Q: Series branch current identity: In a simple series circuit carrying 17 mA, the current through any series element (for example, resistor R1) equals the branch current, i.e., 17 mA. Judge whether this statement is correct.

Introduction / Context: Distinguishing series from parallel behavior is critical. In series connections, the same current flows sequentially through each component. This principle holds regardless of individual resistance values and is the reason voltage, not current, divides among series elements. Given Data / Assumptions: Series circuit with a measured or known branch current of 17 mA. Resistor R1 is one element in that series branch. Ideal wires and components for the conceptual statement. Concept / Approach: By definition, a series connection provides only one path for current. Kirchhoff’s Current Law at the node between series elements states that the current entering equals the current leaving. Therefore, the same 17 mA passes through every series element, including R1. Voltages differ by resistance (V = I * R), but current is identical in series. Step-by-Step Solution: Recognize the topology: one continuous path, no branches.Apply KCL at internal nodes: I_in = I_out.Conclude I_R1 = I_branch = 17 mA.Use V_R1 = 0.017 * R1 for drop if needed. Verification / Alternative check: Measure current with an ammeter inserted at different series points; readings are equal (accounting for meter insertion effects). SPICE simulations confirm equal series currents. Why Other Options Are Wrong: Incorrect / depends on tolerance / only DC: Tolerance affects voltage division; the current equality holds for DC and AC in purely series resistive paths. True only if R1 is smallest: Value does not change current equality; it changes voltage drop. Common Pitfalls: Assuming different currents through different series parts; confusing with parallel branches where current divides. Final Answer: Correct

Q: Wheatstone bridge application: A classic Wheatstone bridge circuit can be used to determine an unknown resistance by balancing the bridge and reading a known ratio. Decide whether this statement correctly describes a common use of the bridge.

Introduction / Context: The Wheatstone bridge is a foundational measurement circuit. By adjusting one known resistor (or using a calibrated ratio), the bridge can be balanced so that the detector indicates zero. At balance, a simple proportion relates the unknown resistor to the known elements, enabling precise resistance measurement without directly reading current or voltage across the unknown at nonzero levels. Given Data / Assumptions: Four-resistor bridge configuration with a sensitive null detector. At balance, no current flows through the detector branch. Resistors are stable during measurement; temperature effects are negligible. Concept / Approach: At balance, the ratio of the two arms is equal: R1/R2 = R_unknown/R3 (labeling may vary). Solving gives R_unknown = (R1/R2) * R3. Because the detector sees near-zero current at balance, measurement error due to detector loading is minimized, making the method precise for a wide range of resistances. Step-by-Step Solution: Arrange the bridge with three known resistors and one unknown.Adjust one known resistor until the detector shows null.Write the ratio equality for the two bridge arms.Solve for the unknown using the known ratio and resistor value. Verification / Alternative check: Compare the result with a direct ohmmeter reading; close agreement validates proper balancing and resistor tolerances. Why Other Options Are Wrong: Incorrect / AC only / only above 1 kΩ / requires digital meter: The bridge works with DC or AC (for resistive cases) and across broad resistance ranges; a simple galvanometer or null detector is sufficient. Common Pitfalls: Miswiring the bridge arms, ignoring lead resistance for low-ohm measurements, or attempting to read the unknown without reaching true null. Final Answer: Correct

Q: Series–parallel in practice: A “loaded voltage divider” (a resistor divider that feeds a finite load) is a common and important application of series–parallel circuits. Determine whether this usage description is accurate.

Introduction / Context: Many real circuits use resistor dividers to derive reference or bias voltages. The moment a load is attached to a divider output, the circuit becomes a series–parallel network: the lower divider resistor is effectively in parallel with the load. Recognizing this is essential for predicting output sag and source current changes. Given Data / Assumptions: Two-resistor divider from a DC supply. Finite load R_L from the divider output to ground. Resistive behavior; frequency effects ignored. Concept / Approach: With a load present, the bottom leg and R_L form a parallel combination. The output voltage becomes V_out = V_in * (R_bottom || R_L) / (R_top + (R_bottom || R_L)). This is a textbook series–parallel transformation and a standard example used to introduce loading effects in basic circuit courses. Step-by-Step Solution: Identify the divider: R_top and R_bottom in series across V_in.Attach R_L across the output node to ground.Compute R_lower_eff = R_bottom || R_L.Use the series formula to compute V_out and confirm series–parallel nature. Verification / Alternative check: Measure V_out with and without the load; the loaded case is lower, matching the series–parallel formula prediction. Why Other Options Are Wrong: Incorrect / only large load / low-voltage only / AC only: The classification as series–parallel does not depend on voltage level or frequency for resistive circuits and holds for any finite load. Common Pitfalls: Designing a divider without considering R_L, leading to excessive voltage sag or wasted current; forgetting to buffer the divider with an op-amp follower when a stiff source is required. Final Answer: Correct

Q: Common faults in circuits: In everyday electronics troubleshooting, the two most typical failures encountered are opens (broken paths) and shorts (unintended low-resistance paths). Decide whether this statement is correct.

Introduction / Context: Technicians often categorize failures broadly as opens or shorts. An open means a conductor or component no longer conducts as intended, breaking the current path. A short means two nodes are unintentionally connected by a very low resistance, often causing excessive current. Recognizing these classes speeds diagnosis across analog and digital systems. Given Data / Assumptions: Applies to discrete-component circuits, PCBs, cables, and connectors. Covers environmental and wear-out mechanisms such as heat, vibration, corrosion, and contamination. No assumption of a specific voltage level or signal type. Concept / Approach: Most electrical issues reduce to a path that is broken (open) or a path that should not exist (short). Examples include cracked solder joints (open), blown fuses (open), carbonized PCB after arcing (short), solder bridges (short), frayed cable shield contacting conductors (short), and corroded connectors (intermittent open). Systematic isolation using continuity tests and resistance measurements quickly confirms which category a fault belongs to. Step-by-Step Solution: Inspect for visible damage and smell for burnt components.Use a DMM continuity or resistance mode to test suspected paths.Measure voltages to find unexpected drops (opens) or collapsed rails (shorts).Isolate sections until the fault location is narrowed down. Verification / Alternative check: Cross-check with current draw: unusually low current suggests an open; abnormally high current or tripping protection suggests a short. Why Other Options Are Wrong: Incorrect / low-voltage only / digital only / wired-only: Opens and shorts occur across all technologies and voltage levels, including high-power systems and high-speed digital PCBs. Common Pitfalls: Overlooking intermittent faults that appear as temperature- or vibration-dependent opens/shorts; assuming IC failure first when a simple connector open is more likely. Final Answer: Correct

Question 1

Series–parallel analysis workflow: when back-solving currents in a series–parallel network, is it good practice to start from the branch farthest from the source and work back toward the source after reducing the network?

Accepted Answer

Introduction / Context: Series–parallel circuits are often simplified by successively combining distant sub-networks into equivalent resistances. Once a single equivalent is found and total current is determined, engineers “back-solve” to recover currents and voltages in original branches. This question examines the recommended direction for that back-solving process. Given Data / Assumptions: Passive DC network with clear series and parallel groupings. Reduction to an equivalent resistance has already been performed. Goal is to determine individual branch currents and drops. Ideal source and ohmic elements (no reactive effects). Concept / Approach: The canonical method is: reduce the network starting from the portions farthest from the source until you obtain a single equivalent. Then compute total current from the source. Next, expand (reverse the reductions) step by step, starting from the farthest equivalent you formed. At each back-step, apply current-divider and voltage-divider rules as appropriate to recover branch values. This “farthest-outward first” expansion ensures consistency with how the equivalents were formed and avoids mixing knowns with unknowns prematurely. Step-by-Step Solution: 1) Combine farthest sub-network into an equivalent resistance.2) Repeat until a single R_eq remains; compute I_total = V_source / R_eq.3) Reverse the last reduction first: split the equivalent back into its components using divider rules to find currents/voltages.4) Continue stepwise toward the source, reconstructing each prior stage’s branch values. Verification / Alternative check: This mirrors standard textbook ladder-network procedures; nodal or mesh analysis would yield identical results but without the intuitive stepwise back-substitution. Why Other Options Are Wrong: “Always start at the source” during back-solving can leave critical branch values unknown because later equivalents have not yet been expanded. “Valid only for AC” or “only if resistors equal” adds constraints that do not apply to the general DC method. Common Pitfalls: Confusing reduction order with back-solving order; forgetting to re-apply correct current- or voltage-divider relations when expanding each equivalent stage. Final Answer: Valid (standard back-solving sequence)

Question 2

Wheatstone bridge property at balance: if a Wheatstone bridge is balanced, what is the voltage between the midpoints A and B relative to the source voltage VS?

Accepted Answer

Introduction / Context: The Wheatstone bridge is a classic four-resistor network used for precise measurement and sensing. A key concept is the “balance” condition, where the ratio of resistances in one branch equals the ratio in the other. This question checks your recall of the bridge’s null property at balance. Given Data / Assumptions: Standard Wheatstone topology: two series legs in parallel across a source VS. Node A is the junction of the upper and lower resistors in the left leg; node B is the junction in the right leg. “Balanced” means R1/R2 = R3/R4, using a consistent top-to-bottom or bottom-to-top ratio convention. Assume ideal meters with infinite input resistance (no loading). Concept / Approach: At balance, the potential divider in the left leg produces the same fraction of source voltage as the right leg. Therefore, the potentials at nodes A and B are equal, making the bridge (detector) voltage VAB = VA − VB equal to zero. This is the basis for using a null detector: the galvanometer reads zero at balance. Step-by-Step Solution: 1) Write left-leg divider: VA = VS * (R2 / (R1 + R2)) with a chosen orientation.2) Write right-leg divider: VB = VS * (R4 / (R3 + R4)).3) Apply balance condition R1/R2 = R3/R4 ⇒ VA = VB.4) Compute VAB = VA − VB = 0. Verification / Alternative check: If the bridge is slightly unbalanced, VAB is small but nonzero; its polarity indicates which ratio is greater. Exactly at balance, the null meter reads zero even though current flows in the legs. Why Other Options Are Wrong: VAB equals VS: impossible since VA and VB are internal divider tap voltages. VAB equals VS/2: only in a special, incidental case and not a balance criterion. “Thevenin voltage of the source”: irrelevant—the null depends on the resistor ratios, not on source reduction. “Undefined without load”: the detector is assumed high impedance; VAB is well-defined. Common Pitfalls: Reversing resistor labeling and ratio orientation; assuming “no current flows anywhere” at balance—current still flows in the legs, but not through the detector. Final Answer: VAB equals 0 (nodes at equal potential)

Question 3

Voltmeter reading in mixed networks: In a series–parallel DC circuit powered by an 18 V source, a correctly connected ideal voltmeter placed directly across the source terminals should indicate 18 V (ignoring wiring drops and meter loading). Decide …

Accepted Answer

Introduction / Context: Knowing what a voltmeter should read in a practical series–parallel circuit is fundamental to troubleshooting. When you connect a voltmeter across the supply terminals, you are measuring the source's terminal voltage. If the DC source is specified as 18 V and loading effects are negligible, the reading should be 18 V. This question checks recognition of proper measurement placement and the definition of terminal voltage. Given Data / Assumptions: DC source nominal value: 18 V. Voltmeter is ideal (very high input resistance), connected directly across the source terminals. Lead resistance and wiring drops are negligible for the concept check. No meter misconfiguration or wrong range selection. Concept / Approach: Voltage is the potential difference between two points. Terminal voltage of a source is the potential difference at its output terminals. An ideal voltmeter has nearly infinite resistance, so it draws negligible current and does not appreciably alter the circuit. Therefore it reports the source terminal voltage, which equals the rated value in this idealized scenario. Step-by-Step Solution: Identify the measurement points: the two source terminals.Recall: an ideal voltmeter measures potential difference without loading.Since the source is 18 V, the expected reading is 18 V.Conclude the statement is accurate under the listed assumptions. Verification / Alternative check: Measure any component in parallel with the source; it will also show approximately 18 V in an ideal model. Bench experiments with a DMM (10 MΩ input) typically read the nominal source value within tolerance. Why Other Options Are Wrong: Incorrect: Conflicts with the definition of terminal voltage. Valid only with an analog meter: Digital or analog does not change the physics. True only if load current is zero: Finite load current does not change an ideal measurement. Depends on lead polarity: Polarity affects sign, not magnitude; meters auto-range absolute value in DC volts mode when leads are swapped. Common Pitfalls: Putting the voltmeter in series (that would be an ammeter position), measuring at the wrong nodes, or expecting internal source resistance drops without modeling them explicitly. Final Answer: Correct

Question 4

Series–parallel reduction rule: For a network where R1 is in series with a parallel group of R2, R3, and R4, the total resistance is given by RT = R1 + (R2 || R3 || R4). Determine whether this formula is valid for that topology.

Accepted Answer

Introduction / Context: Reducing mixed series–parallel resistor networks is a core skill in circuit analysis. Recognizing when a single element is in series with a parallel group allows a quick two-step computation of the equivalent resistance, speeding up hand calculations and sanity checks in design work. Given Data / Assumptions: Topology: R2, R3, and R4 share the same two nodes (are in parallel). R1 is connected in series with that parallel combination. Linear, time-invariant resistors; frequency effects neglected. No dependent sources or bridging connections that would break simple series/parallel relations. Concept / Approach: Series elements carry the same current and their resistances add arithmetically. Parallel elements share the same voltage and their conductances add: R_parallel = 1 / (1/R2 + 1/R3 + 1/R4). Combining both rules yields RT = R1 + R_parallel for this specific topology. Step-by-Step Solution: Compute R_p = (R2 || R3 || R4) = 1 / (1/R2 + 1/R3 + 1/R4).Identify that the same current must pass through R1 and the parallel block in series.Use series rule: RT = R1 + R_p.Conclude the formula in the prompt matches the reduction steps. Verification / Alternative check: Pick numeric values (e.g., R2=300 Ω, R3=600 Ω, R4=600 Ω). Compute R_p=150 Ω, then RT=R1+150 Ω, confirming additivity of the series element with the parallel block. Why Other Options Are Wrong: Incorrect / AC only: The rule is purely topological and applies equally in DC. Valid only when R2=R3=R4 or when R1 is very large: Equality or magnitude ratios are not required for the formula to hold. Common Pitfalls: Accidentally including a bridging element that spoils pure parallel, or adding all four resistors directly as if series. Always confirm node sharing before applying ||. Final Answer: Correct

Question 5

Failure modes of resistors: When a fixed resistor overheats and fails, does it typically fail as a short-circuit path or as an open circuit? Choose the accurate statement about common real-world behavior.

Accepted Answer

Introduction / Context: Diagnosing faults quickly requires knowing typical failure modes. Fixed resistors subjected to excessive power dissipation (heat) usually fracture, crack, or the resistive film breaks, interrupting current flow. Understanding this behavior helps technicians decide where to place the meter probes and what values to expect when equipment fails. Given Data / Assumptions: Component type: fixed carbon film, metal film, or wirewound resistors. Stress: over-power, prolonged overheating, or surge events. No exotic fusible or specially constructed parts designed to fail short. Concept / Approach: Resistors rely on a continuous resistive path. Thermal overstress tends to break or vaporize that path, producing an open circuit (very high resistance). While some rare failure mechanisms can lead to partial shorts (especially with contaminants or carbonization), the overwhelmingly common outcome for standard resistors is an open circuit. Step-by-Step Solution: Consider power P = I^2 * R or V^2 / R; excessive P overheats the element.Heat damages the resistive film or wire, creating breaks.A break implies current cannot flow — an open circuit.Therefore, the typical failure mode is “open,” not “short.” Verification / Alternative check: Field experience: ohmmeter readings on burned resistors trend to very high resistance or infinity. Service manuals often advise checking for opens in suspect resistors after visible scorching. Why Other Options Are Wrong: They usually short: Uncommon for standard parts; shorts are more associated with semiconductors failing catastrophically. Depends on color code: Color code labels value/tolerance, not failure mode. Always perfect conductor/insulator: Real failures vary; “open” means very high resistance, not ideal infinity. Common Pitfalls: Assuming any burnt part is shorted; overlooking that some resistors are intentionally fusible and act like fuses (open by design). Final Answer: Correct — they commonly go open when burned

Question 6

Parallel branch voltage check: If a resistor R3 is connected directly in parallel with an ideal 20 V DC source (i.e., across the same two supply nodes), the voltage across R3 should be 20 V. Decide if this statement is valid.

Accepted Answer

Introduction / Context: Parallel elements share the same voltage because they are connected to the same two nodes. Recognizing this is essential for determining branch currents with Ohm’s law and for verifying measurements made with a voltmeter in complex series–parallel circuits. Given Data / Assumptions: Ideal 20 V DC source. Resistor R3 is connected directly across the source terminals (true parallel). Negligible lead/wiring resistance and no intervening elements. Concept / Approach: By definition of parallel connection, the voltage across each parallel branch equals the source voltage. The resistor’s value affects current (I = V/R) but not the branch voltage in the ideal model. Therefore, any component in parallel with the 20 V source experiences 20 V across it. Step-by-Step Solution: Identify nodes: R3 is tied to the same two nodes as the source.The source enforces a 20 V potential difference between these nodes.Hence V_R3 = 20 V.Conclude the statement is valid for ideal conditions. Verification / Alternative check: Measure with a high-impedance DMM across R3: the reading matches the supply voltage, verifying the rule in practice barring notable wiring drops. Why Other Options Are Wrong: Incorrect / depends on current / valid only for analog meters: Voltage in parallel is independent of resistor value and meter type in the ideal case. True only when R3 = 20 Ω: Resistance value does not set branch voltage in parallel. Common Pitfalls: Confusing series voltage division with parallel voltage equality; inserting a meter incorrectly and loading the circuit, which could change a non-ideal reading. Final Answer: Correct

Question 7

Single-resistor drop equals source voltage: If a 175 V DC source is applied across a lone resistor R1 (no other series elements), the voltage across R1 equals 175 V under ideal conditions. Assess the correctness of this statement.

Accepted Answer

Introduction / Context: A common measurement scenario is applying a source across a single component. In such a case, the entire source voltage appears across that component because there is nowhere else for the drop to occur. This reinforces the basic node-voltage concept used in all series–parallel analyses. Given Data / Assumptions: Ideal 175 V source. Only one element present: resistor R1 across the source terminals. Negligible lead/wiring drops; meter loading ignored. Concept / Approach: With only a single element across the source, KVL dictates that the sum of drops around the loop equals the source voltage. Since R1 is the only drop, V_R1 must equal 175 V. Ohm’s law then determines the current (I = 175 / R1), but the resistor value does not change the magnitude of the voltage imposed by the source across it. Step-by-Step Solution: Draw the loop: source in series with R1.Apply KVL: +175 V + (drop across R1) = 0 around the loop in sign convention.Therefore magnitude V_R1 = 175 V.Use I = V/R if current is needed; voltage remains 175 V. Verification / Alternative check: Bench test with a regulated supply and DMM shows the full source voltage across a single load, subject only to tolerance and negligible lead resistance effects. Why Other Options Are Wrong: Incorrect: Violates KVL. True only if R1 = 175 Ω: Resistance value sets current, not applied voltage. Depends on current range or room temperature: Meter range and temperature do not alter the ideal circuit law. Common Pitfalls: Imagining “voltage division” where no division exists; mixing up current settings on meters with voltage measurements; overlooking that all of the source voltage must appear across the sole component. Final Answer: Correct

Question 8

Loaded voltage divider behavior: When a load resistance is attached to the output node of a DC voltage divider, the source must supply additional current. Evaluate this statement about source current increasing under load.

Accepted Answer

Introduction / Context: Voltage dividers are ubiquitous. However, the ideal unloaded divider assumption breaks once a load is connected. Understanding how loading alters currents and voltages helps in designing dividers with acceptable regulation and selecting buffer stages when needed. Given Data / Assumptions: Two-resistor divider producing an output node. A finite load R_L is connected from the output node to ground (or return). DC steady-state analysis; resistors are linear. Concept / Approach: The source current equals the sum of currents through all paths tied to the source. Without load, current is I_unloaded = V_in / (R_top + R_bottom). With load, the bottom resistor and R_L form a parallel network, decreasing the effective lower-leg resistance and increasing total divider current. Therefore, the source must deliver more current when a load is attached. Step-by-Step Solution: Compute R_lower_eff = R_bottom || R_L.Total divider resistance becomes R_total_loaded = R_top + R_lower_eff.Since R_lower_eff <= R_bottom, R_total_loaded <= R_top + R_bottom.For a fixed V_in, I_loaded = V_in / R_total_loaded ≥ I_unloaded. Verification / Alternative check: Select sample values and calculate currents with and without load; the loaded case always shows higher source current for any finite R_L. Why Other Options Are Wrong: Incorrect / depends on meter type / AC only: The rule follows from resistance reduction; instrument type and DC vs AC (for resistors) do not reverse it. True only if R_L is larger than the bottom resistor: Regardless of R_L magnitude, any finite R_L in parallel reduces the effective resistance and increases current. Common Pitfalls: Ignoring load interaction when specifying divider resistors; forgetting that heavier loading also drops the output voltage below the ideal unloaded value. Final Answer: Correct

Question 9

Series branch current identity: In a simple series circuit carrying 17 mA, the current through any series element (for example, resistor R1) equals the branch current, i.e., 17 mA. Judge whether this statement is correct.

Accepted Answer

Introduction / Context: Distinguishing series from parallel behavior is critical. In series connections, the same current flows sequentially through each component. This principle holds regardless of individual resistance values and is the reason voltage, not current, divides among series elements. Given Data / Assumptions: Series circuit with a measured or known branch current of 17 mA. Resistor R1 is one element in that series branch. Ideal wires and components for the conceptual statement. Concept / Approach: By definition, a series connection provides only one path for current. Kirchhoff’s Current Law at the node between series elements states that the current entering equals the current leaving. Therefore, the same 17 mA passes through every series element, including R1. Voltages differ by resistance (V = I * R), but current is identical in series. Step-by-Step Solution: Recognize the topology: one continuous path, no branches.Apply KCL at internal nodes: I_in = I_out.Conclude I_R1 = I_branch = 17 mA.Use V_R1 = 0.017 * R1 for drop if needed. Verification / Alternative check: Measure current with an ammeter inserted at different series points; readings are equal (accounting for meter insertion effects). SPICE simulations confirm equal series currents. Why Other Options Are Wrong: Incorrect / depends on tolerance / only DC: Tolerance affects voltage division; the current equality holds for DC and AC in purely series resistive paths. True only if R1 is smallest: Value does not change current equality; it changes voltage drop. Common Pitfalls: Assuming different currents through different series parts; confusing with parallel branches where current divides. Final Answer: Correct

Question 10

Wheatstone bridge application: A classic Wheatstone bridge circuit can be used to determine an unknown resistance by balancing the bridge and reading a known ratio. Decide whether this statement correctly describes a common use of the bridge.

Accepted Answer

Introduction / Context: The Wheatstone bridge is a foundational measurement circuit. By adjusting one known resistor (or using a calibrated ratio), the bridge can be balanced so that the detector indicates zero. At balance, a simple proportion relates the unknown resistor to the known elements, enabling precise resistance measurement without directly reading current or voltage across the unknown at nonzero levels. Given Data / Assumptions: Four-resistor bridge configuration with a sensitive null detector. At balance, no current flows through the detector branch. Resistors are stable during measurement; temperature effects are negligible. Concept / Approach: At balance, the ratio of the two arms is equal: R1/R2 = R_unknown/R3 (labeling may vary). Solving gives R_unknown = (R1/R2) * R3. Because the detector sees near-zero current at balance, measurement error due to detector loading is minimized, making the method precise for a wide range of resistances. Step-by-Step Solution: Arrange the bridge with three known resistors and one unknown.Adjust one known resistor until the detector shows null.Write the ratio equality for the two bridge arms.Solve for the unknown using the known ratio and resistor value. Verification / Alternative check: Compare the result with a direct ohmmeter reading; close agreement validates proper balancing and resistor tolerances. Why Other Options Are Wrong: Incorrect / AC only / only above 1 kΩ / requires digital meter: The bridge works with DC or AC (for resistive cases) and across broad resistance ranges; a simple galvanometer or null detector is sufficient. Common Pitfalls: Miswiring the bridge arms, ignoring lead resistance for low-ohm measurements, or attempting to read the unknown without reaching true null. Final Answer: Correct

Question 11

Series–parallel in practice: A “loaded voltage divider” (a resistor divider that feeds a finite load) is a common and important application of series–parallel circuits. Determine whether this usage description is accurate.

Accepted Answer

Introduction / Context: Many real circuits use resistor dividers to derive reference or bias voltages. The moment a load is attached to a divider output, the circuit becomes a series–parallel network: the lower divider resistor is effectively in parallel with the load. Recognizing this is essential for predicting output sag and source current changes. Given Data / Assumptions: Two-resistor divider from a DC supply. Finite load R_L from the divider output to ground. Resistive behavior; frequency effects ignored. Concept / Approach: With a load present, the bottom leg and R_L form a parallel combination. The output voltage becomes V_out = V_in * (R_bottom || R_L) / (R_top + (R_bottom || R_L)). This is a textbook series–parallel transformation and a standard example used to introduce loading effects in basic circuit courses. Step-by-Step Solution: Identify the divider: R_top and R_bottom in series across V_in.Attach R_L across the output node to ground.Compute R_lower_eff = R_bottom || R_L.Use the series formula to compute V_out and confirm series–parallel nature. Verification / Alternative check: Measure V_out with and without the load; the loaded case is lower, matching the series–parallel formula prediction. Why Other Options Are Wrong: Incorrect / only large load / low-voltage only / AC only: The classification as series–parallel does not depend on voltage level or frequency for resistive circuits and holds for any finite load. Common Pitfalls: Designing a divider without considering R_L, leading to excessive voltage sag or wasted current; forgetting to buffer the divider with an op-amp follower when a stiff source is required. Final Answer: Correct

Question 12

Common faults in circuits: In everyday electronics troubleshooting, the two most typical failures encountered are opens (broken paths) and shorts (unintended low-resistance paths). Decide whether this statement is correct.

Accepted Answer

Introduction / Context: Technicians often categorize failures broadly as opens or shorts. An open means a conductor or component no longer conducts as intended, breaking the current path. A short means two nodes are unintentionally connected by a very low resistance, often causing excessive current. Recognizing these classes speeds diagnosis across analog and digital systems. Given Data / Assumptions: Applies to discrete-component circuits, PCBs, cables, and connectors. Covers environmental and wear-out mechanisms such as heat, vibration, corrosion, and contamination. No assumption of a specific voltage level or signal type. Concept / Approach: Most electrical issues reduce to a path that is broken (open) or a path that should not exist (short). Examples include cracked solder joints (open), blown fuses (open), carbonized PCB after arcing (short), solder bridges (short), frayed cable shield contacting conductors (short), and corroded connectors (intermittent open). Systematic isolation using continuity tests and resistance measurements quickly confirms which category a fault belongs to. Step-by-Step Solution: Inspect for visible damage and smell for burnt components.Use a DMM continuity or resistance mode to test suspected paths.Measure voltages to find unexpected drops (opens) or collapsed rails (shorts).Isolate sections until the fault location is narrowed down. Verification / Alternative check: Cross-check with current draw: unusually low current suggests an open; abnormally high current or tripping protection suggests a short. Why Other Options Are Wrong: Incorrect / low-voltage only / digital only / wired-only: Opens and shorts occur across all technologies and voltage levels, including high-power systems and high-speed digital PCBs. Common Pitfalls: Overlooking intermittent faults that appear as temperature- or vibration-dependent opens/shorts; assuming IC failure first when a simple connector open is more likely. Final Answer: Correct

Question 13

Loaded voltage divider — consider attaching different loads to the same divider output. Which statement is more accurate: connecting a 6 kΩ load will cause a smaller drop in the divider's output voltage than connecting a 5 kΩ load (since 6 kΩ loads …

Accepted Answer

Introduction / Context: A voltage divider provides a reference voltage but is sensitive to loading. When an external load is connected, the output falls due to the divider's Thevenin resistance. This question compares two loads (6 kΩ versus 5 kΩ) to evaluate which causes less output sag. Given Data / Assumptions: Same divider and same open-circuit output V_oc for both cases. Load options: 6 kΩ and 5 kΩ connected to the divider output node. Linear, time-invariant resistors; DC or low-frequency behavior; source internal resistance may be modeled as part of the Thevenin resistance R_th. Concept / Approach: The loaded output is V_out = V_oc * (R_L / (R_th + R_L)). For fixed R_th and V_oc, a larger R_L yields a larger fraction and therefore a higher V_out (less droop). Hence, 6 kΩ (larger than 5 kΩ) loads the divider less and produces a smaller decrease from the open-circuit value. Step-by-Step Solution: Model the divider and its source as a Thevenin equivalent: V_oc in series with R_th.For R_L = 6 kΩ: V6 = V_oc * (6k / (R_th + 6k)).For R_L = 5 kΩ: V5 = V_oc * (5k / (R_th + 5k)).Compare: since 6k / (R_th + 6k) > 5k / (R_th + 5k) for R_th ≥ 0, V6 > V5, so the 6 kΩ load causes a smaller drop. Verification / Alternative check: Pick R_th = 2 kΩ and V_oc = 10 V. With 6 kΩ: V_out ≈ 10 * (6000/8000) = 7.5 V. With 5 kΩ: V_out ≈ 10 * (5000/7000) ≈ 7.14 V. The 6 kΩ case has less sag, matching the rule of thumb to use a load at least 10× the source resistance. Why Other Options Are Wrong: Incorrect: Contradicts the divider formula; larger R_L reduces loading. Only true for AC sources: The reasoning is resistive and applies to DC as well. Only true when source resistance is zero: If R_th were zero, both loads would see the same ideal source voltage; the statement remains consistent with the general rule for R_th > 0. Common Pitfalls: Ignoring the source/ divider's Thevenin resistance, or assuming that higher-ohm loads always draw "more" voltage drop in the source. In reality, higher-ohm loads draw less current and thus cause less droop. Final Answer: Correct — a 6 kΩ load produces a smaller output decrease than a 5 kΩ load on the same divider.

Question 14

Thevenin equivalence — the Thevenin equivalent voltage seen at a pair of terminals equals the open-circuit voltage measured at those terminals. Is this statement accurate for linear networks?

Accepted Answer

Introduction / Context: Thevenin’s theorem lets you replace a complicated linear network with a single ideal voltage source in series with a resistance as seen from two terminals. The core quantity is the Thevenin voltage, which many learners confuse. This question asks whether V_th equals the open-circuit voltage at the terminals. Given Data / Assumptions: Linear, bilateral elements (resistors, sources; reactive elements permitted at a fixed frequency for phasors). No load attached when measuring open-circuit voltage. The network is observed from a specified port (two terminals). Concept / Approach: By definition, the Thevenin voltage V_th is the terminal voltage when the output is open-circuited (no load current). With no load current, internal drops through the Thevenin resistance are zero, so the terminal voltage equals the ideal source voltage in the Thevenin model. Therefore, V_th = V_open-circuit is not merely a rule of thumb but the formal definition in Thevenin analysis. Step-by-Step Solution: Identify the output port terminals.Compute or measure V_oc with the load removed (I_load = 0 A).Set V_th = V_oc; find R_th separately (e.g., by zeroing independent sources and measuring resistance, or using short-circuit methods if allowed).Use V_out with any load R_L via V_out = V_th * (R_L / (R_th + R_L)). Verification / Alternative check: Build a resistive divider: with the output open, the node voltage is V_oc from the divider equation. In the Thevenin model, the same node is driven by V_th in series with R_th; with I_load = 0, drops across R_th are zero, so the terminal voltage equals V_th, matching V_oc. Why Other Options Are Wrong: Incorrect: Conflicts with the definition of Thevenin voltage. Only resistive: The statement also holds for reactive networks at a given frequency (phasor steady state). Only when a load is connected: This reverses the concept; V_th is defined with no load connected. Common Pitfalls: Confusing V_th with the loaded output; forgetting that R_th affects the voltage only when load current flows; mixing Norton parameters with Thevenin parameters. Final Answer: Correct — V_th equals the open-circuit terminal voltage.

Question 15

Loaded divider formula — for a Thevenin source of V_th in series with R_th feeding a 10 kΩ load, the output is V_out = V_th * (10 kΩ / (R_th + 10 kΩ)). Is this relationship the correct way to compute the loaded output?

Accepted Answer

Introduction / Context: Any practical voltage source has finite output resistance. The Thevenin model captures this as an ideal source V_th in series with R_th. When you connect a load, the output drops according to a simple divider rule. This item asks you to validate the formula for a 10 kΩ load. Given Data / Assumptions: Thevenin equivalent: V_th in series with R_th. Load: R_L = 10 kΩ. Linear resistive behavior; DC or phasor magnitude for AC at one frequency. Concept / Approach: The output node is the junction between R_th and R_L. By the voltage divider relation: V_out = V_th * (R_L / (R_th + R_L)). Substituting R_L = 10 kΩ gives the stated formula. This is exact for the Thevenin representation and is the standard way to quantify loading effects and set design margins. Step-by-Step Solution: Draw series chain: V_th → R_th → node → R_L (10 kΩ) → return.Apply divider: V_out = V_th * (R_L / (R_th + R_L)).Insert R_L = 10 kΩ: V_out = V_th * (10k / (R_th + 10k)).Analyze sensitivity: increasing R_th lowers V_out; increasing R_L raises V_out toward V_th. Verification / Alternative check: SPICE or bench test with V_th = 5 V, R_th = 1 kΩ: V_out = 5 * (10k / 11k) ≈ 4.545 V, matching simulation and measurement results within tolerance. Why Other Options Are Wrong: Incorrect: The divider formula follows directly from Ohm’s law. Only valid if R_th ≪ 10 kΩ: The formula holds for any values; that inequality only indicates small loading (V_out ≈ V_th). Only valid for AC: The relation is resistive and holds for DC as well. Common Pitfalls: Forgetting the Thevenin resistance, or confusing series and parallel placements of the load; misapplying the parallel formula instead of the series divider. Final Answer: Correct — V_out = V_th * (10 kΩ / (R_th + 10 kΩ)).

Question 16

Ohm’s law in branches — in any series–parallel network, the current through a specific resistor equals the voltage across that resistor divided by its resistance (I = V/R). Does this correctly describe branch current measurement?

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Introduction / Context: Series–parallel circuits can look complicated, but each branch still obeys Ohm’s law locally. Measuring the voltage across one resistor allows you to compute its branch current directly. This question confirms that fundamental relationship for any branch within a linear network. Given Data / Assumptions: Lumped linear elements; constant or slowly varying signals. The resistor's two-terminal voltage can be measured. Temperature and tolerance effects are negligible for the conceptual calculation. Concept / Approach: Ohm’s law states I = V / R for a resistor. Regardless of the complexity of the rest of the network, the local relation between a resistor’s voltage and current is unaffected. Series connections share current; parallel connections share voltage; but within a branch, once V across that element is known, I follows directly via I = V/R. Step-by-Step Solution: Isolate the resistor of interest and identify its two terminals.Measure or compute the voltage V across those terminals.Use Ohm’s law: I_branch = V / R.Report the current with appropriate units (A, mA, µA). Verification / Alternative check: Example: If a 220 Ω branch has 1.70 V across it, I = 1.70 / 220 ≈ 7.73 mA. Measuring with an ammeter in series with that branch should confirm the same value within meter and tolerance limits. Why Other Options Are Wrong: Incorrect: Denies Ohm’s law, which is foundational for linear resistors. Only for series-only circuits: Ohm’s law is local and holds in parallel and mixed networks too. Only if resistance < 10 kΩ: Magnitude does not invalidate the relation; units and accuracy may change but not the law. Common Pitfalls: Confusing the total circuit current with a branch current; mixing up node voltage (common to a parallel group) with element voltage; neglecting meter loading when measuring low-resistance elements. Final Answer: Correct — branch current equals the element’s voltage divided by its resistance.

Question 17

Diagnosing topology — in a series–parallel circuit, components that are in parallel share the same voltage. Therefore, observing that voltages are not shared identifies components that are NOT in parallel. Is this diagnostic statement valid?

Accepted Answer

Introduction / Context: When troubleshooting mixed series–parallel networks, it helps to know which measurements reveal series versus parallel groupings. Parallel elements share the same node pair; consequently, they must have the same voltage across them at a given instant. This item tests whether “non-shared voltages” imply “not parallel.” Given Data / Assumptions: Lumped linear circuit; DC or instantaneous AC measurement. Accurate voltage measurements across candidate components. No significant wiring drops skewing readings between supposed common nodes. Concept / Approach: Parallel definition: components connected across the same two nodes. Voltage is a difference in electric potential between two points. If two components are parallel, the voltage difference across each is identical. Conversely, if two components exhibit different voltages, they cannot share exactly the same two nodes, so they are not in parallel (though they might still be part of a broader parallel path plus series segments). Step-by-Step Solution: Select two candidate elements.Measure V_a and V_b across each with the same polarity reference.If V_a ≠ V_b (beyond tolerance), conclude they are not directly in parallel.If V_a = V_b, they may be parallel; confirm by tracing node connections. Verification / Alternative check: Use a schematic or continuity tester to verify that the terminals of both components connect to the same nodes. Voltage equality is a necessary condition for parallel; coupled with node tracing, it becomes sufficient for diagnosis. Why Other Options Are Wrong: Incorrect: It would imply parallel elements need not share voltage, which contradicts the definition. Valid only for AC / only for equal resistors: Parallel voltage sharing is topology-driven, independent of frequency and resistor equality. Common Pitfalls: Measuring with floating references or swapping meter leads; interpreting small measurement differences due to probe placement or wiring resistance as proof against parallelism without verifying nodes. Final Answer: Correct — if voltages are not shared, the elements are not in parallel.

Question 18

Loaded vs. unloaded divider — if a certain loaded voltage divider produces 12 V at its output, what happens when the load is removed (open-circuit)? Does the output decrease or increase relative to 12 V?

Accepted Answer

Introduction / Context: Voltage dividers sag under load because current through the load creates an additional drop across the source (Thevenin) resistance. Removing the load (open-circuiting the output) eliminates that sag. This question addresses whether the output decreases or increases when the load is taken away. Given Data / Assumptions: A divider or source modeled by V_th in series with R_th. Loaded output measured as 12 V with some R_L attached. Then the load is removed (R_L → ∞), measuring open-circuit output. Concept / Approach: The loaded output is V_load = V_th * (R_L / (R_th + R_L)). When R_L is finite, V_load < V_th (if R_th > 0). As R_L → ∞ (load removed), the divider current into the load goes to zero and the output approaches V_th, which is the open-circuit voltage V_oc. Therefore, the output cannot decrease by removing the load; it increases (or at worst stays the same if R_th = 0). Step-by-Step Solution: Start: V_load = 12 V for some finite R_L.Open-circuit: R_L → ∞ → V_oc = V_th * (∞ / (R_th + ∞)) = V_th.Since V_load = V_th * (fraction < 1), V_oc ≥ V_load.Conclusion: Output increases when the load is removed. Verification / Alternative check: Bench check with V_th = 15 V, R_th = 1 kΩ, R_L = 3 kΩ: V_load = 15 * (3k / 4k) = 11.25 V. Removing the load yields V_oc = 15 V, which is higher than 11.25 V, confirming the principle. Why Other Options Are Wrong: Decreases: Opposite to divider behavior. Drops to 0 V: An open-circuit does not short the source; the node retains the open-circuit voltage. Undefined: It is well-defined; it is the Thevenin (open-circuit) voltage. Common Pitfalls: Confusing open-circuit (infinite resistance) with short-circuit (zero resistance). A short would reduce the output to near 0 V (limited by R_th), but an open makes it rise to V_th. Final Answer: Correct — removing the load raises the output toward the open-circuit value.

Question 19

Voltmeter loading — under what condition can the loading effect of a voltmeter be neglected when measuring the voltage across some resistance in a circuit?

Accepted Answer

Introduction / Context: Real voltmeters are not ideal; they have finite input resistance R_m that loads the circuit and can pull the measured node down. Engineers use a simple criterion to judge whether this loading is negligible: the 10× rule of thumb. Given Data / Assumptions: Voltmeter modeled as R_m in parallel with the circuit resistance across which you are measuring. Quasi-DC or slow AC measurement where reactive effects are minimal. The goal is to keep measurement error small (a few percent or less). Concept / Approach: When you connect a meter across a resistor R_x, the effective resistance becomes R_eff = (R_x * R_m) / (R_x + R_m). If R_m ≫ R_x, then R_eff ≈ R_x and the circuit is barely disturbed. A common practical threshold is R_m ≥ 10 * R_x, which keeps loading error typically under about 10% and often much less in practice, especially with modern DMMs (10 MΩ inputs). Step-by-Step Solution: Model the meter as a parallel resistance R_m.Compute R_eff = (R_x * R_m) / (R_x + R_m).If R_m ≥ 10 * R_x, then R_eff ≈ (R_x * 10R_x) / (11R_x) ≈ 0.909 * R_x, a small perturbation.Conclude loading is negligible at the 10× threshold or higher. Verification / Alternative check: Example: R_x = 100 kΩ; DMM input R_m = 10 MΩ (100×). The equivalent is ~99.01 kΩ, producing <1% error in most divider readings — negligible for many applications. Why Other Options Are Wrong: Very high/very low measured voltage: Voltage magnitude is not the key; the resistance ratio is. Meter resistance 10× less than R_x: That heavily loads the node and severely skews the reading. Common Pitfalls: Confusing meter range with input resistance; some analog meters have low input resistance on low-voltage ranges, causing significant loading errors on high-impedance circuits. Final Answer: Use the 10× rule — R_m should be at least 10 times R_x to ignore loading.

Question 20

Wheatstone bridge behavior — when a Wheatstone bridge is exactly balanced, what is the output (bridge) voltage between the detector nodes?

Accepted Answer

Introduction / Context: The Wheatstone bridge is a classic network used for precision measurement of resistance and for sensing with strain gauges, RTDs, and other transducers. The hallmark of a balanced bridge is a null (zero) indication at the detector. This question checks that fundamental behavior. Given Data / Assumptions: Four resistive arms forming a diamond: R1/R2 in one leg, R3/Rx in the other (for example). Input (excitation) across one diagonal; detector across the other diagonal. Balanced condition satisfied (ratio R1/R2 = R3/Rx). Concept / Approach: At balance, the voltage division in each leg produces identical node potentials at the detector nodes. Since the detector nodes are at the same potential, their difference is zero. The bridge therefore exhibits 0 V across the output diagonal, allowing sensitive null detection and high-resolution measurement by observing deviations from zero when the bridge is slightly unbalanced. Step-by-Step Solution: Apply excitation V_in across the input diagonal.At balance: V_node_top = V_in * (R2 / (R1 + R2)); V_node_bottom = V_in * (Rx / (R3 + Rx)).Set ratio equality: R1/R2 = R3/Rx → the two node voltages are equal.Therefore, V_out = V_node_top − V_node_bottom = 0 V. Verification / Alternative check: Real bridges use galvanometers or instrumentation amplifiers. At balance, the indicator reads zero; minute changes in Rx produce small differential voltages which can be amplified to infer the change precisely. Why Other Options Are Wrong: Same as input / very high: Output is zero at balance, not full-scale or large. Dependent on unknown R: At balance, it is strictly zero regardless of the actual values so long as the ratio condition holds. Common Pitfalls: Confusing bridge balance with maximum sensitivity; maximum sensitivity occurs near but not exactly at null when using certain measurement strategies. Final Answer: Equal to 0 V — a balanced Wheatstone bridge has zero output.

Series-Parallel Circuits Questions