Compound Direct Current (DC) Circuits
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AP Physics 2 › Compound Direct Current (DC) Circuits
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?
The current in the $R_2$ branch is less than the current in the $R_3$ branch.
The current in the $R_2$ branch is zero because $R_1$ limits the current.
The current in the $R_2$ branch is greater than the current in the $R_3$ branch.
The current in the $R_2$ branch equals the current in the $R_3$ branch.
Explanation
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 $18\text{ V}$ 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 $R_3$?
$R_2$ dissipates more power than $R_3$.
Both dissipate zero power because the current is consumed in $R_1$.
$R_2$ dissipates the same power as $R_3$.
$R_2$ dissipates less power than $R_3$.
Explanation
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 $10\text{ V}$ 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 $R_2$?
The current through $R_1$ is less than the current through $R_2$.
The current through $R_1$ is greater than the current through $R_2$.
The current through $R_1$ is zero because $R_3$ is in series after junction K.
The current through $R_1$ equals the current through $R_2$.
Explanation
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 $12\ \text{V}$ 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 $R_2$?
The current through $R_1$ is less than the current through $R_2$.
The current through $R_1$ is greater than the current through $R_2$.
The current through $R_1$ is zero because the junction blocks current.
The current through $R_1$ equals the current through $R_2$.
Explanation
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 $16\ \text{V}$ ideal battery connects to a parallel network between junctions $J$ and $K$, then to a series resistor. Branch 1 between $J$ and $K$ has $R_1=4\ \Omega$; Branch 2 has $R_2=8\ \Omega$. After recombining at $K$, current goes through $R_3=4\ \Omega$ in series back to the battery. Which statement correctly compares the potential differences across $R_1$ and $R_2$?
$\Delta V_{R_1}$ equals $\Delta V_{R_2}$ because they share junctions $J$ and $K$.
$\Delta V_{R_1}$ is greater than $\Delta V_{R_2}$ because $R_1$ is smaller.
$\Delta V_{R_2}$ is zero because the $4\ \Omega$ branch takes all the voltage.
$\Delta V_{R_1}$ is less than $\Delta V_{R_2}$ because $R_2$ is larger.
Explanation
This problem tests understanding of voltage in compound DC circuits with a parallel section followed by series resistance. R₁ = 4 Ω and R₂ = 8 Ω are in parallel between junctions J and K, followed by R₃ = 4 Ω in series. The key principle is that all components connected between the same two junctions (J and K) must have the same potential difference, regardless of their individual resistances. Therefore, the voltage across R₁ equals the voltage across R₂. Choice D incorrectly suggests one branch "takes all the voltage," misunderstanding that parallel branches share the same voltage drop. When analyzing compound circuits, identify junction pairs and remember that all paths between the same junctions have equal potential differences.
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 $R_3$?
The current through $R_2$ equals the current through $R_3$.
The current through $R_2$ is zero because current is used up in $R_1$.
The current through $R_2$ is greater than the current through $R_3$.
The current through $R_2$ is less than the current through $R_3$.
Explanation
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.
A $12.0,\text{V}$ battery connects to resistor $R_1=3.0,\Omega$ in series with a parallel network between junctions $J_1$ and $J_2$. The network has two branches: top branch $R_2=6.0,\Omega$ and bottom branch $R_3=12.0,\Omega$. The branches rejoin at $J_2$ and return to the battery. Assume ideal wires and steady state. Which statement correctly compares the currents in the two branches between $J_1$ and $J_2$?
The current through $R_2$ is zero because $R_1$ uses up the current first.
The current through $R_3$ is twice the current through $R_2$.
The currents through $R_2$ and $R_3$ are equal because they are parallel.
The current through $R_2$ is twice the current through $R_3$.
Explanation
This problem tests understanding of compound DC circuits. In this circuit, R₁ is in series with a parallel combination of R₂ and R₃, meaning all current through R₁ must split between the two parallel branches. Since parallel resistors share the same voltage drop, and current follows Ohm's law (I = V/R), the branch with lower resistance carries more current. With R₂ = 6.0 Ω and R₃ = 12.0 Ω, the current through R₂ is I₂ = V/6 while the current through R₃ is I₃ = V/12, making I₂ = 2I₃. Choice C incorrectly assumes equal currents in parallel branches, reflecting the misconception that parallel elements always have equal currents rather than equal voltages. When analyzing compound circuits, first identify series and parallel regions, then apply the appropriate rules: series elements share current, parallel elements share voltage.
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 $R_3$?
The current through $R_2$ is half the current through $R_3$.
The current through $R_2$ is twice the current through $R_3$.
The current through $R_3$ is greater because it has the larger resistance.
The current through $R_2$ equals the current through $R_3$.
Explanation
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.
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?
More current goes through $R_2$ than through $R_3$.
Equal current goes through $R_2$ and $R_3$ because they are parallel.
More current goes through $R_3$ than through $R_2$.
All current goes through $R_3$ because it has the larger resistance.
Explanation
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.
An ideal $18,\text{V}$ battery is connected to junction J, where the circuit splits into two branches that later rejoin at junction K. Branch 1 contains two series resistors $R_1=3,\Omega$ then $R_2=3,\Omega$. Branch 2 contains a single resistor $R_3=6,\Omega$. Assume steady-state DC. Which statement correctly compares the currents in Branch 1 and Branch 2?
The current in Branch 1 is less than the current in Branch 2.
The current in Branch 1 is greater than the current in Branch 2.
The current in Branch 1 equals the current in Branch 2.
The current in Branch 1 is zero because $R_1$ and $R_2$ consume the current.
Explanation
This problem tests understanding of compound DC circuits. The circuit has two parallel branches between junctions J and K: Branch 1 contains R₁ + R₂ = 3Ω + 3Ω = 6Ω total, while Branch 2 contains R₃ = 6Ω. Since both branches have equal total resistance (6Ω each) and experience the same voltage drop (18V between J and K), they carry equal currents according to Ohm's law. The current in each branch equals 18V/6Ω = 3A. Choice D incorrectly assumes current is "consumed" by resistors, which is a common misconception about current conservation. Strategy: calculate total resistance in each parallel branch before comparing currents.