Question 1

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 2

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

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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 3

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.

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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 4

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 5

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…

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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 6

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…

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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 7

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 8

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 9

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?

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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 10

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?

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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 11

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 12

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 13

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?

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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 14

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 15

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

Question 16

Series circuit calculation (Ohm’s law): three resistors of 220 Ω, 100 Ω, and 180 Ω are connected in series across a 12 V DC source. What is the resulting circuit current (express your choice in appropriate units)?

Accepted Answer

Introduction / Context:When components are connected in series, the same current flows through each element and the total resistance is the arithmetic sum of all series resistances. This problem tests correct application of Ohm’s law and proper unit handling to compute current in a simple DC series network. Given Data / Assumptions: Resistors in series: R1 = 220 ohm, R2 = 100 ohm, R3 = 180 ohm. Applied source: V = 12 V DC (ideal source). Temperature effects and tolerance are ignored; use nominal values. No additional parasitic elements or measurement loading assumed. Concept / Approach:For series connections, total resistance is R_total = R1 + R2 + R3. The circuit current is given by Ohm’s law: I = V / R_total. Keeping track of ohms and volts ensures current is in amperes; convert to milliamperes if useful for readability. Step-by-Step Solution: 1) Compute R_total = 220 + 100 + 180 ohm.2) R_total = 500 ohm.3) Apply Ohm’s law: I = V / R_total = 12 / 500 A.4) I = 0.024 A = 24 mA. Verification / Alternative check:As a quick sense-check, with about 500 ohm across 12 V, current should be a bit under 25 mA (since 12/480 = 25 mA). Our exact 24 mA result is consistent. Why Other Options Are Wrong:41.67 A and 500 A: unrealistically large; would require milliohm-level loads, not hundreds of ohms.120 mA: corresponds to about 100 ohm at 12 V; not our total resistance.2.4 mA: off by a factor of 10; would imply 5 kohm, not 500 ohm. Common Pitfalls:Adding series resistances correctly but then dividing incorrectly; forgetting to convert A to mA; mistakenly using parallel formulas (1/R_total = sum(1/R)). Final Answer:24 mA

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