Q: Single-node reference — node A is wired directly to the positive terminal of an ideal +10 V DC source, while the negative terminal of the source is the reference ground. What is the voltage at node A with respect to ground?

Introduction / Context: Reading node voltages relative to a defined reference (ground) is fundamental in circuit analysis. If a node is tied directly to an ideal source terminal, its potential is fixed by that source, independent of loads elsewhere, assuming ideal connections and no series impedance at the tie point. Given Data / Assumptions: An ideal +10 V DC source with its negative terminal at ground (0 V). Node A is directly connected to the positive terminal (no intervening elements). Wires are ideal with zero drop. Concept / Approach: By definition, the potential difference between the source’s positive and negative terminals is +10 V. Ground is 0 V at the negative terminal. Therefore, any node rigidly connected to the positive terminal must be at +10 V relative to ground. This is a direct application of the definition of ideal sources and reference nodes. Step-by-Step Solution: Set V_ground = 0 V at the source negative terminal.The ideal source enforces V_positive − V_ground = +10 V.Since node A is shorted to the positive terminal, V_A = V_positive.Thus, V_A = +10 V with respect to ground. Verification / Alternative check: Place an ideal voltmeter with the black lead at ground and the red lead at node A; it reads +10 V. In simulation, nodal analysis assigns +10 V to any node directly tied to the source positive terminal. Why Other Options Are Wrong: +9 V or +1 V: would require series impedance or a divider between node A and the source, which is absent.−10 V: would correspond to the opposite polarity connection.0 V: only true if node A were tied to ground, not to the positive terminal. Common Pitfalls: Forgetting to specify the reference node; assuming wire resistance causes drop when the problem states ideal conductors. Final Answer: +10 V

Q: Voltage divider calculation — an ideal 10 V DC source feeds a series divider of R1 = 1 kΩ and R2 = 9 kΩ. What is the voltage drop across R1 (that is, VR1)?

Introduction / Context: Voltage dividers are used to generate a fraction of a supply voltage. The share of the source that appears across each series resistor is set by its resistance relative to the total. This quick exercise reinforces the voltage-division rule for a simple two-resistor chain. Given Data / Assumptions: Ideal source: V_S = 10 V. Series resistors: R1 = 1 kΩ, R2 = 9 kΩ. No loading on the divider output. Concept / Approach: In a series chain with current I, the drop across each resistor is V_i = I * R_i. The common current is I = V_S / (R1 + R2). Therefore, V_R1 = I * R1 = V_S * (R1 / (R1 + R2)). This is the standard voltage-divider expression and is widely used in biasing and sensing circuits. Step-by-Step Solution: Compute the total resistance: R_T = 1 kΩ + 9 kΩ = 10 kΩ.Find the current: I = V_S / R_T = 10 V / 10 kΩ = 1 mA.Compute VR1: VR1 = I * R1 = 1 mA * 1 kΩ = 1 V.Alternatively use divider ratio: VR1 = 10 V * (1 kΩ / 10 kΩ) = 1 V. Verification / Alternative check: Check sum of drops: VR1 + VR2 = 1 V + 9 V = 10 V, which equals the source voltage as required by KVL. Why Other Options Are Wrong: 9 V and 10 V: correspond to the drop on R2 or the full source, not VR1.19 V: exceeds the source; impossible without an active element.0.5 V: would require a different ratio (R1 half of 1 kΩ relative to total). Common Pitfalls: Forgetting units (kΩ vs Ω); mixing up which resistor is being measured; ignoring divider loading when present in real designs. Final Answer: 1 V

Q: Direct measurement across a single resistor — in the network, resistor R3 alone is connected between nodes A and B (no other elements in parallel with it). If the measured voltage between A and B is 9 V, what is the voltage across R3 (VR3)?

Introduction / Context: When a single element is the only component connected between two nodes, the element voltage equals the node-to-node voltage by definition. This is a straightforward but very common situation when taking measurements on boards and in simulations. Given Data / Assumptions: R3 is the only component connected directly between nodes A and B. The measured node voltage difference is V_AB = 9 V. No additional parallel paths exist across A–B. Concept / Approach: Element voltage is the potential difference between its terminals. If those terminals are exactly nodes A and B, then the element voltage equals V_AB. No knowledge of resistance value or currents is needed to equate VR3 with the measured A–B voltage in this configuration. Step-by-Step Solution: Identify the terminals of R3 as node A (one end) and node B (the other end).Recognize VR3 is defined as V_A − V_B (with a chosen polarity).Given V_AB = 9 V, conclude VR3 = 9 V in magnitude with the same polarity convention.No further circuit details are required because R3 spans the measured nodes. Verification / Alternative check: Use an ideal voltmeter directly across R3; it reads the same 9 V as the A–B measurement because it is the same two points. Why Other Options Are Wrong: 8.1 V or 0.9 V: would imply a divider or series sharing not present.“More information is needed”: not in this topology; node-to-node equals element voltage.0 V: would correspond to a short or equal potentials, which contradicts V_AB = 9 V. Common Pitfalls: Misidentifying what nodes an element actually connects to; reversing polarity sign conventions (magnitude remains 9 V either way). Final Answer: 9 V

Q: Open-circuit fault effect between A and B — suppose nodes A and B are connected only through a single resistor R3 (no alternate paths). If R3 fails open, what is the resulting total resistance RT measured between A and B?

Introduction / Context: An open-circuit failure removes a current path. When there is only a single conduction path between two nodes, opening that element isolates the nodes and makes the equivalent resistance between them unbounded (infinite). Recognizing this outcome is vital for fault analysis and troubleshooting. Given Data / Assumptions: Nodes A and B are connected solely through resistor R3. No other series or parallel paths exist between A and B. R3 fails open (break in continuity). Concept / Approach: Resistance between two nodes is defined as the ratio V/I under a small test stimulus. If there is no conductive path, any finite applied voltage produces zero current, so R = V / I → ∞. Therefore, the equivalent resistance becomes infinite when the only branch is open. This is independent of the original resistance value of R3 prior to failure. Step-by-Step Solution: Model the intact circuit as a single branch: RT_intact = R3.Introduce an open in R3: current path between A and B is removed.Apply the definition R = V / I: with I = 0 A for any finite V, R → ∞.Conclude RT_open is infinite (open circuit between A and B). Verification / Alternative check: In measurement, an ohmmeter shows “OL” or very large resistance when probes are placed across A and B after the open occurs. Circuit simulators report no DC path between the nodes. Why Other Options Are Wrong: Finite values like 100 Ω, 110 Ω, 900 Ω, or 10 Ω imply a remaining conductive path, which contradicts the single-path assumption. Common Pitfalls: Assuming other leakage paths exist (they do not in this idealized scenario); confusing “large” with “infinite” — in analysis, we model an ideal open as infinite resistance. Final Answer: Infinite resistance

Question 1

Numerical branch current — if a branch resistor has 5.00 V across it and its resistance is 237 Ω, what is the branch current (choose the closest value)?

Accepted Answer

Introduction / Context: Applying Ohm’s law to compute a branch current from a measured branch voltage is a routine but essential skill. This question provides a realistic resistor value (237 Ω, an E24/E96 series style) and a neat round voltage to check unit handling and arithmetic accuracy. Given Data / Assumptions: Branch voltage V = 5.00 V (across the resistor). Resistance R = 237 Ω. Linear operation; temperature and tolerance effects ignored. Concept / Approach: Use Ohm’s law: I = V / R. Convert the result into an appropriate unit (amperes or milliamperes) and round sensibly to match the options while retaining significant figures consistent with the inputs provided. Step-by-Step Solution: Write Ohm’s law: I = V / R.Substitute: I = 5.00 / 237.Compute: 5.00 / 237 ≈ 0.02110… A.Convert to milliamperes: 0.02110… A ≈ 21.11 mA. Verification / Alternative check: Cross-check by estimating: 5 V across 250 Ω would be 20 mA; since 237 Ω is a bit lower than 250 Ω, current should be a bit higher than 20 mA — around 21 mA. This aligns with the exact calculation 21.11 mA. Why Other Options Are Wrong: 10 mA: Would require about 500 Ω at 5 V; not our case. 90 mA or 100 mA: Would imply ~55 Ω to 50 Ω at 5 V; far from 237 Ω. Common Pitfalls: Mistaking milliamps for amps (writing 0.0211 mA), or rounding too early. Keep unit consistency and convert at the end. Final Answer: 21.11 mA is the correct branch current for 5.00 V across 237 Ω.

Question 2

Series–parallel topology language — complete the statement: “Series components in a series–parallel circuit may be in series with other ______ components, or with other ______ components.”

Accepted Answer

Introduction / Context: Complex networks are built by nesting series and parallel groupings. A precise vocabulary helps when describing how subcircuits are connected. This fill-in statement checks whether you can correctly identify that a series element can continue in series or meet a subcircuit that is arranged in parallel at the next stage of the topology. Given Data / Assumptions: Standard definitions: series elements share the same current path; parallel elements share the same node pair (voltage). Networks are constructed by combining blocks recursively. We are focusing on terminology, not a specific numerical example. Concept / Approach: A series–parallel circuit is formed by connecting components and sub-networks in either series or parallel at each stage. Therefore, a set of series elements may continue in series with other series elements, or that series path may be in series with a block that itself is a parallel combination. The correct completion is “series, parallel.” Step-by-Step Solution: Recall series: one current path; drops add; resistances add.Recall parallel: same node pair; currents add by KCL; conductances add.Recognize nesting: a series chain can precede or follow a parallel block.Fill the blanks: series components may be in series with other series components, or with other parallel components. Verification / Alternative check: Look at a practical RC filter: a series resistor feeding a shunt RC (a parallel block). The series resistor is in series with a parallel network (capacitor || load), matching the statement. Why Other Options Are Wrong: individual, combinations of: Vague and not a recognized pair of topological terms. parallel, series: Reverses the intended order; the first blank must reflect what series components can be with next. shunt, parallel: “Shunt” is another word for parallel, but the pair does not complete the sentence logically. Common Pitfalls: Confusing “in series with a parallel combination” versus “in parallel with a series combination.” The prepositions matter for accurate descriptions. Final Answer: series, parallel — a series path may continue in series or encounter a parallel block.

Question 3

Series–parallel resistance between two nodes A and B  A network consists of two ideal resistors connected in simple series between nodes A and B: a 10 Ω resistor followed by a 1 kΩ (1000 Ω) resistor. For this explicit series connection, what is the …

Accepted Answer

Introduction / Context: This question checks your ability to compute the equivalent resistance of a strictly series connection. Series networks are everywhere in bias chains, current-limiting paths, and sensor dividers. Getting comfortable with series addition is essential before moving to mixed series–parallel reductions and Thevenin/Norton modeling. Given Data / Assumptions: Two ideal resistors are connected in series between nodes A and B. R1 = 10 Ω and R2 = 1 kΩ (1000 Ω). Conductors are ideal; there are no hidden branches or parallel paths. We seek RT as seen between A and B. Concept / Approach: For resistors in series, the equivalent resistance is the arithmetic sum of the individual resistances. The rule is topology-based and independent of excitation (DC or AC) as long as elements are linear resistors. The governing relation is: R_total = R1 + R2 + … + Rn. Once RT is known, any loop current under a given source follows directly from I = V / RT. Step-by-Step Solution: Identify the connection as pure series (one single current path).Write the sum: RT = R1 + R2.Insert values in ohms: RT = 10 + 1000 = 1010 Ω.Report RT to match option formatting: 1,010 Ω. Verification / Alternative check: Use a quick current check. With a 10 V ideal source across A–B, I = 10 / 1010 ≈ 9.90 mA. The drops would be V_R1 ≈ 0.099 V and V_R2 ≈ 9.90 V, which sum back to 10 V, confirming series behavior and the calculated RT. Why Other Options Are Wrong: 100 Ω: would imply a different network (e.g., partial parallel) that is not given.900 Ω: does not equal the sum 10 + 1000.10 Ω: ignores the 1 kΩ element entirely.10,100 Ω: a misplaced digit error; it is a common slip when adding mixed units. Common Pitfalls: Confusing series with parallel (parallel uses reciprocals); mixing kΩ and Ω without converting; accidentally omitting a component or misreading comma formatting in values. Final Answer: 1,010 Ω

Question 4

Loading a voltage divider — when a finite load is connected across the output of an ideal voltage divider, how does the total equivalent resistance seen by the source change?

Accepted Answer

Introduction / Context: Voltage dividers are widely used to scale voltages. However, attaching a load across the divider output alters the circuit seen by the source. Understanding how loading changes the equivalent resistance is crucial for accurate design, source current estimation, and power budgeting. Given Data / Assumptions: An ideal two-resistor divider is driven by a source. A load resistor RL is connected across the divider’s output node and ground. All components are linear, and wires are ideal. Concept / Approach: The upper resistor remains unchanged, but the lower divider resistor now appears in parallel with RL. Since any non-infinite RL in parallel with the lower leg reduces its effective resistance, the total equivalent resistance from the source (upper in series with the new, smaller lower equivalent) must be less than before. In general, R_total_loaded = R_top + (R_bottom // RL), and R_bottom // RL < R_bottom for any finite RL > 0. Step-by-Step Solution: Start from the unloaded divider: R_total_unloaded = R_top + R_bottom.Attach RL across the output: R_bottom_loaded = (R_bottom * RL) / (R_bottom + RL).Compute total: R_total_loaded = R_top + R_bottom_loaded.Conclude: since R_bottom_loaded < R_bottom, R_total_loaded < R_total_unloaded, i.e., it decreases. Verification / Alternative check: Example: R_top = 10 kΩ, R_bottom = 10 kΩ. Unloaded total = 20 kΩ. With RL = 10 kΩ, the lower equivalent is 5 kΩ, so loaded total = 15 kΩ, clearly lower than 20 kΩ. Why Other Options Are Wrong: Increase/Remain/Double: contradict the mathematics of a parallel combination reducing resistance.Become infinite: would require an open circuit, the opposite of adding a load. Common Pitfalls: Forgetting that loading not only changes the source’s seen resistance but also reduces the divider output voltage (divider droop); ignoring the impact on source current and power dissipation. Final Answer: Decrease

Question 5

Mixed series–parallel network with a fault — four identical 10 kΩ resistors are connected in parallel to form a branch, and this parallel branch is in series with a 20 kΩ resistor and an ideal source. If one of the parallel branches develops a hard …

Accepted Answer

Introduction / Context: Diagnosing faults in series–parallel networks requires recognizing how a single failure can change node voltages and element stresses. A short across a parallel group is a classic fault that collapses the node voltage feeding all other parallel elements, affecting both operation and safety margins. Given Data / Assumptions: Four 10 kΩ resistors are in parallel as a single branch. This parallel branch is in series with a 20 kΩ resistor and a source. One branch in the parallel group fails as a near-ideal short (0 Ω) effectively across the entire parallel node pair. Wires and source are otherwise ideal. Concept / Approach: In any parallel network, all components share the same node-to-node voltage. If a hard short is placed across the parallel node pair, the node pair becomes equipotential (no difference), so the voltage across that entire group becomes approximately 0 V. Therefore, every remaining parallel resistor sees zero volts and, consequently, zero current. The series 20 kΩ now carries whatever current is determined by the source and the shorted path, not by the original parallel combination. Step-by-Step Solution: Before the fault, V_parallel has some finite value determined by divider action with 20 kΩ.A short across the parallel node forces V_parallel → 0 V (KVL around the short branch).Since all parallel elements share this node pair, each remaining 10 kΩ now has 0 V across it.Therefore, their currents go to 0 A, and their voltage stress disappears (though the shorted path may overheat). Verification / Alternative check: Replace the entire parallel group with its equivalent. With a short across the output nodes of the group, the equivalent becomes 0 Ω. The series chain reduces to 20 kΩ in series with 0 Ω to ground, making the node feeding the group at ground potential relative to the group—hence 0 V across it. Why Other Options Are Wrong: Increases/Decreases-but-not-zero/Remains: contradict the presence of an ideal short which enforces zero voltage across the shorted terminals.Becomes negative: sign is irrelevant; magnitude becomes zero under an ideal short. Common Pitfalls: Thinking only the shorted branch is affected; in reality, the short collapses the voltage for the entire parallel group. Also, overlooking power implications in the series element and the shorted path. Final Answer: Drops to zero

Question 6

Single-node reference — node A is wired directly to the positive terminal of an ideal +10 V DC source, while the negative terminal of the source is the reference ground. What is the voltage at node A with respect to ground?

Accepted Answer

Introduction / Context: Reading node voltages relative to a defined reference (ground) is fundamental in circuit analysis. If a node is tied directly to an ideal source terminal, its potential is fixed by that source, independent of loads elsewhere, assuming ideal connections and no series impedance at the tie point. Given Data / Assumptions: An ideal +10 V DC source with its negative terminal at ground (0 V). Node A is directly connected to the positive terminal (no intervening elements). Wires are ideal with zero drop. Concept / Approach: By definition, the potential difference between the source’s positive and negative terminals is +10 V. Ground is 0 V at the negative terminal. Therefore, any node rigidly connected to the positive terminal must be at +10 V relative to ground. This is a direct application of the definition of ideal sources and reference nodes. Step-by-Step Solution: Set V_ground = 0 V at the source negative terminal.The ideal source enforces V_positive − V_ground = +10 V.Since node A is shorted to the positive terminal, V_A = V_positive.Thus, V_A = +10 V with respect to ground. Verification / Alternative check: Place an ideal voltmeter with the black lead at ground and the red lead at node A; it reads +10 V. In simulation, nodal analysis assigns +10 V to any node directly tied to the source positive terminal. Why Other Options Are Wrong: +9 V or +1 V: would require series impedance or a divider between node A and the source, which is absent.−10 V: would correspond to the opposite polarity connection.0 V: only true if node A were tied to ground, not to the positive terminal. Common Pitfalls: Forgetting to specify the reference node; assuming wire resistance causes drop when the problem states ideal conductors. Final Answer: +10 V

Question 7

Analyzing circuits with multiple independent sources — which classical network theorem is specifically applied by turning on one source at a time and summing the individual responses?

Accepted Answer

Introduction / Context: Real circuits often contain several independent sources (multiple voltage and/or current sources). Accurately predicting currents and voltages then benefits from a methodical approach that isolates the effect of each source and recombines them. The technique most directly aligned with this idea is the superposition theorem. Given Data / Assumptions: Linear, bilateral components (resistors, linear dependent sources, etc.). Multiple independent sources are present. We seek a principle that evaluates one source at a time. Concept / Approach: The superposition theorem states that in any linear circuit with several independent sources, the response (voltage or current) at any element equals the algebraic sum of the responses caused by each independent source acting alone, with all other independent voltage sources replaced by their internal resistance (ideal: short) and all other independent current sources replaced by their internal resistance (ideal: open). Linearity is essential for superposition to hold. Step-by-Step Solution: Choose a quantity of interest (e.g., current through R or voltage at a node).Deactivate all but one independent source (voltage sources → short, current sources → open).Solve the simplified circuit for the partial response.Repeat for each independent source and algebraically sum the partial responses. Verification / Alternative check: Compare the superposition sum with a single-shot nodal/mesh solution including all sources; they match in linear circuits, validating the theorem. Why Other Options Are Wrong: Thevenin’s theorem: converts a network to a single source and series resistance; useful but different goal.Kirchhoff’s/Ohm’s laws: foundational equations, not a turn-one-source-at-a-time method.Norton’s maximum power rule: not a standard theorem wording; maximum power transfer is a separate condition. Common Pitfalls: Applying superposition to power directly (power is nonlinear in voltage/current); forgetting to properly deactivate sources; using it with nonlinear elements where it does not apply. Final Answer: Superposition

Question 8

Voltage divider calculation — an ideal 10 V DC source feeds a series divider of R1 = 1 kΩ and R2 = 9 kΩ. What is the voltage drop across R1 (that is, VR1)?

Accepted Answer

Introduction / Context: Voltage dividers are used to generate a fraction of a supply voltage. The share of the source that appears across each series resistor is set by its resistance relative to the total. This quick exercise reinforces the voltage-division rule for a simple two-resistor chain. Given Data / Assumptions: Ideal source: V_S = 10 V. Series resistors: R1 = 1 kΩ, R2 = 9 kΩ. No loading on the divider output. Concept / Approach: In a series chain with current I, the drop across each resistor is V_i = I * R_i. The common current is I = V_S / (R1 + R2). Therefore, V_R1 = I * R1 = V_S * (R1 / (R1 + R2)). This is the standard voltage-divider expression and is widely used in biasing and sensing circuits. Step-by-Step Solution: Compute the total resistance: R_T = 1 kΩ + 9 kΩ = 10 kΩ.Find the current: I = V_S / R_T = 10 V / 10 kΩ = 1 mA.Compute VR1: VR1 = I * R1 = 1 mA * 1 kΩ = 1 V.Alternatively use divider ratio: VR1 = 10 V * (1 kΩ / 10 kΩ) = 1 V. Verification / Alternative check: Check sum of drops: VR1 + VR2 = 1 V + 9 V = 10 V, which equals the source voltage as required by KVL. Why Other Options Are Wrong: 9 V and 10 V: correspond to the drop on R2 or the full source, not VR1.19 V: exceeds the source; impossible without an active element.0.5 V: would require a different ratio (R1 half of 1 kΩ relative to total). Common Pitfalls: Forgetting units (kΩ vs Ω); mixing up which resistor is being measured; ignoring divider loading when present in real designs. Final Answer: 1 V

Question 9

Direct measurement across a single resistor — in the network, resistor R3 alone is connected between nodes A and B (no other elements in parallel with it). If the measured voltage between A and B is 9 V, what is the voltage across R3 (VR3)?

Accepted Answer

Introduction / Context: When a single element is the only component connected between two nodes, the element voltage equals the node-to-node voltage by definition. This is a straightforward but very common situation when taking measurements on boards and in simulations. Given Data / Assumptions: R3 is the only component connected directly between nodes A and B. The measured node voltage difference is V_AB = 9 V. No additional parallel paths exist across A–B. Concept / Approach: Element voltage is the potential difference between its terminals. If those terminals are exactly nodes A and B, then the element voltage equals V_AB. No knowledge of resistance value or currents is needed to equate VR3 with the measured A–B voltage in this configuration. Step-by-Step Solution: Identify the terminals of R3 as node A (one end) and node B (the other end).Recognize VR3 is defined as V_A − V_B (with a chosen polarity).Given V_AB = 9 V, conclude VR3 = 9 V in magnitude with the same polarity convention.No further circuit details are required because R3 spans the measured nodes. Verification / Alternative check: Use an ideal voltmeter directly across R3; it reads the same 9 V as the A–B measurement because it is the same two points. Why Other Options Are Wrong: 8.1 V or 0.9 V: would imply a divider or series sharing not present.“More information is needed”: not in this topology; node-to-node equals element voltage.0 V: would correspond to a short or equal potentials, which contradicts V_AB = 9 V. Common Pitfalls: Misidentifying what nodes an element actually connects to; reversing polarity sign conventions (magnitude remains 9 V either way). Final Answer: 9 V

Question 10

Thevenin equivalent form — which option correctly describes the standard Thevenin representation of any linear two-terminal network as seen from a pair of output terminals?

Accepted Answer

Introduction / Context: Thevenin’s theorem is a powerful simplification tool. It allows you to replace any linear, bilateral two-terminal network with an equivalent source and resistance that behave identically from the perspective of the external load. Recognizing the correct canonical form is essential before you compute V_TH and R_TH. Given Data / Assumptions: The original network is linear and bilateral (resistors, independent/dependent sources that are linear). We are looking from a specific pair of output terminals. Frequency effects are ignored unless reactive elements are converted to equivalent impedances (then R_TH becomes Z_TH). Concept / Approach: The Thevenin equivalent is defined as a single ideal voltage source V_TH in series with a single resistance R_TH (or impedance Z_TH). V_TH equals the open-circuit voltage at the terminals. R_TH equals the equivalent resistance seen looking back into the network with independent sources deactivated (voltage sources shorted, current sources opened), or equivalently the ratio of V_TH to I_SC for purely resistive linear circuits. Step-by-Step Solution: Compute V_TH by finding the open-circuit terminal voltage.Compute R_TH by deactivating independent sources and finding the resistance seen at the terminals (or use R_TH = V_TH / I_SC).Construct the equivalent: an ideal V_TH source in series with R_TH.Attach any load RL and analyze easily via a simple series division. Verification / Alternative check: Confirm equivalence by checking that, for any RL, the original network and the Thevenin model produce identical terminal voltage and current. Why Other Options Are Wrong: Current source in series with resistance: that is Norton converted incorrectly; Norton uses a current source in parallel with resistance.Series–parallel ladder or many sources: not a canonical reduced form.Transformer-only model: unrelated to the general linear two-terminal reduction unless the original network is specifically a transformer circuit. Common Pitfalls: Mixing up Norton and Thevenin forms; forgetting to deactivate sources properly when finding R_TH; applying the theorem to nonlinear networks. Final Answer: A voltage source in series with a resistance

Question 11

Open-circuit fault effect between A and B — suppose nodes A and B are connected only through a single resistor R3 (no alternate paths). If R3 fails open, what is the resulting total resistance RT measured between A and B?

Accepted Answer

Introduction / Context: An open-circuit failure removes a current path. When there is only a single conduction path between two nodes, opening that element isolates the nodes and makes the equivalent resistance between them unbounded (infinite). Recognizing this outcome is vital for fault analysis and troubleshooting. Given Data / Assumptions: Nodes A and B are connected solely through resistor R3. No other series or parallel paths exist between A and B. R3 fails open (break in continuity). Concept / Approach: Resistance between two nodes is defined as the ratio V/I under a small test stimulus. If there is no conductive path, any finite applied voltage produces zero current, so R = V / I → ∞. Therefore, the equivalent resistance becomes infinite when the only branch is open. This is independent of the original resistance value of R3 prior to failure. Step-by-Step Solution: Model the intact circuit as a single branch: RT_intact = R3.Introduce an open in R3: current path between A and B is removed.Apply the definition R = V / I: with I = 0 A for any finite V, R → ∞.Conclude RT_open is infinite (open circuit between A and B). Verification / Alternative check: In measurement, an ohmmeter shows “OL” or very large resistance when probes are placed across A and B after the open occurs. Circuit simulators report no DC path between the nodes. Why Other Options Are Wrong: Finite values like 100 Ω, 110 Ω, 900 Ω, or 10 Ω imply a remaining conductive path, which contradicts the single-path assumption. Common Pitfalls: Assuming other leakage paths exist (they do not in this idealized scenario); confusing “large” with “infinite” — in analysis, we model an ideal open as infinite resistance. Final Answer: Infinite resistance

Series-Parallel Circuits Questions