Q: Simple 1:1 divider: A voltage divider uses two equal 68 kΩ resistors across a 24 V source. What is the nominal output voltage measured at the midpoint (no load)?

Introduction / Context: Equal-value voltage dividers are common for halving a supply, establishing bias points, or providing reference levels. With two identical resistors in series across a source, the midpoint sits at exactly half the supply under no-load conditions, assuming ideal components. Given Data / Assumptions: R_top = 68 kΩ, R_bottom = 68 kΩ, connected in series. Source voltage V_in = 24 V (DC). Output is at the junction (midpoint) with no load attached. Concept / Approach: For a two-resistor divider: V_out = V_in * R_bottom / (R_top + R_bottom). With equal values, R_bottom = R_top, the ratio is 1/2. Therefore, V_out equals half of V_in. No load is critical here; any finite load would reduce V_out below this ideal value unless the divider is buffered. Step-by-Step Solution: Substitute into divider formula: V_out = 24 * 68 kΩ / (68 kΩ + 68 kΩ).Compute denominator: 136 kΩ → ratio 68/136 = 1/2.So V_out = 24 * 1/2 = 12 V. Verification / Alternative check: Symmetry argument: Identical resistors share equal voltage. Each drops 12 V, placing the midpoint exactly at 12 V relative to ground. This aligns with KVL and proportional division concepts. Why Other Options Are Wrong: 24 V: That would be the full source at the top node, not the midpoint. 0 V: That is the bottom node reference, not the midpoint. 6 V: Would require unequal resistor values (e.g., a 1:3 ratio), which is not given. Common Pitfalls: Forgetting that the stated output is no-load; a load would alter the midpoint voltage. Final Answer: 12 V

Q: Analog meter sensitivity: For a 30,000 Ω/volt voltmeter set to its 50 V range, what is the instrument's internal resistance (use sensitivity × full-scale voltage)?

Introduction / Context: Traditional analog voltmeters are often specified by their sensitivity in ohms per volt (Ω/V). This figure tells you how much internal resistance the meter presents for each volt of full-scale deflection. Understanding this is essential when estimating meter loading effects on the circuit under test. Given Data / Assumptions: Sensitivity (meter constant): 30,000 Ω/V. Selected meter range: 50 V. We assume an ideal analog meter movement where R_internal = sensitivity × full-scale voltage. Concept / Approach: The internal resistance of an analog voltmeter on a given range equals the product of its Ω/V sensitivity and the full-scale range setting. This arises because the range switch inserts series resistance so that the meter movement sees its required current at full scale while presenting the specified resistance to the circuit. Step-by-Step Solution: Use the relation: R_internal = (Ω/V rating) * (V_range).Substitute values: R_internal = 30,000 Ω/V * 50 V.Compute: R_internal = 1,500,000 Ω = 1.5 MΩ. Verification / Alternative check: If the meter were set to a different range, internal resistance would scale proportionally. For example, at 10 V it would be 300 kΩ; at 100 V it would be 3 MΩ. This linear scaling confirms the formula's consistency. Why Other Options Are Wrong: 15,000 Ω and 150,000 Ω are smaller by factors of 100 and 10 respectively. 15,000,000 Ω corresponds to 300,000 Ω/V on 50 V—far above the stated sensitivity. Common Pitfalls: Confusing the Ω/V rating with the total resistance; forgetting to multiply by the selected range; misreading units (kΩ vs MΩ). Final Answer: 1,500,000 Ω

Q: Current sharing in equal parallel branches: A 470 Ω ‖ 1.5 kΩ network is in series with a parallel group of five 1 kΩ resistors across a 50 V source. What percentage of the total load current flows through any single 1 kΩ branch?

Introduction / Context: In circuits combining series and parallel sections, currents in equal-value parallel branches divide uniformly. This question checks whether you can identify the current through one branch of an equal-parallel group that sits in series with another subnetwork. Given Data / Assumptions: First block: 470 Ω in parallel with 1.5 kΩ. Second block: five 1 kΩ resistors in parallel (equal values). Blocks are in series; source = 50 V. We seek the fraction of current through one 1 kΩ branch relative to the total load current entering the five-branch block. Concept / Approach: In a set of N identical resistors in parallel, the incoming current splits equally: each branch carries 1/N of the block current. Since these five 1 kΩ resistors are identical, each one conducts 1/5 of the current through that parallel group. The series part of the network does not change this equal division within the parallel group. Step-by-Step Solution: Number of equal branches N = 5.Equal-current division → branch current fraction = 1/N = 1/5.Convert to percent: (1/5) * 100% = 20%. Verification / Alternative check: You can compute the equivalent resistance of the five 1 kΩ in parallel (Req = 200 Ω) and confirm that currents into equal arms must be identical by symmetry; the exact magnitude is unnecessary to find the percentage split. Why Other Options Are Wrong: 25% corresponds to four equal branches, not five. 50% would be two branches. 100% would mean only one branch exists, which is false here. Common Pitfalls: Trying to include the series block's value in the division calculation; forgetting that equal-value parallel branches share current equally regardless of other series elements. Final Answer: 20%

Q: Three-tap divider with equal resistors: A 6 V battery feeds three 2.2 kΩ resistors in series to create two output taps. What are the two tap voltages measured from the negative end (assume no load)?

Introduction / Context: Voltage dividers distribute a source voltage across series resistors in proportion to their resistances. With equal-value resistors and no load, the drops are equal, yielding equally spaced output taps. This is a staple concept for bias networks and reference ladders. Given Data / Assumptions: Source: 6 V DC. Three equal resistors: each 2.2 kΩ in series. Outputs: the intermediate node after the first resistor and the node after the second resistor (both referenced to the negative end). No load (open-circuit taps), so divider ratios are ideal. Concept / Approach: Total resistance divides the source proportionally. With equal parts, each section drops V_total/3. Thus the first tap is one section above ground, the second tap is two sections above ground, and the third node is the full source voltage at the top of the ladder. Step-by-Step Solution: Equal segment drop: ΔV = 6 V / 3 = 2 V per resistor.Tap 1 (after first resistor): 2 V.Tap 2 (after second resistor): 2 V + 2 V = 4 V. Verification / Alternative check: KCL/KVL analysis yields the same since the current is uniform and no load draws additional current. Measured voltages stack linearly in equal increments. Why Other Options Are Wrong: 2 V, 2 V ignores the cumulative second drop. 2 V, 6 V and 4 V, 6 V incorrectly include the full source at a tap rather than only at the top of the string. Common Pitfalls: Mixing up node numbering; assuming 6 V appears at an intermediate tap; forgetting that taps are cumulative sums of equal drops. Final Answer: 2 V, 4 V

Q: Mixed series–parallel network: A 12 kΩ, 15 kΩ, and 22 kΩ series string is in series with the parallel of two 10 kΩ resistors. With a 75 V source, what is the approximate current through the 15 kΩ resistor?

Introduction / Context: When multiple series resistors are combined with a parallel subnetwork in series, the same current passes through all series elements. Finding the current requires computing the overall series equivalent first, then applying Ohm’s law to the entire path. This problem reinforces correct series/parallel reduction and unit handling in kilohms. Given Data / Assumptions: Series resistors: 12 kΩ, 15 kΩ, 22 kΩ. Parallel pair: two 10 kΩ → equivalent 5 kΩ. Source voltage: 75 V. Ideal components, DC steady state. Concept / Approach: First reduce the parallel pair (10 kΩ ‖ 10 kΩ = 5 kΩ). Then add all series values to get total resistance. The current through each series element (including the 15 kΩ resistor) is I = V / R_total. No further division is needed because series currents are identical. Step-by-Step Solution: Parallel equivalent: R_p = (10k * 10k) / (10k + 10k) = 5 kΩ.Total series resistance: R_total = 12 kΩ + 15 kΩ + 22 kΩ + 5 kΩ = 54 kΩ.Current: I = 75 V / 54,000 Ω ≈ 0.0013889 A ≈ 1.39 mA ≈ 1.4 mA. Verification / Alternative check: Power sanity: P_total = V * I ≈ 75 * 0.00139 ≈ 0.104 W. Individual drops: V_15k ≈ I * 15k ≈ 20.8 V (consistent). Why Other Options Are Wrong: 14 mA and 50 mA are orders of magnitude too large; 5 mA requires a much smaller total resistance than present. Common Pitfalls: Forgetting to reduce the 10 kΩ pair to 5 kΩ; treating kΩ as Ω; assuming different currents in series elements. Final Answer: 1.4 mA

Question 1

Series network with a shunt across one element: Three 10 kΩ resistors are in series, but one of the 10 kΩ resistors has a 20 kΩ resistor connected in parallel across it. With a 24 V source, what is the total current drawn by the circuit?

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Introduction / Context: This question blends series–parallel reduction with Ohm’s law. Recognizing which elements are in parallel and which are in series is crucial. After simplifying the parallel branch, we add the remaining series resistances and compute the total current from the source voltage. Given Data / Assumptions: Resistors: three of 10 kΩ nominal; across one of these, a 20 kΩ resistor is placed in parallel. Source voltage: 24 V (DC). Ideal components and wiring; ignore tolerances and loading elsewhere. Concept / Approach: First, reduce the parallel pair: R_p = (10 kΩ * 20 kΩ) / (10 kΩ + 20 kΩ) = 200 / 30 kΩ ≈ 6.667 kΩ. Then, since the remaining two 10 kΩ resistors are simply in series with this equivalent, sum them to find the total resistance R_total. Finally, apply I_total = V / R_total and convert to microamperes or milliamperes as appropriate. Step-by-Step Solution: Compute parallel: R_p = (10 kΩ * 20 kΩ) / (30 kΩ) ≈ 6.6667 kΩ.Total series resistance: R_total = 10 kΩ + 6.6667 kΩ + 10 kΩ = 26.6667 kΩ.Total current: I = 24 V / 26.6667 kΩ ≈ 0.0009 A = 900 µA.Thus the source provides approximately 900 µA. Verification / Alternative check: Quick ratio check: 24 V across ~26.7 kΩ should be slightly under 1 mA, consistent with 900 µA. A back-check of power P = V * I ≈ 24 * 0.0009 = 0.0216 W across the network is also reasonable. Why Other Options Are Wrong: 9 mA and 90 mA: Would require total resistances of only a few kilo-ohms or a few hundred ohms, not ≈ 26.7 kΩ. 800 µA: Close but results from rounding the equivalent too aggressively; the correct computed value is ~900 µA. Common Pitfalls: Adding 10 kΩ and 20 kΩ instead of taking their parallel combination. Forgetting to keep units in kΩ consistently during arithmetic. Final Answer: 900 µA

Question 2

Loaded voltage divider analysis: A divider uses two 100 kΩ resistors across a 12 V source. If a 1 MΩ load is connected at the midpoint to ground (i.e., across the lower 100 kΩ), what is the resulting output voltage?

Accepted Answer

Introduction / Context: Real-world voltage dividers are rarely unloaded. When a load is attached to the output node, it forms a parallel path with one of the divider resistors, altering the effective resistance and reducing the output voltage. This is a classic example of source-loading and Thevenin-equivalent thinking. Given Data / Assumptions: Top resistor R_top = 100 kΩ from +12 V to the output node. Bottom resistor R_bottom = 100 kΩ from output node to ground. Load R_load = 1 MΩ from output node to ground (in parallel with R_bottom). All components are ideal; DC operation. Concept / Approach: First, find the equivalent of the bottom leg: R_eq_bottom = (R_bottom || R_load). The output is then V_out = V_in * R_eq_bottom / (R_top + R_eq_bottom). Because R_load is large relative to 100 kΩ, loading is modest but not negligible, pulling the output below the open-circuit 6 V level. Step-by-Step Solution: Compute R_eq_bottom = (100 kΩ * 1 MΩ) / (100 kΩ + 1 MΩ) = (100,000 * 1,000,000) / 1,100,000 ≈ 90.909 kΩ.Compute V_out = 12 V * 90.909 kΩ / (100 kΩ + 90.909 kΩ).Denominator: 190.909 kΩ → Ratio ≈ 0.47619.V_out ≈ 12 * 0.47619 ≈ 5.714 V ≈ 5.7 V. Verification / Alternative check: Open-circuit output would be 6 V. A finite load in parallel with 100 kΩ must reduce the output; 5.7 V is close to, but slightly below, 6 V, which fits expectations. The Thevenin perspective around the midpoint yields the same result. Why Other Options Are Wrong: 6 V: That is the no-load value; it ignores loading. 12 V: Impossible for a passive divider without amplification. 0.57 V: Off by a factor of 10 from the correct loaded output. Common Pitfalls: Treating the load as separate rather than in parallel with the lower resistor. Arithmetic mistakes when handling ratios with kilo-ohm and mega-ohm values. Final Answer: 5.7 V

Question 3

Identifying the branch with maximum current: A 470 Ω || 1.2 kΩ parallel network is in series with a second network formed by three 3 kΩ resistors in parallel. A 200 V source feeds the series combination. Which single resistor carries the greatest cu…

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Introduction / Context: This mixed series–parallel problem asks you to reason about branch currents. Within any parallel group, the branch with smaller resistance carries larger current for the same block voltage. To compare across two parallel blocks in series, find the voltage division between blocks using their equivalent resistances; then compare branch currents inside each block. Given Data / Assumptions: Block A: 470 Ω in parallel with 1.2 kΩ. Block B: three equal 3 kΩ resistors in parallel. Blocks A and B are connected in series to a 200 V source. Ideal components; DC operation. Concept / Approach: First compute each block’s equivalent resistance to find how the 200 V divides between blocks. Then, within each parallel block, branch current I_branch = V_block / R_branch. The highest current will be in the smallest-resistance branch that also sits across the larger block voltage. Step-by-Step Solution: Equivalent of Block A: R_A = (470 * 1200) / (470 + 1200) ≈ 337.8 Ω.Equivalent of Block B: three 3 kΩ in parallel → R_B = 3000 / 3 = 1000 Ω.Voltage split (series): V_A = 200 * R_A / (R_A + R_B) ≈ 200 * 337.8 / 1337.8 ≈ 50.5 V; V_B ≈ 149.5 V.Branch currents: I_470 ≈ 50.5 / 470 ≈ 0.107 A; I_1.2k ≈ 50.5 / 1200 ≈ 0.042 A; I_3k (in Block B) ≈ 149.5 / 3000 ≈ 0.050 A.Largest branch current is through the 470 Ω resistor. Verification / Alternative check: Sanity check: In any given parallel pair at the same voltage, the lower resistance must draw more current. Even after voltage division, the computed numbers confirm the 470 Ω branch dominates. Why Other Options Are Wrong: 1.2 kΩ: Higher resistance in the same parallel block; therefore less current. 3 kΩ: Each 3 kΩ branch carries less current than the 470 Ω branch at their respective block voltages. '470 Ω or 1.2 kΩ': Ambiguous; only 470 Ω is correct. Common Pitfalls: Comparing resistances without accounting for block voltage division. Confusing series and parallel roles when simplifying the network. Final Answer: 470 Ω

Question 4

Balanced Wheatstone bridge computation: In a balanced bridge, RV = 3,500 Ω, R2 = 200 Ω, and R3 = 680 Ω. What is the value of the unknown resistor RUNK that satisfies the balance condition?

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Introduction / Context: The Wheatstone bridge is a precision measurement circuit. Under balance (zero detector current), the ratio of resistances in one leg equals the ratio in the opposite leg. Using this condition, we can compute the unknown resistance from the three known arms without needing the supply voltage or detector characteristics. Given Data / Assumptions: Bridge is balanced (detector at null). RV = 3,500 Ω (variable arm set to this value). R2 = 200 Ω, R3 = 680 Ω. RUNK is the unknown arm opposite RV in the standard ratio relation. Concept / Approach: For a balanced Wheatstone bridge: R_unknown / R2 = RV / R3 (equivalently R_unknown = R2 * RV / R3), depending on arm labeling. This relation arises from equal voltage division in both legs leading to zero potential difference across the detector. Step-by-Step Solution: Write the balance equation: RUNK = R2 * RV / R3.Substitute numbers: RUNK = 200 * 3500 / 680.Compute: 3500 / 680 ≈ 5.14706; multiply by 200 → ≈ 1029.41 Ω.Round appropriately: RUNK ≈ 1,029 Ω. Verification / Alternative check: Check the ratios: RV / R3 ≈ 3500 / 680 ≈ 5.147; RUNK / R2 ≈ 1029 / 200 ≈ 5.145. These are effectively equal, confirming bridge balance within rounding. Why Other Options Are Wrong: 680 Ω, 200 Ω, 880 Ω: Do not satisfy the required ratio equality when paired with the given arms. Common Pitfalls: Missetting the ratio (e.g., inverting R2/R3). Rounding too early and losing precision in the final answer. Final Answer: 1,029 Ω

Question 5

Series–parallel reduction: Two 1.2 kΩ resistors in series form one branch, and this series branch is placed in parallel with a 3.3 kΩ resistor. What is the total equivalent resistance of the network?

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Introduction / Context: Combining series and parallel connections is common in practical circuits. You must correctly identify which resistors add and which combine by reciprocal addition. This exercise walks through a straightforward two-step reduction to arrive at a single equivalent resistance seen by a source. Given Data / Assumptions: Branch A: two resistors of 1.2 kΩ each in series. Branch B: one resistor of 3.3 kΩ. Branches A and B are connected in parallel. Ideal resistors; temperature effects neglected. Concept / Approach: First, add the series pair in Branch A: R_A = 1.2 kΩ + 1.2 kΩ = 2.4 kΩ. Then, compute the parallel combination with Branch B using R_total = (R_A * R_B) / (R_A + R_B). Convert to ohms if needed and simplify to a tidy decimal or rounded whole-number value. Step-by-Step Solution: Series sum: R_A = 2.4 kΩ.Parallel with 3.3 kΩ: R_total = (2.4 kΩ * 3.3 kΩ) / (2.4 kΩ + 3.3 kΩ).Compute numerator: 2.4 * 3.3 = 7.92 (kΩ^2); denominator: 5.7 kΩ.R_total = 7.92 / 5.7 kΩ ≈ 1.38947 kΩ ≈ 1,389 Ω. Verification / Alternative check: Quick bounds: The equivalent must be less than the smaller of 2.4 kΩ and 3.3 kΩ, hence less than 2.4 kΩ. Our result (≈1.389 kΩ) satisfies this and is reasonable. Why Other Options Are Wrong: 138 Ω: An order of magnitude too small; likely from misplacing the decimal. 5,700 Ω: That is the sum, not the parallel result. 880 Ω: Not produced by any correct combination of the given values. Common Pitfalls: Accidentally adding all three values as if all were in series. Forgetting that parallel equivalents are always less than the smallest branch resistance. Final Answer: 1,389 Ω

Question 6

Effect of loading on a voltage divider: The no-load output of a divider is 12 V. After connecting a load across the output, what happens to the output voltage?

Accepted Answer

Introduction / Context: Voltage dividers only hold their ideal output when the output node is unloaded or connected to an infinite resistance (open circuit). Any finite load forms a parallel path with one divider resistor, reducing effective resistance and thereby reducing the output voltage. Recognizing this behavior is fundamental when designing sensor interfaces and bias networks. Given Data / Assumptions: No-load (open-circuit) divider output: 12 V. A load is subsequently connected across the output. Passive resistive divider; no buffering amplifier. Concept / Approach: With a load present, the bottom leg of the divider becomes R_bottom || R_load, which is less than R_bottom alone. The voltage division ratio V_out/V_in = R_eq_bottom / (R_top + R_eq_bottom) decreases because R_eq_bottom decreases. Therefore, V_out must fall below the no-load value. Step-by-Step Solution: Start from V_out(no-load) = V_in * R_bottom / (R_top + R_bottom) = 12 V.Connect load: R_bottom → R_eq_bottom = R_bottom || R_load < R_bottom.Updated ratio: V_out(load) = V_in * R_eq_bottom / (R_top + R_eq_bottom), which is smaller.Hence, the output voltage decreases under load. Verification / Alternative check: A numeric example (e.g., R_top = R_bottom and R_load = 1 MΩ) shows a modest drop from 12 V to a slightly lower value; a smaller load resistance causes a more pronounced drop, confirming the general rule. Why Other Options Are Wrong: increases: Only an active circuit (e.g., amplifier) could raise the output beyond the divider's set ratio. remains the same: True only for infinite load (no load), which is not the case here. becomes zero: Would require a short circuit, not merely a finite load. Common Pitfalls: Forgetting that adding a load changes the divider ratio unless the source has zero output resistance and the load is infinite. Final Answer: decreases

Question 7

Simple 1:1 divider: A voltage divider uses two equal 68 kΩ resistors across a 24 V source. What is the nominal output voltage measured at the midpoint (no load)?

Accepted Answer

Introduction / Context: Equal-value voltage dividers are common for halving a supply, establishing bias points, or providing reference levels. With two identical resistors in series across a source, the midpoint sits at exactly half the supply under no-load conditions, assuming ideal components. Given Data / Assumptions: R_top = 68 kΩ, R_bottom = 68 kΩ, connected in series. Source voltage V_in = 24 V (DC). Output is at the junction (midpoint) with no load attached. Concept / Approach: For a two-resistor divider: V_out = V_in * R_bottom / (R_top + R_bottom). With equal values, R_bottom = R_top, the ratio is 1/2. Therefore, V_out equals half of V_in. No load is critical here; any finite load would reduce V_out below this ideal value unless the divider is buffered. Step-by-Step Solution: Substitute into divider formula: V_out = 24 * 68 kΩ / (68 kΩ + 68 kΩ).Compute denominator: 136 kΩ → ratio 68/136 = 1/2.So V_out = 24 * 1/2 = 12 V. Verification / Alternative check: Symmetry argument: Identical resistors share equal voltage. Each drops 12 V, placing the midpoint exactly at 12 V relative to ground. This aligns with KVL and proportional division concepts. Why Other Options Are Wrong: 24 V: That would be the full source at the top node, not the midpoint. 0 V: That is the bottom node reference, not the midpoint. 6 V: Would require unequal resistor values (e.g., a 1:3 ratio), which is not given. Common Pitfalls: Forgetting that the stated output is no-load; a load would alter the midpoint voltage. Final Answer: 12 V

Question 8

Analog meter sensitivity: For a 30,000 Ω/volt voltmeter set to its 50 V range, what is the instrument's internal resistance (use sensitivity × full-scale voltage)?

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Introduction / Context: Traditional analog voltmeters are often specified by their sensitivity in ohms per volt (Ω/V). This figure tells you how much internal resistance the meter presents for each volt of full-scale deflection. Understanding this is essential when estimating meter loading effects on the circuit under test. Given Data / Assumptions: Sensitivity (meter constant): 30,000 Ω/V. Selected meter range: 50 V. We assume an ideal analog meter movement where R_internal = sensitivity × full-scale voltage. Concept / Approach: The internal resistance of an analog voltmeter on a given range equals the product of its Ω/V sensitivity and the full-scale range setting. This arises because the range switch inserts series resistance so that the meter movement sees its required current at full scale while presenting the specified resistance to the circuit. Step-by-Step Solution: Use the relation: R_internal = (Ω/V rating) * (V_range).Substitute values: R_internal = 30,000 Ω/V * 50 V.Compute: R_internal = 1,500,000 Ω = 1.5 MΩ. Verification / Alternative check: If the meter were set to a different range, internal resistance would scale proportionally. For example, at 10 V it would be 300 kΩ; at 100 V it would be 3 MΩ. This linear scaling confirms the formula's consistency. Why Other Options Are Wrong: 15,000 Ω and 150,000 Ω are smaller by factors of 100 and 10 respectively. 15,000,000 Ω corresponds to 300,000 Ω/V on 50 V—far above the stated sensitivity. Common Pitfalls: Confusing the Ω/V rating with the total resistance; forgetting to multiply by the selected range; misreading units (kΩ vs MΩ). Final Answer: 1,500,000 Ω

Question 9

Current sharing in equal parallel branches: A 470 Ω ‖ 1.5 kΩ network is in series with a parallel group of five 1 kΩ resistors across a 50 V source. What percentage of the total load current flows through any single 1 kΩ branch?

Accepted Answer

Introduction / Context: In circuits combining series and parallel sections, currents in equal-value parallel branches divide uniformly. This question checks whether you can identify the current through one branch of an equal-parallel group that sits in series with another subnetwork. Given Data / Assumptions: First block: 470 Ω in parallel with 1.5 kΩ. Second block: five 1 kΩ resistors in parallel (equal values). Blocks are in series; source = 50 V. We seek the fraction of current through one 1 kΩ branch relative to the total load current entering the five-branch block. Concept / Approach: In a set of N identical resistors in parallel, the incoming current splits equally: each branch carries 1/N of the block current. Since these five 1 kΩ resistors are identical, each one conducts 1/5 of the current through that parallel group. The series part of the network does not change this equal division within the parallel group. Step-by-Step Solution: Number of equal branches N = 5.Equal-current division → branch current fraction = 1/N = 1/5.Convert to percent: (1/5) * 100% = 20%. Verification / Alternative check: You can compute the equivalent resistance of the five 1 kΩ in parallel (Req = 200 Ω) and confirm that currents into equal arms must be identical by symmetry; the exact magnitude is unnecessary to find the percentage split. Why Other Options Are Wrong: 25% corresponds to four equal branches, not five. 50% would be two branches. 100% would mean only one branch exists, which is false here. Common Pitfalls: Trying to include the series block's value in the division calculation; forgetting that equal-value parallel branches share current equally regardless of other series elements. Final Answer: 20%

Question 10

Three-tap divider with equal resistors: A 6 V battery feeds three 2.2 kΩ resistors in series to create two output taps. What are the two tap voltages measured from the negative end (assume no load)?

Accepted Answer

Introduction / Context: Voltage dividers distribute a source voltage across series resistors in proportion to their resistances. With equal-value resistors and no load, the drops are equal, yielding equally spaced output taps. This is a staple concept for bias networks and reference ladders. Given Data / Assumptions: Source: 6 V DC. Three equal resistors: each 2.2 kΩ in series. Outputs: the intermediate node after the first resistor and the node after the second resistor (both referenced to the negative end). No load (open-circuit taps), so divider ratios are ideal. Concept / Approach: Total resistance divides the source proportionally. With equal parts, each section drops V_total/3. Thus the first tap is one section above ground, the second tap is two sections above ground, and the third node is the full source voltage at the top of the ladder. Step-by-Step Solution: Equal segment drop: ΔV = 6 V / 3 = 2 V per resistor.Tap 1 (after first resistor): 2 V.Tap 2 (after second resistor): 2 V + 2 V = 4 V. Verification / Alternative check: KCL/KVL analysis yields the same since the current is uniform and no load draws additional current. Measured voltages stack linearly in equal increments. Why Other Options Are Wrong: 2 V, 2 V ignores the cumulative second drop. 2 V, 6 V and 4 V, 6 V incorrectly include the full source at a tap rather than only at the top of the string. Common Pitfalls: Mixing up node numbering; assuming 6 V appears at an intermediate tap; forgetting that taps are cumulative sums of equal drops. Final Answer: 2 V, 4 V

Question 11

Mixed series–parallel network: A 12 kΩ, 15 kΩ, and 22 kΩ series string is in series with the parallel of two 10 kΩ resistors. With a 75 V source, what is the approximate current through the 15 kΩ resistor?

Accepted Answer

Introduction / Context: When multiple series resistors are combined with a parallel subnetwork in series, the same current passes through all series elements. Finding the current requires computing the overall series equivalent first, then applying Ohm’s law to the entire path. This problem reinforces correct series/parallel reduction and unit handling in kilohms. Given Data / Assumptions: Series resistors: 12 kΩ, 15 kΩ, 22 kΩ. Parallel pair: two 10 kΩ → equivalent 5 kΩ. Source voltage: 75 V. Ideal components, DC steady state. Concept / Approach: First reduce the parallel pair (10 kΩ ‖ 10 kΩ = 5 kΩ). Then add all series values to get total resistance. The current through each series element (including the 15 kΩ resistor) is I = V / R_total. No further division is needed because series currents are identical. Step-by-Step Solution: Parallel equivalent: R_p = (10k * 10k) / (10k + 10k) = 5 kΩ.Total series resistance: R_total = 12 kΩ + 15 kΩ + 22 kΩ + 5 kΩ = 54 kΩ.Current: I = 75 V / 54,000 Ω ≈ 0.0013889 A ≈ 1.39 mA ≈ 1.4 mA. Verification / Alternative check: Power sanity: P_total = V * I ≈ 75 * 0.00139 ≈ 0.104 W. Individual drops: V_15k ≈ I * 15k ≈ 20.8 V (consistent). Why Other Options Are Wrong: 14 mA and 50 mA are orders of magnitude too large; 5 mA requires a much smaller total resistance than present. Common Pitfalls: Forgetting to reduce the 10 kΩ pair to 5 kΩ; treating kΩ as Ω; assuming different currents in series elements. Final Answer: 1.4 mA

Question 12

Loading a two-resistor divider: A voltage divider uses two 12 kΩ resistors in series. Which load resistor connected across the lower 12 kΩ will disturb (change) the divider output the most?

Accepted Answer

Introduction / Context: Voltage dividers provide a fraction of a source voltage, but adding a load across the output forms a parallel combination with the lower leg, changing the effective divider ratio. The smaller the load, the stronger the loading effect and the bigger the output error. This is critical when designing reference taps for sensors or ADCs. Given Data / Assumptions: Divider resistors: 12 kΩ (upper) and 12 kΩ (lower). Load resistor connected across the lower 12 kΩ. We compare candidate load values for how much they alter the output. Concept / Approach: The equivalent lower leg becomes R_eq = (R_lower * R_load) / (R_lower + R_load). A smaller R_load yields a smaller R_eq, pulling the output down more and causing greater error. Therefore, among options, the smallest load value creates the largest disturbance. Step-by-Step Solution: Rank loads by size: 12 kΩ < 18 kΩ < 24 kΩ < 1 MΩ.Smallest load → strongest loading → maximum change in output.Thus, 12 kΩ has the most effect (halving the lower leg to 6 kΩ when paralleled with 12 kΩ). Verification / Alternative check: Compute quickly: 12 kΩ ‖ 12 kΩ = 6 kΩ (output drops to 1/3 of source instead of 1/2). For 24 kΩ, R_eq = 8 kΩ (output = 8/(12+8) = 0.4 of source), smaller disturbance than 1/3 vs 1/2 shift. Why Other Options Are Wrong: Higher load resistances draw less current, hence less loading. 1 MΩ is nearly open, minimally affecting the divider. Common Pitfalls: Assuming equal percent change for any load; ignoring that the output node resistance is the parallel result of the lower leg and load. Final Answer: 12 kΩ

Question 13

Wheatstone bridge balance: Given R1 = 10 kΩ, R2 = 720 Ω, and R4 = 2.4 kΩ in a balanced bridge, compute the unknown resistance using the standard balance relationship.

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Introduction / Context: The Wheatstone bridge is a classic network for precise resistance measurement. At balance (null detector reads zero), the resistor ratios in opposite arms satisfy a fixed relationship, allowing the unknown to be computed from the known arms without measuring current directly. Given Data / Assumptions: A balanced bridge (null condition). R1 = 10 kΩ, R2 = 720 Ω, R4 = 2.4 kΩ. Unknown arm is the remaining resistor (call it Rx). Concept / Approach: At balance, the product of opposite arms is equal. With the common textbook labeling used here, the balance can be expressed as R1 * R2 = Rx * R4 (equivalent to the usual ratio form under appropriate arm assignment). Solving yields Rx = (R1 * R2) / R4. Step-by-Step Solution: Write balance: R1 * R2 = Rx * R4.Rearrange: Rx = (R1 * R2) / R4.Insert values: Rx = (10,000 Ω * 720 Ω) / 2,400 Ω.Compute: Rx = (7,200,000) / 2,400 = 3,000 Ω. Verification / Alternative check: Check with ratios: R1/R4 = R2/Rx → 10,000/2,400 = 720/3,000 → both ≈ 4.1667, confirming balance. Why Other Options Are Wrong: 24 Ω and 2.4 Ω are far too small; 300 Ω is off by a factor of 10. Only 3,000 Ω satisfies the balance relation with the given numbers. Common Pitfalls: Using the wrong pairings in the ratio; mixing kΩ and Ω units; forgetting that balance eliminates the need to know the supply voltage or detector resistance. Final Answer: 3,000 Ω

Question 14

Analog meter loading: For a 20,000 Ω/volt voltmeter set to its 5 V range, determine the internal resistance presented to the circuit.

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Introduction / Context: The Ω/volt rating indicates how much resistance a voltmeter exhibits per volt of full-scale range. This parameter matters because finite internal resistance can load the circuit and alter the measured voltage, especially in high-resistance networks. Given Data / Assumptions: Sensitivity: 20,000 Ω/V. Range: 5 V. Use R_internal = (Ω/V) × (range). Concept / Approach: On each higher voltage range, additional series resistance is introduced so that the meter movement sees the correct full-scale current. Therefore the presented resistance scales linearly with range setting, following the product rule above. Step-by-Step Solution: Compute: R_internal = 20,000 Ω/V * 5 V.Result: R_internal = 100,000 Ω (100 kΩ).Thus the meter loads the node with a 100 kΩ equivalent resistance on the 5 V range. Verification / Alternative check: On a 10 V range, the same instrument would show 200 kΩ; on 1 V, 20 kΩ. The proportionality confirms the logic. Why Other Options Are Wrong: 20,000 Ω corresponds to 1 V range, not 5 V. 200,000 Ω would be the 10 V range value. 1,000,000 Ω implies 200,000 Ω/V sensitivity, not 20,000 Ω/V. Common Pitfalls: Forgetting to multiply by the selected range; confusing Ω/V with kΩ/V; overlooking how meter loading changes with range. Final Answer: 100,000 Ω

Question 15

Total resistance of a mixed network: A 470 Ω ‖ 680 Ω combination is in series with the parallel combination of four 2 kΩ resistors. What is the total equivalent resistance?

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Introduction / Context: Combining series and parallel subnetworks requires careful stepwise reduction. Equivalent resistance calculations are fundamentals used in analyzing bias networks, filters, and power distribution paths. Accuracy in parallel math prevents order-of-magnitude errors. Given Data / Assumptions: First block: 470 Ω in parallel with 680 Ω. Second block: four 2 kΩ resistors in parallel. Blocks are connected in series. Ideal resistors and DC analysis. Concept / Approach: Reduce each parallel subnetwork independently using R_eq = (R1 * R2) / (R1 + R2) for two elements, and R_eq = R/N for N equal elements. Then add the two results because the reduced blocks are in series. Step-by-Step Solution: Two-resistor parallel: R_a = (470 * 680) / (470 + 680) Ω.Compute denominator and result: R_a ≈ 277.9 Ω (≈ 278 Ω).Four equal 2 kΩ in parallel: R_b = 2000 Ω / 4 = 500 Ω.Total series: R_total = R_a + R_b ≈ 277.9 + 500 ≈ 777.9 Ω ≈ 778 Ω. Verification / Alternative check: Check bounds: 470 ‖ 680 must be less than 470 Ω and greater than 0 Ω; 278 Ω satisfies this. Adding 500 Ω yields about 778 Ω, matching expectations. Why Other Options Are Wrong: 1,650 Ω and 1,078 Ω are too high; 77.8 Ω is off by a factor of 10 (likely a decimal slip in the parallel step). Common Pitfalls: Arithmetic slips in reciprocal sums; forgetting that equivalent resistance of parallel elements is less than the smallest branch. Final Answer: 778 Ω

Question 16

Two-resistor parallel solver: The total parallel resistance is 1,403 Ω and one branch is 2 kΩ. What is the value of the other resistor?

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Introduction / Context: Determining an unknown parallel branch from the known total and one branch is a common troubleshooting task. Using the reciprocal form of the parallel resistance formula avoids trial-and-error and gives a quick, exact value. Given Data / Assumptions: Total resistance, R_T = 1,403 Ω. Known branch, R1 = 2,000 Ω. Unknown branch, R2 = ? Only two branches are present. Concept / Approach: For two resistors in parallel, 1/R_T = 1/R1 + 1/R2. Solve for 1/R2 = 1/R_T − 1/R1, then invert to get R2. Keep units consistent in ohms to prevent scaling errors. Step-by-Step Solution: Compute reciprocals: 1/R_T ≈ 1/1403 ≈ 0.00071276.1/R1 = 1/2000 = 0.0005.Find 1/R2 = 0.00071276 − 0.0005 = 0.00021276.Invert: R2 ≈ 1 / 0.00021276 ≈ 4,699 Ω ≈ 4.7 kΩ. Verification / Alternative check: Cross-check with product-over-sum shortcut: R_T = (R1 * R2)/(R1 + R2). Insert R2 = 4.7 kΩ and verify ≈ 1,403 Ω. Why Other Options Are Wrong: 1,403 Ω is the total, not a branch. 2 kΩ is the known branch. 3,403 Ω does not satisfy the parallel equation with the given total. Common Pitfalls: Arithmetic errors in reciprocals; forgetting to invert at the end; mixing kΩ and Ω without conversion. Final Answer: 4.7 kΩ

Question 17

Effect of removing a load: In a simple two-resistor voltage divider, if the load resistance connected to the output is removed (opened), how does the current drawn from the source change?

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Introduction / Context: Voltage dividers are sensitive to loading. Adding a load draws extra current from the source; removing that load reduces the current. Understanding this helps in power budgeting and prevents overloading sources or causing unexpected voltage sag at the divider output. Given Data / Assumptions: A two-resistor divider connected to a source. A load resistor was attached to the output node and is then removed (open circuit). Resistors are ideal; DC conditions. Concept / Approach: Total current from the source equals the sum of the divider's standing current plus any load current. With the load connected, I_source = I_divider + I_load. Removing the load sets I_load to zero, so the source current must decrease to I_divider only. The divider's standing current remains, unless the entire divider is opened (which is not the case here). Step-by-Step Solution: Before removal: I_source,1 = V_source / (R_upper + (R_lower ‖ R_load)).After removal: I_source,2 = V_source / (R_upper + R_lower).Since (R_lower ‖ R_load) < R_lower, the denominator increases after removal, making I_source smaller. Verification / Alternative check: Consider limits: R_load → ∞ (removed) means no output current; only divider current flows. Conversely, a small R_load dramatically increases I_source. These edge cases confirm the trend. Why Other Options Are Wrong: Increases would require adding a load, not removing it. Remains the same is false unless the load current was negligible (not stated). Is cut off implies zero current, which would require opening the divider itself, not just removing the load. Common Pitfalls: Confusing divider current with load current; assuming removing the load opens the entire circuit; overlooking that the divider path still exists. Final Answer: decreases

Question 18

Series–parallel resistive network: Two 3.3 kΩ resistors are connected in series, and this series combination is placed in parallel with a 4.7 kΩ resistor. If the voltage across one 3.3 kΩ resistor is 12 V, what is the voltage across the 4.7 kΩ resis…

Accepted Answer

Introduction / Context: This problem checks understanding of how voltage behaves in series–parallel resistor networks. In DC circuits, series elements share the same current and their voltages add, while parallel elements share the same voltage across the branch terminals. Correctly identifying which components are series and which are parallel is the key to determining unknown voltages. Given Data / Assumptions: A series string of two 3.3 kΩ resistors. This series string is in parallel with a single 4.7 kΩ resistor. The measured voltage across one 3.3 kΩ resistor is 12 V. Ideal resistors, steady-state direct current (no internal source resistance considered). Concept / Approach: In series, voltages add: V_series = V_R1 + V_R2. In parallel, both branches share the same terminal-to-terminal voltage. Therefore, the total voltage across the series pair equals the voltage across the parallel 4.7 kΩ resistor. Step-by-Step Solution: Let each 3.3 kΩ resistor be R.Given V across one R is 12 V.Since the two R are in series with the same current, each drops 12 V (identical value), so V_series = 12 V + 12 V = 24 V.The 4.7 kΩ resistor is connected in parallel with this series string, so V_4.7k = V_series = 24 V. Verification / Alternative check: Ohm’s law on the series branch shows equal-resistance, equal-drop behavior: if current is I, then V_each = I * 3.3 kΩ. Identical resistors in series share equal drops; hence two such drops total 24 V, the same voltage appearing across any parallel branch. Why Other Options Are Wrong: 12 V or 6 V: These ignore that the 4.7 kΩ is across the entire two-resistor series branch, not across a single 3.3 kΩ element. 0 V: Would imply a short circuit across the branch, which is not stated. Common Pitfalls: Mistaking the 4.7 kΩ as being in series with a single 3.3 kΩ rather than in parallel with the pair. Not adding series drops before assigning the parallel voltage. Final Answer: 24 V

Question 19

Instrument loading in measurements: On which selectable voltage range will a voltmeter place the minimum load on the circuit under test (i.e., present the highest internal resistance)?

Accepted Answer

Introduction / Context: When measuring voltage, the voltmeter's internal resistance should be high to minimize loading and avoid altering the circuit's true operating point. Many meters (especially analog and some digital ranges) change their internal resistance when you change the range. Understanding which range yields the highest input resistance helps preserve measurement accuracy. Given Data / Assumptions: A voltmeter with selectable ranges: 1 V, 50 V, 500 V, and 1,000 V. Typical meter behavior: higher voltage ranges correspond to higher internal resistance (or higher ohms-per-volt rating for analog meters). Aim: minimize loading on the circuit under test. Concept / Approach: Circuit loading is minimized when the meter's input resistance is as large as possible compared to the source/network being measured. On most instruments, the highest range provides the highest input resistance, thus the smallest current draw from the circuit. Step-by-Step Solution: Recognize that loading error is inversely related to the voltmeter input resistance.Select the setting that maximizes input resistance.Among the provided choices, the 1,000 V range presents the largest internal resistance and therefore the minimum load. Verification / Alternative check: Analog meters: input resistance ≈ (ohms per volt) * range. Thus, higher range → higher resistance. Many modern DMMs keep roughly constant high input resistance on DC (e.g., 10 MΩ), but on selectable high-voltage ranges certain models effectively apply attenuators, still leading to equal or higher effective input resistance and the least loading at the highest range. Why Other Options Are Wrong: 1 V, 50 V, 500 V: Lower ranges present equal or lower input resistance relative to 1,000 V on many instruments, causing greater loading. Common Pitfalls: Assuming all meters have identical input resistance on every range. Confusing current range behavior (low shunt resistance) with voltage range behavior (high input resistance). Final Answer: 1,000 V

Question 20

Series of two parallel blocks with a 100 V source: The combination (6.8 kΩ ∥ 10 kΩ) is in series with (2.2 kΩ ∥ 1 kΩ). With 100 V across the entire network, which resistor(s) experience the greatest voltage drop?

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Introduction / Context: This question blends series–parallel reduction with voltage division. When parallel blocks are in series, the source voltage divides according to the equivalent resistances of the blocks. Within any one parallel block, each branch shares the same block voltage. We identify which component(s) see the largest voltage by first finding how the total voltage splits between blocks. Given Data / Assumptions: Block A: 6.8 kΩ in parallel with 10 kΩ. Block B: 2.2 kΩ in parallel with 1 kΩ. Blocks A and B are connected in series across 100 V. Ideal resistors, DC operation. Concept / Approach: Compute the equivalent resistance of each block; the larger equivalent takes a larger portion of the total voltage (voltage division). Then, inside each parallel block, all branch resistors see the same block voltage. The branch(es) in the higher-voltage block will therefore have the greatest drop. Step-by-Step Solution: R_A = (6.8k * 10k) / (6.8k + 10k) = 68k / 16.8 ≈ 4.0476 kΩ.R_B = (2.2k * 1k) / (2.2k + 1k) = 2.2k / 3.2 ≈ 0.6875 kΩ.Since R_A > R_B, Block A takes the larger share of the 100 V.Therefore, both resistors in Block A (6.8 kΩ and 10 kΩ) each see the larger block voltage and thus have the greatest voltage drop. Verification / Alternative check: Approximate split: V_A ≈ 100 * 4.0476 / (4.0476 + 0.6875) ≈ 85.5 V; V_B ≈ 14.5 V. Each branch inside A sees ≈ 85.5 V (larger than any drop in B), confirming the selection. Why Other Options Are Wrong: 6.8 kΩ (alone) or 2.2 kΩ or 2.2 kΩ and 1 kΩ: All branches in Block B see the smaller block voltage; none can exceed Block A’s drop. Common Pitfalls: Comparing individual branch resistances without accounting for the block-level voltage division first. Final Answer: 6.8 kΩ and 10 kΩ

Series-Parallel Circuits Questions