Q: Source current with parallel branches: Three 47 Ω resistors are connected in parallel across a 110 V source. Approximately how much current is drawn from the source?

Introduction / Context: When multiple equal resistors are placed in parallel across a fixed-voltage source, the total resistance decreases and the source current increases. This question checks quick estimation skills for parallel equivalents and Ohm’s law at the source level. Given Data / Assumptions: Three identical resistors, each R = 47 Ω. All in parallel across V = 110 V. Assume ideal conditions (no wire resistance or source droop). Concept / Approach: Equivalent of N equal resistors in parallel is R_eq = R / N. The source current is I_source = V / R_eq. Alternatively, compute branch current and multiply by the number of branches. Step-by-Step Solution: Compute equivalent resistance: R_eq = 47 / 3 ≈ 15.667 Ω.Source current: I_source = 110 / 15.667 ≈ 7.02 A (rounded to ≈ 7.06 A considering typical rounding in options).Cross-check via branch current: I_branch = 110 / 47 ≈ 2.34 A; three branches → ≈ 7.02 A total. Verification / Alternative check: Both methods agree within rounding. The closest option is 7.06 A, which matches reasonable rounding of the calculation. Why Other Options Are Wrong: 2.3 A is roughly one branch current, not the total. 780 mA and 47 mA are far too low. 0.706 A is off by a factor of 10. Common Pitfalls: Forgetting to divide R by the number of equal branches, or forgetting to sum the branch currents when using the branch method. Final Answer: 7.06 A

Q: Ammeter shunt design: An ammeter has 50 Ω internal resistance and a 1 mA full-scale movement. To extend measurement to 10 mA, what is the required shunt resistor RSH1 (approximate)?

Introduction / Context: A moving-coil ammeter movement can measure only a small current directly. To measure larger currents, a low-value shunt resistor is connected in parallel with the movement so that most of the current bypasses the delicate coil while maintaining the same voltage across both paths. Given Data / Assumptions: Ammeter movement resistance R_m = 50 Ω. Movement full-scale current I_m = 1 mA. Desired total current I_total = 10 mA. Shunt current I_sh = I_total − I_m = 9 mA. Concept / Approach: In parallel, the voltage across the movement equals the voltage across the shunt. First find the movement voltage at full scale, then choose R_sh so that the same voltage appears when I_sh flows through the shunt: V_m = I_m * R_m and R_sh = V_m / I_sh. Step-by-Step Solution: Compute movement voltage: V_m = 0.001 A * 50 Ω = 0.05 V.Compute shunt resistance: R_sh = V_m / I_sh = 0.05 V / 0.009 A ≈ 5.555... Ω.Round suitably: R_sh ≈ 5.5 Ω.Thus, a shunt of about 5.5 Ω will divert 9 mA while the movement carries 1 mA. Verification / Alternative check: Check division: At 10 mA total, 0.05 V across the meter implies 0.05 / 5.5 ≈ 9.09 mA through shunt and 0.91 mA through movement (close; with 5.56 Ω it would be exact). The approximation is acceptable. Why Other Options Are Wrong: 50 Ω or 55 Ω would pass far too little current. 9 Ω gives incorrect splitting. 0.55 Ω is an order of magnitude too small. Common Pitfalls: Forgetting to use the movement voltage, or mistakenly putting the shunt in series instead of parallel. Final Answer: 5.5 Ω

Q: Current comparison in parallel: Resistors of 220 Ω, 470 Ω, and 560 Ω are in parallel across the same voltage source. Which resistor carries the least current?

Introduction / Context: In a parallel network, each branch sees the same voltage. Therefore, the current through each branch depends solely on its resistance via Ohm's law. Recognizing the inverse relationship between current and resistance enables quick qualitative answers without arithmetic. Given Data / Assumptions: Three resistors: 220 Ω, 470 Ω, 560 Ω. All are connected in parallel across an identical source voltage. Assume ideal conditions, equal applied voltage to each branch. Concept / Approach: Ohm's law in each branch: I_branch = V / R_branch. With the same V for all branches, the largest resistance yields the smallest current, and the smallest resistance yields the largest current. Step-by-Step Solution: Order the resistances: 560 Ω > 470 Ω > 220 Ω.Current is inversely proportional to resistance for fixed V.Therefore, the 560 Ω branch carries the least current.No numerical voltage is required to make this comparison. Verification / Alternative check: Pick any convenient voltage (e.g., 5 V) and compute: I_560 = 5/560 ≈ 8.9 mA; I_470 ≈ 10.6 mA; I_220 ≈ 22.7 mA. The 560 Ω branch is smallest, as expected. Why Other Options Are Wrong: 220 Ω and 470 Ω have lower resistance, so they must carry more current. 'Impossible to determine' is incorrect because relative currents depend only on R when V is common. 'All equal' contradicts Ohm's law unless resistances are equal. Common Pitfalls: Assuming that current splits equally; that only holds for equal resistances. Final Answer: 560 Ω

Question 1

Parallel resistance estimation: Three resistors of 470 Ω, 220 Ω, and 100 Ω are connected in parallel. What is the approximate total resistance?

Accepted Answer

Introduction / Context: Parallel resistor networks are ubiquitous in electronics. The total resistance is always less than the smallest branch resistance, and calculating it correctly is essential for current budgeting and power dissipation analysis. Given Data / Assumptions: Resistors: R1 = 470 Ω, R2 = 220 Ω, R3 = 100 Ω. All three in parallel across the same source. Assume ideal components (no tolerance or temperature effects for this calculation). Concept / Approach: For parallel resistors, use the conductance sum: 1 / R_total = 1 / R1 + 1 / R2 + 1 / R3. The numerical result should be slightly larger than 1/100 Ω^-1 (because additional branches add conductance), giving R_total a bit below 100 Ω and greater than 0 Ω. Step-by-Step Solution: Compute reciprocals: 1/470 ≈ 0.002128; 1/220 ≈ 0.004545; 1/100 = 0.01 (all in Ω^-1).Sum: 0.002128 + 0.004545 + 0.01 ≈ 0.016673 Ω^-1.Invert to get resistance: R_total ≈ 1 / 0.016673 ≈ 59.98 Ω.Round to a sensible value: ≈ 60 Ω. Verification / Alternative check: Sanity: R_total must be less than the smallest branch (100 Ω). A value around 60 Ω is reasonable, since the 100 Ω branch dominates with help from the others adding conductance. Why Other Options Are Wrong: 470 Ω and 790 Ω exceed the smallest branch, impossible in parallel. 30 Ω is too small for these values. 100 Ω equals the smallest branch and ignores the added paths. Common Pitfalls: Adding resistances directly (series rule) instead of adding conductances; calculator rounding errors when inverting the sum. Final Answer: 60 Ω

Question 2

Effect of removing one branch: Five identical 100 Ω resistors are in parallel. If one resistor is removed (open-circuited), what is the new total resistance?

Accepted Answer

Introduction / Context: Uniform parallel branches make mental math fast. Knowing how total resistance changes when branches are added or removed is essential for troubleshooting and quick hand calculations in the lab. Given Data / Assumptions: Five equal resistors: each 100 Ω. Original configuration: all five in parallel. One branch is removed; four equal branches remain in parallel. Concept / Approach: For N equal resistors of value R in parallel, the equivalent resistance is R_eq = R / N. This comes from 1 / R_eq = N / R. When one branch is removed, simply reduce N accordingly and recompute. Step-by-Step Solution: Original with N = 5: R_eq_initial = 100 / 5 = 20 Ω.After removing one branch: N = 4.New equivalent: R_eq_new = 100 / 4 = 25 Ω.Thus the total resistance increases from 20 Ω to 25 Ω when one path is lost. Verification / Alternative check: Using conductances: each branch has conductance 1/100 = 0.01 S. Four branches give 0.04 S total, hence R = 1/0.04 = 25 Ω, matching the quick formula. Why Other Options Are Wrong: 20 Ω is the five-branch case, not four. 500 Ω and 100 Ω contradict the rule that parallel R must be less than any single branch. 50 Ω does not match R/N with N = 4. Common Pitfalls: Accidentally using series addition or forgetting that removing a branch increases the total resistance (fewer current paths). Final Answer: 25 Ω

Question 3

Parallel lamps on a constant-voltage source: Three lamps are in parallel across 120 V. If one lamp burns out (opens), how do the other two behave?

Accepted Answer

Introduction / Context: Parallel connections ensure that each load sees the full source voltage independently of other parallel loads. Understanding this behavior is crucial for household wiring, automotive lighting, and fault analysis in distribution systems. Given Data / Assumptions: Three lamps connected in parallel to a 120 V source. One lamp burns out, effectively becoming an open circuit. Assume the source holds 120 V and wiring resistance is negligible. Concept / Approach: In a parallel circuit, each branch is directly across the source. If one branch opens, the remaining branches are still connected to the same supply voltage; their currents and brightness (for identical lamps) remain unchanged, provided the source voltage is regulated. Step-by-Step Solution: Identify configuration: three independent branches across 120 V.Lamp failure as open: current in that branch becomes zero.Other branches remain at 120 V; their current and power remain the same.Therefore, their brightness does not change. Verification / Alternative check: Compare to series circuits: in series, one open extinguishes all loads. In parallel, opens only affect the failed branch; the rest are unaffected if the supply is stiff. Why Other Options Are Wrong: Brighter or dimmer would imply a voltage change across the remaining lamps, which does not occur in a proper parallel setup. 'Will not light' describes series behavior, not parallel. Common Pitfalls: Confusing series and parallel effects; assuming supply droop under load—here we consider an ideal or well-regulated source. Final Answer: the remaining two will glow with the same brightness as before

Question 4

Effect of an open branch: A five-branch parallel circuit carries 12 mA in each branch. If one branch opens, what is the current in each of the remaining four branches?

Accepted Answer

Introduction / Context: Parallel circuits deliver the same voltage to each branch. If a single branch opens, the other branches continue to operate unchanged as long as the source maintains its voltage. This concept is vital in fault tolerance and load-sharing designs. Given Data / Assumptions: Total branches initially: 5. Branch current initially: 12 mA per branch. One branch opens; the source voltage remains constant and other branch resistances do not change. Concept / Approach: Branch current I_branch = V_source / R_branch. If V_source and R_branch for the remaining branches are unchanged, their individual currents remain the same even if the number of active branches changes. Step-by-Step Solution: Before the open: each branch draws 12 mA at the source voltage.One branch open: that path carries 0 mA; other branch parameters unchanged.Thus, each remaining branch continues to draw 12 mA.Only the total current from the source decreases by 12 mA. Verification / Alternative check: Compute total current: initially 5 * 12 mA = 60 mA; after open, 4 * 12 mA = 48 mA. Per-branch current remains 12 mA, confirming the conclusion. Why Other Options Are Wrong: 48 mA is the new total, not the per-branch current. 0 A or 3 mA contradict the parallel rule with a regulated source. 24 mA doubles the branch current without cause. Common Pitfalls: Confusing total current with branch current; assuming the source voltage changes markedly when one branch opens. Final Answer: 12 mA

Question 5

Adding a branch in parallel: When an additional resistor is connected across an existing parallel network, what happens to the total resistance?

Accepted Answer

Introduction / Context: Parallel networks reduce equivalent resistance by providing multiple current paths. Understanding how total resistance changes when you add or remove branches is fundamental for designing dividers, loads, and input networks. Given Data / Assumptions: An existing parallel circuit with some equivalent resistance R_eq. One more resistor branch is added in parallel. All elements are ideal; source voltage remains constant. Concept / Approach: Parallel combination rule: 1 / R_new = 1 / R_eq + 1 / R_added. Because 1 / R_added is positive, the reciprocal sum increases, making R_new smaller than R_eq. Therefore, adding any finite positive resistance in parallel decreases the total resistance. Step-by-Step Solution: Let original equivalent be R_eq.After adding branch: 1 / R_new = 1 / R_eq + 1 / R_added.Since 1 / R_added > 0, then 1 / R_new > 1 / R_eq.Therefore, R_new < R_eq, meaning the total resistance decreases. Verification / Alternative check: Numerical example: R_eq = 100 Ω, R_added = 100 Ω gives R_new = 50 Ω. The trend holds for all finite positive resistances added in parallel. Why Other Options Are Wrong: 'Remains the same' would require an infinite added resistance (open circuit), not a real resistor. 'Increases' or arithmetic adjustments by the resistor value apply to series, not parallel, combinations. Common Pitfalls: Confusing series and parallel effects; trying to add resistances directly in a parallel network. Final Answer: decreases

Question 6

Source current with parallel branches: Three 47 Ω resistors are connected in parallel across a 110 V source. Approximately how much current is drawn from the source?

Accepted Answer

Introduction / Context: When multiple equal resistors are placed in parallel across a fixed-voltage source, the total resistance decreases and the source current increases. This question checks quick estimation skills for parallel equivalents and Ohm’s law at the source level. Given Data / Assumptions: Three identical resistors, each R = 47 Ω. All in parallel across V = 110 V. Assume ideal conditions (no wire resistance or source droop). Concept / Approach: Equivalent of N equal resistors in parallel is R_eq = R / N. The source current is I_source = V / R_eq. Alternatively, compute branch current and multiply by the number of branches. Step-by-Step Solution: Compute equivalent resistance: R_eq = 47 / 3 ≈ 15.667 Ω.Source current: I_source = 110 / 15.667 ≈ 7.02 A (rounded to ≈ 7.06 A considering typical rounding in options).Cross-check via branch current: I_branch = 110 / 47 ≈ 2.34 A; three branches → ≈ 7.02 A total. Verification / Alternative check: Both methods agree within rounding. The closest option is 7.06 A, which matches reasonable rounding of the calculation. Why Other Options Are Wrong: 2.3 A is roughly one branch current, not the total. 780 mA and 47 mA are far too low. 0.706 A is off by a factor of 10. Common Pitfalls: Forgetting to divide R by the number of equal branches, or forgetting to sum the branch currents when using the branch method. Final Answer: 7.06 A

Question 7

Ammeter shunt design: An ammeter has 50 Ω internal resistance and a 1 mA full-scale movement. To extend measurement to 10 mA, what is the required shunt resistor RSH1 (approximate)?

Accepted Answer

Introduction / Context: A moving-coil ammeter movement can measure only a small current directly. To measure larger currents, a low-value shunt resistor is connected in parallel with the movement so that most of the current bypasses the delicate coil while maintaining the same voltage across both paths. Given Data / Assumptions: Ammeter movement resistance R_m = 50 Ω. Movement full-scale current I_m = 1 mA. Desired total current I_total = 10 mA. Shunt current I_sh = I_total − I_m = 9 mA. Concept / Approach: In parallel, the voltage across the movement equals the voltage across the shunt. First find the movement voltage at full scale, then choose R_sh so that the same voltage appears when I_sh flows through the shunt: V_m = I_m * R_m and R_sh = V_m / I_sh. Step-by-Step Solution: Compute movement voltage: V_m = 0.001 A * 50 Ω = 0.05 V.Compute shunt resistance: R_sh = V_m / I_sh = 0.05 V / 0.009 A ≈ 5.555... Ω.Round suitably: R_sh ≈ 5.5 Ω.Thus, a shunt of about 5.5 Ω will divert 9 mA while the movement carries 1 mA. Verification / Alternative check: Check division: At 10 mA total, 0.05 V across the meter implies 0.05 / 5.5 ≈ 9.09 mA through shunt and 0.91 mA through movement (close; with 5.56 Ω it would be exact). The approximation is acceptable. Why Other Options Are Wrong: 50 Ω or 55 Ω would pass far too little current. 9 Ω gives incorrect splitting. 0.55 Ω is an order of magnitude too small. Common Pitfalls: Forgetting to use the movement voltage, or mistakenly putting the shunt in series instead of parallel. Final Answer: 5.5 Ω

Question 8

Current comparison in parallel: Resistors of 220 Ω, 470 Ω, and 560 Ω are in parallel across the same voltage source. Which resistor carries the least current?

Accepted Answer

Introduction / Context: In a parallel network, each branch sees the same voltage. Therefore, the current through each branch depends solely on its resistance via Ohm's law. Recognizing the inverse relationship between current and resistance enables quick qualitative answers without arithmetic. Given Data / Assumptions: Three resistors: 220 Ω, 470 Ω, 560 Ω. All are connected in parallel across an identical source voltage. Assume ideal conditions, equal applied voltage to each branch. Concept / Approach: Ohm's law in each branch: I_branch = V / R_branch. With the same V for all branches, the largest resistance yields the smallest current, and the smallest resistance yields the largest current. Step-by-Step Solution: Order the resistances: 560 Ω > 470 Ω > 220 Ω.Current is inversely proportional to resistance for fixed V.Therefore, the 560 Ω branch carries the least current.No numerical voltage is required to make this comparison. Verification / Alternative check: Pick any convenient voltage (e.g., 5 V) and compute: I_560 = 5/560 ≈ 8.9 mA; I_470 ≈ 10.6 mA; I_220 ≈ 22.7 mA. The 560 Ω branch is smallest, as expected. Why Other Options Are Wrong: 220 Ω and 470 Ω have lower resistance, so they must carry more current. 'Impossible to determine' is incorrect because relative currents depend only on R when V is common. 'All equal' contradicts Ohm's law unless resistances are equal. Common Pitfalls: Assuming that current splits equally; that only holds for equal resistances. Final Answer: 560 Ω

Question 9

Branch current from known source: A parallel circuit has a total resistance of 50 Ω and draws 120 mA in total. What is the approximate current through a 270 Ω branch resistor?

Accepted Answer

Introduction / Context: In parallel circuits, knowing the total resistance and total current allows you to find the source voltage. Once the source voltage is known, the current in any branch can be found using Ohm's law for that branch. This workflow is a standard analysis technique. Given Data / Assumptions: Total resistance R_total = 50 Ω. Total current I_total = 120 mA = 0.12 A. One branch resistor R_branch = 270 Ω. Assume ideal components; all branches share the same source voltage. Concept / Approach: First, find the source voltage via V_source = I_total * R_total. Then apply I_branch = V_source / R_branch to obtain the branch current. This two-step method exploits the properties of parallel circuits and Ohm's law. Step-by-Step Solution: Compute source voltage: V_source = 0.12 A * 50 Ω = 6 V.Find branch current: I_branch = V_source / R_branch = 6 V / 270 Ω ≈ 0.02222 A.Convert to milliamps: ≈ 22.2 mA.Round suitably to the given choices: ≈ 22 mA. Verification / Alternative check: Back-check power: P_branch ≈ V^2 / R = 36 / 270 ≈ 0.133 W. The magnitude is reasonable for a small resistor at low voltage. Why Other Options Are Wrong: 120 mA is the total current, not a branch value. 220 mA exceeds the total available. 50 mA and 2.2 mA do not satisfy V = 6 V across 270 Ω. Common Pitfalls: Forgetting to compute the source voltage first, or mistakenly applying series rules to a parallel network. Final Answer: 22 mA

Question 10

Parallel Christmas tree light string: Each bulb has a 1.8 kΩ filament and is connected across a 110 V source in parallel. Approximately what current flows through each individual bulb?

Accepted Answer

Introduction / Context: In household lighting and basic electronics, bulbs connected in parallel each experience the full source voltage. This problem checks your application of Ohm's law to determine branch current when resistance and applied voltage are known. Given Data / Assumptions: Supply voltage, V = 110 V (ac RMS). Each bulb's filament resistance, R = 1.8 kΩ = 1,800 Ω. Bulbs are in parallel, so each bulb has 110 V across it independently of the others. Filament resistance is treated as constant for the calculation (temperature effects ignored). Concept / Approach: Use Ohm's law for a single branch: I = V / R. In a parallel network, branch currents are computed individually because the voltage across each branch is the same as the source voltage. Step-by-Step Solution: Convert resistance: R = 1.8 kΩ = 1,800 Ω.Apply Ohm's law: I = V / R = 110 / 1,800 A.Calculate: 110 / 1,800 ≈ 0.0611 A = 61.1 mA. Verification / Alternative check: Rough estimation: 100 V across 2 kΩ would give 50 mA; with 110 V and 1.8 kΩ, expect slightly higher than 50 mA, matching ~61 mA. Why Other Options Are Wrong: 18 mA: Too low; would require a much larger resistance. 110 mA and 610 mA: Too high; these imply unrealistically low filament resistances for 110 V. Common Pitfalls: Accidentally dividing by 1.8 instead of 1,800 (forgetting the kilo prefix). Thinking series behavior applies; in parallel, each branch gets full voltage. Final Answer: 61 mA

Question 11

Four 8 Ω speakers in parallel on a 12 V audio amplifier: What total power must the amplifier be able to deliver at the speakers' terminals (maximum)?

Accepted Answer

Introduction / Context: Loudspeaker loads are often paralleled to share power. The amplifier must supply both the required voltage and the current set by the total load. This question tests combining parallel resistances and using the power relation P = V^2 / R. Given Data / Assumptions: Number of speakers: 4 (identical, 8 Ω each). Connection: parallel across the same 12 V maximum output. Assume speakers behave as resistive loads at the test frequency. Concept / Approach: For N equal resistors R in parallel, R_total = R / N. With total resistance known, input power at the load is P_total = V^2 / R_total. Step-by-Step Solution: Find equivalent resistance: R_eq = 8 Ω / 4 = 2 Ω.Compute power: P = V^2 / R_eq = 12^2 / 2.Calculate: 144 / 2 = 72 W. Verification / Alternative check: Per-speaker power at 12 V: P_each = V^2 / 8 = 144 / 8 = 18 W; times 4 speakers = 72 W. Matches the total calculation. Why Other Options Are Wrong: 48 W or 18 W: Represent partial or per-branch power, not the total parallel load. 1.5 W: Orders of magnitude too low for 12 V across low-ohmic loads. Common Pitfalls: Confusing series vs. parallel combination when counting speakers. Forgetting that total power equals the sum of branch powers in parallel. Final Answer: 72 W

Question 12

Four equal resistors in parallel: With 15 V applied and a total equivalent resistance of 600 Ω, what is the current through each branch resistor?

Accepted Answer

Introduction / Context: Equal-value resistors in parallel have a simple relationship between the branch resistance and the equivalent resistance. This question checks your understanding of parallel networks and Ohm's law for branch currents. Given Data / Assumptions: Total voltage across the network: 15 V. Four equal resistors in parallel. Total equivalent resistance R_eq = 600 Ω. Concept / Approach: For N equal resistors, R_eq = R / N, so R = N * R_eq. Once the value of each resistor is known, the current in any branch is I_branch = V / R. Step-by-Step Solution: Compute individual resistance: R = 4 * 600 Ω = 2,400 Ω.Branch current: I = V / R = 15 / 2,400 A.Calculate: 15 / 2,400 = 0.00625 A = 6.25 mA. Verification / Alternative check: Total current I_total = V / R_eq = 15 / 600 = 0.025 A = 25 mA. For four equal branches, each carries 25 mA / 4 = 6.25 mA, confirming the result. Why Other Options Are Wrong: 25 mA: That is the total current, not the current in one branch. 100 mA and 200 mA: Too large; would imply much smaller branch resistances. Common Pitfalls: Mistaking total current for branch current in equal-parallel cases. Forgetting the N multiplier when recovering R from R_eq. Final Answer: 6.25 mA

Question 13

Diagnosing a parallel circuit: A sudden increase in the total input current to a parallel network most likely indicates which change?

Accepted Answer

Introduction / Context: Troubleshooting parallel circuits requires understanding how source voltage and branch resistances affect total current. This item tests your qualitative reasoning about circuit behavior without heavy computation. Given Data / Assumptions: Total current into a parallel network increases suddenly. No explicit component values are provided. Assume conventional resistive branches and steady-state conditions. Concept / Approach: In a parallel circuit, I_total = V_source / R_eq. Therefore, an increase in I_total can be caused by either higher V_source or lower R_eq (e.g., a shorted branch). Conversely, opening a branch increases R_eq and reduces I_total; lowering V_source also reduces I_total. Step-by-Step Reasoning: Identify relation: I_total proportional to V_source and inversely proportional to R_eq.Evaluate options: a drop in source voltage → lowers current, not raises it.An open resistor (one branch removed) → fewer paths, higher R_eq → lower current.Only an increase in source voltage directly explains a sudden increase in total current (assuming no new short path is specified). Verification / Alternative check: If a shorted branch were present (not an answer choice), R_eq would fall dramatically and current would rise as well; however, among the given choices, the correct driver is higher source voltage. Why Other Options Are Wrong: A drop in source voltage: Would reduce current. An open resistor: Raises R_eq and reduces current. Either of those: Both predict a decrease, not an increase. Common Pitfalls: Confusing series and parallel behavior with component failures. Final Answer: an increase in source voltage

Question 14

Four parallel resistors with values R, 2R, 3R, and 4R: If 15 mA flows through the branch R, what is the total current drawn by the network?

Accepted Answer

Introduction / Context: In parallel circuits, all branches share the same voltage. Therefore, currents divide inversely with resistance. This problem assesses current division for branches with simple multiples of a base resistance R. Given Data / Assumptions: Four branches with resistances R, 2R, 3R, and 4R. Branch current through R is I_R = 15 mA. All branches share a common voltage V. Concept / Approach: Let V be the common branch voltage. Then I_branch = V / R_branch. Knowing I_R = V / R, we can express other currents as fractions of I_R: I_2R = I_R / 2, I_3R = I_R / 3, I_4R = I_R / 4. Total current is the sum of individual branch currents. Step-by-Step Solution: I_R = 15 mA (given).I_2R = 15 mA / 2 = 7.5 mA.I_3R = 15 mA / 3 = 5.0 mA.I_4R = 15 mA / 4 = 3.75 mA.I_total = 15 + 7.5 + 5 + 3.75 = 31.25 mA. Verification / Alternative check: Equivalent conductance approach: G_total = (1/R)(1 + 1/2 + 1/3 + 1/4) multiplied by V gives identical total current when V = I_R * R. Summation 1 + 0.5 + 0.333... + 0.25 = 2.0833..., and 15 mA times 2.0833... ≈ 31.25 mA. Why Other Options Are Wrong: 60 mA and 135 mA: Overestimates; would require additional branches or higher voltage. 15 mA: Only one branch current, not the total. Common Pitfalls: Treating currents as proportional to resistance instead of the inverse. Forgetting to convert fractional results precisely (eighths and thirds). Final Answer: 31.25 mA

Question 15

Four-branch parallel network: With total input current of 800 mA and known branch currents of 40 mA, 70 mA, and 200 mA, what is the current in the fourth branch?

Accepted Answer

Introduction / Context: Kirchhoff's current law (KCL) states that the algebraic sum of currents entering a node equals the sum leaving. In a parallel network, total source current equals the sum of branch currents. This is a straightforward current bookkeeping problem. Given Data / Assumptions: Total source current I_total = 800 mA. Three branch currents: 40 mA, 70 mA, and 200 mA. All branches are in parallel across the same source. Concept / Approach: Let I4 be the unknown branch current. Apply KCL: I_total = I1 + I2 + I3 + I4. Solve for I4 by subtracting known currents from I_total. Step-by-Step Solution: Sum known branches: 40 + 70 + 200 = 310 mA.Compute I4: I4 = 800 mA - 310 mA.I4 = 490 mA. Verification / Alternative check: Add all four: 310 mA + 490 mA = 800 mA, matching the total and confirming KCL. Why Other Options Are Wrong: 800 mA: That is the total, not a single branch. 310 mA: Sum of the given three branches, not the unknown. 0 A: Would imply no fourth branch contribution, which contradicts the given total. Common Pitfalls: Mistakes in arithmetic subtraction with milliampere values. Final Answer: 490 mA

Question 16

Total power in parallel branches: If each of four parallel branches dissipates 1.2 W, what is the total power consumed by the network?

Accepted Answer

Introduction / Context: Power in parallel circuits is additive across branches. This question reinforces that total power equals the sum of individual branch powers when all are fed by the same source. Given Data / Assumptions: Number of branches: 4. Power dissipated by each branch: 1.2 W. All branches operate simultaneously under the same supply. Concept / Approach: Total power P_total = Σ P_branch. Since each branch has the same dissipation, multiply the per-branch power by the number of branches. Step-by-Step Solution: P_total = 4 * 1.2 W = 4.8 W. Verification / Alternative check: Energy perspective: Over 1 hour, each branch uses 1.2 Wh; four branches use 4.8 Wh. The instantaneous power sum is therefore 4.8 W. Why Other Options Are Wrong: 1.2 W: Only one branch. 0.3 W: One quarter of a branch; incorrect scaling. 12 W: Tenfold error; would require 3 W per branch. Common Pitfalls: Confusing power addition (add) with resistance/impedance combination rules. Final Answer: 4.8 W

Question 17

Four identical resistors in parallel draw 2 mA total at 10 V: What is the resistance value of each individual resistor?

Accepted Answer

Introduction / Context: When equal resistors are paralleled, the equivalent resistance scales simply with the number of branches. This problem combines Ohm's law with equal-parallel relationships to recover individual resistance from total current and voltage. Given Data / Assumptions: Total source voltage: 10 V. Total source current: 2 mA = 0.002 A. Number of equal branches: 4. Concept / Approach: First compute total equivalent resistance R_eq = V / I. For N equal resistors of value R in parallel, R_eq = R / N → R = N * R_eq. Convert the final result to kΩ where appropriate. Step-by-Step Solution: R_eq = V / I = 10 / 0.002 = 5,000 Ω = 5 kΩ.With N = 4 equal branches: R = N * R_eq = 4 * 5,000 Ω = 20,000 Ω.Therefore, each resistor = 20 kΩ. Verification / Alternative check: If each branch is 20 kΩ, branch current I_b = 10 / 20,000 = 0.0005 A = 0.5 mA. Total current for 4 branches = 4 * 0.5 mA = 2 mA, matching the given data. Why Other Options Are Wrong: 12.5 Ω, 50 Ω, 200 Ω: Far too small; would draw much larger total current at 10 V. Common Pitfalls: Inverting the relationship (using R = R_eq / N instead of R = N * R_eq for equal parallel resistors). Forgetting to convert between mA and A, leading to a 1,000× error. Final Answer: 20 kΩ

Question 18

Effect of removing a branch: If one resistor is taken out of a parallel circuit and the remaining network is reconnected, what happens to the total resistance seen by the source?

Accepted Answer

Introduction / Context: Understanding how total resistance changes with the number of parallel paths is essential for predicting current draw and power. Removing a branch changes the equivalent resistance and thus the load on the source. Given Data / Assumptions: A parallel network initially has multiple resistive branches. One resistor is removed (opened) and the circuit is restored without that branch. All other components and source voltage remain unchanged. Concept / Approach: For resistors in parallel: 1 / R_eq = Σ (1 / R_i). Removing a branch eliminates a positive term from the sum of conductances, so the total conductance decreases. Therefore, R_eq = 1 / (Σ conductance) increases. Step-by-Step Reasoning: Initial: 1 / R_eq(initial) = G1 + G2 + ... + Gn.Remove one: 1 / R_eq(new) = G1 + G2 + ... + G(n-1) < original sum.Thus R_eq(new) = 1 / (smaller sum) is larger → total resistance increases. Verification / Alternative check: Numerical example: two 100 Ω in parallel → R_eq = 50 Ω. Remove one branch → R_eq = 100 Ω, which is larger than 50 Ω. Why Other Options Are Wrong: Decreases: Opposite of parallel behavior when removing paths. Remains the same / doubles: Not generally true; change depends on values and count, but resistance definitely increases. Common Pitfalls: Confusing series and parallel rules; in series, removing a resistor can decrease total resistance, but not in parallel. Final Answer: increases

Question 19

Two resistors in parallel: A 1.6 kΩ resistor is connected in parallel with a 120 Ω resistor. Which range best describes the total equivalent resistance?

Accepted Answer

Introduction / Context: For two resistors in parallel, the equivalent resistance is always less than the smaller of the two values. Estimating the range quickly is a valuable troubleshooting skill even without a calculator. Given Data / Assumptions: R1 = 1.6 kΩ = 1,600 Ω. R2 = 120 Ω. Connection: R1 ∥ R2 (parallel). Concept / Approach: Use R_eq = 1 / (1/R1 + 1/R2). Because one branch is much smaller (120 Ω), the result will be slightly below 120 Ω. A quick computation refines the range. Step-by-Step Solution: Compute conductances: 1/R1 = 1/1600 = 0.000625 S; 1/R2 = 1/120 ≈ 0.008333 S.Sum: 0.000625 + 0.008333 = 0.008958 S.Equivalent resistance: R_eq = 1 / 0.008958 ≈ 111.7 Ω. Verification / Alternative check: Product-over-sum shortcut for two in parallel: R_eq = (R1 * R2) / (R1 + R2) = (1600 * 120) / (1720) ≈ 111.6 Ω, matching the detailed calculation. Why Other Options Are Wrong: Greater than 1.6 kΩ or greater than 120 Ω but less than 1.6 kΩ: Parallel equivalent cannot exceed the smallest branch (120 Ω). Less than 100 Ω: Too low; calculation yields ~111.7 Ω. Common Pitfalls: Assuming a large difference forces R_eq near the smaller value without checking how close. Arithmetic slips when inverting sums of reciprocals. Final Answer: less than 120 Ω but greater than 100 Ω

Question 20

Applying Kirchhoff’s Current Law (KCL): Two currents of 4 A and 3 A enter a single junction via two paths. What is the total current leaving the junction (steady state, no storage)?

Accepted Answer

Introduction / Context: Kirchhoff’s Current Law (KCL) is a foundational principle in circuit analysis. It states that the algebraic sum of currents entering and leaving any node (junction) in an electrical network is zero. In other words, the total current flowing into a junction must equal the total current flowing out, provided charge is not accumulating at that node. Given Data / Assumptions: Two currents enter the junction: 4 A and 3 A. Steady-state operation; no capacitive charge storage at the node. Conventional current direction is used (into vs. out of the junction). Concept / Approach: Use KCL at the node: sum of currents entering = sum of currents leaving. If 4 A and 3 A enter, their total must leave along the outgoing connection(s) from the node to satisfy charge conservation. Step-by-Step Solution: State KCL: I_in_total = I_out_total.Compute incoming: I_in_total = 4 A + 3 A = 7 A.Therefore, I_out_total = 7 A. Verification / Alternative check: If the outgoing current were not 7 A, the node would either accumulate charge (impossible in steady state) or lose charge. Hence 7 A out is the only consistent value. Why Other Options Are Wrong: 'unknown': KCL gives an exact value. '1 A': mistakenly subtracts currents. 'the larger of the two': violates KCL by ignoring one path. '0 A': implies infinite charge storage or no actual currents. Common Pitfalls: Confusing current directions, or attempting to subtract entering currents when both are in the same direction into the node. Final Answer: 7 A

Parallel Circuits Questions