Question 1

Power fundamentals: Choose the best definition of electrical power in physics and electronics.

Accepted Answer

Introduction / Context: Power connects energy and time. In circuits it determines heating, sizing of supplies, and efficiency. Knowing the correct definition helps relate formulas like P = V * I and P = I^2 * R back to physical meaning. Given Data / Assumptions: Standard physics definitions. Energy measured in joules (J), time in seconds (s), power in watts (W). Electrical work equals charge moved times potential difference. Concept / Approach: Power is the time rate of doing work or transferring energy. Mathematically: P = dW/dt or P = dE/dt. In steady DC circuits, it reduces to algebraic forms like P = V * I because work per unit charge is V and charge per unit time is I. Step-by-Step Solution: Start from definition: P = dE/dt.For electrical systems: E transferred each second by current at voltage V equals V * I.Relate to other forms: P = I^2 * R or P = V^2 / R derived from Ohm’s law. Verification / Alternative check: Unit analysis: V * A = (J/C) * (C/s) = J/s = W, confirming consistency. Why Other Options Are Wrong: Work: That is energy, not its time rate. Conversion of energy: Too vague; power quantifies the rate of conversion. Joules: A unit of energy, not power. Common Pitfalls: Confusing energy with power; using watt-hours (energy) when watts (power) are intended. Final Answer: the rate at which work is done

Question 2

Series drop with a lamp: An 8 Ω resistor is in series with a lamp. The circuit current is 1 A and the applied supply is 20 V DC. What voltage is available across the lamp?

Accepted Answer

Introduction / Context: Series circuits share the same current, but sources divide their voltage among components according to their resistances. This practical problem shows how a series “dropping resistor” reduces the voltage applied to a lamp—useful for quick design checks and troubleshooting. Given Data / Assumptions: Resistor R = 8 Ω in series with a lamp. Current I = 1 A through the series path. Supply voltage V_s = 20 V DC. Lamp modeled as a load that takes the remaining voltage. Concept / Approach: Use Ohm’s law to find the resistor’s voltage drop, then subtract from the source to find the lamp voltage. In any series loop, V_s = V_R + V_lamp. With known I and R, V_R = I * R. Step-by-Step Solution: Compute resistor drop: V_R = I * R = 1 A * 8 Ω = 8 V.Apply KVL: V_lamp = V_s − V_R = 20 V − 8 V.Result: V_lamp = 12 V. Verification / Alternative check: Power check: P_R = I^2 * R = 1^2 * 8 = 8 W; P_lamp = V_lamp * I = 12 * 1 = 12 W; total 8 W + 12 W = 20 W = V_s * I, confirming consistency. Why Other Options Are Wrong: 4 V or 8 V: 8 V is the resistor drop, not the lamp. 4 V is unsupported by the data. 20 V: Would be true only with no series resistor drop (not our case). Common Pitfalls: Forgetting that the same current flows through both elements; mixing up which voltage belongs to which component. Final Answer: 12 V

Question 3

Ideal source behavior: When a battery is connected to a purely series circuit, the current it delivers depends primarily on which single factor?

Accepted Answer

Introduction / Context: Ohm’s law gives a direct relationship among voltage, current, and resistance in simple DC circuits. Recognizing what sets the current helps with power budgeting, fuse sizing, and verifying whether a design will overload a source. Given Data / Assumptions: Ideal battery providing a fixed DC voltage. All elements are in series (single path for current). No internal resistance or nonlinearities considered for the basic principle. Concept / Approach: For an ideal source of voltage V connected to a series resistance R_total, the current is I = V / R_total. Polarity sets direction, not magnitude. Terms such as “primary/secondary” relate to transformers, not DC sources. “Average resistance” is not an electrical parameter; only the algebraic sum matters in series. Step-by-Step Solution: Compute total resistance: R_total = R1 + R2 + ... + Rn.Apply Ohm’s law: I = V / R_total.Conclude that current depends solely on total resistance for a fixed V. Verification / Alternative check: Measure current while adding another series resistor; current decreases according to I_new = V / (R_total + ΔR), confirming dependence on total resistance. Why Other Options Are Wrong: Primary/secondary difference: Not applicable to simple battery circuits. Polarity connections: Affect direction of current, not magnitude for given R_total. Average resistance: Not an electrical standard; magnitude is set by the true total. Common Pitfalls: Ignoring internal battery resistance in real life; while small, it can limit current. The ideal model still teaches the governing dependency. Final Answer: total resistance

Question 4

Ohm’s law in a series circuit: If the total resistance in a series circuit doubles while the applied source voltage remains constant, how will the circuit current change?

Accepted Answer

Introduction / Context: This question tests core understanding of Ohm’s law and how current behaves in a simple series circuit when total resistance changes. In many troubleshooting and design scenarios, recognizing proportional relationships between current, resistance, and voltage is essential for quick, reliable reasoning. Given Data / Assumptions: Total resistance in a series circuit doubles. The source (supply) voltage is constant. Ideal components with no temperature-induced resistance change during comparison. Single-loop series circuit so one current flows through all elements. Concept / Approach: Ohm’s law states I = V / R. If V is constant and R increases, I must decrease proportionally. Specifically, if R becomes 2R, then I becomes V / (2R) = (1/2) * (V / R). Thus current is halved. This inverse proportionality between current and resistance at fixed voltage is a cornerstone of basic circuit analysis. Step-by-Step Solution: Initial current: I1 = V / R_total.After doubling resistance: R_new = 2 * R_total.New current: I2 = V / R_new = V / (2 * R_total) = (1/2) * (V / R_total).Therefore, I2 = I1 / 2, i.e., the current is halved. Verification / Alternative check: Use numerical example. Let V = 10 V and R_total = 100 Ω → I1 = 10 / 100 = 0.1 A. If resistance doubles to 200 Ω → I2 = 10 / 200 = 0.05 A, exactly half. The numerical check confirms the proportional reasoning. Why Other Options Are Wrong: Be the same: would require resistance unchanged or voltage doubled to compensate; not our case.Be doubled: happens if resistance halves at constant voltage, not when it doubles.Reduce source voltage: the source voltage is given as constant; the current change does not “reduce” the source. Common Pitfalls: Confusing proportional and inverse proportional relationships. Remember: at constant V, current varies inversely with resistance (I ∝ 1/R). At constant R, current varies directly with voltage (I ∝ V). Final Answer: be halved

Question 5

Series circuit open-circuit measurement: A 20 V DC source supplies an 8 Ω resistor in series with a lamp. If the lamp is removed (leaving an open circuit at the lamp socket), what voltage will a voltmeter read across the lamp socket terminals?

Accepted Answer

Introduction / Context: Measuring voltages in series circuits requires understanding where drops occur. When a component is removed, the circuit may become open, changing how voltage is distributed. This question reinforces the concept that with zero current, resistive drops vanish and the source appears across the open terminals. Given Data / Assumptions: DC source voltage: 20 V. Series elements: 8 Ω resistor and a lamp originally in series. The lamp is removed, creating an open circuit at the lamp socket. Voltmeter has ideally infinite input resistance and is connected across the lamp socket. Concept / Approach: Ohm’s law dictates the voltage drop across a resistor is V_R = I * R. In an open circuit, the loop current I is zero. Therefore, the drop across the 8 Ω resistor becomes 0 V, and the entire source voltage appears across the open terminals (the lamp socket). Step-by-Step Solution: Lamp removed → circuit open → I = 0 A.Resistor drop: V_R = I * R = 0 * 8 = 0 V.KVL around loop: source 20 V must appear across the only remaining break → lamp socket sees 20 V.Voltmeter reading across lamp socket = 20 V. Verification / Alternative check: Model the lamp socket as two open terminals. With no closed path, current is zero. Any series resistive element has zero drop, leaving the full source across the open—consistent with standard open-circuit behavior. Why Other Options Are Wrong: 0 V: would require a short across the socket, not an open.8 V or 12 V: arbitrary partial drops contradict zero-current condition (no drop across finite resistance with I = 0). Common Pitfalls: Forgetting that voltmeters do not provide a current path. A high-impedance meter does not “close” the circuit; thus no current flows and no series drop occurs. Final Answer: 20 V

Question 6

Flashlight battery polarity: A flashlight uses three 1.5 V cells in series. If one cell is inserted backward (reverse polarity), what happens to the flashlight output?

Accepted Answer

Introduction / Context: Series battery stacks add voltages algebraically, taking polarity into account. Many small flashlights are designed for three 1.5 V cells (nominal 4.5 V) to drive a bulb or LED plus driver. Reversing one cell changes the net series sum and can prevent operation entirely. Given Data / Assumptions: Three cells rated 1.5 V each, intended to be in series and oriented correctly. One cell is reversed, contributing negative polarity relative to the others. Flashlight electronics (bulb/LED driver) expect approximately 4.5 V. Idealized cells assumed for the polarity-sum explanation. Concept / Approach: Series voltages add with sign. With correct orientation, V_total = 1.5 + 1.5 + 1.5 = 4.5 V. With one reversed, V_total = 1.5 + 1.5 − 1.5 = 1.5 V. Most lamps or driver circuits designed for ~4.5 V will not operate properly at ~1.5 V; an incandescent filament may not glow perceptibly, and many LED drivers have undervoltage lockout or cannot maintain current regulation. Step-by-Step Solution: Compute nominal: three cells correct → 4.5 V.Reverse one cell: algebraic sum → 1.5 V total.Compare to required operating voltage: 1.5 V is far below design point.Conclusion: no visible output; flashlight is effectively off. Verification / Alternative check: Practical test: a 4.5 V incandescent bulb at 1.5 V draws much less current and produces negligible light; an LED driver typically shuts down under low input voltage. Both cases appear “off” to a user. Why Other Options Are Wrong: Brighter than normal / the same: contradict the reduced supply voltage.Dimmer than normal: while technically some small glow could occur with simple bulbs, in typical products the result is functionally off; the robust answer is off. Common Pitfalls: Ignoring polarity in series stacks. One reversed cell subtracts from the total, and in some electronics can also stress protection circuitry due to reverse biasing. Final Answer: off

Question 7

Voltage division in series: A 45 V DC source is applied to a series network with total resistance 3300 Ω. If one of the resistors, R3, is 1200 Ω, what is the voltage drop across R3?

Accepted Answer

Introduction / Context: Voltage division in series circuits is a fundamental technique for predicting how a source voltage distributes across individual elements. This problem directly applies the voltage divider rule to determine the drop across one known resistor when total resistance and source voltage are given. Given Data / Assumptions: Total series resistance, R_total = 3300 Ω. Source voltage, V_s = 45 V DC. Target resistor, R3 = 1200 Ω (in series). Ideal resistors; no additional elements. Concept / Approach: For series circuits, current I is the same through all resistors: I = V_s / R_total. The voltage across any resistor R_x is V_x = I * R_x. Combining, V_x = V_s * (R_x / R_total). This is the standard voltage divider formula. Step-by-Step Solution: Compute the ratio: R3 / R_total = 1200 / 3300 = 0.363636…Apply divider: V_R3 = 45 * 0.363636…Multiply: 45 * 0.363636… = 16.3636… V.Round sensibly to two decimals: V_R3 ≈ 16.36 V. Verification / Alternative check: Current method: I = 45 / 3300 = 0.013636… A. Then V_R3 = I * 1200 = 0.013636… * 1200 = 16.3636… V. Both methods agree. Why Other Options Are Wrong: 32.72 V: would correspond to a much larger fraction (about 0.727) not matching 1200/3300.10.90 V: too small; implies R3 is roughly 800 Ω of the total, which it is not.15.00 V: a rounded guess, not the precise divider result. Common Pitfalls: Forgetting that only series elements share the same current; mixing up series and parallel results leads to incorrect voltage predictions. Ensure total resistance includes all series elements when using the divider rule. Final Answer: 16.36 V

Question 8

Series circuits — definition check without a diagram: A correct identification rule is: “Two components are in series if the same current flows through both, sharing a single uninterrupted path with only one common node between them.” Evaluate this …

Accepted Answer

Introduction / Context: Many textbook questions rely on a figure to decide whether components are in series or parallel. Applying the Recovery-First Policy, we remove the dependency on a missing diagram and test the core definition instead. Knowing the precise topological rule lets you identify series connections in any circuit drawing. Given Data / Assumptions: Ideal wires and lumped components. No branching between the two devices under test. Current continuity is the criterion, not physical proximity. Concept / Approach: Components are in series if the same current must pass through them sequentially because there is only one path for current flow. This means the junction between them connects to nothing else (no additional branches). Parallel elements instead share the same two nodes and therefore the same voltage, not necessarily the same current. The series criterion is independent of component values and source type (DC or AC). Step-by-Step Solution: Check for branching between the two components—there should be none.Confirm that the current entering the first must exit through the second.Verify that they do not share the same two nodes (that would be parallel).Conclude the provided rule correctly defines a series connection. Verification / Alternative check: Apply Kirchhoff’s laws: the same branch current flows through series components; voltages add according to Ohm’s law, consistent with the definition. Why Other Options Are Wrong: Incorrect: Conflicts with the standard topological definition. Applies only to DC: AC behavior still follows the same series topology. Valid only if resistances equal: Series status does not depend on values. Depends on supply polarity: Polarity does not change network topology. Common Pitfalls: Judging by physical layout instead of connectivity; mistaking a node with hidden branches for a pure series junction; assuming equal current implies equal voltage (it does not). Final Answer: Correct

Question 9

Measurement reference — definition of ground: In practical electronic circuits, “ground” (or reference) is the node against which all other voltages are measured. Decide whether this statement accurately describes ground.

Accepted Answer

Introduction / Context: Voltage is inherently a difference in electric potential between two points. To report a single number for “the voltage” at a node, engineers choose a reference node—commonly called ground (GND). Understanding that ground is a reference, not always literally the Earth, is essential for measurement, safety, and PCB design. Given Data / Assumptions: Ground refers to the designated reference node in the circuit schematic. Measuring instruments report V_node = V(node) - V(ground). Earth bonding may or may not be used depending on system class. Concept / Approach: Ground provides a common return path and a reference potential for signals and supplies. In many systems there are multiple grounds (analog, digital, chassis) that are tied at a single point. Whether connected to Earth or not, the chosen ground serves as the zero-volt reference for measurements and specifications throughout the circuit. Step-by-Step Solution: Identify the reference node on the schematic (usually the ground symbol).Understand that meter black lead connects to ground to measure node voltages.Recognize that absolute potential is not required; only differences matter.Conclude that defining ground as the measurement reference is accurate. Verification / Alternative check: Oscilloscope probes and DMMs default to ground-referenced measurements; datasheets specify output “with respect to GND,” confirming the definition in practice. Why Other Options Are Wrong: Incorrect: Ground is indeed the reference node by definition. True only for earth-grounded systems / AC power circuits / metal chassis: Ground as a reference applies irrespective of earthing method or enclosure material. Common Pitfalls: Assuming ground always equals Earth potential; mixing up safety ground, chassis ground, and signal return; creating ground loops by multiple bonding points. Final Answer: Correct

Question 10

Series resistors — quick sum check: If 7.3 kΩ, 1.8 kΩ, and 4.9 kΩ resistors are connected in series, is the total resistance equal to 14 kΩ? Verify the claim using series-resistance rules.

Accepted Answer

Introduction / Context: Being able to sanity-check series resistor totals quickly is invaluable for design and troubleshooting. Series combinations add directly because the same current flows through each resistor and each contributes its share of voltage drop. This problem asks you to confirm a specific numerical claim for three common E24 values. Given Data / Assumptions: Resistor values: 7.3 kΩ, 1.8 kΩ, 4.9 kΩ. Series connection: same current through all three resistors. Ideal components; tolerance effects ignored for the conceptual check. Concept / Approach: For series resistances, the equivalent resistance is the arithmetic sum: R_total = R1 + R2 + R3. Units should be consistent; since all are in kΩ, we can add directly in kΩ or convert to Ω first and convert back later. The calculation is straightforward but must be performed carefully to avoid simple arithmetic slips. Step-by-Step Solution: Write the series formula: R_total = R1 + R2 + R3.Substitute values in kΩ: R_total = 7.3 + 1.8 + 4.9 kΩ.Compute partial sum: 7.3 + 1.8 = 9.1 kΩ.Add the third value: 9.1 + 4.9 = 14.0 kΩ. Verification / Alternative check: Convert to ohms for confirmation: 7300 + 1800 + 4900 = 14,000 Ω, which equals 14 kΩ. Either way, the arithmetic agrees with the claim. Why Other Options Are Wrong: Incorrect: The sum is exactly 14 kΩ under ideal arithmetic. Valid only at 25 °C / analog meter only: Temperature and instrument type do not change the ideal series rule; tolerances might slightly vary measured values but not the nominal total. Cannot be determined without current: Equivalent resistance is independent of current magnitude. Common Pitfalls: Mistyping values (e.g., 7.3 as 7.03), mixing units (Ω vs kΩ), or mistakenly averaging instead of summing for series networks (averaging applies to some parallel intuition, not series). Final Answer: Correct

Question 11

Series circuits — assess the statement: “The total resistance of a series circuit equals the average (arithmetic mean) of all individual resistance values.” Is this claim valid for ideal series connections?

Accepted Answer

Introduction / Context: Understanding how resistances combine is foundational for circuit analysis. In a series circuit, there is a single path for current, and each resistor shares the same current. This question challenges a common misconception: that the total series resistance equals the average of the resistor values, rather than the sum. Clarifying this point prevents systematic errors in design and troubleshooting. Given Data / Assumptions: Ideal lumped components (no parasitic effects). Resistors R1, R2, …, Rn connected in series (one current path). Goal: compare “sum” versus “average” for total resistance. Concept / Approach: Kirchhoff’s Voltage Law (KVL) states that the sum of voltage drops around a loop equals the source voltage. With a common current I in a series chain, the total voltage drop is I*R1 + I*R2 + … + I*Rn. Equating with V = I*R_total directly yields R_total = R1 + R2 + … + Rn. An average would be (R1 + … + Rn) / n, which underestimates the real series total whenever n > 1 (unless all but one are zero, which is not a useful physical case). Step-by-Step Solution: Let series resistors be R1..Rn with current I.By Ohm’s law: V_drop,k = I * Rk for each k.By KVL: V_source = I*(R1 + R2 + … + Rn).Hence R_total = (V_source / I) = R1 + R2 + … + Rn (sum, not average). Verification / Alternative check: Take a numeric example: R1 = 10 Ω, R2 = 20 Ω. Sum = 30 Ω. Average = 15 Ω. Measuring current with a fixed source shows the 30 Ω result matches I = V / 30, not V / 15, confirming that average is incorrect. Why Other Options Are Wrong: Correct: Conflicts with R_total = sum of series resistances. Applies only when all resistors are equal: Even then, the total is n*R, not the average R. Applies only below 1 kΩ: Resistance combination rules are not value-range dependent. Common Pitfalls: Confusing series (resistances add) with parallel (conductances add); thinking “average” because of evenly shared current. Series totals always add arithmetically. Final Answer: Incorrect — total series resistance is the sum, not the average.

Question 12

Series-resistance rule — for ideal resistors connected in series, how is the total resistance calculated?

Accepted Answer

Introduction / Context: Series connections appear everywhere in electronics, from simple dividers to bias networks. Knowing how to combine resistances in series enables quick current estimates and power budgeting. This item asks for the correct method to find the equivalent resistance of several resistors in series. Given Data / Assumptions: Ideal resistors R1, R2, …, Rn in series. Single current flows through all series components. Standard Ohm’s law and Kirchhoff’s laws apply. Concept / Approach: With one common current I, each resistor drops V_k = I*R_k. The source must supply the sum of these drops, so V_total = I*(R1 + R2 + … + Rn). Therefore, the equivalent resistance is R_total = V_total / I = R1 + R2 + … + Rn. This is a direct consequence of KVL and Ohm’s law in a single-loop circuit. Step-by-Step Solution: Model the series chain with current I.Write drops: V1 = I*R1, V2 = I*R2, etc.Apply KVL: V_total = V1 + V2 + … + Vn.Factor I: V_total = I*(R1 + R2 + … + Rn) → R_total = sum of R_k. Verification / Alternative check: Measure with a bench supply: doubling one resistor doubles its portion of the drop, and the net acts like one larger resistor equal to the arithmetic sum, matching calculations and meter readings. Why Other Options Are Wrong: Product over sum: That is the two-resistor parallel shortcut, not series. Average: Underestimates the true series equivalent. Reciprocal of sum of reciprocals: That is the general parallel formula. Common Pitfalls: Memorizing the parallel formula and misapplying it to series, or assuming averages apply because the current is common. In series, resistances add directly. Final Answer: Correct — add individual resistances to obtain the total series resistance.

Question 13

KVL in series networks — the sum of the individual voltage drops across series resistors must equal the total applied source voltage. Is this fundamental statement accurate?

Accepted Answer

Introduction / Context: This statement encapsulates Kirchhoff’s Voltage Law (KVL) applied to a series resistive circuit. KVL is one of the two cornerstone conservation laws for lumped circuits, ensuring energy consistency around closed loops. The item tests whether you can connect KVL to observable behavior in series resistor chains. Given Data / Assumptions: Ideal source and ideal resistors in series. Closed loop such that KVL applies. Steady-state conditions (but the principle also holds instant-by-instant in AC). Concept / Approach: KVL states the algebraic sum of voltages around any closed loop is zero. Traversing a loop with a source and series drops yields V_source − V1 − V2 − … − Vn = 0, or equivalently V_source = V1 + V2 + … + Vn. Each drop V_k follows Ohm’s law V_k = I*R_k with common current I in series. Thus the sum of the drops equals the applied source voltage regardless of individual resistor values. Step-by-Step Solution: Assume series chain R1..Rn with loop current I.Compute each drop: V_k = I*R_k.Apply KVL: V_source = Σ V_k.Therefore, the statement holds for all series networks. Verification / Alternative check: Bench test with a DMM across each resistor and across the source shows that measured drops sum to the source voltage within meter tolerance, validating KVL experimentally for DC and snapshot values in AC. Why Other Options Are Wrong: Incorrect: Violates energy conservation around the loop. Only true for DC: KVL holds for AC as an instantaneous law in lumped circuits. Only when resistors are equal: Equality is irrelevant; drops scale with R. Common Pitfalls: Sign convention mistakes when summing rises and drops; forgetting that meter polarity affects the sign, not the magnitude relation that the algebraic sum around the loop is zero. Final Answer: Correct — by KVL, individual drops sum to the applied source voltage.

Question 14

Series-aiding sources — four 9 V batteries connected in series and aiding each other produce a total of 36 V. Is this description correct for ideal sources?

Accepted Answer

Introduction / Context: Stacking voltage sources is a common way to reach higher supply rails in portable systems. When sources are connected “series-aiding,” their voltages add algebraically with the same polarity orientation. This question checks your understanding by using four identical 9 V batteries as an example. Given Data / Assumptions: Four ideal 9 V sources. Series-aiding orientation (polarity aligned head-to-tail). Neglect internal resistance for the basic calculation. Concept / Approach: For ideal voltage sources connected in series, the net voltage equals the signed sum of individual voltages. “Aiding” means all polarities line up to produce a larger resultant. Therefore, 9 V + 9 V + 9 V + 9 V = 36 V. In practice, small internal resistances cause voltage sag under load, but the open-circuit sum remains 36 V and loaded values are close if currents are modest. Step-by-Step Solution: Identify orientation: series-aiding (same polarity).Add ideal source voltages: 9 + 9 + 9 + 9 = 36 V.Conclude total: V_total = 36 V (ideal).Note real-world caveat: internal resistance introduces droop under load. Verification / Alternative check: Measure open-circuit with a DMM: approximately 36 V. Under load, compute V_load = 36 V − I * (Σ r_internal) to estimate any droop; for light loads the reading remains near 36 V. Why Other Options Are Wrong: Incorrect: Opposes the fundamental series addition rule. Correct only if internal resistance is zero: Internal resistance affects load voltage, not the ideal series summation principle. Voltages never add: They do add algebraically when sources are in series. Common Pitfalls: Confusing series-aiding with series-opposing; reversing one cell flips the algebraic sum and reduces the total rather than increasing it. Final Answer: Correct — four 9 V sources in series-aiding give 36 V (ideal).

Question 15

Faults in series chains — a short circuit across one element in a series network will cause the total circuit current to decrease. Is this claim accurate for an ideal voltage source?

Accepted Answer

Introduction / Context: Troubleshooting often hinges on predicting how faults affect current. In a series circuit powered by an ideal voltage source, replacing a finite resistor with a short (near zero ohms) changes the equivalent resistance. This question tests whether you know in which direction the total current moves under such a fault. Given Data / Assumptions: Ideal voltage source of fixed value. Series network of resistors. Fault: one resistor becomes a short (its resistance → ~0 Ω). Concept / Approach: Total series resistance is the sum of individual resistances. If one resistance collapses to nearly zero, the sum decreases. Ohm’s law for the whole circuit is I_total = V_source / R_total. With a smaller denominator (R_total), the current increases, not decreases. Only non-idealities like source current limits, fuses, or saturation would reduce current, but those are separate protective effects, not ideal circuit behavior. Step-by-Step Solution: Original: R_total(original) = R1 + R2 + … + Rk + … + Rn.Fault: Rk → 0 → R_total(fault) = R_total(original) − Rk.Since Rk > 0 originally, R_total decreases.Therefore, I_total(fault) = V_source / R_total(fault) increases. Verification / Alternative check: Numerical example: V = 12 V, series 3 Ω + 3 Ω. Normal I = 12 / 6 = 2 A. Short one resistor to 0 Ω → R_total = 3 Ω → I = 12 / 3 = 4 A (increased). Why Other Options Are Wrong: Correct: Opposite to Ohm’s law with a fixed source voltage. Only correct with current sources: A current source fixes current; voltage then changes across the network, but that is not the scenario here. Only correct if other resistors are large: Magnitude changes, not the direction of the change, depend on values. Common Pitfalls: Associating the word “short” with “less current.” A short reduces resistance; with a voltage source, current rises and may blow protective devices. Final Answer: Incorrect — a short reduces total resistance and increases total current (ideal source).

Question 16

Open-circuit in series — consider an ideal series circuit that develops an open in one component. Which statement best reflects the condition across the open break and the source?

Accepted Answer

Introduction / Context: Open-circuit faults are common in connectors, fuses, and fractured solder joints. Recognizing the electrical signature of an open in a series path is critical for diagnostics. This item asks you to identify what happens to current and voltages when a series path is interrupted by an open break. Given Data / Assumptions: Ideal voltage source. Series chain of components with one section “open” (infinite resistance). No alternative paths bypass the open (single path only). Concept / Approach: With a series open, the circuit is incomplete; therefore, current is zero everywhere in the loop: I_total = 0 A. Without current, the normal I*R drops on intact resistors disappear (each finite resistor drop is 0 V). The source still maintains its terminal voltage, and by KVL the entire source voltage appears across the open gap (the element or connection where the break occurred). This is why an open fuse often has the full supply voltage across it during operation. Step-by-Step Solution: Series path interrupted → I_total = 0 A.Finite resistors: V_k = I*R_k = 0 V.KVL: V_source = V_open + Σ V_resistors = V_open + 0.Therefore, V_open = V_source appears across the open. Verification / Alternative check: Measure with a voltmeter: across any intact resistor you will read ~0 V; across the open terminals you will read near the source voltage. This is the classic signature of an open series component. Why Other Options Are Wrong: Current increases: Impossible without a closed loop. Open sees smallest drop: Incorrect; it sees the full source voltage. Source collapses to 0 V: Ideal sources maintain their set voltage; collapse suggests current limiting or failure, not the ideal model. Common Pitfalls: Confusing a short with an open; assuming all components still drop voltage despite zero current. In series, no current means no I*R drops on finite resistors. Final Answer: Correct — current is zero and the full source voltage appears across the open break.

Question 17

Series-opposing sources — when two voltage sources are series-opposing, should their voltages be added arithmetically to get the total? Evaluate this statement.

Accepted Answer

Introduction / Context: Multiple sources in series combine according to their polarities. “Series-aiding” means voltages add; “series-opposing” means they subtract. This question checks whether you recognize that opposing sources reduce the net voltage instead of increasing it by addition. Given Data / Assumptions: Two ideal voltage sources in series. Opposing polarity orientation (one source fights the other). Linear, lumped circuit model. Concept / Approach: The correct way to combine series sources is to take their algebraic sum with sign, not a blind arithmetic sum. If V1 and V2 are oriented so that one is a rise while the other is a drop in the same loop traversal, the net is V_net = V1 − V2 (magnitude per polarity). Only when the sources aid each other do they add: V_net = V1 + V2. Therefore, the statement “series-opposing sources are added” is wrong. Step-by-Step Solution: Assign loop direction and source polarities.Write KVL including algebraic signs for rises/drops.For opposing orientation: V_net = |V1 − V2| (with sign by chosen direction).Conclude: They do not add; they subtract. Verification / Alternative check: Example: 12 V source opposing a 9 V source → net 3 V. A DMM across the pair confirms ~3 V with polarity set by the larger source orientation, matching algebraic summation, not addition. Why Other Options Are Wrong: Correct: Contradicts basic source combination rules. Only if identical / Only for AC: Irrelevant; algebraic sum applies regardless of equality or AC/DC as long as phasor polarities are considered for AC. Common Pitfalls: Ignoring polarity arrows and simply summing magnitudes. Always mark polarities and signs before applying KVL. Final Answer: Incorrect — series-opposing sources subtract; use the algebraic sum with signs.

Question 18

Dependence of total series resistance — does the equivalent resistance of a series circuit “always depend only on the highest-value resistor” present in that circuit?

Accepted Answer

Introduction / Context: Designers sometimes use rules of thumb when one element dominates. However, rigorous analysis of series circuits requires summation of all resistances. This item probes whether you know that every series element contributes to the total, not just the largest one, even if approximation may be acceptable in special contexts. Given Data / Assumptions: Series chain of N resistors. Ideal lumped model. No element is bypassed or shorted. Concept / Approach: The series formula is exact: R_total = R1 + R2 + … + Rn. While a large resistor may dominate numerically, the exact total still includes the smaller ones. Approximations (e.g., neglecting 1 Ω alongside 10 kΩ) are engineering conveniences, not the governing law. Therefore, the claim that total resistance “always depends only” on the largest value is false. Step-by-Step Solution: Write exact relation: R_total = Σ R_k for k = 1..n.Evaluate contribution: Each R_k adds positively to R_total.Consider dominance: If R_max ≫ others, R_total ≈ R_max + small terms, but not solely R_max.Conclude the statement is incorrect literally. Verification / Alternative check: Example: 9,900 Ω + 100 Ω = 10,000 Ω. The 100 Ω contributes 1% and cannot be ignored for precise work. Bench measurements reflect the full sum within tolerances. Why Other Options Are Wrong: Correct: Misstates the series rule. Correct only if >10×: Even then, the smaller parts still add; the claim “only” remains wrong. Temperature stability: Unrelated to the additive rule; it affects value drift, not whether each resistor counts. Common Pitfalls: Mistaking a convenient approximation for an exact law; overlooking that billing, power, and tolerance stacks rely on the exact sum, not a dominant-single-value simplification. Final Answer: Incorrect — the total series resistance is the sum of all resistors, not merely the largest one.

Question 19

Applied example — three resistors of 5 Ω, 10 Ω, and 20 Ω are connected in series. Is the total (equivalent) resistance equal to 35 Ω under ideal conditions?

Accepted Answer

Introduction / Context: Concrete numerical examples reinforce the series-resistance rule. Here, you are asked to validate a straightforward calculation for three resistors connected in series. Such checks are essential when sizing supplies and estimating current and power distribution in a chain. Given Data / Assumptions: Resistors: 5 Ω, 10 Ω, 20 Ω. Series connection (single current path). Ideal components; ignore tolerance and temperature coefficients. Concept / Approach: The series combination rule says: R_total = R1 + R2 + R3. Since the current is common to all three, the voltage drops add to the source voltage, and the resistances add arithmetically to give the equivalent seen by the source. No reciprocals are used for series addition; those apply to parallel networks. Step-by-Step Solution: List values: R1 = 5 Ω, R2 = 10 Ω, R3 = 20 Ω.Compute sum: R_total = 5 + 10 + 20 = 35 Ω.Therefore, the statement that the total is 35 Ω is correct.If a source V is applied, current would be I = V / 35 (for completeness). Verification / Alternative check: Measure with an ohmmeter across the three in series. Within tolerance, the reading will approach 35 Ω. Any deviation comes from component tolerances and lead/contact resistances, not from the combination rule. Why Other Options Are Wrong: Incorrect: Conflicts with the series sum rule. Correct only at 25 °C: Temperature affects value slightly but not the arithmetic rule. Series uses reciprocals: Reciprocals are for parallel, not series. Common Pitfalls: Accidentally applying the parallel formula to a series problem or overlooking unit symbols (Ω vs kΩ) when summing values. Final Answer: Correct — 5 Ω + 10 Ω + 20 Ω = 35 Ω in series.

Question 20

Power accounting — in a purely resistive series circuit, does the total power dissipated equal the sum of the powers dissipated by each resistor, or does it exceed that sum?

Accepted Answer

Introduction / Context: Energy conservation in circuits dictates that the power delivered by sources is absorbed by elements as heat or stored energy. In steady-state resistive networks, all source power ends up as heat in the resistors. This question checks whether you understand that total power is an additive quantity across elements, not something that mysteriously exceeds the parts’ contributions. Given Data / Assumptions: Purely resistive series network driven by a source. Steady state (no reactive storage). Ideal interconnects without parasitic losses. Concept / Approach: Use P = V*I at the source, and P_k = I^2 * R_k (or V_k^2 / R_k) at each resistor. Since the same current I flows in series, the total resistor power is Σ (I^2 * R_k) = I^2 * Σ R_k. The equivalent resistance is R_total = Σ R_k, so the power seen from the source is P_total = I^2 * R_total = I^2 * Σ R_k, exactly the same as the sum of the individual powers. There is no extra “hidden” power term in an ideal resistive loop. Step-by-Step Solution: Write element powers: P_k = I^2 * R_k.Sum them: Σ P_k = I^2 * Σ R_k.Compute source power: P_source = I^2 * R_total with R_total = Σ R_k.Therefore, P_source = Σ P_k (equality, not excess). Verification / Alternative check: Measure current I and each drop V_k, compute P_k = V_k * I. Summing P_k matches the source measurement V_source * I within meter tolerance, confirming conservation of energy. Why Other Options Are Wrong: Exceeds the sum: Violates conservation of energy in an ideal resistive loop. Less than the sum: Also violates conservation; nothing else absorbs the “missing” power. Only for equal resistors: Equality of values is irrelevant to the additivity of power. Common Pitfalls: Confusing instantaneous versus average power, or mixing AC reactive elements (which store and return energy) with purely resistive networks. In a DC or resistive AC case, powers simply add. Final Answer: Correct — total power equals the sum of the individual resistor powers in a series resistive circuit.

Series Circuits Questions