Q: 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?

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 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.

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

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.

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

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.

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

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

Question 1

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.

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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 2

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.

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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 3

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.

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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 4

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.

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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 5

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 …

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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 6

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?

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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 7

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?

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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 8

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 9

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 10

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 11

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 12

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.

Question 13

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

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

Question 14

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

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

Question 15

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

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

Question 16

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

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

Question 17

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

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

Question 18

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

Accepted Answer

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

Question 19

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

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

Question 20

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

Accepted Answer

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

Series-Parallel Circuits Questions