Simple Circuits - AP Physics 2

1. A closed single-loop circuit contains an ideal $5.0\ \text{V}$ battery and one resistor. When the battery is replaced with an ideal $2.5\ \text{V}$ battery, the same resistor remains in the loop. Which statement best describes the resistance of the resistor after the battery replacement?
   Answer: It stays the same because resistance is a property of the resistor, not the battery. This question tests understanding of simple circuits. Resistance is an intrinsic property of a resistor that depends on its material, dimensions, and temperature, not on the applied voltage or resulting current. When the battery voltage changes from 5.0 V to 2.5 V, the resistor's resistance remains unchanged. The current will decrease proportionally (halving when voltage halves), but the resistance stays constant. Choice A incorrectly suggests voltage determines resistance, revealing the misconception that circuit components change their properties based on applied voltages. In a single-loop circuit, a resistor's resistance value is independent of the battery voltage.
2. A closed single-loop circuit has a $3.0\ \text{V}$ ideal battery and one resistor. If the resistor is replaced with another resistor of half the resistance, which statement best describes the current?
   Answer: It doubles because $I=\dfrac{V}{R}$ and $R$ is halved. This question tests understanding of simple circuits. According to Ohm's law, current I = V/R in a single-loop circuit. If the original resistance is R and it's replaced with R/2, the new current becomes I = V/(R/2) = 2V/R, which is twice the original current. With the 3.0 V battery voltage constant and resistance halved, the current must double due to the inverse relationship between current and resistance. Choice D incorrectly suggests current splits, which only occurs in parallel circuits, not in a single-loop series circuit. When resistance is halved in a simple circuit with fixed voltage, the current always doubles.
3. A closed single-loop circuit contains an ideal $8.0\,\text{V}$ battery and two identical resistors in series. Compared to a circuit with only one of those resistors, which statement best describes the current?
   Answer: The current is halved because the total resistance doubles This question tests understanding of simple circuits. In a single-loop circuit with resistors in series, the total resistance is the sum of individual resistances. With two identical resistors, the total resistance doubles compared to having just one resistor. Since I = V/R and the voltage remains constant at 8.0 V while resistance doubles, the current must be halved. Choice B incorrectly thinks that having more resistors somehow increases the current, not understanding that resistors oppose current flow. In a single-loop circuit, adding resistors in series always increases total resistance and decreases current.
4. A closed single-loop circuit contains an ideal $10\ \text{V}$ battery and one resistor. The resistor is replaced by a different single resistor, and the measured current increases. Which statement best describes the resistance of the new resistor compared to the original?
   Answer: It is smaller because higher current at the same voltage implies less resistance. This question tests understanding of simple circuits. In a single-loop circuit, Ohm's law states I = V/R, meaning current and resistance are inversely related when voltage is constant. Since the battery voltage remains at 10 V and the current increases after replacing the resistor, the new resistance must be smaller than the original. This follows from rearranging Ohm's law: R = V/I, where larger current means smaller resistance. Choice A incorrectly suggests higher current requires more resistance, revealing the misconception that current and resistance are directly proportional. In a single-loop circuit with constant voltage, higher current always indicates lower resistance.
5. A closed single-loop circuit has an ideal $12\ \text{V}$ battery and a single resistor. The current is $3.0\ \text{A}$. The resistor is replaced with another resistor that causes the current to become $1.5\ \text{A}$ with the same battery. Which statement best describes the new resistance compared to the original?
   Answer: It is twice as large because the current is one-half as large at the same voltage. This question tests understanding of simple circuits. In a single-loop circuit with constant voltage, Ohm's law shows that current and resistance are inversely related: I = V/R. The original resistance is R₁ = 12 V / 3.0 A = 4.0 Ω. With the new current of 1.5 A (half the original), the new resistance is R₂ = 12 V / 1.5 A = 8.0 Ω, which is twice the original resistance. When current halves at constant voltage, resistance must double. Choice D incorrectly mentions current splitting, revealing the misconception that current can divide even in a single continuous loop. In a single-loop circuit, halving the current at constant voltage means doubling the resistance.
6. A closed single-loop circuit contains an ideal $9.0\,\text{V}$ battery and one resistor. If the resistor's resistance increases while the battery voltage stays the same, which statement best describes the current in the circuit?
   Answer: The current decreases because $I=\dfrac{V}{R}$ This question tests understanding of simple circuits. In a single-loop circuit with one battery and one resistor, Ohm's law states that current I = V/R, where V is the battery voltage and R is the resistance. Since the battery voltage remains constant at 9.0 V and the resistance increases, the current must decrease according to this inverse relationship. Choice B incorrectly assumes batteries supply constant current rather than constant voltage, which is a common misconception about how batteries work. In a single-loop circuit, apply Ohm's law directly: when resistance increases with constant voltage, current must decrease.
7. A closed single-loop circuit contains an ideal battery and two resistors in series, $2.0\ \Omega$ and $4.0\ \Omega$. Which statement best describes the current in the $2.0\ \Omega$ resistor compared to the $4.0\ \Omega$ resistor?
   Answer: It is the same in both resistors because they are in series in a single loop. This question tests understanding of simple circuits. In a series circuit, components are connected end-to-end forming a single path for current flow. By the principle of charge conservation, the same amount of charge must flow through each component per unit time, meaning the current is identical through both resistors. The different resistances affect the voltage drop across each resistor (V = IR), not the current through them. Choice A incorrectly applies the relationship between resistance and current as if the resistors were in parallel. In any series connection within a single loop, the current through all components is always the same.
8. A closed single-loop circuit contains an ideal battery and a single resistor. The measured current is $0.50\ \text{A}$ when the battery voltage is $6.0\ \text{V}$. Which statement best describes the resistance of the resistor?
   Answer: It is $12\ \Omega$ because $R=V/I$. This question tests understanding of simple circuits. To find resistance when current and voltage are known, we rearrange Ohm's law to R = V/I. With a voltage of 6.0 V and current of 0.50 A, the resistance is R = 6.0 V / 0.50 A = 12 Ω. This is the fundamental relationship between voltage, current, and resistance in any circuit element. Choice A incorrectly multiplies current and voltage (R = IV), which would give units of power (watts), not resistance (ohms). When solving circuit problems, always check that your formula gives the correct units for the quantity you're finding.
9. A closed single-loop circuit has an ideal $1.5\ \text{V}$ battery and a $3.0\ \Omega$ resistor. The battery is replaced by an ideal $3.0\ \text{V}$ battery, with the resistor unchanged. Which statement best describes the current in the circuit after the replacement?
   Answer: It doubles because the voltage doubles while $R$ stays the same. This question tests understanding of simple circuits. In a single-loop circuit, the current is determined by Ohm's law: I = V/R. Initially, with a 1.5 V battery and 3.0 Ω resistor, the current is 0.5 A. When the battery is replaced with a 3.0 V battery (doubling the voltage) while keeping the same 3.0 Ω resistor, the current doubles to 1.0 A. This follows directly from I = V/R: doubling the numerator doubles the result. Choice B incorrectly suggests that higher voltage "uses up" charge faster, misunderstanding that current is charge flow rate, not charge depletion. In simple circuits, current is directly proportional to voltage when resistance is constant.
10. A closed single-loop circuit has an ideal $6.0\,\text{V}$ battery and a single resistor $R$. If $R$ increases, which statement best describes the circuit current?
    Answer: It decreases because $I=\dfrac{V}{R}$ for the loop. This question tests understanding of simple circuits. In a single-loop circuit, the current is determined by Ohm's law: I = V/R, where V is the battery voltage and R is the total resistance. Since the battery voltage remains constant at 6.0 V, increasing R causes the current to decrease proportionally. Choice A incorrectly suggests that batteries can somehow "push harder" to maintain current, reflecting a misconception about how batteries work. Always apply Ohm's law to the entire loop: when resistance increases, current must decrease if voltage stays constant.
11. A closed single-loop circuit contains an ideal $6.0\ \text{V}$ battery and a single resistor. The resistor's value increases due to heating while the battery voltage stays constant. Which statement best describes the current in the circuit?
    Answer: It decreases because the resistance increase reduces $I=V/R$. This question tests understanding of simple circuits. In a single-loop circuit with an ideal battery and a resistor, the current follows Ohm's law: I = V/R. When the resistor heats up and its resistance increases while the 6.0 V battery voltage remains constant, the current must decrease according to the inverse relationship in Ohm's law. As resistance increases in the denominator, the current decreases proportionally. Choice C incorrectly assumes ideal batteries supply constant current, but ideal batteries actually maintain constant voltage. When analyzing simple circuits, always apply I = V/R with the battery providing fixed voltage.
12. A closed single-loop circuit contains an ideal battery and two resistors in series, $R_1$ and $R_2$. If $R_2$ is increased while $R_1$ and the battery remain unchanged, which statement best describes the current in the circuit?
    Answer: It decreases because the total series resistance increases. This question tests understanding of simple circuits. In a single-loop circuit with resistors in series, the total resistance is the sum of individual resistances: R_total = R₁ + R₂. When R₂ increases while R₁ and the battery voltage remain constant, the total resistance increases. By Ohm's law (I = V/R), increasing the denominator while keeping the numerator constant causes the current to decrease. Choice A incorrectly suggests larger resistors "pull" more current, when actually they oppose current flow. Remember that in series circuits, increasing any resistance decreases the current through all components.
13. A closed single-loop circuit has an ideal $3.0\,\text{V}$ battery and a $6.0\,\Omega$ resistor. Which statement best describes the current in the circuit?
    Answer: The current is $0.50\,\text{A}$ everywhere in the loop. This question tests understanding of simple circuits. In a single-loop circuit, the current is determined by Ohm's law: I = V/R = 3.0 V / 6.0 Ω = 0.50 A. This current is the same everywhere in the loop because charge cannot accumulate at any point. Choice D reflects the misconception that current varies with position relative to battery terminals, ignoring the fundamental principle of current conservation in circuits. For any single-loop circuit, calculate current using total voltage divided by total resistance.
14. A closed single-loop circuit uses an ideal $12\,\text{V}$ battery and two resistors, $2.0\,\Omega$ and $4.0\,\Omega$, in series. Which statement best describes the current through the $4.0\,\Omega$ resistor?
    Answer: It equals the current through the $2.0\,\Omega$ resistor. This question tests understanding of simple circuits. In a series circuit, all components share the same current because there is only one path for charge flow. The total resistance is 2.0 Ω + 4.0 Ω = 6.0 Ω, so the current is I = V/R = 12 V / 6.0 Ω = 2.0 A through both resistors. Choice C reflects the misconception that voltage gets "used up" by the first resistor, leaving none for subsequent components. Remember that in series circuits, current is constant but voltage divides among components.
15. A closed single-loop circuit uses an ideal $12\,\text{V}$ battery and a single unknown resistor. The measured current is $3.0\,\text{A}$. Which statement best describes the resistance?
    Answer: It is $4.0\,\Omega$ because $R=\dfrac{V}{I}$. This question tests understanding of simple circuits. Using Ohm's law rearranged to solve for resistance: R = V/I = 12 V / 3.0 A = 4.0 Ω. This is the fundamental relationship between voltage, current, and resistance in any circuit element. Choice C incorrectly inverts the formula (R = I/V), which would give units of A/V instead of ohms, reflecting confusion about the mathematical relationship. When finding unknown resistance, always use R = V/I, ensuring voltage is across the resistor and current flows through it.
16. A closed single-loop circuit contains a $12\ \text{V}$ ideal battery and two identical resistors in series. Which statement best describes the current through each resistor?
    Answer: It is the same through both resistors because there is only one path. This question tests understanding of simple circuits. In a single-loop circuit, there is only one path for current to flow, so the same current must pass through every component in the circuit. This is a consequence of charge conservation - charge cannot accumulate or disappear at any point in the circuit. The current that leaves the battery must flow through the first resistor, then the second resistor, and return to the battery. Choice C incorrectly assumes current splits in a series circuit, which only happens in parallel circuits. In any single-loop circuit, remember that the current is the same through all components because there's only one path.
17. A single-loop circuit contains an ideal $9.0\ \text{V}$ battery and a resistor, but the switch is open so the circuit is open. Which statement best describes the current in the resistor while the switch remains open?
    Answer: It is zero because there is no complete path for charge flow. This question tests understanding of simple circuits. In any circuit, current requires a complete conducting path from the battery's positive terminal through the circuit and back to the negative terminal. With the switch open, this path is broken, preventing any charge flow, so the current is zero everywhere in the circuit. The battery voltage still exists across the open switch, but no current flows. Choice B incorrectly assumes batteries supply constant current regardless of circuit conditions, revealing the misconception that batteries are current sources rather than voltage sources. In a single-loop circuit, an open switch always results in zero current throughout the entire loop.
18. A closed single-loop circuit contains an ideal $4.5\ \text{V}$ battery and a single resistor. The current in the circuit is $0.30\ \text{A}$. Which statement best describes the voltage across the resistor?
    Answer: It is $4.5\ \text{V}$ because the only resistor has the full battery voltage. This question tests understanding of simple circuits. In a single-loop circuit with an ideal battery and one resistor, Kirchhoff's voltage law requires that the voltage across the resistor equals the battery voltage. Since there's only one resistor in the loop, it must have the full 4.5 V across it to complete the circuit. This can be verified using Ohm's law: V = IR = 0.30 A × R, and since the battery provides 4.5 V, the resistor must have 4.5 V across it. Choice B incorrectly suggests voltage is "used up" by current, confusing voltage (potential difference) with energy. In simple circuits, the sum of voltage drops must equal the source voltage.
19. A closed single-loop circuit contains an ideal $10\,\text{V}$ battery and a single resistor carrying $2.0\,\text{A}$. Which statement best describes the resistance of the resistor?
    Answer: $R=5.0\,\Omega$ because $R=\dfrac{V}{I}$ This question tests understanding of simple circuits. In a single-loop circuit, Ohm's law can be rearranged to find resistance: R = V/I. With a 10 V battery and 2.0 A of current flowing through the circuit, the resistance is R = 10V/2.0A = 5.0 Ω. This resistance opposes the flow of current and determines how much current flows for a given voltage. Choice A incorrectly multiplies voltage and current (R = VI), which would give units of power (watts) rather than resistance (ohms). In a single-loop circuit, use R = V/I to find resistance when you know the voltage and current.
20. A closed single-loop circuit includes an ideal battery and one resistor. If the circuit is opened by a switch, which statement best describes the voltage across the resistor immediately after opening?
    Answer: It becomes $0\ \text{V}$ because no current flows through the resistor in an open circuit. This question tests understanding of simple circuits. When a switch opens a previously closed circuit, current immediately stops flowing because there's no longer a complete path. According to Ohm's law (V = IR), the voltage across a resistor equals the product of current and resistance. With zero current flowing through the resistor in an open circuit, the voltage across it must be 0 V regardless of the resistor's value. Choice B incorrectly assumes voltage is the same everywhere, confusing voltage (potential difference) with electric potential. In an open circuit, no current flows through any component, so the voltage across resistors is always zero.

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