Question 1

Series network of discrete resistors: three resistors 120 Ω, 270 Ω, and 330 Ω are connected in series. What is the total equivalent resistance of the series combination?

Accepted Answer

Introduction / Context: In a series circuit, resistances add directly because the same current flows through each element and voltage drops sum. Recognizing this rule is foundational for quickly computing totals and distributing voltages with the voltage-divider principle. Given Data / Assumptions: R₁ = 120 Ω, R₂ = 270 Ω, R₃ = 330 Ω. Series connection (same current through all). We seek R_total. Concept / Approach: For series resistors, R_total = R₁ + R₂ + R₃. There is no averaging or product; simple arithmetic addition applies. The result must be greater than any individual resistor value. Step-by-Step Solution: Add values: 120 + 270 + 330 = 720 Ω.Therefore, R_total = 720 Ω. Verification / Alternative check: Since each term is positive, the total exceeds the largest single resistor (330 Ω), which is consistent with series addition. Why Other Options Are Wrong: Less than 120 Ω contradicts series addition. The average is not a series rule. 120 Ω is only one component, not the total. Common Pitfalls: Confusing series with parallel (where reciprocals add and the total is less than any individual branch); mis-adding values. Final Answer: 720 Ω

Question 2

Voltage division in series: a 5.6 kΩ resistor and a 4.7 kΩ resistor are in series. If the voltage across the 5.6 kΩ resistor is 10 V, what is the voltage across the 4.7 kΩ resistor (assume the same series current)?

Accepted Answer

Introduction / Context: In a series network, the same current flows through all resistors. Voltage drops are proportional to resistance values, a principle known as the voltage divider. If one drop is known, the other can be found using ratios without knowing the total source voltage explicitly. Given Data / Assumptions: Series resistors: R₁ = 5.6 kΩ, R₂ = 4.7 kΩ. Measured V₁ across 5.6 kΩ is 10 V. Ideal components and steady-state DC. Concept / Approach: In series: I is common. Therefore V ∝ R. The voltage across the second resistor is V₂ = V₁ * (R₂ / R₁). No need to know the source voltage; the ratio suffices. Step-by-Step Solution: Compute ratio R₂ / R₁ = 4.7 / 5.6.Multiply: V₂ = 10 * (4.7 / 5.6) ≈ 10 * 0.8392857 ≈ 8.392857 V.Rounded to two decimals: ≈ 8.39 V. Verification / Alternative check: If I is common, V₁/V₂ = R₁/R₂ = 5.6/4.7 ≈ 1.1915; with V₁ = 10 V, V₂ ≈ 10 / 1.1915 ≈ 8.39 V, confirming the result. Why Other Options Are Wrong: 10 V would imply equal resistances. 2.32 V is far too small for nearly equal-kilo-ohm values. 0 V is impossible with nonzero resistance and nonzero current. Common Pitfalls: Using R₁ + R₂ incorrectly; forgetting that the series current is the same and that voltages divide in proportion to resistances. Final Answer: 8.39 V

Question 3

Series-aiding sources with series resistors: Two 6 V batteries are connected in series aiding and applied across two 1.2 kΩ resistors connected in series. What is the current through each resistor (steady-state DC)?

Accepted Answer

Introduction / Context: Series circuits are foundational in basic electrical engineering. When sources are connected series aiding, their voltages algebraically add. In a pure series path, the same current flows through every element. This question reinforces applying Ohm's law to a composite of series sources and series resistors, and it checks that you remember current is identical through each series component. Given Data / Assumptions: Batteries: two sources of 6 V each, connected series aiding. Resistors: two resistors of 1.2 kΩ each, connected in series. Assume ideal components (no internal resistance), steady DC conditions. Concept / Approach: Combine series sources by addition and combine series resistors by summation. Then apply Ohm's law I = V / R for the total loop. In a series circuit, that same current passes through each resistor; therefore the computed loop current is the current through each resistor as well. Step-by-Step Solution: Total source voltage: V_total = 6 V + 6 V = 12 V.Total resistance: R_total = 1.2 kΩ + 1.2 kΩ = 2.4 kΩ = 2400 Ω.Loop current: I = V_total / R_total = 12 / 2400 A = 0.005 A = 5 mA.Since series current is the same everywhere, each resistor carries 5 mA. Verification / Alternative check: Voltage drops: V1 = I * 1.2 kΩ = 5 mA * 1200 Ω = 6 V; V2 = 6 V. Sum is 12 V, matching the source addition, so KVL is satisfied. Why Other Options Are Wrong: 10 mA: Would require R_total = 1.2 kΩ, not 2.4 kΩ. 0 A: Only true for an open circuit, which is not the case here. 2.5 mA: Implies an incorrect total resistance or voltage sum. Common Pitfalls: Forgetting that series aiding sources add and series resistors sum. Applying the current to only one resistor rather than the entire loop. Final Answer: 5 mA

Question 4

Series-resistance concept check: When a fourth resistor is connected in series with an existing string of three resistors, how does the total circuit resistance change?

Accepted Answer

Introduction / Context: Understanding how resistances combine is a core skill in circuit analysis. In series networks, resistances simply add. This conceptual item tests whether you recall that adding any positive resistance in series increases total resistance, thus reducing current for a fixed source voltage. Given Data / Assumptions: Three resistors are already connected in series. A fourth resistor (with positive resistance) is added in series. No specific values are given; we reason qualitatively. Concept / Approach: The equivalent resistance of series elements is R_eq = R1 + R2 + R3 (+ R4 if added). Adding a positive term to a sum must increase the total. The phrase “increases by one-fourth” would only apply to a special case (all four resistors equal), which is not guaranteed here. Step-by-Step Solution: Start with R_eq(initial) = R1 + R2 + R3.Add R4 in series: R_eq(final) = R1 + R2 + R3 + R4.Since R4 > 0, R_eq(final) > R_eq(initial).Therefore, the total resistance increases (strictly). Verification / Alternative check: Consider a numeric example: 10 Ω + 10 Ω + 10 Ω = 30 Ω. Add 5 Ω → 35 Ω, which is larger. With a fixed voltage V, the current I = V / R_eq decreases accordingly, confirming the qualitative conclusion. Why Other Options Are Wrong: Increases by one-fourth: Not necessarily; only true if R4 equals the average of the existing three in a specific configuration (e.g., all equal). Decreases: Impossible when adding a positive series resistance. Remains the same: Also impossible unless R4 = 0 Ω (not stated). Common Pitfalls: Confusing series with parallel rules (where adding a branch can reduce total resistance). Assuming equal values without explicit information. Final Answer: increases

Question 5

Equal resistors in series with a 12 V battery: Four identical resistors in series draw 13.63 mA. What is the resistance value of each individual resistor?

Accepted Answer

Introduction / Context: Determining component values from measured current and source voltage is a routine application of Ohm's law. With identical series resistors, the total resistance is an integer multiple of one resistor, simplifying calculations and offering a quick consistency check. Given Data / Assumptions: Battery voltage V = 12 V (DC). Total current I = 13.63 mA = 0.01363 A. Four equal-value resistors in series; call each R. Concept / Approach: Total resistance is R_total = 4R. Use R_total = V / I to find the combined resistance, then divide by 4 to obtain each resistor's value. Keep sufficient significant figures before rounding to the nearest offered option. Step-by-Step Solution: Compute total resistance: R_total = V / I = 12 / 0.01363 ≈ 880.26 Ω.Individual resistor: R = R_total / 4 ≈ 880.26 / 4 ≈ 220.07 Ω.Round to practical value: ≈ 220 Ω. Verification / Alternative check: Back-calculate current using 4 × 220 Ω = 880 Ω: I = 12 / 880 = 0.013636… A = 13.636 mA, which matches the given 13.63 mA within rounding. Power per resistor P = I^2 * R ≈ (0.01363)^2 * 220 ≈ 0.0409 W, safe for 1/8 W parts. Why Other Options Are Wrong: 22 Ω or 88 Ω: Would give total resistance far too small, producing much larger currents. 880 Ω: That is the total, not each resistor. Choosing this confuses per-element value with the overall series sum. Common Pitfalls: Forgetting to divide the total by the number of equal resistors. Rounding intermediate values too early, causing noticeable discrepancy. Final Answer: 220 Ω

Question 6

Mixed series string and current through one element: Resistors of 470 Ω, 680 Ω, 1 kΩ, and 1.2 kΩ are in series on a 20 V source. What is the approximate current through the 680 Ω resistor?

Accepted Answer

Introduction / Context: In a DC series circuit, the current is identical through every element. Therefore, to find the current through any one resistor, we can compute the total series resistance and apply Ohm's law once for the entire loop. This checks fluency with combining varied resistor values and converting the result to milliamperes. Given Data / Assumptions: Series resistors: 470 Ω, 680 Ω, 1 kΩ (1000 Ω), and 1.2 kΩ (1200 Ω). Source voltage: 20 V (DC). Ideal conditions (no internal source resistance). Concept / Approach: Total series resistance is the arithmetic sum. One loop current flows, so that same current passes through the 680 Ω resistor. Apply I = V / R_total and express the answer in mA for readability. Step-by-Step Solution: Sum resistances: R_total = 470 + 680 + 1000 + 1200 = 3350 Ω.Compute loop current: I = 20 / 3350 A ≈ 0.00597 A.Convert to milliamperes: 0.00597 A ≈ 5.97 mA ≈ 6 mA.Thus, current through the 680 Ω resistor is ≈ 6 mA (same as the loop current). Verification / Alternative check: Voltage on the 680 Ω element: V_680 = I * 680 ≈ 0.00597 * 680 ≈ 4.06 V. Summing drops across all four parts returns approximately 20 V, confirming KVL consistency within rounding. Why Other Options Are Wrong: 60 mA and 300 mA: Would require much smaller total resistance than 3.35 kΩ. 30 mA: Still too large; would imply R_total ≈ 667 Ω, not the case here. Common Pitfalls: Treating current through one series element as different from the loop current. Arithmetic mistakes when summing mixed kΩ and Ω values. Final Answer: 6 mA

Question 7

Uniform series current: Five resistors are connected in one series path. If 3 A enters the first resistor, how much current enters the second resistor in the same circuit?

Accepted Answer

Introduction / Context: A hallmark of series circuits is that the same current flows through each component because there is only one path for charge. This quick concept check ensures you can reason about current distribution without unnecessary calculations. Given Data / Assumptions: Five resistors connected strictly in series. Measured current entering the first resistor: 3 A. Ideal wires and elements; DC steady state. Concept / Approach: In series, current continuity applies: the current that enters one element must leave it and enter the next, as there is no branching. Therefore, the current into every resistor is identical to the loop current. Step-by-Step Solution: Recognize a single-path circuit: one loop implies identical current through all elements.Given I_loop = 3 A at the first element, so the current into the second element is also 3 A.No numeric computation is needed beyond this rule. Verification / Alternative check: Kirchhoff’s current law at the node between the first and second resistors: the current leaving the first equals the current entering the second, hence 3 A continues unchanged. Why Other Options Are Wrong: 1 A, 0.3 A: Would require a branch or time-varying storage element changing current, not present here. 4 A: Would violate current continuity in a single series path. Common Pitfalls: Confusing series circuits with parallel circuits where current divides. Assuming current diminishes across elements; voltage divides, not current, in series. Final Answer: 3 A

Question 8

Voltage division insight: A series circuit has resistors of 4.7 kΩ, 12 kΩ, and 2.2 kΩ. Which resistor experiences the greatest voltage drop when powered by a DC source?

Accepted Answer

Introduction / Context: In series networks, the supply voltage divides in proportion to resistance values. Identifying which component has the largest voltage drop helps with power rating choices and with understanding bias networks and sensor dividers. Given Data / Assumptions: Three resistors in series: 4.7 kΩ, 12 kΩ, and 2.2 kΩ. Same current flows through each resistor (series circuit). DC, linear resistor behavior assumed. Concept / Approach: The drop across each resistor is V_i = I * R_i. Since the current I is identical for all three, the voltage drop is directly proportional to the resistance value. The largest resistance therefore drops the largest voltage. Step-by-Step Solution: Compare magnitudes: 12 kΩ > 4.7 kΩ > 2.2 kΩ.Thus, V_12k = I * 12 kΩ is the greatest among the three.Therefore, the 12 kΩ resistor has the most voltage drop. Verification / Alternative check: Pick a notional current (e.g., 1 mA). Then drops would be: 2.2 V, 4.7 V, and 12 V respectively, confirming the ranking with any positive current value. Why Other Options Are Wrong: the 2.2 kΩ / the 4.7 kΩ: Smaller resistances produce proportionally smaller drops for the same current. impossible to determine: False; proportionality makes the determination straightforward without knowing the exact current. Common Pitfalls: Assuming voltage divides equally; that only occurs when resistors are equal. Mixing series and parallel intuitions; in series, current is the same and voltage divides. Final Answer: the 12 kΩ

Question 9

Find the missing series element: A series circuit has three resistors. Two resistors are 1.2 kΩ each, and the total resistance is 12 kΩ. What is the value of the third resistor?

Accepted Answer

Introduction / Context: Solving for an unknown in a series network is a direct application of the additive property of resistances. This is common when back-calculating a missing component from a measured total or when designing a target equivalent resistance from available parts. Given Data / Assumptions: Total series resistance: 12 kΩ. Two known resistors: 1.2 kΩ and 1.2 kΩ. All are simple linear resistors in series. Concept / Approach: For series resistors, R_total = R1 + R2 + R3. Rearrange to find the unknown: R3 = R_total − (R1 + R2). Keep units consistent, then convert back to a convenient expression (ohms or kilo-ohms) as needed. Step-by-Step Solution: Sum the knowns: R1 + R2 = 1.2 kΩ + 1.2 kΩ = 2.4 kΩ.Compute the unknown: R3 = 12 kΩ − 2.4 kΩ = 9.6 kΩ.Express in ohms if preferred: 9.6 kΩ = 9600 Ω. Verification / Alternative check: Add them back: 1.2 kΩ + 1.2 kΩ + 9.6 kΩ = 12.0 kΩ exactly. The arithmetic closes perfectly, confirming the result. Why Other Options Are Wrong: is 960 Ω or is 1.2 kΩ: Far too small; their sum with the two 1.2 kΩ parts would not reach 12 kΩ. cannot be determined: Incorrect; the series sum uniquely determines the missing value. Common Pitfalls: Mixing kilo-ohms and ohms without conversion, causing factor-of-1000 mistakes. Subtracting only one of the known resistors from the total instead of both. Final Answer: is 9,600 Ω

Question 10

Power sharing in equal series resistors: A circuit dissipates a total of 12 W across four equal-value resistors in series. How much power does each resistor dissipate?

Accepted Answer

Introduction / Context: Knowing how power divides among components helps you select appropriate wattage ratings and avoid overheating. For equal resistors in series, voltage divides equally, and with the same current through each, power splits evenly as well. This question makes you apply those ideas quantitatively to a total power figure. Given Data / Assumptions: Total power P_total = 12 W. Four resistors of equal resistance, connected in series. Steady DC, ideal components. Concept / Approach: In series, the same current flows in all resistors. For equal resistances, each drops the same voltage. Power in each is P_i = I^2 * R (same I and same R), hence equal for all. Therefore, each resistor gets an equal fraction of the total power: P_total / 4. Step-by-Step Solution: Compute per-resistor power: P_each = P_total / 4 = 12 W / 4 = 3 W.No additional calculation is necessary because resistances are equal.Thus, each resistor dissipates 3 W. Verification / Alternative check: Let each resistor be R and total be 4R. If source voltage is V, then I = V / (4R). Power in one: P1 = I^2 * R = (V^2 / (16R^2)) * R = V^2 / (16R). Total: P_total = 4 * P1 = V^2 / (4R). Solving gives P1 = P_total / 4, matching the simple division used above. Why Other Options Are Wrong: 12 W: That would award the whole power to one resistor. 48 W: Exceeds the total power; impossible. 8 W: Would imply unequal sharing, which contradicts equal resistances in series. Common Pitfalls: Forgetting that equal R in series split voltage equally, not current (current is already equal). Mixing series and parallel intuitions about power distribution. Final Answer: 3 W

Question 11

Effect of removing a series element: When one of three series resistors is removed and the circuit is reconnected across the same source, how does the current change?

Accepted Answer

Introduction / Context: Current in a circuit depends on total resistance for a fixed source voltage. Removing a series element reduces total resistance, which in turn increases current according to Ohm's law. This qualitative question probes your grasp of that inverse relationship without requiring specific numbers. Given Data / Assumptions: Original circuit: three resistors in series. Modified circuit: one of the three series resistors is removed; the remaining two are still in series with the same source. Source voltage remains unchanged; ideal conditions. Concept / Approach: Total resistance for three in series is R_total(3) = R1 + R2 + R3. After removing one, R_total(2) = R1 + R2 (or similar). Since R_total(2) < R_total(3), Ohm’s law dictates I_new = V / R_total(2) > V / R_total(3) = I_old. The exact percentage change depends on the actual values, not necessarily “one-third.” Step-by-Step Solution: Recognize that series removal lowers total resistance.With fixed V, current varies inversely with total resistance.Thus, the loop current must increase after removing one resistor. Verification / Alternative check: Numeric example: Let R1 = R2 = R3 = 100 Ω. Initially R_total = 300 Ω, I_old = V/300. After removing one, R_total = 200 Ω, I_new = V/200 = 1.5 * I_old. That is an increase, not necessarily “by one-third.” Why Other Options Are Wrong: Increases by one-third: The change depends on actual values; equal resistors give a 50% increase, not 33%. Decreases by one-third: Removing a series resistor cannot decrease current for a fixed source. Decreases by the amount of current through the removed resistor: Nonsense in a series path; current is the same through all elements and does not “subtract.” Common Pitfalls: Assuming a fixed percentage change without numeric context. Confusing current division (a parallel concept) with series behavior. Final Answer: increases

Question 12

Series-opposing sources: If a 24 V battery and a 6 V battery are connected in series but opposing each other, what is the resulting total voltage across the pair?

Accepted Answer

Introduction / Context: When multiple sources are placed in series, their voltages add algebraically, taking polarity into account. In a series-opposing configuration, one source subtracts from the other. This knowledge is important in battery pack design and when analyzing multi-source circuits for net driving voltage. Given Data / Assumptions: Source 1: 24 V. Source 2: 6 V. They are connected in series opposing (polarities opposed). Concept / Approach: For series-aiding, voltages add; for series-opposing, the net is the difference between magnitudes if they are aligned head-to-head. The effective polarity follows the larger-magnitude source. Therefore, compute V_net = 24 V − 6 V = 18 V in the direction of the 24 V source. Step-by-Step Solution: Identify opposing polarities from the description.Subtract magnitudes: V_net = 24 − 6 = 18 V.Conclude the net voltage equals 18 V, oriented with the stronger source. Verification / Alternative check: Draw a quick polarity sketch: placing the 6 V reversed relative to the 24 V shows two opposing EMFs. KVL around the loop yields the same algebraic difference of 18 V. Why Other Options Are Wrong: 30 V: That would be series aiding, not opposing. 24 V: Ignores the opposing 6 V source. 0 V: Only true if the magnitudes matched exactly (e.g., 12 V opposing 12 V). Common Pitfalls: Failing to account for polarity, assuming all series sources add. Sign mistakes when writing KVL, which can flip the result. Final Answer: 18 V

Question 13

Series combination calculation: eight resistors each of 5.6 kΩ are connected in series in a DC circuit. What is the total equivalent resistance in ohms (Ω)?

Accepted Answer

Introduction / Context: In basic circuit theory, resistors in series simply add. This question tests your ability to convert between kilo-ohms (kΩ) and ohms (Ω) and to compute the total resistance for a string of identical resistors connected end-to-end. Given Data / Assumptions: Number of resistors in series: 8. Each resistor value: 5.6 kΩ. Ideal components; no temperature or tolerance effects considered. Concept / Approach: For series resistors, the equivalent resistance is the arithmetic sum: R_total = R1 + R2 + … + Rn. Unit consistency matters; convert kΩ to Ω if the answers are expressed in Ω. Recall 1 kΩ = 1,000 Ω. Step-by-Step Solution: Convert a single 5.6 kΩ resistor: 5.6 kΩ = 5,600 Ω.Compute total for eight in series: R_total = 8 * 5,600 Ω = 44,800 Ω.Express result with thousands separator for clarity: 44,800 Ω. Verification / Alternative check: Sum directly in kΩ: 8 * 5.6 kΩ = 44.8 kΩ, which equals 44,800 Ω. Both routes agree exactly. Why Other Options Are Wrong: 8 Ω confuses multiplication by count with unit scaling. 5,600 Ω is only one resistor. 4,480 Ω is off by a factor of 10 (missing the kilo-to-ohm conversion). Common Pitfalls: Forgetting to multiply by the number of resistors; mixing kΩ with Ω and dropping the 10^3 factor. Final Answer: 44,800 Ω

Question 14

Power dissipation across a fixed 120 V source: which series combination of 220 Ω resistors (one, two, three, or four in series) dissipates the greatest total power?

Accepted Answer

Introduction / Context: Total power drawn from a constant-voltage source depends on the equivalent resistance. This question checks your understanding of how power changes when series resistance increases while the supply voltage is held constant. Given Data / Assumptions: Source voltage: 120 V (constant). Each resistor value: 220 Ω. Series connection only; no tolerance or heating effects considered. Concept / Approach: For a fixed voltage source, total power P_total equals V^2 / R_total. Therefore, smaller R_total yields larger P_total. In series, resistances add: R_total = n * 220 Ω, where n is the number of resistors in series. Step-by-Step Solution: Compute R_total for each choice: 1→220 Ω; 2→440 Ω; 3→660 Ω; 4→880 Ω.Use P_total = V^2 / R_total. The smallest R_total (220 Ω) gives the largest power.Hence, one 220 Ω resistor dissipates the most total power when connected across 120 V. Verification / Alternative check: Numerical example: P_1 = 120^2 / 220 ≈ 65.45 W; P_2 ≈ 32.73 W; P_3 ≈ 24.55 W; P_4 ≈ 16.36 W. The trend confirms the reasoning. Why Other Options Are Wrong: Adding more resistors increases R_total and therefore reduces total power drawn from the same source voltage. Common Pitfalls: Confusing constant-current with constant-voltage cases; incorrectly averaging resistances instead of summing for series. Final Answer: one 220 Ω resistor

Question 15

Voltage division percentage: in a 680 Ω total series circuit, what percentage of the total source voltage appears across a 47 Ω series resistor?

Accepted Answer

Introduction / Context: Voltage division states that in a series circuit the voltage across a resistor is proportional to its resistance. Expressing this as a percentage helps in quick mental checks and design estimates for dividers and sensor interfaces. Given Data / Assumptions: Total series resistance: 680 Ω. One resistor: 47 Ω in series. All components are ideal; DC steady state. Concept / Approach: Voltage across a series element: V_i = (R_i / R_total) * V_total. The percentage share of total voltage is simply 100 * R_i / R_total, independent of the actual source voltage value. Step-by-Step Solution: Compute fraction: 47 / 680 ≈ 0.0691176.Convert to percentage: 0.0691176 * 100 ≈ 6.91176%.Rounded to two decimals: ≈ 6.91%. Verification / Alternative check: Sanity check: 47 Ω is much smaller than 680 Ω, so the percentage should be well below 10%, matching 6.91%. Why Other Options Are Wrong: 68% and 69% would require a resistor close to two-thirds of the total. 47% incorrectly equates resistance value with percentage without dividing by the total. Common Pitfalls: Forgetting to divide by total series resistance; confusing absolute resistance with percentage of voltage. Final Answer: 6.91%

Question 16

Series opposing sources: two 1.5 V cells are connected in series opposition across two 100 Ω resistors in series. What is the total current in the circuit?

Accepted Answer

Introduction / Context: Battery sources can be connected in series aiding (voltages add) or series opposing (voltages subtract). Understanding net source voltage is essential before applying Ohm’s law to find current in a simple series network. Given Data / Assumptions: Two cells: each 1.5 V. Connection: series opposing (opposite polarity). Load: two 100 Ω resistors in series (total 200 Ω). Ideal sources and resistors (no internal resistance considered). Concept / Approach: In series opposition, the net voltage equals the difference of the two cell voltages. Since both are 1.5 V, the net source becomes zero volts. With zero applied voltage across a finite resistance, circuit current is zero amperes. Step-by-Step Solution: Compute net voltage: V_net = 1.5 V − 1.5 V = 0 V.Total resistance: R_total = 100 Ω + 100 Ω = 200 Ω.Apply Ohm’s law: I = V_net / R_total = 0 / 200 = 0 A. Verification / Alternative check: Kirchhoff’s voltage law perspective: opposing sources cancel, leaving no driving emf, hence no current in a closed loop of finite resistance. Why Other Options Are Wrong: Any nonzero current would require a nonzero net voltage or a short (zero resistance), neither is present per the data. Common Pitfalls: Accidentally adding the voltages as if they were aiding; forgetting that equal and opposite sources fully cancel. Final Answer: 0 A

Question 17

Fault diagnosis in a series chain: five resistors are in series across a 6 V battery. Measured voltage is zero across all resistors except R3. What voltage appears across R3?

Accepted Answer

Introduction / Context: In a healthy series circuit with current flowing, voltages divide among all resistors proportionally. If a meter shows zero volts across every element except one, this indicates a fault such as an open circuit. This question tests reasoning about abnormal measurements and where the full source voltage appears under an open condition. Given Data / Assumptions: Series chain: five resistors labeled R1–R5. Source: 6 V battery. Measurements: 0 V across all except R3; nonzero only across R3. Assume ideal meter (no loading). Concept / Approach: If a resistor is open, current in the series loop becomes zero. With zero current, there is no voltage drop across intact resistors (V = I * R = 0). The open element takes the entire source voltage because it is the interruption point separating the two battery terminals via the series chain. Step-by-Step Solution: Interpret zero drops elsewhere: I_series = 0 A (no current).Open suspected at R3; the potential difference from the left side of the string to the right must appear across that break.Therefore, V_R3 = full source voltage = 6 V. Verification / Alternative check: Kirchhoff’s laws: with I = 0, V across each intact resistor is 0; loop equation reduces to V_source = V_open. Only the open bears the entire 6 V. Why Other Options Are Wrong: 1.2 V or 0.6 V imply current and normal division; 0 V contradicts the observation that only across R3 a voltage is present. Common Pitfalls: Assuming nonzero current despite the readings; forgetting that an open in series halts current everywhere. Final Answer: 6 V

Question 18

Where to place an ammeter: to measure the current leaving the second resistor in a four-resistor series circuit, where can the ammeter be inserted without changing the reading?

Accepted Answer

Introduction / Context: In a pure series circuit, the same current flows through every element because there is only one path for charge flow. This concept underpins ammeter placement and ensures consistent current readings regardless of location along the loop. Given Data / Assumptions: Four resistors connected in series. Goal: measure current leaving the second resistor. Assume an ideal ammeter (very low internal resistance) placed in series. Concept / Approach: Series circuits have identical current at all points: I1 = I2 = I3 = I4. Hence, inserting an ammeter anywhere in series yields the same current measurement, provided the meter is correctly placed in series (not in parallel) and does not significantly alter the circuit. Step-by-Step Solution: Recognize the single-loop path: current is uniform along the series chain.Place the ammeter at any break in the series (e.g., at the source return, or between any pair of resistors).The reading will be identical regardless of specific insertion point. Verification / Alternative check: Kirchhoff’s current law: in a series loop with no branches, current continuity ensures equal current everywhere. Why Other Options Are Wrong: Specifying one segment suggests different currents in different segments, which is false for series circuits. All positions yield the same reading. Common Pitfalls: Placing an ammeter in parallel (shorting part of the circuit) or assuming current changes between series elements. Final Answer: at any point in the circuit

Question 19

Series aiding sources: if 6 V and 9 V supplies are connected in series aiding (same polarity), what is the total resulting voltage across the load?

Accepted Answer

Introduction / Context: Combining sources in series requires attention to polarity. With series aiding orientation, voltages add algebraically to increase the total emf. This principle is frequently used to obtain higher voltages from multiple cells or batteries. Given Data / Assumptions: Two sources: 6 V and 9 V. Connection: series aiding (positive to negative, same polarity aiding). Ideal sources with negligible internal resistance. Concept / Approach: For series aiding, V_total = V1 + V2. Ensure polarities are aligned so that the individual emfs reinforce rather than oppose each other. No further calculation is needed. Step-by-Step Solution: Compute: V_total = 6 V + 9 V = 15 V.State the result: 15 V across the load. Verification / Alternative check: Polarity diagram confirms both sources push current in the same direction, so their voltages add. Why Other Options Are Wrong: 6 V or 9 V ignores addition; 3 V would suggest opposing orientation with a 9−6 difference, not aiding. Common Pitfalls: Misreading orientation and subtracting instead of adding; overlooking internal resistance (not relevant here). Final Answer: 15 V

Question 20

Find current through a series string: a 12 V battery is connected across resistors of 68 Ω, 47 Ω, 220 Ω, and 33 Ω in series. What is the circuit current in milliamperes (mA)?

Accepted Answer

Introduction / Context: Determining series current requires computing the total equivalent resistance and then applying Ohm’s law. This is fundamental in sizing sources and predicting voltage drops for component networks in DC circuits. Given Data / Assumptions: Source: 12 V battery. Resistors in series: 68 Ω, 47 Ω, 220 Ω, 33 Ω. Ideal components; DC steady state. Concept / Approach: Total series resistance is the sum. Current I = V / R_total. Convert the final answer to mA for readability if needed (1 A = 1000 mA). Step-by-Step Solution: Sum resistances: R_total = 68 + 47 + 220 + 33 = 368 Ω.Apply Ohm’s law: I = 12 V / 368 Ω ≈ 0.0326087 A.Convert to mA: 0.0326087 A × 1000 ≈ 32.6 mA. Verification / Alternative check: Power sanity check: P = V * I ≈ 12 * 0.0326 ≈ 0.391 W total, plausible for small resistors. Why Other Options Are Wrong: 16.3 mA would require roughly double the resistance. 163 mA or 326 mA imply much smaller total resistance than 368 Ω. Common Pitfalls: Mistakes in arithmetic while summing resistors; forgetting to convert A to mA for the final expression. Final Answer: 32.6 mA

Series Circuits Questions