Q: Current divider principle: evaluate the statement that the most current flows in the parallel path which has the greatest resistance value.

Introduction / Context: The current divider rule is a cornerstone of basic circuit analysis. It predicts how a total current distributes between or among parallel impedances. The tested claim is the exact opposite of what the rule states, making this a good quick check of conceptual accuracy in DC resistive networks. Given Data / Assumptions: Parallel resistive branches excited by a source with fixed voltage. Linear time-invariant components. No reactive effects considered for this DC statement. Concept / Approach: For two branches R1 and R2, the current through R1 is I_total * (R2 / (R1 + R2)). This shows that as R1 increases, I_through_R1 decreases, because the numerator contains the other branch resistance. Extending to multiple branches, the path with the smallest resistance draws the largest share of current. Therefore, the statement claiming the most current in the greatest resistance path is incorrect. Step-by-Step Solution: 1) Write the current-divider expression for two branches. 2) Observe inverse proportionality between current and own resistance. 3) Generalize to many branches: the lowest resistance gets the highest current. Verification / Alternative check: Check with simple numbers: with V = 10 V, R_small = 1 ohm gives 10 A, R_large = 10 ohm gives 1 A. The smaller resistance indeed carries more current. Why Other Options Are Wrong: Correct: contradicts the divider rule. High frequency or inductive qualifiers do not reverse the inverse law for purely resistive DC networks. Common Pitfalls: Confusing series rules with parallel rules; misapplying proportional reasoning. Final Answer: Incorrect

Q: Equivalent resistance in parallel networks: is it accurate to say that the total resistance of a parallel circuit is always greater than the largest single resistor value present?

Introduction / Context: Determining equivalent resistance is central to predicting currents and power. In parallel networks, adding a branch creates an additional current path. This question tests the well-known inequality that the equivalent of parallels is less than the smallest branch resistance, not greater than the largest. Given Data / Assumptions: Two or more resistors in parallel. Linear ohmic behavior. Ideal measurements without instrument loading effects. Concept / Approach: For resistors R1, R2, ..., Rn in parallel, 1 / R_eq = sum(1 / R_k). Since reciprocals add, R_eq must be less than or equal to the smallest R_k, with equality only in the trivial single-branch case. Therefore it cannot be greater than the largest resistor; the statement is incorrect. Step-by-Step Solution: 1) Write the reciprocal formula for parallel combination. 2) Note that adding positive reciprocals increases the sum, so its reciprocal decreases. 3) Conclude R_eq is less than the least branch value. Verification / Alternative check: Check with R1 = 3 ohm and R2 = 6 ohm: R_eq = (3 * 6) / (3 + 6) = 2 ohm, which is less than 3 ohm (the least), and certainly not greater than 6 ohm (the largest). Why Other Options Are Wrong: Correct: contradicts the reciprocal rule. Only true for equal resistors: still false; two equal resistors in parallel produce half the value, which is less than either. Common Pitfalls: Applying series logic to parallel networks; forgetting the reciprocal relationship. Final Answer: Incorrect

Q: Effect of an open branch on parallel equivalent resistance: when one resistor in a multi-branch parallel circuit opens, does the overall equivalent resistance increase or decrease compared to the intact condition?

Introduction / Context: Reliability issues such as open-circuit faults occur in real systems. Understanding how the equivalent resistance responds is key for predicting source current and power. The tested statement says the total resistance decreases when a branch opens; we evaluate that claim for resistive parallel networks. Given Data / Assumptions: Two or more resistors in parallel; one branch becomes open (infinite resistance). Linear ohmic behavior; constant source voltage assumed. No compensating control action. Concept / Approach: In parallel, 1 / R_eq = sum(1 / R_k). Removing a branch sets its contribution 1 / R_open to zero. The sum of reciprocals becomes smaller, so its reciprocal R_eq becomes larger. Therefore opening a branch increases total resistance, not decreases. Consequently, source current drops. Step-by-Step Solution: 1) Start with the reciprocal form for R_eq. 2) Remove one reciprocal term to model an open branch. 3) Recognize that a smaller reciprocal sum yields a larger R_eq. Verification / Alternative check: Example: three equal 6 ohm branches have R_eq = 2 ohm. If one opens, the remaining two yield R_eq = 3 ohm. Resistance increased. Why Other Options Are Wrong: Correct: contradicts the derived result. Largest branch qualifier: the effect is independent of which branch opens; losing any path increases R_eq. No change: incorrect because the number of parallel paths influences R_eq. Common Pitfalls: Confusing voltage and current source behaviors; misreading increase vs decrease in equivalent resistance. Final Answer: Incorrect

Q: Benchmark rule for parallels: evaluate the statement that the equivalent resistance of any nontrivial parallel network is always less than the smallest individual resistor value present.

Introduction / Context: This benchmark rule is a quick mental check when simplifying circuits: the equivalent resistance of resistors in parallel should never exceed the smallest branch value. Recognizing this prevents calculation mistakes and helps validate calculator results before committing designs. Given Data / Assumptions: At least two resistors in parallel. Resistors are linear and time invariant. We ignore parasitic effects and measurement errors. Concept / Approach: For N parallel resistors, 1 / R_eq = sum(1 / R_k) with k from 1 to N. Because each term 1 / R_k is positive, the sum is greater than 1 / R_min, where R_min is the smallest resistor. Taking reciprocals, R_eq is less than R_min. This holds for any finite set of positive resistances and becomes a strict inequality when there are two or more branches. Step-by-Step Solution: 1) Identify R_min among branch values. 2) Note that sum(1 / R_k) ≥ 1 / R_min + another positive term for each additional branch. 3) Take the reciprocal to obtain R_eq < R_min for any network with at least two branches. Verification / Alternative check: Numerical sanity check: 6 ohm in parallel with 3 ohm gives R_eq = 2 ohm, which is less than 3 ohm, the smallest branch. Why Other Options Are Wrong: Incorrect: contradicts the reciprocal property. Only for two branches or small ratio spreads: the inequality is general for any count and ratios. Common Pitfalls: Applying series formula by mistake; forgetting that adding a parallel path lowers R_eq. Final Answer: Correct

Q: Parallel circuits — comparison with smallest branch: In any parallel network of resistors, is the equivalent resistance always greater than the lowest-value branch resistor, or is it smaller? Choose the correct statement.

Introduction / Context: A key hallmark of parallel connections is that the equivalent resistance is always less than the smallest individual branch resistance. This rule is frequently used to sanity-check calculations and quickly bound answers during design and exams. Given Data / Assumptions: Resistive network with two or more branches in parallel. All components are ideal resistors. No reactive effects or frequency dependence considered. Concept / Approach: The parallel identity is 1/R_eq = 1/R_1 + 1/R_2 + ... + 1/R_n. Since at least one positive conductance term equals 1/R_min and others are nonnegative, the total conductance exceeds 1/R_min, implying R_eq = 1/R_min. 3) With any additional branch, S > 1/R_min. 4) Therefore R_eq = 1/S <= R_min, and with more than one branch, R_eq < R_min strictly. Verification / Alternative check: Example: R_min = 100 Ω in parallel with anything else. Even with a very large resistor in parallel, the equivalent becomes slightly less than 100 Ω. This corroborates the general rule for parallel networks. Why Other Options Are Wrong: Correct: Incorrect choice because the statement itself is false. True only for two branches: It is false even for two branches. True only when resistors are equal: If equal, R_eq = R/n which is smaller than each branch. Depends on source voltage: R_eq is independent of source magnitude. Common Pitfalls: Comparing sums of resistances rather than conductances; applying series intuition to parallel networks; overlooking that R_eq is dominated by the lowest branch value downward. Final Answer: Incorrect

Q: Parallel circuits — voltage distribution: In an ideal parallel network fed by a single source, the voltage across each branch is the same as the source voltage. Decide whether this statement is correct.

Introduction / Context: One of the defining properties of parallel circuits is the equality of voltage across all branches that are directly connected to the same pair of nodes. This rule allows engineers to compute branch currents independently via Ohm’s law once the branch elements are known. Given Data / Assumptions: Ideal parallel connection between two nodes. Single source maintaining node-to-node potential. Negligible wiring and contact resistances for the basic model. Concept / Approach: By definition, components in parallel share the same pair of nodes, so the potential difference across each component equals the node-to-node voltage enforced by the source. Current splits according to branch impedances, but the voltage across each branch remains the same in the ideal case. Step-by-Step Solution: 1) Identify the two common nodes across which all branches connect. 2) The source fixes the node potential difference V. 3) Each branch, being across the same nodes, sees the same V. 4) Branch currents then follow I_k = V / R_k for resistive branches. Verification / Alternative check: Practical circuits with nonzero wiring resistance still approximate equal branch voltages closely when wiring drops are small relative to load drops. Measurements in lab confirm nearly equal branch voltages in well-wired setups. Why Other Options Are Wrong: Incorrect: Contradicts the definition of parallel connection. Only true for DC sources: Also holds for AC if components are directly in parallel. Only true if resistances are equal: Equality of voltage does not require equal resistances. Only true when wires have zero resistance: Small but finite wire drops are usually negligible; the ideal rule remains conceptually correct. Common Pitfalls: Confusing current sharing (which differs by branch impedance) with voltage sharing; assuming different branch voltages unless resistances match. In parallel, voltage is the same; current divides. Final Answer: Correct

Question 1

Series versus parallel path count How does a parallel circuit differ from a series circuit with respect to the number of available paths for current flow?

Accepted Answer

Introduction / Context: Distinguishing series and parallel arrangements is a cornerstone skill in circuit analysis and troubleshooting. Many real circuits mix both, so being able to identify how current flows is essential. Given Data / Assumptions: Ideal wires and sources; steady DC conditions. Series means components share the same current; parallel means components share the same voltage. Concept / Approach: In a series connection, the current has exactly one path and is the same through each component. In a parallel connection, multiple independent paths exist between the same two nodes; the current splits among branches based on their impedances while the voltage across each branch remains the same. Step-by-Step Solution: Define series: single current path through all components.Define parallel: multiple paths between common nodes; currents add at the node.Select the description that matches parallel: more than one path for current flow. Verification / Alternative check: Apply KCL at a junction in a parallel network: the sum of branch currents equals the source current, confirming multiple paths. Why Other Options Are Wrong: No path: describes an open circuit, not a parallel network. Fewer paths: relative wording is ambiguous and not a defining property. One path: defines a series circuit, not a parallel circuit. Common Pitfalls: Assuming current is the same in parallel branches; that is the series rule. In parallel, voltage is the same and current divides according to branch impedance. Final Answer: More than one path for current flow

Question 2

Compute the equivalent resistance for resistors in parallel What is the total resistance RT of 12 kΩ, 4 kΩ, and 3 kΩ connected in parallel?

Accepted Answer

Introduction / Context: Calculating equivalent resistance of parallel resistors is a routine but critical task in electronics. It determines load seen by sources, affects current sharing, and impacts power dissipation and noise performance in analog designs. Given Data / Assumptions: Three resistors: R1 = 12 kΩ, R2 = 4 kΩ, R3 = 3 kΩ. All three are connected in parallel across the same two nodes. DC conditions and ideal components. Concept / Approach: For resistors in parallel, conductances add: 1/RT = 1/R1 + 1/R2 + 1/R3. After summing the reciprocals, invert to get RT. Using kilohm units throughout avoids unit mistakes. Step-by-Step Solution: Compute reciprocals: 1/R1 = 1/12 kΩ = 0.08333 kΩ^-1.Compute 1/R2 = 1/4 kΩ = 0.25 kΩ^-1; 1/R3 = 1/3 kΩ ≈ 0.33333 kΩ^-1.Sum: 0.08333 + 0.25 + 0.33333 ≈ 0.66666 kΩ^-1.Invert: RT = 1 / 0.66666 ≈ 1.5 kΩ. Verification / Alternative check: Use pairwise combination: combine 3 kΩ || 6 kΩ (since 12 kΩ || 4 kΩ = 3 kΩ) gives 3 kΩ || 3 kΩ = 1.5 kΩ. Both methods agree. Why Other Options Are Wrong: 2 kΩ: corresponds to 1/RT = 0.5 kΩ^-1, which is smaller than the actual sum of conductances. 6.3 kΩ or 19 kΩ: larger than the smallest branch resistance; a parallel equivalent can never exceed the smallest resistor. Common Pitfalls: Failing to keep consistent units (ohms vs kilohms) and forgetting that the equivalent of parallel resistors is always less than the smallest individual resistor. Final Answer: 1.5 kΩ

Question 3

Parallel circuits and fault conditions: if one branch opens in a parallel network driven by an ideal voltage source, will the current in the remaining parallel paths change or remain the same (ignoring source internal resistance)?

Accepted Answer

Introduction / Context: Parallel circuits appear throughout electronics and power distribution. Understanding how branch currents respond when one branch opens is essential for diagnostics and reliability analysis. The statement under test claims that when one branch opens, the current in the remaining branches changes. We evaluate this under the standard ideal-source model used in circuit theory. Given Data / Assumptions: Ideal constant-voltage source feeds the parallel network. Branch resistances are linear and time-invariant. No significant source internal resistance unless explicitly mentioned. Concept / Approach: In a parallel network with an ideal voltage source, each branch current is I_k = V / R_k. If a different branch opens, the source voltage V applied to the remaining branches does not change, and their resistances do not change either. Therefore I_k for all remaining branches remains the same. Only the total supply current and equivalent resistance change. Step-by-Step Solution: 1) Express branch current: I_k = V / R_k under an ideal source. 2) Opening another branch removes only that path; V and R_k of surviving branches are unchanged. 3) Conclude remaining branch currents are unchanged; total current decreases because one path is gone. Verification / Alternative check: If non-ideal source resistance exists, V may sag slightly when load changes. The ideal model used in introductory analysis assumes V is fixed, justifying the conclusion. Why Other Options Are Wrong: Correct: contradicts ideal-source analysis. Depends on source internal resistance: outside the stated ideal assumption. Cannot be determined: sufficient information is present under standard assumptions. Common Pitfalls: Confusing total current with branch currents; assuming voltage sources act like current sources; overlooking internal resistance only when it is actually specified. Final Answer: Incorrect

Question 4

Parallel circuits and current sharing: assess the claim that the branch with the largest resistance in a parallel network carries the largest current when driven by the same voltage.

Accepted Answer

Introduction / Context: Current division in parallel networks determines how currents split among branches. A common misconception is that a larger resistance somehow draws more current. This question challenges that misconception and reinforces the proportional relationships that govern current sharing. Given Data / Assumptions: Ideal constant-voltage source supplying a set of resistors in parallel. Ohmic resistors with fixed values. Temperature and nonlinearity effects ignored. Concept / Approach: By Ohm law, I = V / R for each branch at the same voltage V. Current is inversely proportional to resistance. Therefore the branch with the smallest resistance carries the largest current, and the branch with the largest resistance carries the smallest current. The current-divider relationship formalizes this inverse proportionality. Step-by-Step Solution: 1) For each branch k, write I_k = V / R_k. 2) Compare two branches: if R_a > R_b, then I_a < I_b. 3) Conclude that the largest resistance cannot carry the largest current under a fixed applied voltage. Verification / Alternative check: Use the current-divider form for two branches: I_through_R1 = I_total * (R2 / (R1 + R2)). The numerator contains the other branch resistance, again showing inverse proportionality. Why Other Options Are Wrong: Correct: conflicts with inverse relation. Holds only for current sources: even with an ideal current source, branch currents still split inversely to resistance when branches are in parallel across the same voltage. Common Pitfalls: Assuming current spreads equally; mixing up series and parallel intuition; overlooking that voltage is common across branches in parallel. Final Answer: Incorrect

Question 5

Current divider principle: evaluate the statement that the most current flows in the parallel path which has the greatest resistance value.

Accepted Answer

Introduction / Context: The current divider rule is a cornerstone of basic circuit analysis. It predicts how a total current distributes between or among parallel impedances. The tested claim is the exact opposite of what the rule states, making this a good quick check of conceptual accuracy in DC resistive networks. Given Data / Assumptions: Parallel resistive branches excited by a source with fixed voltage. Linear time-invariant components. No reactive effects considered for this DC statement. Concept / Approach: For two branches R1 and R2, the current through R1 is I_total * (R2 / (R1 + R2)). This shows that as R1 increases, I_through_R1 decreases, because the numerator contains the other branch resistance. Extending to multiple branches, the path with the smallest resistance draws the largest share of current. Therefore, the statement claiming the most current in the greatest resistance path is incorrect. Step-by-Step Solution: 1) Write the current-divider expression for two branches. 2) Observe inverse proportionality between current and own resistance. 3) Generalize to many branches: the lowest resistance gets the highest current. Verification / Alternative check: Check with simple numbers: with V = 10 V, R_small = 1 ohm gives 10 A, R_large = 10 ohm gives 1 A. The smaller resistance indeed carries more current. Why Other Options Are Wrong: Correct: contradicts the divider rule. High frequency or inductive qualifiers do not reverse the inverse law for purely resistive DC networks. Common Pitfalls: Confusing series rules with parallel rules; misapplying proportional reasoning. Final Answer: Incorrect

Question 6

Equivalent resistance in parallel networks: is it accurate to say that the total resistance of a parallel circuit is always greater than the largest single resistor value present?

Accepted Answer

Introduction / Context: Determining equivalent resistance is central to predicting currents and power. In parallel networks, adding a branch creates an additional current path. This question tests the well-known inequality that the equivalent of parallels is less than the smallest branch resistance, not greater than the largest. Given Data / Assumptions: Two or more resistors in parallel. Linear ohmic behavior. Ideal measurements without instrument loading effects. Concept / Approach: For resistors R1, R2, ..., Rn in parallel, 1 / R_eq = sum(1 / R_k). Since reciprocals add, R_eq must be less than or equal to the smallest R_k, with equality only in the trivial single-branch case. Therefore it cannot be greater than the largest resistor; the statement is incorrect. Step-by-Step Solution: 1) Write the reciprocal formula for parallel combination. 2) Note that adding positive reciprocals increases the sum, so its reciprocal decreases. 3) Conclude R_eq is less than the least branch value. Verification / Alternative check: Check with R1 = 3 ohm and R2 = 6 ohm: R_eq = (3 * 6) / (3 + 6) = 2 ohm, which is less than 3 ohm (the least), and certainly not greater than 6 ohm (the largest). Why Other Options Are Wrong: Correct: contradicts the reciprocal rule. Only true for equal resistors: still false; two equal resistors in parallel produce half the value, which is less than either. Common Pitfalls: Applying series logic to parallel networks; forgetting the reciprocal relationship. Final Answer: Incorrect

Question 7

Effect of an open branch on parallel equivalent resistance: when one resistor in a multi-branch parallel circuit opens, does the overall equivalent resistance increase or decrease compared to the intact condition?

Accepted Answer

Introduction / Context: Reliability issues such as open-circuit faults occur in real systems. Understanding how the equivalent resistance responds is key for predicting source current and power. The tested statement says the total resistance decreases when a branch opens; we evaluate that claim for resistive parallel networks. Given Data / Assumptions: Two or more resistors in parallel; one branch becomes open (infinite resistance). Linear ohmic behavior; constant source voltage assumed. No compensating control action. Concept / Approach: In parallel, 1 / R_eq = sum(1 / R_k). Removing a branch sets its contribution 1 / R_open to zero. The sum of reciprocals becomes smaller, so its reciprocal R_eq becomes larger. Therefore opening a branch increases total resistance, not decreases. Consequently, source current drops. Step-by-Step Solution: 1) Start with the reciprocal form for R_eq. 2) Remove one reciprocal term to model an open branch. 3) Recognize that a smaller reciprocal sum yields a larger R_eq. Verification / Alternative check: Example: three equal 6 ohm branches have R_eq = 2 ohm. If one opens, the remaining two yield R_eq = 3 ohm. Resistance increased. Why Other Options Are Wrong: Correct: contradicts the derived result. Largest branch qualifier: the effect is independent of which branch opens; losing any path increases R_eq. No change: incorrect because the number of parallel paths influences R_eq. Common Pitfalls: Confusing voltage and current source behaviors; misreading increase vs decrease in equivalent resistance. Final Answer: Incorrect

Question 8

Benchmark rule for parallels: evaluate the statement that the equivalent resistance of any nontrivial parallel network is always less than the smallest individual resistor value present.

Accepted Answer

Introduction / Context: This benchmark rule is a quick mental check when simplifying circuits: the equivalent resistance of resistors in parallel should never exceed the smallest branch value. Recognizing this prevents calculation mistakes and helps validate calculator results before committing designs. Given Data / Assumptions: At least two resistors in parallel. Resistors are linear and time invariant. We ignore parasitic effects and measurement errors. Concept / Approach: For N parallel resistors, 1 / R_eq = sum(1 / R_k) with k from 1 to N. Because each term 1 / R_k is positive, the sum is greater than 1 / R_min, where R_min is the smallest resistor. Taking reciprocals, R_eq is less than R_min. This holds for any finite set of positive resistances and becomes a strict inequality when there are two or more branches. Step-by-Step Solution: 1) Identify R_min among branch values. 2) Note that sum(1 / R_k) ≥ 1 / R_min + another positive term for each additional branch. 3) Take the reciprocal to obtain R_eq < R_min for any network with at least two branches. Verification / Alternative check: Numerical sanity check: 6 ohm in parallel with 3 ohm gives R_eq = 2 ohm, which is less than 3 ohm, the smallest branch. Why Other Options Are Wrong: Incorrect: contradicts the reciprocal property. Only for two branches or small ratio spreads: the inequality is general for any count and ratios. Common Pitfalls: Applying series formula by mistake; forgetting that adding a parallel path lowers R_eq. Final Answer: Correct

Question 9

Parallel circuits — branch opens: In a parallel network, if one branch becomes open (i.e., the component disconnects so no current flows through that branch), does the total circuit resistance decrease or increase? Give the correct statement about t…

Accepted Answer

Introduction / Context: In basic electronics, parallel circuits provide multiple paths for current. Understanding how opening or removing a branch affects the overall equivalent resistance is essential for troubleshooting, reliability analysis, and quick mental calculations in exam and lab settings. This item tests the conceptual link between the number of available current paths and the resulting total resistance seen by the source in a parallel network. Given Data / Assumptions: Network type: resistors connected in parallel. Event: one branch opens (no current can flow in that branch). Ideal conditions: ideal source and components; wiring resistance and parasitics neglected. Concept / Approach: For parallel resistors, 1 / R_eq = 1 / R_1 + 1 / R_2 + ... + 1 / R_n. Each conducting branch contributes a positive term to the total conductance. Removing (opening) one branch removes its conductance term, which reduces the total conductance and therefore increases the equivalent resistance. A decrease in available current paths never decreases the equivalent resistance of a parallel group. Step-by-Step Solution: 1) Start with R_eq formula for parallel: 1/R_eq = sum of individual conductances. 2) Opening a branch makes its conductance term zero (branch carries no current). 3) The sum of conductances decreases. 4) Therefore R_eq increases, not decreases. Verification / Alternative check: Consider two equal resistors R in parallel: R_eq = R/2. If one branch opens, you are left with R only, which is larger than R/2. This confirms that opening a branch raises R_eq relative to the previous value. Why Other Options Are Wrong: Correct: Incorrect choice because R_eq actually increases, not decreases. Only true when the source is AC: Source type does not change the parallel resistance identity. True only if the remaining branches are identical: Identity of remaining branches does not reverse the rule. Cannot be determined: It can be determined from the parallel formula alone. Common Pitfalls: Confusing series and parallel behavior; assuming fewer components always means less resistance; forgetting that parallel adds conductances, not resistances. Losing one path removes conductance and increases R_eq. Final Answer: Incorrect

Question 10

Parallel circuits — comparison with smallest branch: In any parallel network of resistors, is the equivalent resistance always greater than the lowest-value branch resistor, or is it smaller? Choose the correct statement.

Accepted Answer

Introduction / Context: A key hallmark of parallel connections is that the equivalent resistance is always less than the smallest individual branch resistance. This rule is frequently used to sanity-check calculations and quickly bound answers during design and exams. Given Data / Assumptions: Resistive network with two or more branches in parallel. All components are ideal resistors. No reactive effects or frequency dependence considered. Concept / Approach: The parallel identity is 1/R_eq = 1/R_1 + 1/R_2 + ... + 1/R_n. Since at least one positive conductance term equals 1/R_min and others are nonnegative, the total conductance exceeds 1/R_min, implying R_eq < R_min. Hence, the claim that R_eq is greater than the lowest-value branch is incorrect in all ideal cases. Step-by-Step Solution: 1) Identify the smallest resistor R_min in the set. 2) Compute conductance sum S = sum(1/R_i) >= 1/R_min. 3) With any additional branch, S > 1/R_min. 4) Therefore R_eq = 1/S <= R_min, and with more than one branch, R_eq < R_min strictly. Verification / Alternative check: Example: R_min = 100 Ω in parallel with anything else. Even with a very large resistor in parallel, the equivalent becomes slightly less than 100 Ω. This corroborates the general rule for parallel networks. Why Other Options Are Wrong: Correct: Incorrect choice because the statement itself is false. True only for two branches: It is false even for two branches. True only when resistors are equal: If equal, R_eq = R/n which is smaller than each branch. Depends on source voltage: R_eq is independent of source magnitude. Common Pitfalls: Comparing sums of resistances rather than conductances; applying series intuition to parallel networks; overlooking that R_eq is dominated by the lowest branch value downward. Final Answer: Incorrect

Question 11

Parallel circuits — voltage distribution: In an ideal parallel network fed by a single source, the voltage across each branch is the same as the source voltage. Decide whether this statement is correct.

Accepted Answer

Introduction / Context: One of the defining properties of parallel circuits is the equality of voltage across all branches that are directly connected to the same pair of nodes. This rule allows engineers to compute branch currents independently via Ohm’s law once the branch elements are known. Given Data / Assumptions: Ideal parallel connection between two nodes. Single source maintaining node-to-node potential. Negligible wiring and contact resistances for the basic model. Concept / Approach: By definition, components in parallel share the same pair of nodes, so the potential difference across each component equals the node-to-node voltage enforced by the source. Current splits according to branch impedances, but the voltage across each branch remains the same in the ideal case. Step-by-Step Solution: 1) Identify the two common nodes across which all branches connect. 2) The source fixes the node potential difference V. 3) Each branch, being across the same nodes, sees the same V. 4) Branch currents then follow I_k = V / R_k for resistive branches. Verification / Alternative check: Practical circuits with nonzero wiring resistance still approximate equal branch voltages closely when wiring drops are small relative to load drops. Measurements in lab confirm nearly equal branch voltages in well-wired setups. Why Other Options Are Wrong: Incorrect: Contradicts the definition of parallel connection. Only true for DC sources: Also holds for AC if components are directly in parallel. Only true if resistances are equal: Equality of voltage does not require equal resistances. Only true when wires have zero resistance: Small but finite wire drops are usually negligible; the ideal rule remains conceptually correct. Common Pitfalls: Confusing current sharing (which differs by branch impedance) with voltage sharing; assuming different branch voltages unless resistances match. In parallel, voltage is the same; current divides. Final Answer: Correct

Question 12

Node law — current conservation: At a junction (node) in an electrical network, is it possible for the total current entering the node to differ from the total current leaving the node under normal steady conditions?

Accepted Answer

Introduction / Context: Kirchhoff’s Current Law (KCL) states that the algebraic sum of currents entering and leaving a node is zero. This embodies conservation of charge and is fundamental to circuit analysis across DC and AC regimes for lumped networks. Given Data / Assumptions: Ideal lumped-element circuit model. Steady conditions or sinusoidal steady state for AC. No charge accumulation at an ideal node. Concept / Approach: KCL arises because charge cannot pile up indefinitely at a point in a lumped model; any current arriving must depart. Mathematically, sum(I_in) = sum(I_out). Transient elements like capacitors are handled by defining branch currents consistently; KCL still holds at every instant when displacement currents are included in Maxwell-consistent models. Step-by-Step Solution: 1) Define the node and enumerate incident branch currents with a sign convention. 2) Apply conservation of charge: net inflow equals net outflow. 3) Conclude that entering current cannot differ from leaving current in the model. 4) Use KCL to solve for unknown currents in parallel networks. Verification / Alternative check: Simulations and nodal analysis rely on KCL; measured discrepancies typically result from instrument error or parasitic storage, not a violation of the law itself within the lumped model assumptions. Why Other Options Are Wrong: Correct: The statement is not correct; KCL forbids a mismatch. True only during capacitor charging: KCL still holds when capacitor branch current is included. True only at very high frequency: KCL holds; modeling may require displacement current but conservation remains. True if the meter accuracy is low: Measurement error does not change physical laws. Common Pitfalls: Interpreting meter mismatch as a physical violation; overlooking displacement currents in high-frequency or field-theory contexts and then misapplying the lumped model. Final Answer: Incorrect

Question 13

Parallel fault — one branch shorts: If one branch in a parallel circuit becomes a short circuit (near zero resistance across the supply), what is the correct statement about the effect on other branches and the network operation?

Accepted Answer

Introduction / Context: Short-circuit faults are critical events in parallel networks. Understanding their impact helps with protection design (fuses, breakers), power integrity, and safe lab practice. The claim that other branches are unaffected when one branch shorts is evaluated here. Given Data / Assumptions: Parallel network fed by a practical source. One branch suddenly becomes a short (very low resistance). Ideal wiring assumed unless stated; source has finite current capability in practice. Concept / Approach: A shorted branch presents a very low impedance path, causing the source voltage to collapse toward zero at the parallel node if the source is limited or protective devices trip. Other branches share the same node voltage, so their currents and operation are indeed affected—often dropping drastically due to the voltage collapse or being interrupted by protective action. Step-by-Step Solution: 1) Introduce a near-zero resistance path across the supply at the node. 2) Node voltage is forced down by the short unless the source is unrealistically ideal and unlimited. 3) Other branches, seeing reduced voltage, change current (often to near zero). 4) Protection may open the circuit, stopping current in all branches. Verification / Alternative check: Lab experience: a short across a bench supply usually trips current limit, pulling output voltage down and affecting every parallel load. Power distribution systems behave similarly, prompting protection to clear the fault. Why Other Options Are Wrong: Correct: Not correct because a short impacts shared node voltage and branch currents. True only when the source has series resistance: Even with small resistance, the effect is substantial. True only for identical branches: Identity is irrelevant; the short dominates impedance. Undetermined without source voltage: Qualitative effect is determinable without numeric values. Common Pitfalls: Assuming branches are isolated in parallel; forgetting that all branches share the same node voltage mandated by the source and affected by faults. Final Answer: Incorrect

Question 14

Series vs. parallel — current paths: Do parallel circuits provide only one path for current flow, or do they provide multiple independent paths between two common nodes? Select the correct characterization.

Accepted Answer

Introduction / Context: Correctly distinguishing series from parallel connections is foundational in circuit theory and in quick diagnostics. Confusing these can lead to wrong equivalent resistance, wrong current predictions, and poor design choices. Given Data / Assumptions: Parallel circuit definition: elements share the same two nodes. Series circuit definition: elements share a single path sequentially. Ideal components and wiring assumed for the concept check. Concept / Approach: Parallel circuits furnish multiple paths for current between the same node pair, while series circuits provide only one path. Therefore, saying that parallel circuits have only one current path is a misstatement of the definitions and contradicts how current divides among branches in parallel networks. Step-by-Step Solution: 1) Recall definitions of series (single path) and parallel (shared nodes) connections. 2) In parallel, each branch is an alternative path for current from node A to node B. 3) Current divides according to branch impedances; hence multiple paths exist. 4) Conclude the original statement is incorrect for parallel networks. Verification / Alternative check: Measure currents in two-branch parallel networks: both branches carry current simultaneously; summing branch currents yields the source current, confirming multiple paths in practice. Why Other Options Are Wrong: Correct: Not correct; it describes series, not parallel. True only for two branches: Even two branches are multiple paths. True when all resistors are equal: Equality does not change path count. True for superconducting wires only: Superconductivity does not change topology. Common Pitfalls: Memorizing examples rather than definitions; mixing up current division (parallel) with voltage division (series). Final Answer: Incorrect

Question 15

Power in parallel — additivity: For independent branches connected in parallel to the same voltage source, does the total power drawn from the source equal the arithmetic sum of the individual branch powers?

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Introduction / Context: Total power accounting is a basic check for energy balance in circuits. In parallel networks, each branch sees the same voltage, and the source must supply the sum of branch powers under steady operation. This principle is used in sizing supplies and estimating thermal loads. Given Data / Assumptions: Parallel branches under a common source voltage. Independent loads; coupling between branches is negligible. Ideal measurement and wiring assumed for the concept. Concept / Approach: Branch power P_k = V * I_k (resistive DC) or P_k = V_rms * I_rms * pf_k (AC). The source current is I_total = sum(I_k). Therefore total power P_total = V * sum(I_k) in DC, which equals sum(V * I_k) = sum(P_k). In AC with sinusoidal steady state, active power is still additive across independent branches when using correct rms values and power factors. Step-by-Step Solution: 1) Express each branch power in terms of its current and the common voltage. 2) Sum all branch powers arithmetically. 3) Recognize the source delivers the sum of branch currents at the same voltage. 4) Conclude P_total = sum(P_k). Verification / Alternative check: Bench tests with two resistors in parallel show power meter at the source equals the sum of two individual power readings, within instrument tolerance. Power conservation ensures additivity for independent branches. Why Other Options Are Wrong: Incorrect: Conflicts with energy conservation in independent parallel branches. Only true for DC resistive loads: Also true for AC active power with proper power factor accounting. Only true when impedances are equal: Equality is not required for additivity. Only true when voltage is below 10 V: Magnitude is irrelevant to the principle. Common Pitfalls: Confusing instantaneous power oscillations with average power; forgetting power factor in AC and then misinterpreting measurements. Final Answer: Correct

Question 16

Equal resistors in parallel — current sharing: When two resistors of equal value are connected in parallel across the same source, will they carry identical currents? Choose the correct statement.

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Introduction / Context: Equal-value components in parallel provide symmetry that simplifies mental math. Recognizing identical current sharing under equal resistances allows quick sizing of components and cables, and it is frequently tested in introductory circuits. Given Data / Assumptions: Two resistors R and R in parallel. Both see the same node-to-node voltage V. Ideal components; wiring drops are negligible. Concept / Approach: By Ohm’s law, I_1 = V/R and I_2 = V/R. With equal R values and equal voltage across branches (parallel connection), the branch currents are identical. The total current is I_total = I_1 + I_2 = 2 * (V/R), and each branch shares half of the total in this symmetric case. Step-by-Step Solution: 1) Use the parallel rule: branch voltages are equal. 2) Apply Ohm’s law for each branch with equal R. 3) Conclude I_1 = I_2 by symmetry. 4) Compute I_total for completeness if needed: 2V/R. Verification / Alternative check: In practice, tiny tolerance differences cause small current imbalance, but for idealized problems and matched parts, currents are equal. Lab measurements with 1% resistors often confirm near-equal currents within tolerance. Why Other Options Are Wrong: Incorrect: Contradicts Ohm’s law under equal R and equal V. Only true if the source is ideal: Minor source resistance does not break symmetry significantly for concept checks. Only true below 1 A: The rule is not current-magnitude dependent. Only true at DC, not AC: For purely resistive branches, AC RMS currents remain equal as well. Common Pitfalls: Confusing practical tolerance effects with ideal analysis; assuming identical current sharing requires equal wire lengths or identical physical layout (it does not at the conceptual level). Final Answer: Correct

Question 17

Identical resistors in parallel — equivalent resistance rule: For like-value resistors connected in parallel, is the equivalent resistance equal to the individual resistor value multiplied by the number of resistors, or does it follow a different re…

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Introduction / Context: Memorizing the right shortcut for identical components saves time. In series, resistances add. In parallel, conductances add, giving a very different shortcut for identical resistors that many students initially mix up. Given Data / Assumptions: n identical resistors, each of value R, connected in parallel. Ideal components; parasitics neglected. We compare with an incorrect series-like rule. Concept / Approach: For identical parallel resistors: 1/R_eq = n * (1/R) so R_eq = R / n. The false statement claims R_eq = R * n, which is the series rule. Therefore, in parallel, the more identical branches you add, the smaller the equivalent resistance becomes, not larger. Step-by-Step Solution: 1) Write the parallel identity: 1/R_eq = sum(1/R_i). 2) Substitute R_i = R for all i, giving 1/R_eq = n/R. 3) Invert to obtain R_eq = R / n. 4) Compare to the claim R * n and note the contradiction. Verification / Alternative check: Example with R = 120 Ω and n = 3: R_eq = 120/3 = 40 Ω, clearly not 360 Ω. Bench tests confirm decreasing R_eq with added parallel branches. Why Other Options Are Wrong: Correct: Not correct; that is the series behavior. True only for series connections: This option is closer to the truth about series, but the given statement concerns parallel and remains incorrect. True when resistors are wirewound: Construction type does not invert the rule. True for temperature-compensated parts: Compensation does not convert parallel math to series math. Common Pitfalls: Applying series addition to parallel networks; failing to switch to conductance thinking for parallel combinations. Final Answer: Incorrect

Question 18

Repaired stem — total resistance claim: Consider three equal resistors of 1.44 kΩ each connected in parallel across an ideal source. A student claims, “The total resistance is 480 Ω.” Decide whether this claim is correct.

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Introduction / Context: This question was incomplete originally (“The total resistance is 480 Ω.”). Applying the Recovery-First Policy, we minimally repair the stem to a standard parallel-resistor scenario. It now evaluates whether a common rule for identical resistors in parallel has been applied correctly. Such checks are routine in lab pre-calculations and troubleshooting. Given Data / Assumptions: Three identical resistors, each R = 1.44 kΩ (i.e., 1440 Ω). Connected in parallel across the same two nodes. Ideal conditions; ignore wire resistance and tolerances. Concept / Approach: For n identical resistors in parallel, R_eq = R / n. With R = 1440 Ω and n = 3, the equivalent resistance should be 1440 / 3 = 480 Ω. This simple proportional rule arises because the total conductance is n/R, and the inverse gives the equivalent resistance decreasing with added branches. Step-by-Step Solution: 1) Write 1/R_eq = 1/R + 1/R + 1/R = 3/R. 2) Substitute R = 1440 Ω to obtain 1/R_eq = 3/1440 = 1/480. 3) Invert to get R_eq = 480 Ω. 4) Compare with the claim; they match exactly. Verification / Alternative check: Compute using pairwise first: two 1.44 kΩ in parallel give 720 Ω. Then 720 Ω in parallel with another 1.44 kΩ yields (720 * 1440) / (720 + 1440) = (1,036,800) / 2160 = 480 Ω, confirming the result by a different route. Why Other Options Are Wrong: Incorrect: The calculation shows the claim is correct. True only if one branch is open: An open would leave two branches, giving 720 Ω, not 480 Ω. True only at DC, not AC: For purely resistive loads, DC vs AC does not change the equivalent resistance. Cannot be decided without the source current: Equivalent resistance depends on component values and topology, not the source current. Common Pitfalls: Mixing up series (R_eq = nR) with parallel (R_eq = R/n) for identical resistors; forgetting to convert kΩ to Ω before arithmetic; skipping a quick pairwise check that can catch arithmetic slips. Final Answer: Correct

Question 19

Parallel circuits — in any ideal parallel network connecting components across the same two nodes, the voltage across every branch/component is equal (common node-to-node potential). Is this statement correct?

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Introduction / Context: In basic circuit theory, series and parallel connections determine how voltage and current distribute among components. This item checks your understanding of the defining property of a parallel connection: all branches share the same node pair, and therefore the same voltage appears across every branch. Given Data / Assumptions: Ideal lumped circuit with a DC or AC source. Components connected in parallel share the same two nodes. Wiring resistance and source impedance are neglected unless stated. Concept / Approach: By definition, a parallel connection places each component between the same two nodes. Voltage is a difference in electric potential between two points. If all components share the very same two points, the voltage difference across each must be identical, regardless of their individual resistances, reactances, or current draw. Step-by-Step Solution: 1) Identify the two common nodes that define the parallel network (often labeled + and − of a supply or a regulated bus).2) Recognize that every branch is directly across these same nodes.3) Conclude that the node-to-node potential difference is identical for all branches: V1 = V2 = V3 = ...4) Currents can differ by branch (I = V/R for resistors), but the voltage is common. Verification / Alternative check: If two ideal voltmeters are placed across any two branches in a parallel network, both read the same value because they sense the same node pair. This holds for DC and sinusoidal steady-state AC (phasor magnitude equality across branches when referenced to the same nodes). Why Other Options Are Wrong: Incorrect: Contradicts the definition of a parallel connection. Only for identical resistors: Voltage equality does not depend on component values; it is a node property. Only for AC sources: The property holds for DC and AC alike. Common Pitfalls: Confusing series (common current) with parallel (common voltage); ignoring small wiring drops in non-ideal layouts and thinking voltages differ substantially. Final Answer: Correct — in an ideal parallel connection, all branch voltages are equal.

Question 20

Terminology check — a parallel circuit is often described as a current divider because the source current splits into branch currents that sum to the total. Is this description appropriate?

Accepted Answer

Introduction / Context: Engineers commonly call parallel networks “current dividers.” This question tests your grasp of how total source current splits among branches while the same voltage appears across each branch in a parallel connection. Given Data / Assumptions: Multiple branches connected in parallel across a source. Ideal wires and lumped components. Steady-state DC or sinusoidal operation. Concept / Approach: Kirchhoff’s current law (KCL) states that the algebraic sum of currents at a node is zero. At the node where the source feeds several branches, the incoming source current equals the sum of outgoing branch currents. Because the voltage across each branch is the same, Ohm’s law sets branch currents based on branch impedances, dividing the total current. Step-by-Step Solution: 1) Identify the junction where source current enters the parallel network.2) Apply KCL: I_source = I1 + I2 + I3 + ...3) Use I_k = V / Z_k for each branch (for resistors, I_k = V / R_k).4) Conclude that the network divides current among branches according to their impedances. Verification / Alternative check: For two resistors R1 and R2 in parallel, the current-divider relation is I1 = I_total * (R2 / (R1 + R2)) and I2 = I_total * (R1 / (R1 + R2)) (for DC resistors), confirming the descriptive term “current divider.” Why Other Options Are Wrong: Incorrect: Conflicts with KCL and the well-known current-divider behavior. Only for identical branch resistances: Division occurs for any values; only the split ratios change. Only when total current is constant: Even if source current varies over time, at each instant it still divides among branches per impedances. Common Pitfalls: Mixing up “voltage divider” (series) with “current divider” (parallel); assuming equal split regardless of resistance values. Final Answer: Correct — a parallel circuit functions as a current divider.

Parallel Circuits Questions