is incomplete, making current essentially zero. The correct answer is based on the requirement for a closed path in steady DC current flow. A common distractor might incorrectly suggest I increases, confusing open with short circuits. Teaching strategies include using circuit simulations to show open vs. closed states and reinforcing with hands-on switch experiments.
7. In a series string of resistors, what happens to $\vec{I}$ when another resistor is added?
   Answer: $\vec{I}$ decreases because total $\vec{R}$ increases at fixed source $\vec{V}$ This question tests understanding of electric current in DC circuits, specifically current response to added series resistance. 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 a resistor increases total R, decreasing I at fixed V. The correct answer is based on the inverse proportionality in Ohm's Law. A common distractor might suggest I stays the same, confusing with parallel additions. Teaching strategies include simulations of series strings and hands-on resistor addition experiments.
8. A wire carries a steady current of $2.5 \text{ A}$. If the cross-sectional area of the wire is doubled while keeping the same material and length, and the voltage across the wire remains constant, what will be the new current?
   Answer: $5.0 \text{ A}$ Resistance of a wire is $R = \rho \frac{L}{A}$ where $\rho$ is resistivity, $L$ is length, and $A$ is cross-sectional area. When area doubles, resistance becomes $R' = \rho \frac{L}{2A} = \frac{R}{2}$. Since voltage remains constant and $I = \frac{V}{R}$, when resistance halves, current doubles: $I' = \frac{V}{R'} = \frac{V}{R/2} = 2 \times \frac{V}{R} = 2I = 2 \times 2.5 = 5.0 \text{ A}$. Choice A incorrectly assumes current decreases. Choice B assumes no change. Choice D incorrectly applies the area doubling directly to current without considering the resistance relationship.
9. Refer to the circuit diagram. What is the current through the $6.0 \, \Omega$ resistor?
   Answer: $1.5 \text{ A}$ The circuit shows a $9.0 \text{ V}$ battery connected to a $6.0 \, \Omega$ resistor in parallel with a $3.0 \, \Omega$ resistor. In parallel, each resistor experiences the full battery voltage. For the $6.0 \, \Omega$ resistor: $I = \frac{V}{R} = \frac{9.0}{6.0} = 1.5 \text{ A}$. Choice A uses the parallel equivalent resistance, choice B uses an averaging approach, choice D uses the $3.0 \, \Omega$ resistance value, and choice E uses the $3.0 \, \Omega$ resistor's current.
10. A student measures the current in a circuit and finds that when a $6.0 \text{ Ω}$ resistor is added in series, the current decreases by $25\%$. What was the original resistance of the circuit before adding the resistor?
    Answer: $18 \text{ Ω}$ Let the original resistance be $R_0$ and voltage be $V$. Original current: $I_0 = \frac{V}{R_0}$. After adding $6.0 \text{ Ω}$: $I_1 = \frac{V}{R_0 + 6.0}$. Given that current decreases by $25\%$: $I_1 = 0.75 I_0$. Therefore: $\frac{V}{R_0 + 6.0} = 0.75 \times \frac{V}{R_0}$. Simplifying: $\frac{R_0}{R_0 + 6.0} = 0.75$, so $R_0 = 0.75(R_0 + 6.0) = 0.75 R_0 + 4.5$. Solving: $0.25 R_0 = 4.5$, so $R_0 = 18 \text{ Ω}$. Choice A assumes the added resistance equals original resistance. Choice B uses incorrect fractional relationship. Choice D comes from misapplying the percentage change.
11. A student connects three identical batteries in parallel, each with emf $1.5 \text{ V}$ and internal resistance $0.8 \text{ Ω}$. This battery pack is then connected to an external resistor. If the terminal voltage of the battery pack drops to $1.2 \text{ V}$ under load, what is the current being drawn from the external circuit?
    Answer: $1.1 \text{ A}$ When batteries are connected in parallel, the total emf remains $1.5 \text{ V}$, but the internal resistance becomes $\frac{0.8}{3} = 0.27 \text{ Ω}$. The voltage drop across the internal resistance is $\mathcal{E} - V_{terminal} = 1.5 - 1.2 = 0.3 \text{ V}$. Using Ohm's law for the internal resistance: $I = \frac{V_{drop}}{r_{total}} = \frac{0.3}{0.27} = 1.11 \text{ A} ≈ 1.1 \text{ A}$. Choice A uses the individual battery internal resistance instead of parallel equivalent. Choice B miscalculates the parallel internal resistance. Choice D uses incorrect voltage relationships.
12. 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?
    Answer: $7.5 \text{ A}$ 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.
13. 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?
    Answer: $2.5 \text{ A}$ 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.
14. In a series circuit, what happens to $\vec{I}$ when total $\vec{R}$ increases?
    Answer: $\vec{I}$ decreases for fixed source $\vec{V}$, consistent with Ohm's law 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.
15. 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?
    Answer: $0.9 \text{ A}$ 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 $1.5 \text{ V}$ to $4.5 \text{ V}$) triples the current (from $0.3 \text{ A}$ to $0.9 \text{ A}$) 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.
16. 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?
    Answer: $4.0 \text{ A}$ 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.
17. 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?
    Answer: $4.0 \text{ A}$ 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.
18. 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?
    Answer: It decreases by $33\%$ 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.
19. In a simple DC loop, what causes a voltage drop across a resistor?
    Answer: Electrical energy per charge is dissipated as thermal energy in the resistor This question tests understanding of electric current in DC circuits, specifically voltage drops in resistors. 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. Voltage drop dissipates electrical energy as thermal in the resistor. The correct answer is based on energy conversion in resistors. A common distractor might suggest charge accumulation, ignoring flow. Teaching strategies include energy balance discussions and joule heating demonstrations.
20. 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?
    Answer: $1.0 \text{ A}$ 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.

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