Question 1

Series circuits — as additional resistors are added in series (one after another in a single path), the total equivalent resistance of the circuit increases. Evaluate this statement for ideal resistors.

Accepted Answer

Introduction / Context: Understanding how total resistance changes when components are added is foundational for circuit design and troubleshooting. In a series connection, components share exactly one continuous path for current. Knowing how equivalent resistance behaves helps predict changes in current, voltage division, and power dissipation when adding or removing parts. Given Data / Assumptions: Ideal, linear resistors are connected strictly in series. Temperature effects and tolerance are ignored. Single DC source and steady-state conditions are assumed for intuition, though the rule is topology-based and source-agnostic. Concept / Approach: For resistors in series, the equivalent resistance is the arithmetic sum: R_total = R1 + R2 + … + Rn. Each added resistor contributes a positive amount of ohmic opposition to current. Therefore, adding any additional positive resistance increases R_total, which for a fixed source voltage reduces circuit current via Ohm’s law I = V/R_total. Step-by-Step Solution: Model the original circuit with R_total1 = ΣRi.Add one more series resistor Rx > 0.Compute the new equivalent: R_total2 = R_total1 + Rx.Since Rx > 0, R_total2 > R_total1, confirming the statement. Verification / Alternative check: Try numbers: Two 1 kΩ resistors in series → 2 kΩ. Add 470 Ω → 2.47 kΩ. With a 10 V source, current drops from 10/2000 = 5 mA to 10/2470 ≈ 4.05 mA, consistent with increased total resistance. Why Other Options Are Wrong: Incorrect: contradicts R_total = ΣRi for series networks.“Only AC” or “only equal values”: series summation holds regardless of frequency or equality of values (for ideal resistors).“Only with a capacitor”: passive capacitors are irrelevant to the algebraic sum of series resistances. Common Pitfalls: Confusing series with parallel (where adding a branch can reduce R_total); assuming physical proximity implies series—node analysis must confirm a single current path. Final Answer: Correct

Question 2

Series-path definition — a “series circuit” between two nodes provides only one continuous path for current flow between those points. Assess this definition.

Accepted Answer

Introduction / Context: Series and parallel are topological relationships that dictate how voltage and current distribute in circuits. Correctly identifying a series path is crucial before applying Ohm’s law, voltage division, and Kirchhoff’s laws. Misclassification leads to wrong calculations of currents, voltages, and power ratings. Given Data / Assumptions: The statement defines a series circuit as a single current path between two nodes. Components are ideal and connections are perfect conductors. Steady-state analysis is considered for simplicity. Concept / Approach: In a strictly series connection, every electron that leaves the source must pass through each element sequentially and return—there is no branching node that creates alternative paths. Consequently, the same current flows through all series elements, while voltages across them generally differ according to their individual impedances (V_i = I * R_i for resistors). Step-by-Step Solution: Mark the two nodes under consideration (start and return).Trace the conductive path and check for any node where current could split.If no branching exists, all elements on that path are in series and share identical current.Apply voltage division if needed: V_i = (R_i / ΣR) * V_source for resistive series networks. Verification / Alternative check: Use Kirchhoff’s Current Law at each intermediate node; with one path only, the current in equals the current out and equals the same I everywhere. Measurements with an ammeter in multiple series locations will read the same current (within instrument tolerance). Why Other Options Are Wrong: “Incorrect”: conflicts with the standard definition found in textbooks and datasheets.AC-only / identical resistors / low-current caveats: series topology is independent of frequency, value equality, and magnitude of current. Common Pitfalls: Assuming components are in series just because they are drawn in a line—node labeling must confirm a single path; ignoring hidden branches like measurement shunts or protection paths. Final Answer: Correct

Question 3

Series current continuity — in an ideal series circuit at steady state, the current leaving a resistor is equal to the current entering that resistor. Evaluate this statement.

Accepted Answer

Introduction / Context: Current continuity is a direct consequence of charge conservation. In a series circuit, every element experiences the same current because there is only one path for charges to move. Recognizing this allows you to simplify analysis and avoid mistakes such as summing currents in series networks (which is unnecessary and incorrect). Given Data / Assumptions: Ideal conductors and resistors. Steady-state DC (though the principle applies to AC instantaneously as well). No parallel leakage paths or measurement shunts. Concept / Approach: Kirchhoff’s Current Law (KCL) states that the algebraic sum of currents at a node is zero. In series, a node exists between elements but does not branch; thus, current entering that node must equal current leaving it. Consequently, current in any series element is identical to current in every other series element. Step-by-Step Solution: Select any resistor in a series chain.Label the entering current as I_in and the leaving current as I_out.Apply KCL to the node on either side of the resistor: ΣI = 0 implies I_in = I_out.Conclude that the statement is correct for ideal series circuits. Verification / Alternative check: Measure with an ammeter placed in different series positions; readings are the same. Simulations (SPICE) confirm identical branch currents when no parallel branches exist. Why Other Options Are Wrong: “Incorrect” and the conditional options contradict KCL and the nature of a single, unbranched path. Common Pitfalls: Confusing series with parallel (in parallel, currents differ by branch conductance); overlooking small unintended branches like meter shunts that slightly alter currents. Final Answer: Correct

Question 4

Single-resistor series example — consider a simple circuit with one ideal resistor R1 connected directly across an ideal 10 V DC source. What is the voltage across R1?

Accepted Answer

Introduction / Context: This problem reinforces the idea that in a one-element series circuit, the entire source voltage appears across that element. It is a direct application of Kirchhoff’s Voltage Law (KVL) and Ohm’s law and sets the stage for understanding voltage division when more than one element is present. Given Data / Assumptions: Ideal 10 V DC source. Only one component in the loop: resistor R1. Ideal wires, no internal source resistance. Concept / Approach: KVL states that the algebraic sum of voltages around any closed loop is zero. With a single resistor in the loop, the resistor must drop the full source voltage to satisfy KVL. Ohm’s law relates current and resistance, but the drop is set by the source magnitude because there are no other elements to share it. Step-by-Step Solution: Write KVL: +V_source − V_R1 = 0.Substitute the given source: +10 − V_R1 = 0.Solve for V_R1: V_R1 = 10 V.Note: Current I = 10 / R1, but the exact R1 value does not change the fact that V_R1 = 10 V. Verification / Alternative check: Measure with an ideal voltmeter connected across R1; it reads the node-to-node source voltage because the resistor is directly across the source terminals. Why Other Options Are Wrong: 5 V: would require a second equal drop in series; not present.“Depends only on resistance” or “more information required”: the lone element must take the entire 10 V.0 V: implies a short circuit or zero source, which contradicts the setup. Common Pitfalls: Assuming “voltage division” always applies; it only applies when multiple series elements share the source voltage. Final Answer: 10 V

Question 5

Kirchhoff’s Voltage Law (KVL) in series loops — the sum of all individual voltage drops around a series circuit equals the source voltage. Assess this statement for ideal components.

Accepted Answer

Introduction / Context: Kirchhoff’s Voltage Law is a cornerstone of circuit analysis. It expresses energy conservation around a closed path: the total rise from sources equals the total drop across elements. In series circuits, this means the source voltage is apportioned among the series elements according to their impedances. Given Data / Assumptions: Ideal wiring and components (no parasitic EMF sources). Closed single-loop series circuit. Steady-state analysis for simplicity; KVL is general and also applies instantaneously in AC. Concept / Approach: KVL states ΣV_rises − ΣV_drops = 0. For a single source V_S and series elements with drops V1, V2, …, Vn, we have V_S = V1 + V2 + … + Vn. This enables voltage-divider design and power budgeting in series networks. Step-by-Step Solution: Define loop direction and label polarities using the passive sign convention.Write KVL: +V_S − V1 − V2 − … − Vn = 0.Rearrange: V_S = V1 + V2 + … + Vn.Use Ohm’s law as needed: Vi = I * Ri, where I is the common series current. Verification / Alternative check: Measure all individual drops and sum them; the total equals the measured source voltage within meter tolerance. SPICE simulations enforce KVL numerically, providing another check. Why Other Options Are Wrong: “Incorrect” and conditional options contradict a fundamental conservation law; equality of resistances, frequency, or waveform type does not invalidate KVL for lumped circuits. Common Pitfalls: Mixing source internal resistance into “external” drops without accounting for it; mislabeling polarities and signs; assuming wiring is ideal while significant wiring drops exist in high-current layouts. Final Answer: Correct

Question 6

Voltage divider check — a series circuit with three resistors R1, R2, and R3 across a 55 V ideal source is measured to have drops VR1 = 10 V, VR2 = 15 V, and VR3 = 30 V. Do these readings satisfy Kirchhoff’s Voltage Law for the loop?

Accepted Answer

Introduction / Context: Voltage-divider measurements in series circuits must obey Kirchhoff’s Voltage Law (KVL): the sum of drops equals the source. This problem checks whether provided voltage readings are self-consistent for a loop driven by an ideal 55 V source. Given Data / Assumptions: Ideal source of 55 V and ideal wiring. Three resistors in strict series; no parallel branches. Measured drops: 10 V, 15 V, 30 V across R1, R2, R3. Concept / Approach: Apply KVL around the loop. If the algebraic sum of measured drops equals the source magnitude, the readings are consistent with KVL. Individual resistor values are not necessary to verify the law; only the sum matters for the loop equation. Step-by-Step Solution: Compute the sum of measured drops: 10 + 15 + 30 = 55 V.Compare with source: V_S = 55 V.Since ΣV_drops = V_S, KVL is satisfied.Therefore, the measurement set is self-consistent for an ideal series circuit. Verification / Alternative check: Using voltage division, Vi = (Ri / ΣR) * V_S. For any positive resistances that sum to ΣR, the drops will sum to V_S by construction. The exact Ri values are not needed to validate KVL compliance. Why Other Options Are Wrong: “KVL violated” contradicts the demonstrated equality 10 + 15 + 30 = 55.“Only if identical” and “only AC” are irrelevant to KVL in lumped circuits.“Cannot be judged without resistor values” is incorrect; KVL concerns loop sums, not individual resistances. Common Pitfalls: Forgetting to include the source’s internal resistance drop when modeling non-ideal sources; misreading meter polarities leading to sign errors. Final Answer: Correct — the readings satisfy KVL

Question 7

Total current in a series circuit — should you add individual branch currents to obtain the total current in a single-path series network?

Accepted Answer

Introduction / Context: Students often confuse rules for series and parallel networks. In a single-path series circuit, current does not split; thus, “adding currents” has no meaning because there is only one current everywhere in the loop. Recognizing this prevents calculation errors and misinterpretation of ammeter readings. Given Data / Assumptions: A single closed series path (no branches). Ideal components and wiring. Steady-state operation. Concept / Approach: By definition, series elements carry identical current. KCL at each internal node implies current entering equals current leaving; with no branching nodes, the same current traverses all components. Therefore, the total current drawn from the source is simply that one current value, obtained via I = V_source / R_total for resistive networks, not a sum of separate branch currents (as would be the case for parallel networks). Step-by-Step Solution: Compute equivalent resistance: R_total = ΣRi for series resistors.Apply Ohm’s law: I = V_source / R_total.Recognize that this I is the same through each resistor: I1 = I2 = … = I.Conclude there is no summation of currents in series; there is only one current. Verification / Alternative check: Insert an ammeter at different series points; readings are identical, confirming that current is not distributed among branches but remains singular in magnitude and direction (for DC). Why Other Options Are Wrong: “Add the currents”: applies to parallel branches, not to a single series path.Frequency, value equality, or capacitors do not change the series current principle for the same instantaneous current in each element. Common Pitfalls: Confusing series and parallel behaviors; overlooking small leakage or measurement branches that can create unintended parallel paths. Final Answer: Incorrect — the same current flows through every series element

Question 8

Voltage division rule — in a pure series resistive circuit carrying a common current, larger resistance values develop larger voltage drops across them. Evaluate this statement.

Accepted Answer

Introduction / Context: Voltage division in series circuits is a standard tool for biasing, sensing, and signal scaling. It relies on the fact that the same current flows through all series elements, so each element’s voltage drop is proportional to its resistance. This question checks understanding of that proportionality. Given Data / Assumptions: Purely resistive series network. Common current I flows through all resistors. Ideal conditions (no parasitics). Concept / Approach: Ohm’s law says V_i = I * R_i. Since I is common to all series resistors, the element with the larger R_i will have the larger V_i. The voltage-divider expression quantifies this: V_k = (R_k / ΣR) * V_source. Hence, larger R_k yields a larger fraction of V_source, provided all resistances are positive and finite. Step-by-Step Solution: Identify series network and common current I.Use V_i = I * R_i to express each drop.Compare two resistors, Ra and Rb: if Ra > Rb then V_a > V_b.Conclude the statement is correct for ideal resistive series circuits. Verification / Alternative check: Example: Two resistors 1 kΩ and 3 kΩ in series on 8 V give drops 2 V and 6 V respectively, confirming that the larger resistor gets the larger drop in proportion to its value. Why Other Options Are Wrong: Conditional options about voltage level, tolerance, or AC do not alter the fundamental proportionality in ideal, linear resistive networks. Common Pitfalls: Forgetting internal source resistance or measurement loading; confusing series with parallel where voltage across branches is equal rather than divided. Final Answer: Correct

Question 9

Recovered data for solvability — a series network has three resistors R1, R2, and R3 with total equivalent resistance R_total = 600 kΩ. If R1 = 100 kΩ and R3 = 300 kΩ, what is the value of R2?

Accepted Answer

Introduction / Context: When analyzing series circuits, the equivalent resistance is simply the sum of individual resistances. If the total is known along with all but one element, you can directly compute the missing value. This mirrors many troubleshooting scenarios where you measure total and some components but one reading is unavailable. Given Data / Assumptions: Series network: R_total = R1 + R2 + R3. R_total = 600 kΩ, R1 = 100 kΩ, R3 = 300 kΩ. Ideal resistors; temperature effects ignored. Concept / Approach: Use the series-sum formula and solve for the unknown. This is an algebraic rearrangement requiring no advanced methods. Ensuring units are consistent (all in kΩ) simplifies the arithmetic and avoids conversion errors. Step-by-Step Solution: Write the sum: R_total = R1 + R2 + R3.Substitute known values: 600 = 100 + R2 + 300 (all in kΩ).Combine known terms: 600 = R2 + 400.Solve for the unknown: R2 = 600 − 400 = 200 kΩ. Verification / Alternative check: Back-substitute: 100 + 200 + 300 = 600 kΩ, which matches the specified total, confirming the calculation. Why Other Options Are Wrong: 100 kΩ or 300 kΩ: would sum to 500 kΩ or 700 kΩ, inconsistent with the given total.“More information is needed”: not true; series-sum provides a unique solution.50 kΩ: totals 450 kΩ, not 600 kΩ. Common Pitfalls: Mishandling units (Ω vs kΩ); forgetting that only series, not parallel, uses direct addition; accidentally subtracting the wrong way around. Final Answer: 200 kΩ

Question 10

Passive sign convention in resistors — for a resistor carrying conventional current from one terminal to the other, the terminal where current enters is labeled positive and the terminal where current exits is labeled negative. Choose the correct po…

Accepted Answer

Introduction / Context: Correct polarity labeling is essential for applying Kirchhoff’s laws and for calculating power in passive components. The passive sign convention standardizes how we assign voltage polarities relative to current direction so that power absorbed by a passive element is positive when current enters its positive-labeled terminal. Given Data / Assumptions: A single resistor with conventional current defined from the entry terminal to the exit terminal. Ideal resistor behavior (V = I * R). Steady-state consideration, though the convention also applies instantaneously in AC. Concept / Approach: Under the passive sign convention, voltage polarity across a passive element is assigned such that the terminal where current enters is marked positive. For a resistor this ensures that power P = V * I is positive when the resistor dissipates energy as heat: with V defined positive at the entry terminal and current I defined entering that terminal, P becomes positive, matching physical intuition. Step-by-Step Solution: Assume conventional current flows into terminal A and out of terminal B.Label terminal A as “+” and terminal B as “−”.Relate voltage to current: V_AB = I * R with I entering terminal A.Compute power: P = V_AB * I ≥ 0, indicating power absorbed by the resistor. Verification / Alternative check: Measure the voltage with a meter whose red lead is at the entry side; a positive reading corresponds to a drop from entry to exit consistent with energy dissipation. Reversing the leads or current direction flips signs consistently while preserving the convention. Why Other Options Are Wrong: Negative at entry / positive at exit: this violates the passive sign convention for the given current direction.Same polarity at both terminals: impossible because a voltage is a potential difference between two points.“Cannot be defined”: polarity can always be defined once current direction is chosen. Common Pitfalls: Mixing up active versus passive sign conventions; forgetting to keep current and voltage reference directions consistent, which leads to wrong signs in KVL/KCL equations and power calculations. Final Answer: Positive at entry, negative at exit

Question 11

Series-circuit reasoning (insufficient data case): In a resistor network that includes R3, what is the voltage drop across R3? Assume only that elements are passive and linear, with a single DC source; no numerical values or configuration are provid…

Accepted Answer

Introduction / Context: Finding a voltage drop across an individual resistor requires knowing how the resistor is connected (series, parallel, or mixed), the source value(s), and the relevant component values. Without this information, any numeric answer is a guess. This question trains you to recognize when data are insufficient and to ask for the missing parameters before solving. Given Data / Assumptions: No circuit diagram is supplied in the stem. The resistor is labeled R3 but its value and connection are not provided. Only the general context of DC passive linear components is implied. Concept / Approach: Use Kirchhoff’s laws and Ohm’s law. To compute a drop v_R3, you need either the branch current through R3 (then v_R3 = I_R3 * R3) or the node voltages at each end of R3 (then v_R3 = V_node1 − V_node2). Both require actual component values and topology. Without them, there is no unique solution. Step-by-Step Solution: Recognize required inputs: R3 value plus the current through it or node voltages around it.Check the stem: none of these are given, and the configuration (series/parallel) is unknown.Conclude: the problem is underdetermined; many different circuits could use an “R3” label with different outcomes. Verification / Alternative check: If a series chain of known resistors and a known source were provided, you would compute total current I = V_total / R_total and then v_R3 = I * R3. In a parallel branch, you would first find the node voltage across that branch, then multiply by current or use divider relations. In all cases, missing data prevent calculation here. Why Other Options Are Wrong: Numeric values such as 1.5 V, 3 V, 6 V, or 9 V imply a specific circuit that is not supplied. Selecting them assumes facts not in evidence. Common Pitfalls: Memorizing a single number for “typical” drops. Voltage division depends on actual ratios and connections; never assume. Final Answer: Cannot be determined without the circuit configuration and values

Question 12

Series resistance addition (basic recall): If a 47 Ω resistor is connected in series with a 68 Ω resistor, what is the total equivalent resistance of the series combination?

Accepted Answer

Introduction / Context: In a series circuit, resistances add directly because the same current must pass through each element. This question checks your ability to compute an equivalent resistance quickly and to distinguish series behavior from parallel behavior, where the rule is different. Given Data / Assumptions: Two resistors: 47 Ω and 68 Ω. They are connected in series across some source. Ideal components, DC conditions. Concept / Approach: For series elements, use the formula R_total = R1 + R2 + ... because the same current flows and voltage drops add. There is no current splitting in series paths, unlike in parallel networks. Step-by-Step Solution: Identify series connection: one path for current through 47 Ω then 68 Ω.Apply series rule: R_total = 47 + 68.Compute: 47 + 68 = 115 Ω.Report the equivalent: 115 Ω. Verification / Alternative check: If these were in parallel, the equivalent would be less than the smallest resistor. Since our answer is greater than either alone, that is consistent with series behavior. Why Other Options Are Wrong: 3,196 Ω is the product (47*68) and not applicable in series. “Less than 47 Ω/68 Ω” would describe a parallel equivalent, not series. 1,115 Ω is a miscalculation by adding an extra digit. Common Pitfalls: Mixing the series and parallel formulas, or mistakenly using the product-over-sum rule (valid only for two resistors in parallel). Final Answer: 115 Ω

Question 13

Voltage division in a series network: Resistors of 3.9 kΩ, 7.5 kΩ, and 5.6 kΩ are connected in series across a 34 V DC source. What is the voltage drop across the 7.5 kΩ resistor?

Accepted Answer

Introduction / Context: Voltage division is a cornerstone concept in series circuits. When several resistors share a series current, each drop is proportional to its resistance. This problem exercises the divider formula and careful arithmetic in kilo-ohm units under a DC supply. Given Data / Assumptions: Series resistors: 3.9 kΩ, 7.5 kΩ, 5.6 kΩ. Supply voltage: 34 V DC. Ideal resistors and wires (no extra drops). Concept / Approach: For a series string, the same current I flows through all resistors. The voltage drop across any one resistor R_i is v_i = V_total * (R_i / R_total). First compute the total resistance, then apply the proportion for the 7.5 kΩ element. Step-by-Step Solution: Compute R_total = 3.9k + 7.5k + 5.6k = 17.0 kΩ.Apply divider: v_7.5k = 34 * (7.5 / 17.0).Calculate: 34 * 7.5 = 255; 255 / 17 = 15 V.Therefore, the 7.5 kΩ resistor drops 15 V. Verification / Alternative check: Find current I = V / R_total = 34 / 17k = 2 mA. Then v_7.5k = I * 7.5k = 0.002 * 7500 = 15 V. Both methods agree. Why Other Options Are Wrong: 34 V is the whole supply, not a single drop. 7.8 V and 11.2 V correspond to other ratios in the network, not 7.5 kΩ. 2.9 V is not consistent with the series proportion. Common Pitfalls: Forgetting units (kΩ vs Ω) and mixing formulas; double-check arithmetic when dividing by totals. Final Answer: 15 V

Question 14

Series aiding/opposing sources (algebraic sum): Three DC sources of −1.2 V, +15 V, and −6 V are connected in series. What is the magnitude of the total resultant voltage?

Accepted Answer

Introduction / Context: When DC sources are connected in series, their voltages add algebraically, taking polarity into account. Aiding sources add; opposing sources subtract. The final question often asks either for signed result or its magnitude. Here, the magnitude is requested. Given Data / Assumptions: Three ideal sources: −1.2 V, +15 V, −6 V. Series connection so voltages sum. The question asks for magnitude only (absolute value). Concept / Approach: Use algebraic addition: V_total = V1 + V2 + V3. After computing the signed total, take the absolute value to get the magnitude. Be careful with negative signs representing opposing polarities. Step-by-Step Solution: Sum signed voltages: V_total = −1.2 + 15 − 6.Compute: −1.2 + 15 = 13.8; 13.8 − 6 = 7.8 V.Magnitude: |7.8 V| = 7.8 V.Answer: 7.8 V. Verification / Alternative check: Think in terms of aiding/opposing: the +15 V aids; the −1.2 V and −6 V oppose, netting 15 − (1.2 + 6) = 15 − 7.2 = 7.8 V. Same result. Why Other Options Are Wrong: 10.2 V and 22.2 V come from incorrect addition/misplacing signs. 1.3 V is unrelated to the provided numbers. 9.6 V would be 15 − 5.4, which is not our opposing sum. Common Pitfalls: Dropping a sign, or forgetting that the question asks for magnitude, not signed polarity. Final Answer: 7.8 V

Question 15

Definition check: A voltage divider (two or more resistors establishing a fraction of a source) is always which type of network in its simplest, canonical form?

Accepted Answer

Introduction / Context: A basic voltage divider is the standard way to derive a lower voltage from a higher DC source using resistors. In its canonical form, it comprises two resistors in series, with the output taken from the junction. Understanding this prevents confusion with other topologies that incorporate loading or feedback elements. Given Data / Assumptions: Unloaded divider unless otherwise stated. Two or more resistors connected end-to-end. Output is the node between two series elements. Concept / Approach: The defining trait of a voltage divider is a simple series chain that splits the voltage proportionally to resistance values. Loading by a subsequent stage forms a series-parallel system, but the divider itself remains a series element pair. Hence, the primary classification is “series circuit.” Step-by-Step Solution: Identify the divider: R_top and R_bottom in series across the source.Output is taken at the midpoint node, yielding V_out = V_in * (R_bottom / (R_top + R_bottom)).Classify the topology: series.Therefore, select “Series circuit.” Verification / Alternative check: Consider loading: adding a load in parallel with the lower resistor creates a series-parallel network, but the divider core remains a series pair. The question asks for “always” in the canonical sense. Why Other Options Are Wrong: Series-parallel and bridge circuits describe more complex connections. A pure parallel circuit would not divide voltage proportionally. A delta network is a three-branch form not required for a basic divider. Common Pitfalls: Confusing the loaded divider (series-parallel system) with the intrinsic divider definition. Final Answer: Series circuit

Question 16

Topology identification rule: Which fundamental aspect determines whether interconnected resistors form a series, parallel, or series–parallel circuit in DC analysis?

Accepted Answer

Introduction / Context: Recognizing circuit topology quickly is essential for applying the correct solving technique. The essence of “series” and “parallel” lies in how current can flow: one path or multiple paths. This conceptual clarity avoids mistakes when redrawing or simplifying networks. Given Data / Assumptions: Standard DC circuits with ideal wires and components. Definitions: series = one current path; parallel = multiple current paths between the same two nodes. No reactive effects considered. Concept / Approach: The classification hinges on current paths. If there is only one path for current through components sequentially, they are in series. If components share both terminal nodes (same two nodes), they are in parallel, providing multiple current paths. Mixed combinations create series–parallel networks. Step-by-Step Solution: Inspect the nodes: do components share both nodes (parallel) or connect end-to-end (series)?Count paths: one path implies series; two or more paths imply parallel.Classify accordingly and simplify stepwise if mixed.Choose the determinant: current path count. Verification / Alternative check: Redraw the circuit using node labeling. Components that bridge the same node pair are parallel; elements chained between successive nodes are in series. This method consistently matches the “current paths” criterion. Why Other Options Are Wrong: Source type, “voltage flow,” wattage rating, or physical spacing do not define topology. Only connection nodes and consequent current paths do. Common Pitfalls: Letting physical drawing layout (left/right, up/down) mislead you; rely on node connections, not artwork. Final Answer: Current flow (number of available paths)

Question 17

Underdetermined value in a labeled network: A resistor R1 is referenced in a circuit, but no component values or node voltages are supplied. What is R1 equal to under these conditions?

Accepted Answer

Introduction / Context: Component values cannot be inferred from labels alone. Without measurement data, design constraints, or relationships from the rest of the circuit, a label like “R1” is just an identifier. This question reinforces the need for complete specifications before calculating a numerical value. Given Data / Assumptions: No diagram or numerical constraints are included. R1 is referenced but not defined by any equations in the stem. Standard passive linear component behavior is assumed. Concept / Approach: To solve for an unknown resistance, you typically need Ohm’s law with known voltage and current, a divider/bridge relationship, or network equations (KCL/KVL). Without such information, the value is indeterminate; many different values could satisfy an unspecified circuit. Step-by-Step Solution: Look for given voltages/currents: none provided.Look for ratios (e.g., bridge balance): none provided.Conclude the system is underdetermined; R1 cannot be solved numerically.Therefore, select the non-numeric option indicating insufficiency of data. Verification / Alternative check: If even a single additional relation were provided (e.g., total current and total voltage in a simple series), you could compute R_total and potentially deduce R1 in combination with other known resistors. The absence of such data confirms the conclusion. Why Other Options Are Wrong: Specific numbers (100 kΩ, 200 kΩ, 300 kΩ, 50 kΩ) are arbitrary without supporting data. Common Pitfalls: Guessing “typical” values; always demand the equations or measurements that define an unknown. Final Answer: Cannot be determined from the information provided

Question 18

Power in a circuit (insufficient data recognition): Without any given voltages, currents, or resistances for the referenced circuit, what can be concluded about the total power dissipated by the circuit?

Accepted Answer

Introduction / Context: Total power in a circuit equals the sum of individual element powers. To compute it, you need at least one of the following for each element or for the whole network: voltage and current, resistance and current, or voltage and resistance. If the stem mentions a “given circuit” but supplies no values, the problem is underdetermined, and no unique numeric answer exists. Given Data / Assumptions: No source magnitude, no element values, and no measured currents are provided. We are asked for the total power dissipated. Passive linear elements are assumed (typical resistive dissipation). Concept / Approach: Select the appropriate power expression: P = V * I, or P = I^2 * R, or P = V^2 / R. Each requires specific numeric inputs. In network form, P_total can also be derived from source power if the source voltage and current are known. Lacking all such data, a numeric result is impossible. Step-by-Step Solution: Identify missing information: V, I, or R are not specified.Recognize dependency: every power formula needs at least two of these quantities.Conclude: there is insufficient information to compute total power.Select the only logically correct choice expressing indeterminacy. Verification / Alternative check: Even if a schematic were referenced, without numerical labels or measured values you cannot evaluate power. Any proposed number would be speculative. Why Other Options Are Wrong: 64 µW, 4 µW, 0 W, and 16 × 10^−12 W are arbitrary in the absence of circuit data. “0 W” might be true only for an open-circuited, unpowered network—an assumption not given. Common Pitfalls: Picking a small number thinking it is “safe.” Correct engineering requires evidence-based calculation, not guessing. Final Answer: Cannot be determined from the information provided

Question 19

Finding each resistor from total series resistance: Five equal resistors are in series with a 20 V DC source. The measured circuit current is 400 µA. What is the resistance of each individual resistor?

Accepted Answer

Introduction / Context: When equal resistors are in series, the total resistance is simply N times one resistor’s value. Measuring the current with a known source quickly yields total resistance, and division by N gives each individual resistance. This is a standard application of Ohm’s law and series rules. Given Data / Assumptions: Number of equal resistors N = 5 in series. Source voltage V = 20 V DC. Measured current I = 400 µA = 0.0004 A. Concept / Approach: Ohm’s law: R_total = V / I. For equal series parts, R_each = R_total / N. Unit handling is critical: microampere to ampere conversion must be correct to avoid a 10^6 error factor. Step-by-Step Solution: Compute total resistance: R_total = 20 / 0.0004 = 50,000 Ω = 50 kΩ.Divide by number of equal parts: R_each = 50 kΩ / 5 = 10 kΩ.State the result: each resistor equals 10 kΩ.Quick check: The series string is 50 kΩ; at 20 V the current is 0.4 mA, consistent with the measurement. Verification / Alternative check: Reverse-calculate current using R_each = 10 kΩ: total is 5 * 10 kΩ = 50 kΩ; I = 20 / 50,000 = 0.0004 A = 400 µA. Match confirmed. Why Other Options Are Wrong: 50 kΩ is the total, not each. 125 Ω and 20 Ω are far too small; they would produce much larger currents. 2 kΩ would make the total 10 kΩ, implying 2 mA, not the given 400 µA. Common Pitfalls: Forgetting to convert microamperes to amperes; using parallel formulas instead of series addition for equal parts. Final Answer: 10 kΩ

Question 20

Series current from total resistance: Four resistors each of 90 Ω are connected in series across an 18 V DC source. What is the circuit current?

Accepted Answer

Introduction / Context: Calculating current in a simple series network is a direct application of Ohm’s law once the total resistance is known. Series connection increases resistance by summation; hence current decreases relative to a single element for the same source voltage. Given Data / Assumptions: Resistors: four identical units of 90 Ω each, in series. Source: 18 V DC. Ideal components with negligible lead resistance. Concept / Approach: Compute total resistance, then apply I = V / R_total. Because resistors are in series, R_total = 90 + 90 + 90 + 90. Compare the resulting current to answer choices, and convert to milliamperes for readability. Step-by-Step Solution: Sum resistances: R_total = 4 * 90 Ω = 360 Ω.Apply Ohm’s law: I = V / R = 18 / 360 A.Compute: 18 / 360 = 0.05 A.Convert units: 0.05 A = 50 mA. Verification / Alternative check: If a single 90 Ω were across 18 V, current would be 0.2 A. In series with four such resistors, current is reduced by a factor of 4, giving 0.05 A. This matches the calculation. Why Other Options Are Wrong: 0.2 A is the single-resistor case. 5 A and 20 A are unrealistic for 18 V across hundreds of ohms. 9 mA would imply 2 kΩ total, not 360 Ω. Common Pitfalls: Forgetting to sum series resistances or mishandling unit conversions between amperes and milliamperes. Final Answer: 50 mA

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