Compound Direct Current (DC) Circuits - AP Physics 2

A $6\text{ V}$ ideal battery feeds $R_1=1\,\Omega$ to junction J. From J to junction K are two branches: branch A has two series resistors $R_2=2\,\Omega$ then $R_3=2\,\Omega$ ; branch B has a single resistor $R_4=4\,\Omega$ . The branches rejoin at K and return. Which statement correctly compares the currents in branch A and branch B?
Answer: Branch A current equals branch B current. This problem tests understanding of compound DC circuits. Branch A contains R₂ and R₃ in series (total 4Ω), while branch B contains only R₄ (4Ω). Since both branches connect the same two junctions (J and K), they experience the same voltage drop. With equal total resistance (4Ω each) and equal voltage, the branches carry equal current according to Ohm's law (I = V/R). Choice D incorrectly assumes that having two resistors prevents current flow, misunderstanding that resistors in series simply add their resistances. To analyze parallel branches, first find the total resistance of each branch, then apply Ohm's law using the common voltage across the branches.
A $15\ \text{V}$ ideal battery is connected to a network where $R_1=5\ \Omega$ is in series with a parallel pair between junctions $J$ and $K$ : Branch 1 has $R_2=10\ \Omega$ , Branch 2 has $R_3=20\ \Omega$ . The branches recombine at $K$ and return to the battery. Which statement correctly compares the potential differences across $R_2$ and $R_3$ ?
Answer: $\Delta V_{R_2}$ equals $\Delta V_{R_3}$ because they are in parallel. This question tests understanding of voltage in compound DC circuits. The circuit consists of R₁ in series with a parallel combination of R₂ and R₃ between junctions J and K. In parallel connections, all branches between the same two junctions must have the same potential difference, regardless of the individual resistance values. Therefore, the voltage across R₂ equals the voltage across R₃, even though R₂ = 10 Ω and R₃ = 20 Ω are different. Choice D incorrectly assumes that R₁ "uses up" most of the voltage first, reflecting a misconception about voltage distribution in series-parallel combinations. To analyze voltage in compound circuits, remember that parallel elements share the same voltage drop, while series elements share the same current.
An ideal $20\,\text{V}$ source drives $R_1=10\,\Omega$ in series with a parallel network between junctions $C$ and $D$ . The top branch contains $R_2=10\,\Omega$ and the bottom branch contains $R_3=30\,\Omega$ ; the branches recombine at $D$ . At junction $C$ , which statement correctly compares the branch currents?
Answer: The current in the $R_2$ branch is greater than the current in the $R_3$ branch. This problem tests understanding of compound DC circuits. At junction C, current from R₁ splits between parallel branches containing R₂ (10 Ω) and R₃ (30 Ω). Since parallel resistors share the same voltage and follow Ohm's law (I = V/R), the branch with lower resistance carries more current. With R₂ having one-third the resistance of R₃, it carries three times the current: I₂ = V/10 while I₃ = V/30, making I₂ = 3I₃. Choice A incorrectly reverses this relationship, suggesting higher resistance means more current, a common misconception about Ohm's law. When current splits at a junction in compound circuits, always remember that more current flows through the path of least resistance.
A circuit has an ideal $10\,\text{V}$ battery and resistor $R_1=5.0\,\Omega$ in series with a parallel network between junctions $M$ and $N$ . The upper branch contains $R_2=10\,\Omega$ and the lower branch contains $R_3=20\,\Omega$ ; branches rejoin at $N$ . Which statement correctly compares the currents through $R_2$ and $R_3$ ?
Answer: The current through $R_2$ is twice the current through $R_3$ . This problem tests understanding of compound DC circuits. Between junctions M and N, resistors R₂ (10 Ω) and R₃ (20 Ω) are in parallel, sharing the same voltage drop. By Ohm's law (I = V/R), current through each branch is inversely proportional to its resistance. Since R₃ has twice the resistance of R₂, it carries half the current: I₂ = V/10 and I₃ = V/20, making I₂ = 2I₃. Choice D incorrectly suggests larger resistance means more current, reversing the actual relationship between resistance and current in parallel circuits. When solving compound circuits, remember that in parallel branches with the same voltage, lower resistance always means higher current.
An ideal $5\text{ V}$ battery connects to $R_1=1\,\Omega$ in series, then reaches junction J. From J to junction K, two parallel branches contain $R_2=1\,\Omega$ and $R_3=2\,\Omega$ . The branches rejoin at K and return. Which statement correctly compares the charge per second entering junction J to the total leaving J?
Answer: The charge per second entering J equals the total charge per second leaving J. This problem tests understanding of compound DC circuits. At any junction in a circuit, charge conservation requires that charge per second entering equals charge per second leaving—this is Kirchhoff's current law. Current (charge per second) entering junction J through R₁ must equal the sum of currents leaving through R₂ and R₃, as charge cannot accumulate or disappear at a junction. Choice B incorrectly claims resistors "consume" charge, confusing energy dissipation with charge flow—resistors transform electrical energy to heat but don't destroy charge carriers. When analyzing junctions, always apply conservation of charge: the sum of currents entering any junction equals the sum of currents leaving.
A circuit has an ideal $18\text{ V}$ battery, then $R_1=6\,\Omega$ to junction J. From J, two branches run to junction K: branch A has $R_2=6\,\Omega$ ; branch B has $R_3=12\,\Omega$ . The branches rejoin at K and return to the battery. Which statement correctly compares the power dissipated in $R_2$ and $R_3$ ?
Answer: $R_2$ dissipates more power than $R_3$ . This problem tests understanding of compound DC circuits. R₁ (6Ω) is in series with parallel branches containing R₂ (6Ω) and R₃ (12Ω). In parallel branches, voltage is the same across each resistor, but current divides inversely with resistance—R₂ gets twice the current of R₃ since it has half the resistance. Power dissipation follows P = V²/R for parallel resistors with the same voltage, so lower resistance means higher power. Since R₂ (6Ω) has half the resistance of R₃ (12Ω), it dissipates twice the power. Choice D incorrectly assumes current is "consumed" in R₁, violating conservation of charge. To analyze power in compound circuits, first determine whether resistors are in series (same current) or parallel (same voltage), then apply the appropriate power formula.
An ideal $18\,\text{V}$ source powers $R_1=2.0\,\Omega$ in series with a parallel network between junctions $X$ and $Y$ . One branch contains $R_2=3.0\,\Omega$ and the other contains $R_3=6.0\,\Omega$ ; the branches recombine at $Y$ . Which statement correctly compares the voltage drops across $R_2$ and $R_3$ ?
Answer: The voltage across $R_2$ equals the voltage across $R_3$ because they share junctions $X$ and $Y$ . This problem tests understanding of compound DC circuits. In this configuration, R₁ is in series with a parallel network of R₂ and R₃ between junctions X and Y. The key principle for parallel resistors is that they share the same voltage drop because they connect to the same two points. Regardless of their individual resistances (R₂ = 3.0 Ω and R₃ = 6.0 Ω), both resistors experience the same potential difference between X and Y. Choice A incorrectly assumes voltage depends on resistance in parallel circuits, confusing the current-resistance relationship with voltage behavior. When analyzing compound circuits, remember that parallel components always have equal voltages across them, while their currents depend inversely on their resistances.
A $6\ \text{V}$ ideal battery connects to $R_1=1\ \Omega$ in series, then reaches junction $J$ where it splits into two branches that rejoin at $K$ . Branch 1 has $R_2=2\ \Omega$ ; Branch 2 has $R_3=6\ \Omega$ . Which statement correctly compares the currents through $R_2$ and $R_1$ ?
Answer: The current through $R_2$ is less than the current through $R_1$ . This problem examines current relationships in compound DC circuits. In this configuration, R₁ carries the total circuit current before it splits at junction J into two parallel branches containing R₂ and R₃. By Kirchhoff's current law, the current through R₁ equals the sum of currents through R₂ and R₃, making the current through R₁ greater than the current through either parallel branch alone. Since current divides between parallel branches, the current through R₂ must be less than the current through R₁. Choice D incorrectly suggests current is "consumed" when it splits, misunderstanding that current is conserved at junctions. When analyzing compound circuits, trace the current path and remember that series elements carry the same current while parallel branches divide the total current.
A $12\ \text{V}$ ideal battery is connected to resistor $R_1=4\ \Omega$ in series with a parallel network. At junction $J$ , the circuit splits into two branches: Branch 1 contains $R_2=6\ \Omega$ and Branch 2 contains $R_3=3\ \Omega$ ; the branches rejoin at junction $K$ and return to the battery. Which statement correctly compares the currents in $R_2$ and $R_3$ ?
Answer: The current through $R_2$ is less than the current through $R_3$ . This problem tests understanding of compound DC circuits. In this circuit, R₁ is in series with a parallel combination of R₂ and R₃, where the parallel branches rejoin before returning to the battery. In parallel branches, the voltage across each branch is the same, but current divides inversely proportional to resistance—more current flows through the smaller resistance. Since R₃ = 3 Ω is smaller than R₂ = 6 Ω, more current flows through R₃ than through R₂. Choice D incorrectly assumes current is "used up" in R₁, reflecting the misconception that current is consumed rather than conserved. To solve compound circuit problems, first identify series and parallel regions, then apply the appropriate rules for current and voltage distribution in each region.
An ideal battery is connected to resistor $R_1$ in series, then junction J. Between J and K are two parallel branches: Branch 1 contains $R_2$ only; Branch 2 contains $R_3$ only. The branches rejoin at K. Which statement correctly compares the voltage across $R_2$ and the voltage across $R_3$ ?
Answer: The voltage across $R_2$ equals the voltage across $R_3$ . This problem tests understanding of compound DC circuits. Between junctions J and K, R₂ and R₃ are in parallel, meaning they connect to the same two junction points. The fundamental principle of parallel circuits is that all elements between the same junctions experience identical voltage drops, regardless of their individual resistances or the currents they carry. This voltage equality holds whether R₂ and R₃ have equal or different resistance values. Choice D incorrectly assumes voltage becomes zero when current splits at a junction, confusing current division with voltage behavior. Key strategy: recognize that parallel elements always share the same voltage across their terminals.
An ideal battery is connected to $R_1=5\,\Omega$ in series, then junction J. Between J and K, two resistors are in parallel: Branch 1 has $R_2=10\,\Omega$ and Branch 2 has $R_3=10\,\Omega$ . The branches rejoin at K and return to the battery. Which statement correctly compares the currents through $R_2$ and $R_3$ ?
Answer: The current through $R_2$ equals the current through $R_3$ . This problem tests understanding of compound DC circuits. The circuit has R₁ in series with a parallel combination of R₂ and R₃ between junctions J and K. Since R₂ = 10Ω and R₃ = 10Ω have identical resistance values and are in parallel, they experience the same voltage drop and carry equal currents. When parallel resistors have equal resistance, current divides equally between them regardless of other circuit elements. Choice D incorrectly assumes R₁ "uses up" current, reflecting the misconception that current is consumed rather than conserved in circuits. Key principle: equal resistances in parallel always carry equal currents.
An ideal $10\text{ V}$ battery connects to a parallel pair between junctions J and K: branch 1 has $R_1=5\,\Omega$ , branch 2 has $R_2=10\,\Omega$ . After junction K, the circuit continues through $R_3=5\,\Omega$ in series back to the battery. Which statement correctly compares the currents through $R_1$ and $R_2$ ?
Answer: The current through $R_1$ is greater than the current through $R_2$ . This problem tests understanding of compound DC circuits. The parallel combination of R₁ (5Ω) and R₂ (10Ω) is in series with R₃ (5Ω). In parallel branches, current divides inversely with resistance—since R₁ has half the resistance of R₂, it carries twice the current of R₂. The total current from the parallel section then flows through R₃, which equals the sum of currents through R₁ and R₂. Therefore, the current through R₁ is greater than the current through R₂. Choice D incorrectly assumes that having R₃ in series somehow blocks current through R₁, misunderstanding how series and parallel combinations work. When solving compound circuits, trace the current path and apply junction rules: current entering a junction equals current leaving.
A circuit has an ideal $6\,\text{V}$ battery, followed by a resistor $R_1=1\,\Omega$ in series, then junction J. From J, Branch 1 contains $R_2=2\,\Omega$ and Branch 2 contains $R_3=4\,\Omega$ ; the branches rejoin at junction K. Which statement correctly compares the current through $R_1$ to the current through $R_2$ ?
Answer: The current through $R_1$ is greater than the current through $R_2$ . This problem tests understanding of compound DC circuits. In this configuration, R₁ is in series with the entire parallel combination of R₂ and R₃, meaning all current through R₁ must split between the two parallel branches at junction J. By Kirchhoff's current law, the current through R₁ equals the sum of currents through R₂ and R₃, making I₁ greater than I₂ alone. Since R₂ = 2Ω is less than R₃ = 4Ω, more current flows through R₂ than R₃, but both branch currents are less than the total current through R₁. Choice D incorrectly assumes current becomes zero when splitting, violating current conservation. Remember: series elements before a junction carry the total current that later divides among parallel branches.
An ideal $15\text{ V}$ battery is connected to $R_1=3\,\Omega$ in series with a parallel section between junctions J and K. The parallel section contains $R_2=3\,\Omega$ in one branch and $R_3=6\,\Omega$ in the other. The branches rejoin at K and return. Which statement correctly compares the current through $R_1$ to the current through $R_2$ ?
Answer: The current through $R_1$ is greater than the current through $R_2$ . This problem tests understanding of compound DC circuits. R₁ (3Ω) is in series with the entire circuit, so it carries the total current. The parallel section has R₂ (3Ω) and R₃ (6Ω), where current divides inversely with resistance—R₂ gets twice the current of R₃ since it has half the resistance. The current through R₁ equals the sum of currents through R₂ and R₃, making it greater than the current through R₂ alone. Choice D incorrectly assumes current splitting at a junction means no current flows through R₁, violating current conservation. To solve compound circuits, remember that series elements carry the total current while parallel elements divide the current based on their resistance ratios.
A $9\,\text{V}$ ideal battery drives a circuit where $R_1=3\,\Omega$ is in series with a parallel network between junctions J and K. In Branch 1, $R_2=2\,\Omega$ ; in Branch 2, $R_3=5\,\Omega$ . The branches rejoin at K and return to the battery. Which statement correctly compares the voltage drops across $R_2$ and $R_3$ ?
Answer: The voltage across $R_2$ equals the voltage across $R_3$ . This problem tests understanding of compound DC circuits. The circuit has R₁ in series with a parallel network of R₂ and R₃ between junctions J and K. In parallel branches, the key principle is that voltage across parallel elements is always equal, regardless of their individual resistances. Both R₂ and R₃ experience the same voltage drop between junctions J and K, even though R₂ = 2Ω and R₃ = 5Ω have different resistances. Choice D incorrectly assumes voltage becomes zero when current splits, reflecting a fundamental misunderstanding of how voltage behaves in parallel circuits. Always remember: parallel elements share the same voltage, while series elements share the same current.
A $10\ \text{V}$ ideal battery powers a series resistor $R_1=5\ \Omega$ followed by a parallel network between junctions $J$ and $K$ . Branch 1 contains $R_2=5\ \Omega$ ; Branch 2 contains $R_3=5\ \Omega$ . Which statement correctly compares the currents in $R_2$ and $R_3$ ?
Answer: The current through $R_2$ equals the current through $R_3$ . This question tests understanding of current in compound DC circuits with identical parallel resistors. The circuit has R₁ in series with two parallel branches containing identical resistors R₂ = R₃ = 5 Ω. When parallel branches have equal resistance, the current divides equally between them—this is a special case of the inverse proportionality rule. Therefore, the current through R₂ equals the current through R₃, with each carrying half of the total current that flows through R₁. Choice D incorrectly claims voltage divides equally in parallel, confusing the behavior of current (which divides based on resistance) with voltage (which is always equal across parallel branches). To solve compound circuits efficiently, recognize special cases like identical parallel resistors where current splits equally.
A $12\ \text{V}$ ideal battery supplies a circuit where $R_1=2\ \Omega$ is in series with a parallel section between junctions $J$ and $K$ . Branch 1 contains $R_2=8\ \Omega$ ; Branch 2 contains $R_3=8\ \Omega$ . Which statement correctly compares the current through $R_1$ to the current through $R_2$ ?
Answer: The current through $R_1$ is greater than the current through $R_2$ . This question examines current relationships in compound DC circuits where a series resistor precedes identical parallel branches. R₁ is in series with the entire circuit, so it carries the total current before it splits at junction J into two equal branches (R₂ = R₃ = 8 Ω). Since the parallel branches have equal resistance, the current divides equally between them, with each branch carrying half the total current. Therefore, the current through R₁ (total current) is greater than the current through R₂ (half the total current). Choice D incorrectly claims junctions block current, misunderstanding that junctions are simply connection points where current conservation applies. To analyze compound circuits systematically, identify which components carry total current versus partial current based on their position relative to junctions.
A $6.0\,\text{V}$ ideal battery is connected to two resistors in series: $R_1=2.0\,\Omega$ followed by a parallel section between junctions $J$ and $K$ . The parallel branches are $R_2=4.0\,\Omega$ (top) and $R_3=12.0\,\Omega$ (bottom), rejoining at $K$ . At junction $J$ , which statement correctly describes how the current splits?
Answer: More current goes through $R_2$ than through $R_3$ . This problem tests understanding of compound DC circuits. At junction J, the total current from R₁ splits between parallel branches R₂ (4.0 Ω) and R₃ (12.0 Ω). Since parallel resistors share the same voltage and follow Ohm's law (I = V/R), current divides inversely with resistance. With R₂ having one-third the resistance of R₃, it carries three times the current: if I₃ = I, then I₂ = 3I. Choice C incorrectly assumes equal currents in parallel branches, a common misconception that ignores how resistance affects current distribution. To analyze current splitting in compound circuits, apply the principle that lower resistance paths carry proportionally more current when voltage is constant.
A $12\,\text{V}$ ideal battery connects to resistor $R_1=4\,\Omega$ in series, then reaches junction J where the circuit splits into two branches: Branch 1 has $R_2=6\,\Omega$ ; Branch 2 has $R_3=12\,\Omega$ . The branches rejoin at junction K and return to the battery. Assume steady-state DC and ideal wires. Which statement correctly compares the currents in $R_2$ and $R_3$ ?
Answer: The current through $R_2$ is greater than the current through $R_3$ . This problem tests understanding of compound DC circuits. In this circuit, R₁ is in series with a parallel combination of R₂ and R₃, where R₂ = 6Ω and R₃ = 12Ω are connected between junctions J and K. In a parallel connection, both branches experience the same voltage drop, but current divides inversely with resistance according to Ohm's law (I = V/R). Since R₂ has lower resistance than R₃, more current flows through R₂ than through R₃. Choice D incorrectly assumes current is "used up" in R₁, which reflects the misconception that current is consumed as it passes through circuit elements. Remember: identify series and parallel regions before reasoning about current or voltage distribution.
A $15\,\text{V}$ ideal source is connected to $R_1=5.0\,\Omega$ in series with a parallel network between junctions $S$ and $T$ . The parallel branches are $R_2=5.0\,\Omega$ (top) and $R_3=10\,\Omega$ (bottom), rejoining at $T$ . Which statement correctly compares the voltage across $R_2$ and the voltage across $R_3$ ?
Answer: The voltage across $R_2$ equals the voltage across $R_3$ because both connect between $S$ and $T$ . This problem tests understanding of compound DC circuits. Between junctions S and T, resistors R₂ (5.0 Ω) and R₃ (10 Ω) form a parallel network. The fundamental property of parallel resistors is that they share the same voltage drop because they connect between the same two points. Despite having different resistances, both R₂ and R₃ experience the identical potential difference between S and T. Choice A incorrectly relates resistance to voltage in parallel circuits, confusing this with how current varies with resistance. When analyzing compound circuits, remember the key rule: parallel components always have equal voltages across them, regardless of their individual resistances or the currents flowing through them.

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