Q: Current division in parallel circuits — does total current split directly proportional to resistance values?

Introduction / Context: Understanding how current divides among parallel branches is essential for sizing components and predicting power dissipation. This item targets the proportionality rule for current division. Given Data / Assumptions: Two or more resistors connected in parallel across a common voltage source. Ohmic behavior (linear resistors) and steady-state conditions. Concept / Approach: In parallel, branch voltages are equal. Using Ohm’s law, branch current Ik = V / Rk. Therefore, current in a branch is inversely proportional to its resistance. Lower resistance draws more current, higher resistance draws less. The current divider formula for two branches is I1 = Itot * (R2 / (R1 + R2)), I2 = Itot * (R1 / (R1 + R2)). Note that each branch current multiplies by the opposite branch resistance in the numerator, embodying inverse proportionality. Step-by-Step Solution: Given V across all branches: I1 = V / R1, I2 = V / R2.Take ratios: I1 / I2 = (V / R1) / (V / R2) = R2 / R1.Thus, Ik ∝ 1 / Rk, not Rk.Apply current divider to compute each branch from total current if desired. Verification / Alternative check: Example: V = 10 V, R1 = 5 Ω, R2 = 10 Ω → I1 = 2 A, I2 = 1 A. The branch with half the resistance draws twice the current, consistent with inverse proportionality. Why Other Options Are Wrong: “True” would claim larger resistors get more current, which contradicts Ohm’s law under equal voltage. Common Pitfalls: Memorizing formulas without noting symmetry: the “opposite resistance” in the numerator is a cue that current is inversely related to resistance. Final Answer: False

Q: Kirchhoff’s Current Law (KCL) — at a junction, do sums of entering and leaving currents match?

Introduction / Context: Kirchhoff’s Current Law is one of the two foundational network laws used in circuit analysis. It enforces charge conservation at a node (junction) and underpins nodal analysis, Thevenin/Norton conversions, and network simulators. Given Data / Assumptions: Ideal lumped circuit model with clearly defined nodes. No net charge accumulation at a node in steady state. Currents defined as positive into or out of the node with a consistent sign convention. Concept / Approach: KCL states that the algebraic sum of currents at a node equals zero, which is equivalent to “sum of currents entering equals sum of currents leaving.” This is a direct consequence of charge conservation: a node cannot store net charge over time in the lumped-parameter assumption, hence what flows in must flow out. Step-by-Step Solution: Choose a sign convention: currents entering positive, leaving negative.Write KCL: Σ I_k = 0 across all connected branches.Rearrange: Σ I_in = Σ I_out, which matches the statement.Apply to example node with three branches to verify numerically. Verification / Alternative check: Use nodal analysis on a simple parallel network; solving produces branch currents whose sum equals the source current, confirming KCL at the supply node. Why Other Options Are Wrong: “False” would contradict charge conservation and invalidate standard analytical methods. Common Pitfalls: Inconsistent current directions leading to sign errors; always define arrows and stick to them to avoid apparent “violations” of KCL. Final Answer: True

Question 1

Finding unknown parallel resistors: Two resistors are connected in parallel with total equivalent resistance RT = 8 Ω. One resistor has double the resistance of the other. What are the individual resistor values?

Accepted Answer

Introduction / Context: Determining branch values from the equivalent parallel resistance is a classic algebraic exercise. It appears when replacing networks with standard part values or diagnosing mismatched components. Given Data / Assumptions: Two unknown resistors in parallel. RT = 8 Ω. One resistor is exactly double the other. Concept / Approach: Let the smaller resistor be R and the larger be 2R. For two resistors in parallel, RT = (R * 2R) / (R + 2R) = (2R^2) / (3R) = (2/3) R. Solve algebraically for R, then list both values. Step-by-Step Solution: Set 8 = (2/3) R.Solve for R: R = 8 * 3 / 2 = 12 Ω.Therefore, the other resistor is 2R = 24 Ω. Verification / Alternative check: Compute equivalent: (12 * 24) / (12 + 24) = 288 / 36 = 8 Ω. Checks out exactly. Why Other Options Are Wrong: Other pairs do not satisfy both conditions (parallel equivalent of 8 Ω and a 2:1 ratio). For instance, 10 Ω and 20 Ω in parallel yield 6.67 Ω, not 8 Ω. Common Pitfalls: Using series formulas or mixing up which number is larger (the 2R must exceed R). Final Answer: 12 Ω and 24 Ω

Question 2

Parallel circuits in basic electronics Which quantity is considered the common reference (i.e., the same across every branch) in any parallel circuit?

Accepted Answer

Introduction / Context: In electrical engineering, recognizing what is common and what divides in parallel circuits is fundamental for troubleshooting and design. This concept underpins how we size components, measure quantities, and predict behavior when loads change. Given Data / Assumptions: The network is an ideal parallel circuit with ideal conductors. All branches are connected directly across the same two nodes of a voltage source. Component non-idealities like lead resistance are neglected for clarity. Concept / Approach: By definition, elements are in parallel if both of their terminals are connected to the same two nodes. That guarantees the same potential difference across each branch. While the branch currents may differ (depending on branch resistance or impedance), the voltage across each branch equals the source voltage at those nodes. Step-by-Step Solution: Identify the two common nodes in a parallel network.State the parallel property: the potential difference across each branch equals the node-to-node voltage.Conclude that voltage is the common reference quantity in parallel circuits. Verification / Alternative check: Use Ohm’s law on two different branches: I1 = V/R1 and I2 = V/R2. The same V appears in both relations; only currents vary with resistance. Measurements with a voltmeter across any branch show the same reading (within tolerances). Why Other Options Are Wrong: Current: divides among branches according to their impedances. Power: differs per branch because P = V^2/R or P = I^2*R. Resistance: each branch has its own value; the equivalent resistance is the parallel combination, not a common per-branch value. Common Pitfalls: Confusing series and parallel rules; in series, current is common and voltage divides, whereas in parallel, voltage is common and current divides. Final Answer: Voltage

Question 3

Power dissipation in unequal parallel resistors When three different resistor values are connected in parallel, which branch experiences the greatest power loss (for the same applied voltage)?

Accepted Answer

Introduction / Context: Understanding how power distributes among branches in a parallel network is essential for thermal design and component selection. Overheating typically occurs in the branch with the highest dissipation, so predicting which branch that is prevents premature failures. Given Data / Assumptions: Three resistors of unequal value connected in parallel across the same source. Ideal conditions: wires and source internal resistance neglected. Steady DC operation. Concept / Approach: In parallel circuits, the voltage across each resistor is identical. Power in a resistor can be computed as P = V^2/R. With a common V, the branch power is inversely proportional to resistance. Therefore, the smallest resistance draws the largest current and dissipates the most power. Step-by-Step Solution: Write branch power: P_i = V^2 / R_i for each branch.Compare P values: smaller R_i yields larger P_i.Conclude that the smallest resistance dissipates the greatest power. Verification / Alternative check: Using P = I^2 * R with I = V/R in parallel gives P = (V/R)^2 * R = V^2 / R, the same conclusion: smallest R gives largest power. Why Other Options Are Wrong: Largest resistance: has the lowest current and lowest P = V^2/R. Same power loss: only true if all resistances were equal. Need more values: not necessary; the relationship holds for any positive unequal resistances. Common Pitfalls: Applying the series-circuit intuition where current is common; in parallel, voltage is common, changing the power relationship with resistance. Final Answer: The smallest resistance

Question 4

Troubleshooting parallel circuits Which instruments are best for isolating a suspected open component within a parallel circuit during testing and maintenance?

Accepted Answer

Introduction / Context: Opens in parallel networks can be tricky to find because other branches may keep the circuit functioning partially. Selecting the right instrument speeds up diagnostics and reduces downtime in labs and field service work. Given Data / Assumptions: DC parallel circuit with accessible nodes. Component suspected of being open (infinite resistance). Safe procedures followed for live and de-energized tests. Concept / Approach: A voltmeter (high input impedance) across a suspect branch under power will show full source voltage if the branch is open (no drop elsewhere), while a good branch in parallel will show normal voltage but carries current. An ohmmeter (with power off and component isolated as needed) directly measures very high resistance for an open. These two instruments are the primary tools for detecting opens. Step-by-Step Solution: With power on, place a voltmeter across the suspect component: an open typically reads near source voltage.Power down and disconnect at least one lead, then measure with an ohmmeter: an open reads extremely high or out of range.Confirm by comparing with a known-good component or branch. Verification / Alternative check: Ammeter testing is less direct in parallel because current simply diverts to other branches; zero branch current may be masked unless you break into that branch. A wattmeter is unnecessary and less informative for simple continuity faults. Why Other Options Are Wrong: Neither an ammeter nor a voltmeter: voltmeters are highly effective at detecting opens. Wattmeter or voltmeter: wattmeters complicate measurement and are not needed for opens. Ammeter or ohmmeter: ammeters require series insertion and can be impractical in multi-branch circuits without rewiring. Common Pitfalls: Measuring resistance with power applied can damage the meter; always de-energize for ohmmeter tests. Final Answer: A voltmeter or an ohmmeter

Question 5

Topology vocabulary in circuit analysis When components are connected in parallel between the same two nodes, they form what are commonly called:

Accepted Answer

Introduction / Context: Clear terminology helps in reading schematics and describing circuit topology. In networks, elements connected between a common pair of nodes are described using a specific term that highlights current pathways. Given Data / Assumptions: Parallel connection implies both ends of each element share the same two nodes. No faults (opens or shorts) are assumed unless stated. Concept / Approach: Each separate path that connects the pair of nodes is called a branch. Multiple branches in parallel share the same voltage but carry different currents based on their impedance. Recognizing branches is central to applying Kirchhoff’s Current Law at a node where branch currents sum. Step-by-Step Solution: Identify the two nodes that define the parallel connection.Count each independent path between these nodes; each is a branch.Conclude that parallel elements constitute branches in the circuit. Verification / Alternative check: Apply KCL at one node: the algebraic sum of all branch currents leaving and entering the node is zero, confirming the branch construct. Why Other Options Are Wrong: Open circuits: indicate a broken path, not a normal parallel connection. Short circuits: imply near-zero impedance path bypassing other elements; not the definition of parallel branches. Voltage divider: typically a series connection used to divide voltage, not a parallel structure. Common Pitfalls: Calling every path a loop; loops include series and return paths, whereas branches specifically refer to parallel paths between common nodes. Final Answer: Branches

Question 6

Series versus parallel path count How does a parallel circuit differ from a series circuit with respect to the number of available paths for current flow?

Accepted Answer

Introduction / Context: Distinguishing series and parallel arrangements is a cornerstone skill in circuit analysis and troubleshooting. Many real circuits mix both, so being able to identify how current flows is essential. Given Data / Assumptions: Ideal wires and sources; steady DC conditions. Series means components share the same current; parallel means components share the same voltage. Concept / Approach: In a series connection, the current has exactly one path and is the same through each component. In a parallel connection, multiple independent paths exist between the same two nodes; the current splits among branches based on their impedances while the voltage across each branch remains the same. Step-by-Step Solution: Define series: single current path through all components.Define parallel: multiple paths between common nodes; currents add at the node.Select the description that matches parallel: more than one path for current flow. Verification / Alternative check: Apply KCL at a junction in a parallel network: the sum of branch currents equals the source current, confirming multiple paths. Why Other Options Are Wrong: No path: describes an open circuit, not a parallel network. Fewer paths: relative wording is ambiguous and not a defining property. One path: defines a series circuit, not a parallel circuit. Common Pitfalls: Assuming current is the same in parallel branches; that is the series rule. In parallel, voltage is the same and current divides according to branch impedance. Final Answer: More than one path for current flow

Question 7

Compute the equivalent resistance for resistors in parallel What is the total resistance RT of 12 kΩ, 4 kΩ, and 3 kΩ connected in parallel?

Accepted Answer

Introduction / Context: Calculating equivalent resistance of parallel resistors is a routine but critical task in electronics. It determines load seen by sources, affects current sharing, and impacts power dissipation and noise performance in analog designs. Given Data / Assumptions: Three resistors: R1 = 12 kΩ, R2 = 4 kΩ, R3 = 3 kΩ. All three are connected in parallel across the same two nodes. DC conditions and ideal components. Concept / Approach: For resistors in parallel, conductances add: 1/RT = 1/R1 + 1/R2 + 1/R3. After summing the reciprocals, invert to get RT. Using kilohm units throughout avoids unit mistakes. Step-by-Step Solution: Compute reciprocals: 1/R1 = 1/12 kΩ = 0.08333 kΩ^-1.Compute 1/R2 = 1/4 kΩ = 0.25 kΩ^-1; 1/R3 = 1/3 kΩ ≈ 0.33333 kΩ^-1.Sum: 0.08333 + 0.25 + 0.33333 ≈ 0.66666 kΩ^-1.Invert: RT = 1 / 0.66666 ≈ 1.5 kΩ. Verification / Alternative check: Use pairwise combination: combine 3 kΩ || 6 kΩ (since 12 kΩ || 4 kΩ = 3 kΩ) gives 3 kΩ || 3 kΩ = 1.5 kΩ. Both methods agree. Why Other Options Are Wrong: 2 kΩ: corresponds to 1/RT = 0.5 kΩ^-1, which is smaller than the actual sum of conductances. 6.3 kΩ or 19 kΩ: larger than the smallest branch resistance; a parallel equivalent can never exceed the smallest resistor. Common Pitfalls: Failing to keep consistent units (ohms vs kilohms) and forgetting that the equivalent of parallel resistors is always less than the smallest individual resistor. Final Answer: 1.5 kΩ

Question 8

Parallel networks — if you add more resistors in parallel, do total resistance and total current behave as stated? Claim: Adding resistors in parallel increases total resistance and decreases total current.

Accepted Answer

Introduction / Context: This statement probes the hallmark property of parallel circuits: how total or equivalent resistance changes as additional branches are added, and how that affects the current drawn from a fixed-voltage source. Given Data / Assumptions: Resistors connected in parallel between the same two nodes of an ideal voltage source. Ohm’s law applies; source voltage remains constant. No wire resistance or parasitic effects. Concept / Approach: The equivalent resistance of resistors in parallel is found from conductances: 1/Req = 1/R1 + 1/R2 + … . Adding another parallel branch adds positive conductance, so Req must decrease. With a fixed source voltage, the total current Itot = V / Req therefore increases when Req decreases. Thus the claim in the stem is the opposite of the true behavior. Step-by-Step Solution: Start with two resistors: 1/Req = 1/R1 + 1/R2 → Req = (R1 * R2) / (R1 + R2) < min(R1, R2).Add a third resistor R3: new 1/Req' = 1/Req + 1/R3 → Req' < Req.For fixed V: Itot = V / Req increases as Req decreases.Therefore, adding resistors in parallel decreases total resistance and increases total current drawn. Verification / Alternative check: Numerical example: V = 12 V; R1 = 12 Ω gives Itot = 1 A. Add R2 = 12 Ω in parallel: Req = 6 Ω, Itot = 2 A. Add R3 = 12 Ω: Req = 4 Ω, Itot = 3 A. The pattern confirms the rule. Why Other Options Are Wrong: Choosing “True” confuses parallel with series behavior. Series connections increase resistance as more resistors are added; parallel decreases it. Common Pitfalls: Using arithmetic averages or adding magnitudes directly; always combine parallel resistances via reciprocals or conductances. Final Answer: False

Question 9

Current division in parallel circuits — does total current split directly proportional to resistance values?

Accepted Answer

Introduction / Context: Understanding how current divides among parallel branches is essential for sizing components and predicting power dissipation. This item targets the proportionality rule for current division. Given Data / Assumptions: Two or more resistors connected in parallel across a common voltage source. Ohmic behavior (linear resistors) and steady-state conditions. Concept / Approach: In parallel, branch voltages are equal. Using Ohm’s law, branch current Ik = V / Rk. Therefore, current in a branch is inversely proportional to its resistance. Lower resistance draws more current, higher resistance draws less. The current divider formula for two branches is I1 = Itot * (R2 / (R1 + R2)), I2 = Itot * (R1 / (R1 + R2)). Note that each branch current multiplies by the opposite branch resistance in the numerator, embodying inverse proportionality. Step-by-Step Solution: Given V across all branches: I1 = V / R1, I2 = V / R2.Take ratios: I1 / I2 = (V / R1) / (V / R2) = R2 / R1.Thus, Ik ∝ 1 / Rk, not Rk.Apply current divider to compute each branch from total current if desired. Verification / Alternative check: Example: V = 10 V, R1 = 5 Ω, R2 = 10 Ω → I1 = 2 A, I2 = 1 A. The branch with half the resistance draws twice the current, consistent with inverse proportionality. Why Other Options Are Wrong: “True” would claim larger resistors get more current, which contradicts Ohm’s law under equal voltage. Common Pitfalls: Memorizing formulas without noting symmetry: the “opposite resistance” in the numerator is a cue that current is inversely related to resistance. Final Answer: False

Question 10

Kirchhoff’s Current Law (KCL) — at a junction, do sums of entering and leaving currents match?

Accepted Answer

Introduction / Context: Kirchhoff’s Current Law is one of the two foundational network laws used in circuit analysis. It enforces charge conservation at a node (junction) and underpins nodal analysis, Thevenin/Norton conversions, and network simulators. Given Data / Assumptions: Ideal lumped circuit model with clearly defined nodes. No net charge accumulation at a node in steady state. Currents defined as positive into or out of the node with a consistent sign convention. Concept / Approach: KCL states that the algebraic sum of currents at a node equals zero, which is equivalent to “sum of currents entering equals sum of currents leaving.” This is a direct consequence of charge conservation: a node cannot store net charge over time in the lumped-parameter assumption, hence what flows in must flow out. Step-by-Step Solution: Choose a sign convention: currents entering positive, leaving negative.Write KCL: Σ I_k = 0 across all connected branches.Rearrange: Σ I_in = Σ I_out, which matches the statement.Apply to example node with three branches to verify numerically. Verification / Alternative check: Use nodal analysis on a simple parallel network; solving produces branch currents whose sum equals the source current, confirming KCL at the supply node. Why Other Options Are Wrong: “False” would contradict charge conservation and invalidate standard analytical methods. Common Pitfalls: Inconsistent current directions leading to sign errors; always define arrows and stick to them to avoid apparent “violations” of KCL. Final Answer: True

Question 11

Parallel circuits — do they function as current dividers by definition?

Accepted Answer

Introduction / Context: Recognizing series vs. parallel behaviors helps you quickly analyze networks. This question focuses on the current-divider role of a parallel connection and how current distributes based on branch impedances. Given Data / Assumptions: Multiple resistive branches connected between the same two nodes of a source. Linear, time-invariant behavior under steady-state operation. Concept / Approach: In parallel, all branches share the same voltage. Currents in the branches depend on their individual resistances or impedances: Ik = V / Rk for resistors. The total source current is the sum of branch currents, so the parallel network divides the incoming current among the available paths according to admittance ratios. This is precisely the definition and practical utility of a current divider. Step-by-Step Solution: Write total current: Itot = Σ (V / Rk) = V * Σ (1 / Rk).Branch current for two-branch case: I1 = Itot * (R2 / (R1 + R2)), I2 = Itot * (R1 / (R1 + R2)).Observe inverse dependence: smaller resistance pulls a larger share of the current.Extend to impedances for AC: use complex values and the same divider principle with admittances. Verification / Alternative check: Measure with an ammeter in each branch and compare with calculated divider ratios; experimental values match predictions within component tolerances. Why Other Options Are Wrong: “False” would deny the fundamental current-splitting property that distinguishes parallel from series, where voltage divides instead. Common Pitfalls: Swapping voltage and current divider formulas; remember: series → voltage divider, parallel → current divider. Final Answer: True

Question 12

Definition check — are parallel resistors connected between the same two nodes (points) in a circuit?

Accepted Answer

Introduction / Context: Properly identifying parallel connections is foundational for applying current divider rules and computing equivalent resistance. This item verifies the formal definition of parallel components in circuit topology. Given Data / Assumptions: Ideal lumped-element schematic with distinct nodes. No ambiguity in connection points or shared elements. Concept / Approach: Two (or more) components are in parallel if they share the same two nodes, meaning both terminals of each element connect to the same pair of nodes. As a consequence, the voltage across each parallel element is identical. This is distinct from series connection, where elements share exactly one node and the same current flows through each element in sequence. Step-by-Step Solution: Define nodes A and B as the pair connecting all resistor ends.Because each resistor has one terminal at A and the other at B, V_AB is common to every branch.Thus, by definition, they are parallel and experience the same voltage, while currents add.Use parallel formulas to compute Req via conductance summation if needed. Verification / Alternative check: A quick check is to “collapse” wires in your mind: if every element’s terminals can be traced directly to the exact same two nodes without intermediate components, those elements are parallel. Why Other Options Are Wrong: “False” would contradict the universally accepted definition used in textbooks and standards. Common Pitfalls: Confusing networks that have shared segments but are not strictly connected to the same two nodes; use a node-labeling approach to avoid mistakes. Final Answer: True

Question 13

Two-resistor parallel formula — is the total resistance simply the arithmetic sum of the two resistors?

Accepted Answer

Introduction / Context: Computing equivalent resistance for two resistors in parallel is a frequent task in circuit design. This question challenges a common misconception that parallel totals add like series totals. Given Data / Assumptions: Two resistors R1 and R2 connected in parallel across the same source. Resistors are linear and temperature effects are neglected. Concept / Approach: For two resistors in parallel, the correct equivalent resistance is Req = (R1 * R2) / (R1 + R2). More generally, 1/Req = 1/R1 + 1/R2. The arithmetic sum R1 + R2 applies to series connections, not parallel. Because parallel adds conductances, the final equivalent must be less than the smallest individual resistor value. Step-by-Step Solution: Start from conductance: G_eq = 1/R1 + 1/R2.Invert to get resistance: Req = 1 / G_eq = (R1 * R2) / (R1 + R2).Check inequality: Req < min(R1, R2), which contradicts any claim of a sum.If R1 = R2 = R, then Req = R/2, a useful quick check. Verification / Alternative check: Numerical example: R1 = 6 Ω, R2 = 3 Ω → Req = (6 * 3) / (6 + 3) = 18/9 = 2 Ω; clearly not 9 Ω. Why Other Options Are Wrong: “True” incorrectly transfers the series rule to a parallel situation and would drastically overestimate current. Common Pitfalls: Forgetting to invert sums when switching between resistance and conductance; to avoid errors, compute with admittances in parallel networks. Final Answer: False

Question 14

Topology insight — in a purely resistive parallel circuit, does the number of current paths equal the number of resistors?

Accepted Answer

Introduction / Context: Counting current paths helps visualize how current splits and how removing a branch changes overall behavior. This question formalizes the relationship between branches and current paths in a simple parallel network. Given Data / Assumptions: Each resistor forms its own branch directly between the same two nodes. No branching within a branch (i.e., no resistor networks nested inside a single branch). Concept / Approach: In a basic parallel circuit, each distinct resistor connected across the same two nodes provides an independent path for current. Hence, the number of independent current paths equals the number of parallel resistors. This is the reason why adding another parallel resistor increases total conductance and decreases overall resistance: more paths enable more current for a given voltage. Step-by-Step Solution: Let N resistors be connected in parallel between nodes A and B.Each resistor offers one distinct branch path for charge flow from A to B (or B to A, depending on polarity).Therefore, total current Itot equals the sum of branch currents across these N paths.Conclusion: number of current paths = N (the number of resistors), subject to the stated assumptions. Verification / Alternative check: Draw a node-branch graph: nodes are vertices; each resistor is an edge between the same two vertices; each edge represents one path carrying branch current. Why Other Options Are Wrong: “False” would only be correct if branches shared internal elements or if some resistors were in series inside a branch, which violates the simple “each element is a separate branch” assumption. Common Pitfalls: Misidentifying series-parallel combinations as pure parallel; always label nodes to see whether every element truly spans the same two nodes. Final Answer: True

Question 15

Fault behavior in parallel networks — if one branch opens, does the total (equivalent) resistance decrease?

Accepted Answer

Introduction / Context: Understanding how faults affect equivalent resistance is key to troubleshooting. An open circuit in one branch of a parallel network removes that path entirely. This item asks whether the overall resistance goes down or up when a branch opens. Given Data / Assumptions: Several resistors connected in parallel across a source. One branch becomes open (infinite resistance), eliminating its conductance contribution. Source voltage remains constant and other branches are unaffected. Concept / Approach: Parallel equivalent resistance is governed by conductance addition: G_eq = Σ (1/Rk). Opening a branch sets 1/R_open = 0, removing that term. Therefore total conductance decreases, and equivalently, total resistance Req = 1/G_eq increases. With higher Req at a fixed voltage, total current drawn from the source decreases. Hence, the statement that total resistance decreases is incorrect; it increases. Step-by-Step Solution: Initial: G_eq = 1/R1 + 1/R2 + … + 1/Rn.Fault: branch j opens → remove 1/Rj term → G_eq_fault < G_eq.Therefore Req_fault = 1 / G_eq_fault > 1 / G_eq = Req.Consequently, Itot_fault = V / Req_fault is smaller than before. Verification / Alternative check: Example: R1 = R2 = 10 Ω in parallel → Req = 5 Ω. Open R2 → only R1 remains → Req = 10 Ω, which is larger, not smaller. Why Other Options Are Wrong: Answering “True” would imply that removing a current path somehow lowers resistance, contradicting the physics of fewer available conduction paths. Common Pitfalls: Confusing series and parallel responses to opens: an open in series breaks the circuit and sends resistance to infinity; an open in one parallel branch simply removes that branch while leaving other paths intact but increasing overall resistance. Final Answer: False

Question 16

Power accounting in parallel networks: Is the total power in a parallel circuit obtained by summing the power dissipated in each individual resistor?  Assume a steady-state circuit with an ideal voltage source feeding multiple resistive branches in …

Accepted Answer

Introduction / Context: Understanding how power combines in circuits is essential for thermal design, fuse sizing, and efficiency analysis. In a parallel circuit, each branch has the same applied voltage, but different currents may flow depending on branch resistances. The question asks whether total power equals the sum of individual branch powers. Given Data / Assumptions: Ideal voltage source maintains the same voltage across all parallel branches. Elements are linear resistors (no active sources within branches). Steady-state operation (DC or sinusoidal AC with RMS values). Concept / Approach: Power conservation states that input electrical power from the source equals the total power consumed by all loads. In parallel, each resistor k dissipates Pk = V^2 / Rk when supplied by a common voltage V. The source supplies the algebraic sum of these powers, so P_total = ΣPk. Step-by-Step Solution: Let the supply voltage across all branches be V.For branch k with resistance Rk: Ik = V / Rk and Pk = V * Ik = V^2 / Rk.Sum over all branches: P_total = Σ(V^2 / Rk) = ΣPk.Therefore, the total power equals the sum of individual branch powers. Verification / Alternative check: Using currents: the source current I_source = ΣIk. Source power is P_source = V * I_source = V * Σ(V / Rk) which simplifies to Σ(V^2 / Rk) = ΣPk, confirming additivity. Why Other Options Are Wrong: Equal resistors are unnecessary; additivity holds for any set of resistances. In AC, using RMS values yields the same result for pure resistors. Finite source resistance can be modeled as an additional element whose power simply adds to the total dissipation. Common Pitfalls: Confusing voltage addition (which does not occur in parallel across the same two nodes) with power addition. Also, forgetting to use RMS values in AC analysis can lead to incorrect power calculations. Final Answer: True

Question 17

Current distribution in parallel circuits: Does the branch with the lowest resistance carry the greatest current when branches share the same voltage?

Accepted Answer

Introduction / Context: Parallel circuits place components across the same two nodes, so each branch experiences the same voltage. This item checks the fundamental current-voltage-resistance relationship governing how current divides among parallel branches. Given Data / Assumptions: Ideal voltage source maintains node-to-node voltage across all branches. Branches contain resistors (linear, time-invariant). Steady-state DC or AC using RMS values for magnitudes. Concept / Approach: Ohm’s law states I = V / R. In a parallel network, each branch current Ik equals V / Rk. If two branches share the same V, the smaller Rk produces a larger Ik. Thus, the branch with the lowest resistance draws the most current. Step-by-Step Solution: All branches have the same applied voltage V.For branch k: Ik = V / Rk.If Ri < Rj then Ii = V / Ri > V / Rj = Ij.Therefore, the lowest-resistance branch carries the maximum current. Verification / Alternative check: Numerical example: V = 12 V; R1 = 3 Ω, R2 = 6 Ω. I1 = 4 A, I2 = 2 A. The lower resistance branch indeed carries twice the current. Why Other Options Are Wrong: No need for other branches to be open; current division works with any number of branches. In AC, the same principle holds for resistive branches using RMS values. Power rating does not determine current; resistance and voltage do. Common Pitfalls: Confusing resistance with power rating or physical size. Also, treating parallel branches as if currents must be equal—this only occurs when the resistances are equal. Final Answer: True

Question 18

Parallel circuits and fault conditions: if one branch opens in a parallel network driven by an ideal voltage source, will the current in the remaining parallel paths change or remain the same (ignoring source internal resistance)?

Accepted Answer

Introduction / Context: Parallel circuits appear throughout electronics and power distribution. Understanding how branch currents respond when one branch opens is essential for diagnostics and reliability analysis. The statement under test claims that when one branch opens, the current in the remaining branches changes. We evaluate this under the standard ideal-source model used in circuit theory. Given Data / Assumptions: Ideal constant-voltage source feeds the parallel network. Branch resistances are linear and time-invariant. No significant source internal resistance unless explicitly mentioned. Concept / Approach: In a parallel network with an ideal voltage source, each branch current is I_k = V / R_k. If a different branch opens, the source voltage V applied to the remaining branches does not change, and their resistances do not change either. Therefore I_k for all remaining branches remains the same. Only the total supply current and equivalent resistance change. Step-by-Step Solution: 1) Express branch current: I_k = V / R_k under an ideal source. 2) Opening another branch removes only that path; V and R_k of surviving branches are unchanged. 3) Conclude remaining branch currents are unchanged; total current decreases because one path is gone. Verification / Alternative check: If non-ideal source resistance exists, V may sag slightly when load changes. The ideal model used in introductory analysis assumes V is fixed, justifying the conclusion. Why Other Options Are Wrong: Correct: contradicts ideal-source analysis. Depends on source internal resistance: outside the stated ideal assumption. Cannot be determined: sufficient information is present under standard assumptions. Common Pitfalls: Confusing total current with branch currents; assuming voltage sources act like current sources; overlooking internal resistance only when it is actually specified. Final Answer: Incorrect

Question 19

Parallel circuits and current sharing: assess the claim that the branch with the largest resistance in a parallel network carries the largest current when driven by the same voltage.

Accepted Answer

Introduction / Context: Current division in parallel networks determines how currents split among branches. A common misconception is that a larger resistance somehow draws more current. This question challenges that misconception and reinforces the proportional relationships that govern current sharing. Given Data / Assumptions: Ideal constant-voltage source supplying a set of resistors in parallel. Ohmic resistors with fixed values. Temperature and nonlinearity effects ignored. Concept / Approach: By Ohm law, I = V / R for each branch at the same voltage V. Current is inversely proportional to resistance. Therefore the branch with the smallest resistance carries the largest current, and the branch with the largest resistance carries the smallest current. The current-divider relationship formalizes this inverse proportionality. Step-by-Step Solution: 1) For each branch k, write I_k = V / R_k. 2) Compare two branches: if R_a > R_b, then I_a < I_b. 3) Conclude that the largest resistance cannot carry the largest current under a fixed applied voltage. Verification / Alternative check: Use the current-divider form for two branches: I_through_R1 = I_total * (R2 / (R1 + R2)). The numerator contains the other branch resistance, again showing inverse proportionality. Why Other Options Are Wrong: Correct: conflicts with inverse relation. Holds only for current sources: even with an ideal current source, branch currents still split inversely to resistance when branches are in parallel across the same voltage. Common Pitfalls: Assuming current spreads equally; mixing up series and parallel intuition; overlooking that voltage is common across branches in parallel. Final Answer: Incorrect

Question 20

Current divider principle: evaluate the statement that the most current flows in the parallel path which has the greatest resistance value.

Accepted Answer

Introduction / Context: The current divider rule is a cornerstone of basic circuit analysis. It predicts how a total current distributes between or among parallel impedances. The tested claim is the exact opposite of what the rule states, making this a good quick check of conceptual accuracy in DC resistive networks. Given Data / Assumptions: Parallel resistive branches excited by a source with fixed voltage. Linear time-invariant components. No reactive effects considered for this DC statement. Concept / Approach: For two branches R1 and R2, the current through R1 is I_total * (R2 / (R1 + R2)). This shows that as R1 increases, I_through_R1 decreases, because the numerator contains the other branch resistance. Extending to multiple branches, the path with the smallest resistance draws the largest share of current. Therefore, the statement claiming the most current in the greatest resistance path is incorrect. Step-by-Step Solution: 1) Write the current-divider expression for two branches. 2) Observe inverse proportionality between current and own resistance. 3) Generalize to many branches: the lowest resistance gets the highest current. Verification / Alternative check: Check with simple numbers: with V = 10 V, R_small = 1 ohm gives 10 A, R_large = 10 ohm gives 1 A. The smaller resistance indeed carries more current. Why Other Options Are Wrong: Correct: contradicts the divider rule. High frequency or inductive qualifiers do not reverse the inverse law for purely resistive DC networks. Common Pitfalls: Confusing series rules with parallel rules; misapplying proportional reasoning. Final Answer: Incorrect

Parallel Circuits Questions