Question 1

The total current in the given circuit equals _____.

Accepted Answer

170 mA

Question 2

Effect of adding more branches — as additional resistors are connected in parallel to an existing network, how does the total (equivalent) resistance of the circuit change?

Accepted Answer

Introduction / Context:Understanding how equivalent resistance changes with circuit topology is essential for sizing power supplies and estimating currents. This item examines the well-known behavior of parallel combinations when more branches are added. Given Data / Assumptions: One or more resistors are already in parallel. An additional resistor is connected in parallel across the same two nodes. Ideal lumped components and wiring. Concept / Approach:The reciprocal relation for parallel resistances is 1 / R_eq = 1 / R1 + 1 / R2 + ... Adding another positive term (1 / R_new) to the right-hand side increases the total conductance, which means R_eq (the reciprocal of conductance) must decrease. Step-by-Step Solution: 1) Start with 1 / R_eq(old) = 1 / R1 + 1 / R2 + ...2) Add a new branch: 1 / R_eq(new) = 1 / R_eq(old) + 1 / R_new.3) Since 1 / R_new is positive, the right side increases.4) Therefore R_eq(new) = 1 / (larger positive number) is strictly smaller than R_eq(old). Verification / Alternative check:For two equal resistors R in parallel, R_eq = R / 2; adding a third equal branch makes R_eq = R / 3 — a clear decrease with each added branch. Why Other Options Are Wrong: Increase: Opposite to the conductance sum rule. Increase by the value of the resistor being connected: That describes series, not parallel. Remain unchanged: Only true if the added branch has infinite resistance (i.e., not actually connected). Common Pitfalls:Confusing series (resistances add) with parallel (conductances add); forgetting that any finite added branch lowers R_eq. Final Answer:Total resistance decreases when more resistors are added in parallel.

Question 3

Aging effects — if the resistance value of one branch in a parallel network drifts with age, which electrical quantity in that branch changes as a direct consequence (assuming the supply voltage remains regulated)?

Accepted Answer

Introduction / Context:Real components age, and resistors can drift upward or downward in value. In a parallel network supplied by a regulated source, it is important to predict which quantities change when one branch’s resistance changes over time. Given Data / Assumptions: Parallel branches share the same two nodes (common voltage). Supply is ideal or well-regulated; bus voltage stays approximately constant. Only one resistor’s value drifts; others remain nominal. Concept / Approach:In parallel, the voltage across each branch equals the supply bus voltage. Ohm’s law in the drifting branch is I_branch = V_bus / R_branch. If R_branch changes while V_bus is held constant, then I_branch must change inversely with R_branch, while the branch voltage remains essentially the same as the bus voltage. Step-by-Step Solution: 1) Recognize V_branch = V_bus for all parallel branches.2) Write I_branch = V_bus / R_branch.3) If R_branch increases with age, I_branch decreases proportionally; if R_branch decreases, I_branch increases.4) V_branch remains at the bus value; other branches’ currents are unaffected except for small supply-regulation effects. Verification / Alternative check:You can model the network as a Norton equivalent seen by the branch. With a stiff source (low source resistance), branch voltage is fixed and branch current follows Ohm’s law with the changed resistance. Why Other Options Are Wrong: Branch voltage: In parallel with a regulated source, voltage is essentially constant. Both branch current and branch voltage: Voltage is stable; only current changes markedly. Power supply voltage: A regulated source maintains voltage independent of single-branch drift. Common Pitfalls:Assuming that any change in a component forces the supply to change voltage; with proper regulation, current changes while voltage stays fixed. Final Answer:Branch current changes when a parallel resistor’s value drifts (with the bus voltage held constant).

Question 4

Parallel circuits — a circuit is said to be parallel between two nodes if more than one independent current path exists between those two points (and all such branches share the same node-to-node voltage).

Accepted Answer

Introduction / Context:A “parallel” circuit describes how elements connect between the same pair of nodes. Recognizing parallel topology is foundational for applying Ohm’s law and Kirchhoff’s laws in analysis and design. Given Data / Assumptions: Two specific nodes exist in the circuit. We classify the connection as series or parallel based on current paths and node sharing. Ideal wires and components are assumed. Concept / Approach:Elements are in parallel if they connect across the same two nodes; this guarantees equal voltage across each branch while currents may differ. Multiple, independent paths imply branch currents can split and recombine according to branch impedances. Step-by-Step Solution: 1) Identify the two nodes of interest.2) Check if components’ terminals land on the same two nodes.3) If ≥2 distinct paths exist between the nodes, the connection is parallel.4) Conclude that branch voltages are equal, currents divide by conductance. Verification / Alternative check:Apply Kirchhoff’s Voltage Law around each branch: each branch spans the same node-to-node voltage; apply Kirchhoff’s Current Law at the node to confirm current splitting and recombination. Why Other Options Are Wrong: Voltage divides: that is a series property.Equal resistors: not required for parallel.Only one current path: defines series, not parallel.Source necessarily AC: topology is independent of source type. Common Pitfalls:Confusing equal voltage (parallel) with equal current (series); relying on physical drawing proximity instead of node identity. Final Answer:More than one current path exists between the two nodes

Question 5

Parallel network measurement — if an ideal voltmeter indicates 12 V across the source in a simple parallel circuit, what is the voltage across branch resistor R2?

Accepted Answer

Introduction / Context:In a parallel circuit all branches share the same two nodes, so the branch voltage equals the source (node-to-node) voltage. This property simplifies measurement and analysis. Given Data / Assumptions: Ideal voltmeter reads 12 V across the source terminals. R2 is connected in parallel across the same source nodes. Wiring and instruments are ideal (no wiring drop, no meter loading). Concept / Approach:By definition of parallel, the voltage across each branch element equals the node-to-node voltage supplied. Ohm’s law then determines each branch current independently from the common voltage and its resistance. Step-by-Step Solution: 1) Identify that R2 spans the same two nodes as the source.2) Therefore VR2 = VS by parallel definition.3) Given VS = 12 V, set VR2 = 12 V.4) Branch current can be computed separately: IR2 = 12 V / R2. Verification / Alternative check:Measure voltage across any other branch (e.g., R1). It will also read 12 V, confirming equal branch voltages in parallel. Why Other Options Are Wrong: VS − 12 V: nonsensical here because VS is 12 V.0 V: only if R2 were shorted out by nodes at the same potential, which contradicts parallel to source.More information / depends on R2: resistance value affects current, not the parallel branch voltage. Common Pitfalls:Transferring series “voltage division” intuition to parallel networks; misplacing meter leads to unintended nodes. Final Answer:12 V

Question 6

Current division insight — in a parallel circuit, the branch having the largest resistance will have which relative current compared with other branches?

Accepted Answer

Introduction / Context:Current division governs how source current splits among parallel branches. Since all branches share the same voltage, the branch current depends inversely on its resistance (Ohm’s law). Given Data / Assumptions: Ideal DC conditions, lumped components. Branches are purely resistive for this question. Each branch sees the same node-to-node voltage. Concept / Approach:Ohm’s law: I = V/R. With equal V across branches, higher R implies lower I. Thus, the largest resistance carries the smallest current. Voltage across each branch is equal and equals the source voltage. Step-by-Step Solution: 1) Express branch currents: I1 = V/R1, I2 = V/R2, …2) Compare for Rlargest: Ilargest = V/Rlargest is the smallest magnitude.3) Conclude: the branch with the largest resistance has the least current.4) Total current equals the sum of branch currents per Kirchhoff’s Current Law. Verification / Alternative check:Compute a quick example (V = 12 V; R = 6 kΩ vs 12 kΩ). Currents are 2 mA and 1 mA respectively, confirming the inverse relation. Why Other Options Are Wrong: Most current: occurs for the smallest resistance, not the largest.Least/Most voltage: all branches have the same voltage in parallel.Same current: only if all resistances are equal. Common Pitfalls:Confusing series/parallel attributes; thinking voltage divides in parallel (it does not). Final Answer:Least current

Question 7

Instrument placement — to measure a branch current, an ammeter is placed in series with that branch; to measure the total current drawn from the supply in a parallel circuit, the ammeter is placed in series with the source lead.

Accepted Answer

Introduction / Context:Ammeter insertion must not change the circuit’s node-to-node voltages significantly. Ideal ammeters have very low resistance and are inserted in the path of the current being measured. Given Data / Assumptions: Ideal ammeter (negligible internal resistance). Parallel circuit with multiple branches. Need to measure branch and total currents. Concept / Approach:Current is the same at every point in a series path; hence an ammeter must be placed in series with the intended path. For total current, insert the ammeter in series with one supply lead so all branch currents pass through it. Step-by-Step Solution: 1) Break the branch conductor and insert the ammeter in series to read branch current.2) Break the supply lead and insert the ammeter in series to read total current.3) Verify low burden so readings do not disturb the circuit. Verification / Alternative check:KCL confirms the source current equals the sum of branch currents; the series supply ammeter thus reads the total. Why Other Options Are Wrong: Parallel placements measure voltage (that is a voltmeter behavior), not current.Mixing series/parallel incorrectly yields meaningless or dangerous readings. Common Pitfalls:Placing ammeters across a source (short circuit risk); confusing series vs parallel measurement rules. Final Answer:Series, series

Question 8

Fault effect — in a parallel resistive network, opening (going open-circuit) one branch resistor will cause the total equivalent resistance seen by the source to _____ .

Accepted Answer

Introduction / Context:Understanding how faults affect equivalent resistance helps with diagnostics. In parallel networks, each branch adds conductance (1/R). Removing a branch by opening it reduces total conductance. Given Data / Assumptions: Purely resistive DC network. One branch goes open (infinite resistance). All other branches remain unchanged. Concept / Approach:Total conductance G_total = Σ(1/Ri). If one branch opens, its 1/R term becomes 0, so G_total decreases, meaning R_total = 1/G_total increases. Source current drops accordingly for a fixed source voltage. Step-by-Step Solution: 1) Express R_total via conductance sum.2) Remove one positive term from the sum due to open branch.3) Conclude G_total decreases → R_total increases.4) Expect lower total current and power draw. Verification / Alternative check:Try numbers: parallel of 6 kΩ and 12 kΩ is 4 kΩ; opening the 12 kΩ branch leaves 6 kΩ, which is larger than 4 kΩ. Why Other Options Are Wrong: Decrease/zero/negative: contradict conductance arithmetic and physical realizability.Remain same: only true if the opened branch carried negligible current (infinite R), which is the definition of already-open. Common Pitfalls:Thinking of voltage division (series behavior) instead of conductance addition in parallel. Final Answer:Increase

Question 9

Instrument placement (repeat) — to measure a branch current, an ammeter is placed in series with that branch; to measure the total current drawn from the supply, the ammeter is placed in series with the source lead.

Accepted Answer

Introduction / Context:This is the same principle as earlier: current measurement requires insertion into the current path with minimal burden. The ammeter must be in series with the path whose current is measured. Given Data / Assumptions: Ideal meter, negligible resistance. Parallel circuit drawing total current from a source. Concept / Approach:Series placement ensures the measured path’s current flows through the meter. For total current, place it in series with the source so every branch current sums through the meter. Step-by-Step Solution: 1) Break the branch lead → insert meter for branch current.2) Break the source lead → insert meter for total current.3) Confirm readings with KCL at the supply node. Verification / Alternative check:Sum of branch readings equals the total reading within meter tolerance. Why Other Options Are Wrong: Any “parallel” placement is a voltmeter-like connection and is incorrect for current. Common Pitfalls:Accidentally shorting a source by putting an ammeter across it. Final Answer:Series, series

Question 10

Repair for solvability — two resistors of 12 kΩ and 8 kΩ are connected in parallel. What is the total (equivalent) resistance of the circuit?

Accepted Answer

Introduction / Context:When resistors are in parallel, their equivalent resistance is less than the smallest branch resistance. Calculating with the reciprocal formula gives precise values used in design checks. Given Data / Assumptions: R1 = 12 kΩ, R2 = 8 kΩ, connected in parallel. Ideal resistors and wiring. Concept / Approach:Use the parallel formula: 1/Req = 1/R1 + 1/R2. For two branches it is often convenient to use the product-over-sum shortcut: Req = (R1 * R2) / (R1 + R2). Step-by-Step Solution: 1) Compute numerator: R1 * R2 = 12 kΩ * 8 kΩ = 96 (kΩ)^2.2) Compute denominator: R1 + R2 = 20 kΩ.3) Divide: Req = 96 / 20 = 4.8 kΩ.4) Sanity check: 4.8 kΩ is less than 8 kΩ, as required for parallel. Verification / Alternative check:Reciprocal method: 1/Req = 1/12 + 1/8 = 0.0833 + 0.125 = 0.2083 (1/kΩ) → Req ≈ 4.8 kΩ. Why Other Options Are Wrong: Greater than 8 kΩ / greater than 30 kΩ / 58 kΩ: contradict the rule that parallel equivalent is less than the smallest branch.2.0 kΩ: much too low for these values. Common Pitfalls:Adding resistances directly (series rule) instead of using reciprocal sum for parallel. Final Answer:4.8 kΩ

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