Question 1

Parallel lighting calculation: Five identical 200 W bulbs are connected in parallel to a 110 V supply. What is the approximate current through each bulb?

Accepted Answer

Introduction / Context: In parallel circuits, each branch receives the full supply voltage. For a rated appliance such as a light bulb, the current through that branch is determined by its power rating and the applied voltage. This is a standard power calculation useful for household and laboratory circuits. Given Data / Assumptions: Supply voltage V = 110 V (constant). Each bulb rated P = 200 W. Five bulbs are in parallel (but question asks per-bulb current). Assume bulbs operate at nominal ratings. Concept / Approach: Use P = V * I for each bulb. Since each bulb in parallel sees 110 V, the branch current is I = P / V. The number of bulbs does not change the current in each branch; it changes the total current drawn from the source. Step-by-Step Solution: Compute single-bulb current: I_bulb = 200 W / 110 V.I_bulb ≈ 1.818 A.Rounded to the nearest option: ≈ 1.8 A. Verification / Alternative check: Total current with five bulbs would be 5 * 1.818 A ≈ 9.09 A, which matches another option but refers to total, not per bulb. The question explicitly asks per-bulb current. Why Other Options Are Wrong: 2.2 A overestimates current. 137 mA is far too small for 200 W at 110 V. 9.09 A is the approximate total for five bulbs, not per bulb. Common Pitfalls: Confusing branch (per bulb) current with total source current in parallel networks. Final Answer: 1.8 A

Question 2

Reasoning about parallel networks: Six resistors are connected in parallel. The two smallest values are both 1.2 kΩ. What can you conclude about the total equivalent resistance?

Accepted Answer

Introduction / Context: Understanding limits on equivalent resistance helps you quickly sanity-check parallel networks without full calculation. With multiple branches, the equivalent is strongly influenced by the smallest resistances because conductances add in parallel. Given Data / Assumptions: Six resistors in parallel (values unspecified except two are 1.2 kΩ). All branches connect across the same voltage. Ideal components (no parasitics) for reasoning. Concept / Approach: For any parallel network, R_eq is less than or equal to the smallest branch resistance, and adding any additional finite branch strictly decreases R_eq. Two equal resistors R in parallel give R/2. Additional branches further decrease R_eq. Step-by-Step Solution: Equivalent of the two 1.2 kΩ branches alone: R_2 = 1.2 kΩ / 2 = 600 Ω.Adding four more branches in parallel increases total conductance.Therefore, R_total must be less than 600 Ω. Verification / Alternative check: Even if the remaining four resistors were very large (but finite), they would still reduce the equivalent below 600 Ω, though perhaps only slightly. Why Other Options Are Wrong: 'is less than 1.2 kΩ' is true but not the strongest conclusion. 'is greater than 1.2 kΩ' and 'equals 1.2 kΩ' contradict parallel rules. 'is less than 6 kΩ' is trivially true and uninformative compared with the tighter bound 600 Ω. Common Pitfalls: Adding resistances instead of conductances; forgetting that even very large added resistances still reduce R_eq (though by a small amount). Final Answer: is less than 600 Ω

Question 3

Branch current completion using KCL: A total of 120 mA enters a three-branch parallel network. Two branch currents are 40 mA and 10 mA. What is the third branch current?

Accepted Answer

Introduction / Context: Kirchhoff’s Current Law (KCL) enables quick determination of unknown branch currents when the total entering or leaving a node and other branch currents are known. This is a common troubleshooting step in parallel circuits. Given Data / Assumptions: Total incoming current I_total = 120 mA. Known branch currents: 40 mA and 10 mA (leaving the node). Steady-state; no charge storage at the node. Concept / Approach: By KCL, the sum of branch currents leaving the node must equal the current entering. Therefore, the third branch current equals the difference between total incoming current and the sum of the known branch currents. Step-by-Step Solution: Sum known leaves: 40 mA + 10 mA = 50 mA.Compute unknown: I_3 = I_total − 50 mA = 120 mA − 50 mA = 70 mA.Hence the third branch carries 70 mA. Verification / Alternative check: Check: 40 + 10 + 70 = 120 mA, which matches the total entering current. KCL satisfied. Why Other Options Are Wrong: 50 mA confuses sum with difference. 120 mA ignores existing branches. 40 mA duplicates a known branch, not the computed one. Common Pitfalls: Sign errors (entering vs. leaving), and forgetting to subtract all known branch currents. Final Answer: 70 mA

Question 4

Diagnosing a parallel network from measured total current: Three-branch circuit with branch currents R1 = 12 mA, R2 = 15 mA, R3 = 25 mA. If the measured total is only 27 mA, what conclusion can you draw?

Accepted Answer

Introduction / Context: Troubleshooting often involves comparing expected totals from branch data with measured totals. In a parallel circuit, the total current equals the sum of all branch currents. A discrepancy can immediately suggest an open branch or other fault. Given Data / Assumptions: Expected branch currents: 12 mA, 15 mA, 25 mA. Measured total current: 27 mA. Assume the given branch values represent the normal operating currents. Concept / Approach: Compute the expected total and compare to the measured value. If the measured value equals the total minus exactly one branch, that branch is likely open (carrying 0 mA). Step-by-Step Solution: Expected total: 12 + 15 + 25 = 52 mA.Difference: 52 mA − 27 mA = 25 mA.The missing 25 mA matches branch R3. Therefore, R3 is open (no current). Verification / Alternative check: If R3 were open, the total would be 12 + 15 = 27 mA, which matches the measured value. This confirms the diagnosis. Why Other Options Are Wrong: R1 open would give 15 + 25 = 40 mA. R2 open would give 12 + 25 = 37 mA. 'Operating properly' contradicts the mismatch. 'Multiple branches are shorted' would raise, not reduce, current. Common Pitfalls: Arithmetic slips adding branch currents; overlooking that an open branch contributes zero current. Final Answer: R3 is open

Question 5

Node inflow by KCL: In a three-branch parallel circuit, currents of 200 mA, 340 mA, and 700 mA are measured in the same direction into a junction. What is the current into the junction?

Accepted Answer

Introduction / Context: Kirchhoff’s Current Law (KCL) also applies when multiple branch currents converge into a node. The total current into the node is the sum of all branch currents directed into that node at that instant. Given Data / Assumptions: Three measured currents entering the node: 200 mA, 340 mA, 700 mA. All measurements are referenced with the same sign convention (into the node). Steady-state analysis. Concept / Approach: Add currents that share the same direction into the node to find the total inflow. This is a direct application of current conservation. Step-by-Step Solution: Sum: 200 mA + 340 mA + 700 mA = 1,240 mA.Convert to amperes: 1,240 mA = 1.24 A.Thus, the node inflow is 1.24 A. Verification / Alternative check: If the node had only 900 mA entering, the missing 340 mA would violate KCL unless there were an unaccounted source/sink, which the problem does not imply. Why Other Options Are Wrong: 900 mA and 540 mA are partial sums. 200 mA is just one branch. 0 A contradicts the measurements. Common Pitfalls: Forgetting unit conversion mA to A, or accidentally subtracting when all currents share the same direction. Final Answer: 1.24 A

Question 6

Parallel circuits — effect of an open branch on total resistance In a circuit containing multiple resistors in parallel, what happens to the overall (equivalent) resistance if one of the parallel resistors opens (becomes an open circuit)?

Accepted Answer

Introduction / Context: Understanding how faults affect network behavior is essential for troubleshooting. In parallel resistor networks, an “open” means one branch no longer conducts. Knowing how this changes the equivalent resistance helps predict current changes and diagnose failures. Given Data / Assumptions: Two or more resistors are connected in parallel. One resistor becomes open (its resistance effectively becomes infinite). Supply voltage remains unchanged. Concept / Approach: The equivalent resistance of parallel branches is determined by the sum of branch conductances: 1/Req = 1/R1 + 1/R2 + ... . Removing a branch by opening it eliminates its conductance contribution (1/Rbranch → 0). With fewer conducting paths, the total conductance decreases, so the equivalent resistance increases. Step-by-Step Solution: Start with: 1/Req = Σ(1/Ri) over all parallel branches.When a branch opens: its term 1/Ropen = 0 (since Ropen → ∞), reducing Σ(1/Ri).With smaller Σ(1/Ri), the reciprocal Req = 1 / Σ(1/Ri) becomes larger.Therefore, the total (equivalent) resistance increases when any conducting branch opens. Verification / Alternative check: Example: two 100 Ω resistors in parallel yield Req = (100*100)/(100+100) = 50 Ω. If one opens, only one 100 Ω path remains, so Req jumps from 50 Ω to 100 Ω, clearly an increase. Why Other Options Are Wrong: It halves: halving happens only under specific numeric changes (e.g., adding identical parallel branches), not when removing one. It remains the same: losing a path always changes the conductance sum. It decreases: adding branches decreases Req; removing branches increases Req. Common Pitfalls: Confusing series and parallel behavior; in series, opening a component breaks the circuit and current falls to zero, but in parallel, other paths still conduct, albeit with higher overall resistance. Final Answer: It increases.

Question 7

Quick parallel combination — product-over-sum shortcut Using the product-over-sum rule for two parallel resistances, what is the combined value of 150 Ω and 6800 Ω?

Accepted Answer

Introduction / Context: Technicians routinely combine two resistors in parallel using the product-over-sum shortcut. This fast mental or calculator method avoids repeatedly applying the reciprocal formula and is ideal for quick checks during design or troubleshooting. Given Data / Assumptions: Two resistors: R1 = 150 Ω and R2 = 6800 Ω. They are connected in parallel. We seek the equivalent resistance. Concept / Approach: For two parallel resistors, the equivalent resistance is given by either the reciprocal form or the product-over-sum shortcut: Req = (R1 * R2) / (R1 + R2). This works because 1/Req = 1/R1 + 1/R2 = (R1 + R2) / (R1 * R2). Step-by-Step Solution: Use product-over-sum: Req = (R1 * R2) / (R1 + R2).Compute product: R1 * R2 = 150 * 6800 = 1,020,000.Compute sum: R1 + R2 = 150 + 6800 = 6950.Divide: Req = 1,020,000 / 6950 ≈ 146.762... Ω.Round to one decimal place commonly used in options: 146.7 Ω. Verification / Alternative check: Check reasonableness: the equivalent of parallel resistors must be less than the smallest branch (150 Ω). Our result (~146.8 Ω) is slightly less than 150 Ω, which is consistent. Also, because 6800 Ω is much larger than 150 Ω, the equivalent should be only slightly below 150 Ω, matching the computed value. Why Other Options Are Wrong: 150: equals the smaller resistor and ignores the small parallel effect of 6800 Ω. 0.006: not a plausible resistance in this context; likely a units mix-up. 6800: equals the larger resistor; parallel cannot exceed the smaller branch. Common Pitfalls: Forgetting to divide by the sum after multiplying (using only the product). Rounding before the final division, which can skew the result. Final Answer: 146.7

Question 8

Kirchhoff’s Current Law (KCL) — node balance calculation If 550 mA of current leaves a node and one branch provides 250 mA entering that node, how much current must enter from the other branch to satisfy KCL?

Accepted Answer

Introduction / Context: Kirchhoff’s Current Law (KCL) is fundamental for analyzing circuits. It states that the algebraic sum of currents at a node is zero: total entering current equals total leaving current. Applying KCL quickly reveals unknown branch currents and validates measurements. Given Data / Assumptions: Total current leaving the node: 550 mA. Known entering current from one branch: 250 mA. All currents are steady-state direct currents; directions are defined as entering or leaving the node. Concept / Approach: KCL requires Σ I(in) = Σ I(out). If we know the total leaving current and part of the entering current, the remaining entering current is the difference needed to balance the equality. Step-by-Step Solution: Write KCL: I_in,total = I_out,total.Given: I_out,total = 550 mA.Known entering branch: 250 mA.Let the unknown entering current be Ix. Then 250 mA + Ix = 550 mA.Solve: Ix = 550 mA − 250 mA = 300 mA. Verification / Alternative check: Check balance: entering currents sum to 250 mA + 300 mA = 550 mA, which equals the leaving current; KCL satisfied. Why Other Options Are Wrong: 250 mA: would make total entering 500 mA, not equal to 550 mA. 550 mA: would make total entering 800 mA, violating KCL. 800 mA: would make entering far exceed leaving; incorrect. Common Pitfalls: Mixing signs and directions; always define “entering” and “leaving” clearly. Adding currents without respecting direction, which breaks KCL. Final Answer: 300 mA

Question 9

Parallel resistors – total equivalent resistance calculation: What is the total resistance of four identical resistors, each 1 kΩ, when they are connected in parallel?

Accepted Answer

Introduction / Context: Calculating the equivalent resistance of parallel-connected resistors is a foundational skill in circuit analysis. It is frequently used in designing bias networks, sensor interfaces, and power distribution where lower equivalent resistance or higher current capability is needed. Given Data / Assumptions: There are 4 identical resistors. Each resistor value R = 1 kΩ. Connection is purely parallel; leads and source resistances are neglected. Concept / Approach: For n identical resistors in parallel, the equivalent resistance is R_eq = R / n. More generally, for parallel branches the conductances add: 1 / R_eq = 1 / R_1 + 1 / R_2 + ... + 1 / R_n. Step-by-Step Solution: Identify n = 4, R = 1 kΩ.Use identical-parallel shortcut: R_eq = R / n.Compute: R_eq = 1 kΩ / 4 = 0.25 kΩ.Convert to ohms: 0.25 kΩ = 250 Ω. Verification / Alternative check: Using the full formula: 1/R_eq = 1/1000 + 1/1000 + 1/1000 + 1/1000 = 4/1000 = 0.004; hence R_eq = 1 / 0.004 = 250 Ω. Why Other Options Are Wrong: 200 Ω and 400 Ω: do not match R/4 for identical 1 kΩ branches. 4 kΩ or 1 kΩ: these are typical of series or single-branch values, not four-way parallel. Common Pitfalls: Mixing series and parallel rules, or forgetting that equivalent resistance in parallel must be less than the smallest branch resistance. Final Answer: 250 ohms

Question 10

Safe troubleshooting with meters: When using an ammeter or voltmeter to troubleshoot an unknown circuit, which range-selection procedure should you follow first?

Accepted Answer

Introduction / Context: Proper range selection prevents damage to test equipment and the circuit under test. This is especially important for analog meters and for digital meters with limited input protection at lower ranges. Given Data / Assumptions: The magnitude of the circuit quantity (voltage or current) is initially unknown. A standard handheld meter with selectable ranges is available. Goal is safety and reliability of measurement. Concept / Approach: Starting at the highest range protects the meter from overload. Once a safe reading is observed, step down ranges to improve resolution while ensuring the reading remains within limits. For current measurements, breaking the circuit and inserting the ammeter in series must be done carefully. Step-by-Step Solution: Set the meter to the highest appropriate measurement type and range.Connect probes correctly (parallel for voltage, series for current).If the reading is small relative to full scale, step down one range and observe again.Repeat until you obtain a readable value near mid-scale without exceeding limits. Verification / Alternative check: Compare manufacturer recommendations: they universally advise starting high to avoid input protection tripping or fuse blowing on ammeter inputs. Why Other Options Are Wrong: Short the leads and adjust: can damage the meter and offers no useful information. Check external power: many meters are battery powered; this is good maintenance but not the immediate range strategy. Start with the lowest scale: risks overload and meter damage. Mid-scale fixed: offers no protection if the value is very large. Common Pitfalls: Leaving the meter on the current jack and then measuring voltage; always verify lead placement and meter mode before probing. Final Answer: Start with the highest scale and adjust down to a lower scale.

Question 11

Stereo amplifier power question with incomplete load data: Channel 1 of a stereo amplifier shows 12 V at the speaker terminals. How much total power is the amplifier delivering to the speakers?

Accepted Answer

Introduction / Context: Power calculations in audio systems require both voltage and load characteristics. Without the speaker impedance and whether the stated voltage is RMS, peak, or peak-to-peak, any wattage result would be speculative. Given Data / Assumptions: Reported value: 12 V at the speaker terminals of Channel 1. No information provided on speaker impedance (e.g., 4 Ω, 8 Ω) or wiring (single speaker, multiple in parallel/series). Voltage type (RMS vs. peak vs. peak-to-peak) is unspecified. Concept / Approach: Electrical power for a resistive load is computed from any of these equivalent forms: P = V * I, P = V^2 / R, or P = I^2 * R. To use them, you must know at least two quantities among V, I, R and ensure consistent definitions (e.g., RMS values for AC power). Step-by-Step Solution: Recognize that P = V^2 / R requires the load resistance R.The current I is also unknown; thus P = V * I cannot be applied.Voltage type uncertainty (RMS vs. peak) further prevents a correct calculation.Hence, additional information (speaker impedance and whether 12 V is RMS) is mandatory. Verification / Alternative check: Illustration: If R = 8 Ω and V_RMS = 12 V, then P = 12^2 / 8 = 18 W. If R = 4 Ω, P = 36 W. This spread shows the answer depends on R. Why Other Options Are Wrong: 0 W: incorrect unless muted or open circuit; not stated. 18 W and 36 W: each corresponds to a specific assumed impedance (8 Ω or 4 Ω), which is not given. 1.5 W: arbitrary without supporting data. Common Pitfalls: Using peak or peak-to-peak voltages directly in P = V^2 / R as if they were RMS; this overestimates power by factors of 2 or 4. Final Answer: More information is needed to find the total power delivered to the speakers.

Question 12

Power in parallel branches: Two resistors are connected in parallel and dissipate 6 W and 10 W respectively. What is the total power loss in the network?

Accepted Answer

Introduction / Context: Power bookkeeping is essential for thermal design and source sizing. In any circuit, the total power delivered by the source equals the sum of the powers dissipated in all elements (ignoring storage in ideal reactive components at steady state). Given Data / Assumptions: Two parallel branches with measured or calculated dissipations: P1 = 6 W, P2 = 10 W. DC or steady-state AC with resistive loads so real power adds arithmetically. Concept / Approach: Conservation of energy requires that the source supplies the sum of branch powers. In parallel circuits, voltage across each branch is common, but power still adds linearly: P_total = Σ P_branch. Step-by-Step Solution: Write: P_total = P1 + P2.Substitute: P_total = 6 + 10.Compute: P_total = 16 W. Verification / Alternative check: If branch currents are I1 and I2 at common voltage V, then P_total = V * (I1 + I2) = V * I_total = V * I1 + V * I2 = P1 + P2, confirming additivity. Why Other Options Are Wrong: 3.75 W and 4 W: do not reflect power addition; possibly confused with equivalent resistance ideas. 60 W: gross overestimate without basis. 6 W: ignores the second branch. Common Pitfalls: Mixing current-division or resistance calculations with power addition; power does not divide inversely like resistance. Final Answer: 16 watts

Question 13

Understanding parallel resistive circuits: Which statement correctly describes a property of a parallel resistive network?

Accepted Answer

Introduction / Context: Recognizing the defining characteristics of parallel circuits helps in predicting behavior such as current sharing, voltage distribution, and fault tolerance in power and signal networks. Given Data / Assumptions: Purely resistive branches. Ideal connections without parasitic effects. Concept / Approach: In a parallel network, all branch elements share the same two nodes; therefore, the voltage across each branch is the same, and the total current is the sum of the branch currents. Multiple current paths exist between the nodes, which is the hallmark of a parallel connection. Step-by-Step Reasoning: Identify the defining feature: common node pair for all branches.Therefore, multiple current paths exist, and voltage is equal across branches.Select the option that states the existence of more than one current path. Verification / Alternative check: Apply Kirchhoff's Current Law at the node: I_total = Σ I_branch, confirming that currents split among several paths while node voltage remains common. Why Other Options Are Wrong: Voltage divides between branches: false for parallel; voltage is common. Division of voltage applies to series circuits. Total branch power exceeds source power: violates energy conservation; source power equals sum of branch powers. Total conductance less than smallest branch conductance: opposite of reality; conductances add, so total conductance is greater than any single branch conductance. Equivalent resistance greater than largest branch resistance: for parallel it is always less than the smallest branch resistance. Common Pitfalls: Confusing series and parallel rules (voltage division vs. current division) and misinterpreting conductance behavior. Final Answer: there is more than one current path between two points

Question 14

Parallel circuits as dividers: Besides being widely used in power distribution, a parallel circuit is also used as a practical divider for which electrical quantity?

Accepted Answer

Introduction / Context: Just as series networks are used for voltage division, parallel networks are the standard approach for splitting current among multiple branches. Understanding this guides design of sensing shunts, bias networks, and load sharing. Given Data / Assumptions: Resistive branches at steady state. Single source feeding multiple parallel branches. Concept / Approach: In a parallel circuit, the voltage across each branch is identical, but currents split according to branch conductances. The current divider rule describes the portion of total current in any branch. Step-by-Step Solution (Current Divider Rule): For two branches with resistances R1 and R2 at source voltage V:I_total = V * (1/R1 + 1/R2).Branch currents: I1 = V / R1 and I2 = V / R2.Equivalently: I1 = I_total * (R2 / (R1 + R2)) for two-branch case. Verification / Alternative check: Measure currents with an ammeter in each branch; their sum equals source current by Kirchhoff's Current Law, confirming current division. Why Other Options Are Wrong: Voltage: divided by series networks, not parallel. Conductance: it adds in parallel; we do not use a parallel circuit as a “conductance divider.” Power: total power is the sum of branch powers; “dividing power” is not a standard design objective. Frequency: passive DC resistive networks do not divide frequency. Common Pitfalls: Assuming voltage divides in parallel or misapplying the current divider formula by swapping resistances in the ratio. Final Answer: current

Question 15

Parallel resistors and total power: Two resistors of 2 kΩ and 1 kΩ are connected in parallel. The measured total current into the parallel network is IT = 3 mA. What is the total power dissipated by the parallel combination?

Accepted Answer

Introduction / Context: This problem tests core DC circuit analysis for parallel resistors: combining equivalent resistance, using the given total current, and computing total power. Such steps appear routinely in electronics troubleshooting and power budgeting. Given Data / Assumptions: R1 = 2 kΩ, R2 = 1 kΩ, connected in parallel. Total current into the parallel pair: IT = 3 mA. Use ideal components and steady-state DC. Concept / Approach: For a parallel pair, compute equivalent resistance Req, then use P_total = IT^2 * Req. Alternatively, find the common voltage V = IT * Req and compute P_total = V^2 / Req; both are equivalent and consistent with power conservation. Step-by-Step Solution: Compute Req = (R1 * R2) / (R1 + R2) = (2000 * 1000) / (2000 + 1000) = 2,000,000 / 3000 = 666.666… Ω.Compute power using total current: P_total = IT^2 * Req = (0.003)^2 * 666.666… = 9e-6 * 666.666… ≈ 0.006 W.Convert to milliwatts: 0.006 W = 6 mW. Verification / Alternative check: Find branch currents: I1 = V/2000, I2 = V/1000 with V = IT * Req = 0.003 * 666.666… ≈ 2 V. Then I1 ≈ 1 mA, I2 ≈ 2 mA, and P_total = V * IT = 2 V * 3 mA = 6 mW. Matches the result. Why Other Options Are Wrong: 6 µW and 36 µW: off by factors of 1000 due to unit mistakes. Common Pitfalls: Confusing series vs parallel formulas, or using P = I^2 * R with a branch current but the total resistance, which mixes totals and branches incorrectly. Final Answer: 6 mW

Question 16

Power in a parallel network: A 5 kΩ resistor and a 25 kΩ resistor are connected in parallel across a 10 V DC supply. What is the total power dissipated by the two-resistor network?

Accepted Answer

Introduction / Context: Evaluating power in parallel circuits is a staple skill for electronics. Here, we use voltage known across each parallel branch to compute total power cleanly using conductances or individual branch powers. Given Data / Assumptions: R1 = 5 kΩ, R2 = 25 kΩ in parallel. Supply voltage V = 10 V across both resistors. Ideal DC conditions. Concept / Approach: Total power for parallel resistors can be found via P_total = V^2 * (1/R1 + 1/R2). Since voltage is the same on each branch, this approach is direct and avoids finding branch currents separately. Step-by-Step Solution: Compute equivalent conductance: 1/R1 + 1/R2 = 1/5000 + 1/25000 = 0.0002 + 0.00004 = 0.00024 S.Compute total power: P_total = V^2 * (sum conductance) = 10^2 * 0.00024 = 100 * 0.00024 = 0.024 W.Convert to milliwatts: 0.024 W = 24 mW. Verification / Alternative check: Branch powers: P1 = V^2/R1 = 100/5000 = 0.02 W (20 mW); P2 = 100/25000 = 0.004 W (4 mW). Sum = 24 mW. Confirms the result. Why Other Options Are Wrong: 2.4 mW and 3.3 mW are too small; 33 mW is too large relative to the calculated conductance. Common Pitfalls: Using series formulas by mistake or forgetting that both branches see the full 10 V in parallel. Final Answer: 24 mW

Question 17

Parallel resistors with a fixed supply: Two resistors 1 kΩ and 2 kΩ are connected in parallel across a 12 V source. What is the current through the 2 kΩ branch?

Accepted Answer

Introduction / Context: Branch currents in parallel circuits are straightforward because each branch experiences the same applied voltage. This is a core concept in Ohm’s law applications and nodal analysis. Given Data / Assumptions: R_branch = 2 kΩ. Parallel connection to a 12 V DC source. Ideal components, steady state. Concept / Approach: For any branch in parallel, I_branch = V / R_branch. No need to compute the total equivalent resistance unless total current is required. Step-by-Step Solution: Use Ohm’s law on the 2 kΩ branch: I_2k = 12 V / 2000 Ω = 0.006 A.Convert to milliamps: 0.006 A = 6 mA. Verification / Alternative check: Compute the 1 kΩ branch current (12 mA) and note total current would be 18 mA. Power check: P_total = V * I_total = 12 V * 0.018 A = 0.216 W; equals P1 + P2, confirming consistency. Why Other Options Are Wrong: 4 mA and 8 mA are incorrect divisions of 12 V by 2 kΩ; 12 mA corresponds to the 1 kΩ branch, not the 2 kΩ branch. Common Pitfalls: Accidentally using series current rules or confusing branch labels, leading to swapping the 1 kΩ and 2 kΩ results. Final Answer: 6 mA

Question 18

Many-equal-resistor parallel network: Fifteen resistors, each of value 2 MΩ, are all connected in parallel. What is the total equivalent resistance RT of the combination?

Accepted Answer

Introduction / Context: Equal-value resistors in parallel reduce to a simple formula that is frequently used in bias networks and sensor arrays. This question reinforces the shortcut and careful unit handling from megaohms to kiloohms. Given Data / Assumptions: N = 15 identical resistors. Each resistor R = 2 MΩ. Ideal DC behavior. Concept / Approach: For N equal resistors in parallel, RT = R / N. This follows from adding conductances, since total conductance G_total = N * (1/R). Step-by-Step Solution: RT = R / N = 2 MΩ / 15.Compute numeric value: 2,000,000 Ω / 15 ≈ 133,333.33 Ω.Express in kiloohms: ≈ 133 kΩ. Verification / Alternative check: Check conductance: each is 0.5 µS (since 1 / 2 MΩ). Times 15 gives 7.5 µS. Equivalent resistance is 1 / 7.5 µS ≈ 133.3 kΩ. Matches the result. Why Other Options Are Wrong: 300 kΩ and 750 kΩ correspond to incorrect division or mixing units; 30 MΩ is larger than a single branch and cannot be the parallel result. Common Pitfalls: Forgetting that parallel decreases resistance; mis-converting megaohms to kiloohms leading to order-of-magnitude errors. Final Answer: 133 kΩ

Question 19

Kirchhoff’s Current Law (KCL) in parallel circuits: Which statement correctly expresses the junction rule for currents at a node?

Accepted Answer

Introduction / Context: Kirchhoff’s Current Law (KCL) is fundamental for analyzing parallel circuits, nodal voltages, and complex networks. It embodies charge conservation at a node in steady state. Given Data / Assumptions: Steady-state DC (no charge accumulating at the node). Ideal wires and components; currents are algebraic with sign conventions. Concept / Approach: KCL states that the algebraic sum of currents at a node is zero. Equivalently, the sum of currents entering a node equals the sum of currents leaving that node. This is independent of the number of branches or their resistances. Step-by-Step Reasoning: Define current directions with signs (into node positive, out negative).Apply conservation of charge: Σ I_into − Σ I_out = 0 ⇒ Σ I_into = Σ I_out.This directly matches the correct response. Verification / Alternative check: In nodal analysis, each node equation enforces KCL. Solutions that violate Σ I_into = Σ I_out indicate modeling or calculation errors. Why Other Options Are Wrong: Option (a) resembles Kirchhoff’s Voltage Law (KVL), not KCL. Option (b) is a property of parallel resistors, not a current law. Option (c) misstates the equality with a “difference.” Common Pitfalls: Mixing KCL and KVL statements or forgetting to keep a consistent current sign convention. Final Answer: sum of the total currents flowing out of a junction equals the sum of the total currents flowing into that junction

Question 20

Voltage distribution in parallel circuits: What statement best describes the voltage across any branch of a parallel circuit relative to the source?

Accepted Answer

Introduction / Context: In a parallel circuit, each branch is connected directly across the source. Recognizing that the same voltage appears across every branch is foundational for calculating branch currents and powers. Given Data / Assumptions: Ideal DC source and wiring. Multiple branches connected in parallel to the same two nodes. Concept / Approach: The defining property of parallel connection is common voltage. Regardless of branch resistance or conductance, the branch endpoints share the source nodes, so the branch voltage equals the source voltage. Step-by-Step Reasoning: Identify the two nodes of the source.Each branch connects between the same two nodes, thus V_branch = V_source.Branch currents then follow I_branch = V_source / R_branch, differing according to resistance. Verification / Alternative check: Using nodal analysis with the source set as the reference (or applied) potential, every branch sees the same node-to-node voltage difference. Why Other Options Are Wrong: (a) and (b) relate voltage to total current or total resistance, which is a series-circuit viewpoint. (d) describes series voltage division, not parallel circuits. Common Pitfalls: Confusing series and parallel behaviors, especially assuming voltage divides among branches in parallel (it does not). Final Answer: is equally applied to all branch conductances

Parallel Circuits Questions