College Physics Quiz: Kirchhoffs Loop Rule
20 questions · exam conditions
0:00
Kirchhoffs Loop RuleQuestion 1 of 20

A student applies Kirchhoff's loop rule to a single-loop circuit and writes: 9+2I+4I−6=09 + 2I + 4I - 6 = 0. If this equation is correct, which statement best describes the circuit configuration the student analyzed?

A 9V battery opposing a 6V battery, with a 2Ω resistor and 4Ω resistor in series, traversed clockwise
A 9V battery aiding a 6V battery, with a 2Ω resistor and 4Ω resistor in series, traversed counterclockwise
A 9V battery opposing a 6V battery, with a 2Ω resistor and 4Ω resistor in series, traversed counterclockwise
A 9V battery aiding a 6V battery, with a 2Ω resistor and 4Ω resistor in parallel, traversed clockwise
Two 9V batteries in parallel opposing one 6V battery, with resistors in series
← Back to quizzes

College Physics Quiz

College Physics Quiz: Kirchhoffs Loop Rule

Practice Kirchhoffs Loop Rule in College Physics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Kirchhoffs Loop Rule, giving you a quick way to practice the rules, question types, and explanations that matter most for College Physics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A student applies Kirchhoff's loop rule to a single-loop circuit and writes: 9+2I+4I−6=09 + 2I + 4I - 6 = 0. If this equation is correct, which statement best describes the circuit configuration the student analyzed?

  1. A 9V battery opposing a 6V battery, with a 2Ω resistor and 4Ω resistor in series, traversed clockwise
  2. A 9V battery aiding a 6V battery, with a 2Ω resistor and 4Ω resistor in series, traversed counterclockwise
  3. A 9V battery opposing a 6V battery, with a 2Ω resistor and 4Ω resistor in series, traversed counterclockwise (correct answer)
  4. A 9V battery aiding a 6V battery, with a 2Ω resistor and 4Ω resistor in parallel, traversed clockwise
  5. Two 9V batteries in parallel opposing one 6V battery, with resistors in series
Explanation: When analyzing Kirchhoff's loop rule problems, you need to interpret the signs in the equation to understand both the circuit configuration and the direction of traversal. The key is recognizing that positive terms indicate voltage rises (moving from negative to positive terminal of a battery) while negative terms indicate voltage drops (resistors or moving from positive to negative terminal of a battery). Looking at the equation 9+2I+4I−6=09 + 2I + 4I - 6 = 0, the +9 term indicates a voltage rise from a 9V battery, the +2I and +4I terms represent voltage drops across 2Ω and 4Ω resistors (which appear positive because you're traversing against the current direction), and the -6 term indicates a voltage drop from a 6V battery. This pattern means you're moving counterclockwise: entering the 9V battery at its negative terminal (voltage rise), then passing through resistors against current flow, and finally entering the 6V battery at its positive terminal (voltage drop). The batteries oppose each other since one provides a rise and the other a drop. Choice A incorrectly suggests clockwise traversal. Choice B has the batteries aiding rather than opposing each other - if they aided, both would contribute voltage rises or both would contribute drops. Choice D incorrectly describes parallel resistors, but parallel resistors would require separate current terms and a more complex analysis. The correct answer is C: opposing batteries with series resistors traversed counterclockwise. Study tip: Always trace the loop direction carefully and remember that the sign of each term reveals whether you're experiencing a voltage rise or drop at that component.

Question 2

A student writes the loop equation 24−8I−12I+16=024 - 8I - 12I + 16 = 0 for a single loop. After solving, they get I=2I = 2 A. However, when they check their answer by calculating power, they find that the batteries are delivering negative power. What is the most likely explanation?

  1. The student made an algebraic error when solving for current I
  2. The student used the wrong sign convention when traversing the batteries
  3. The current direction assumption was opposite to the actual current flow (correct answer)
  4. The student incorrectly combined the series resistances before writing the equation
  5. One of the batteries is actually a load, not a source, in this circuit
Explanation: When analyzing DC circuits using loop equations, you need to understand the relationship between assumed current direction and the signs in your equation. The student's algebra appears correct: solving 24−8I−12I+16=024 - 8I - 12I + 16 = 0 gives 40=20I40 = 20I, so I=2I = 2 A. The key insight comes from the power check. When batteries deliver negative power, it means they're actually absorbing power (being charged) rather than supplying it. This happens when the actual current flows opposite to your assumed direction. The student assumed current flowed in one direction around the loop, but the positive value of 2 A combined with negative battery power indicates the current actually flows the opposite way. Option A is incorrect because the algebra 40=20I40 = 20I yielding I=2I = 2 A is mathematically sound. Option B is wrong because sign convention errors would typically prevent you from getting a sensible solution altogether, not give you the right magnitude with wrong interpretation. Option D is incorrect because the equation 24−8I−12I+16=024 - 8I - 12I + 16 = 0 properly shows individual resistance terms rather than an incorrectly combined resistance. The solution I=2I = 2 A represents the magnitude of current, but the negative power indicates it flows opposite to the assumed direction. In circuit problems, this means the actual current is -2 A in the assumed direction, or equivalently, 2 A in the opposite direction. Study tip: Always check your circuit solutions using power calculations. If batteries show negative power, your current direction assumption was backwards—flip it and reinterpret your results accordingly.

Question 3

In a two-loop circuit, a student obtains the system: Loop 1: 15−5I1−10(I1−I2)=015 - 5I_1 - 10(I_1 - I_2) = 0 and Loop 2: −10(I2−I1)−8I2+12=0-10(I_2 - I_1) - 8I_2 + 12 = 0. After solving, they get I1=1.4I_1 = 1.4 A and I2=0.8I_2 = 0.8 A. What is the actual current through the 10Ω resistor?

  1. 0.6 A in the direction of I1I_1 (correct answer)
  2. 0.6 A in the direction of I2I_2
  3. 2.2 A in the direction of I1I_1
  4. 1.4 A in the direction of I1I_1
  5. 0.8 A in the direction of I2I_2
Explanation: When analyzing multi-loop circuits, the key insight is understanding that currents through shared components depend on how you define your loop currents and their directions. The 10Ω resistor sits between the two loops, so its actual current is determined by the difference between the loop currents. From the given equations, you can see that the 10Ω resistor appears as 10(I1−I2)10(I_1 - I_2) in Loop 1 and −10(I2−I1)-10(I_2 - I_1) in Loop 2. This tells us the actual current through the 10Ω resistor is I1−I2=1.4−0.8=0.6I_1 - I_2 = 1.4 - 0.8 = 0.6 A. Since this value is positive, the current flows in the direction of I1I_1 through this resistor. Looking at the wrong answers: Choice B incorrectly assumes the current flows in the direction of I2I_2, which would happen if I2>I1I_2 > I_1, but that's not the case here. Choice C gives 2.2 A, which might result from adding the currents (I1+I2I_1 + I_2) instead of finding their difference—a common error when students confuse how currents combine in shared branches. Choice D simply uses I1I_1 alone, ignoring that I2I_2 also contributes to the current through the shared resistor. The correct answer is A: 0.6 A in the direction of I1I_1. Study tip: In multi-loop problems, always remember that current through a shared component equals the difference between the relevant loop currents, and the direction depends on which loop current is larger. The algebra in your loop equations will guide you to the right setup.

Question 4

A student writes three loop equations for a circuit: (1) 24−4I1−8(I1−I2)=024 - 4I_1 - 8(I_1 - I_2) = 0, (2) −8(I2−I1)−6I2−12(I2−I3)=0-8(I_2 - I_1) - 6I_2 - 12(I_2 - I_3) = 0, (3) −12(I3−I2)−10I3+18=0-12(I_3 - I_2) - 10I_3 + 18 = 0. When solving this system, which approach will be most efficient?

  1. Substitute equation (3) into equation (2), then solve the resulting two-equation system
  2. Add all three equations together to eliminate the shared resistor terms immediately
  3. Solve equation (1) for I1I_1 in terms of I2I_2, then substitute into equation (2) (correct answer)
  4. Use matrix methods since the system has three equations in three unknowns
  5. Rearrange equation (2) to eliminate all shared resistor terms before proceeding
Explanation: When solving systems of loop equations in circuit analysis, you want to choose the approach that minimizes algebraic complexity while maintaining accuracy. The key is identifying which equation can be most easily manipulated to create useful substitutions. Looking at equation (1): 24−4I1−8(I1−I2)=024 - 4I_1 - 8(I_1 - I_2) = 0, you can simplify this to 24−12I1+8I2=024 - 12I_1 + 8I_2 = 0, which easily rearranges to I1=24+8I212=2+2I23I_1 = \frac{24 + 8I_2}{12} = 2 + \frac{2I_2}{3}. This gives you a clean expression for I1I_1 in terms of I2I_2. Substituting this into equation (2) eliminates I1I_1 and creates a two-variable system that's much simpler to solve. Option A is inefficient because substituting equation (3) into (2) creates a complex expression involving both I2I_2 and the messy terms from equation (3). Option B seems appealing, but adding all three equations doesn't actually eliminate the shared resistor terms—you still get a complicated single equation with all three currents. Option D suggests matrix methods, but while mathematically valid, it's overkill for a system where simple substitution works so elegantly. The winning strategy here is recognizing that equation (1) has the simplest structure and can be solved for I1I_1 with minimal algebra. This creates the cleanest substitution path forward. Study tip: When facing systems of circuit equations, always scan for the equation with the fewest terms or the one that can be most easily solved for a single variable. Start with the simplest manipulation to reduce the system's complexity efficiently.

Question 5

In a circuit with four meshes, a student needs to write three independent loop equations. After writing two equations, they consider several options for the third equation. Which choice would definitely result in a linearly dependent equation?

  1. Writing an equation for the outer perimeter of the entire circuit
  2. Writing an equation for any loop that shares exactly one resistor with each of the first two loops
  3. Writing an equation that can be obtained by adding or subtracting the first two equations (correct answer)
  4. Writing an equation for the smallest loop that hasn't been used yet
  5. Writing an equation for any loop that contains all the batteries in the circuit
Explanation: When analyzing circuits with multiple meshes, you need to understand linear independence in loop equations. For a circuit with four meshes, you need exactly three independent equations because the number of independent loop equations equals the number of meshes minus one. The key insight is that any loop equation must be linearly independent from the others you've already written. Linear dependence occurs when one equation can be derived by combining existing equations, meaning it provides no new information about the circuit. Choice C is correct because if an equation can be obtained by adding or subtracting your first two equations, it's mathematically redundant. This new equation would be a linear combination of existing equations, violating the independence requirement. It won't help you solve for unknown currents because it doesn't provide any new constraints on the system. Choice A is wrong because the outer perimeter loop often provides independent information, especially if your first two equations used internal loops. Choice B is incorrect because sharing exactly one resistor with each previous loop typically indicates the new loop traverses different circuit elements, likely making it independent. Choice D is wrong because the size of a loop doesn't determine its independence—a small unused loop could easily be independent of larger loops you've already analyzed. Remember this pattern: when writing mesh equations, avoid any loop whose equation you can create by algebraically manipulating your existing equations. Always check that each new loop introduces fresh circuit relationships.

Question 6

When solving a three-loop circuit, a student obtains the currents I1=2.5I_1 = 2.5 A, I2=−1.8I_2 = -1.8 A, and I3=0.7I_3 = 0.7 A. The negative value for I2I_2 indicates that:

  1. There is an error in the loop equations because currents cannot be negative
  2. The assumed direction for I2I_2 was opposite to the actual current flow in that loop (correct answer)
  3. Loop 2 contains a battery that is being charged rather than delivering power
  4. The resistances in loop 2 are larger than those in the other loops
  5. Loop 2 is electrically isolated from loops 1 and 3
Explanation: When analyzing multi-loop circuits using Kirchhoff's laws, you must assign assumed directions for each current before writing your equations. The sign of your calculated results tells you whether your assumptions were correct. A negative current value simply means the actual current flows opposite to the direction you initially assumed. Since I2=−1.8I_2 = -1.8 A, the current in that branch is actually 1.8 A flowing in the direction opposite to what you drew on your circuit diagram. This is completely normal and expected in circuit analysis. Let's examine why the other options are incorrect. Choice A is wrong because negative currents are mathematically valid and physically meaningful—they indicate direction, not an error in your equations. Choice C misinterprets what the negative sign means; it has nothing to do with whether batteries are charging or discharging, only with the direction you initially chose for that current. Choice D incorrectly connects the sign of current to resistance values. While resistance affects current magnitude, it doesn't determine the sign of your calculated current—only your initial directional assumption does that. The correct answer is B because when you solve circuit equations and get a negative current, it always means your assumed direction was backwards from the actual current flow. Study tip: When setting up circuit problems, don't worry about guessing current directions correctly. Pick any direction for each current, solve the equations, and let the math tell you the actual directions through the signs of your answers.

Question 7

When writing loop equations for a multi-loop circuit, a student writes: Loop 1: 12−3I1−6I1=012 - 3I_1 - 6I_1 = 0 and Loop 2: −6I2−4I2+8=0-6I_2 - 4I_2 + 8 = 0. What error has the student most likely made?

  1. Used the wrong sign convention for one of the batteries in the circuit
  2. Failed to account for shared currents in resistors common to both loops (correct answer)
  3. Applied Kirchhoff's current law instead of loop rule to one of the equations
  4. Incorrectly combined series resistances before writing the loop equations
  5. Traversed both loops in opposite directions, causing sign inconsistencies
Explanation: When analyzing multi-loop circuits using Kirchhoff's voltage law, you must carefully track how currents flow through resistors that are shared between loops. Looking at these equations, each loop appears to be treated as completely independent, but this is rarely the case in actual circuit problems. The correct answer is B because the student has failed to account for shared currents in resistors common to both loops. In most multi-loop circuits, some resistors lie along the boundary between loops, meaning they carry current from both loops simultaneously. When writing loop equations, you must add or subtract these shared currents depending on their relative directions. For example, if a 5Ω resistor is shared between loops with currents I1I_1 and I2I_2 flowing in opposite directions, the voltage drop would be 5(I1−I2)5(I_1 - I_2), not simply 5I15I_1 in one equation and 5I25I_2 in another. Answer A is wrong because sign convention errors would typically show up as incorrect signs on the battery terms, but the battery polarities (12V positive, 8V positive) seem consistent. Answer C is incorrect because both equations follow the proper loop rule format (sum of voltage drops equals zero), not current law. Answer D doesn't apply since the issue isn't about combining resistances incorrectly, but about treating currents independently. Study tip: When writing loop equations, always identify which resistors are shared between loops first. Then carefully determine whether the loop currents flow in the same or opposite directions through those shared elements before writing your voltage terms.

Question 8

A student analyzing a three-loop circuit writes: Loop A: 12−4I1−2(I1−I2)=012 - 4I_1 - 2(I_1 - I_2) = 0, Loop B: −2(I2−I1)−6I2+8=0-2(I_2 - I_1) - 6I_2 + 8 = 0, Loop C: −3I3+6−9I3=0-3I_3 + 6 - 9I_3 = 0. What can you conclude about the circuit topology?

  1. All three loops share common resistors with each adjacent loop
  2. Loops A and B share a resistor, but loop C is electrically isolated from the others (correct answer)
  3. Loop C contains two resistors in series with a battery between them
  4. The circuit contains exactly five resistors and three batteries total
  5. Loops A and C share a resistor, while loop B is connected to both through junction points
Explanation: When analyzing circuit topology from Kirchhoff's voltage law equations, look for shared terms between loops - these reveal which components connect different parts of the circuit. In Loop A's equation, the term −2(I1−I2)-2(I_1 - I_2) represents a 2Ω resistor with current flowing from loop A to loop B. Loop B's equation contains −2(I2−I1)-2(I_2 - I_1), which is the same resistor viewed from the opposite direction. This shared term proves loops A and B are connected through this common 2Ω resistor. However, Loop C's equation −3I3+6−9I3=0-3I_3 + 6 - 9I_3 = 0 contains only I3I_3 terms. There are no I1I_1 or I2I_2 terms, and loops A and B contain no I3I_3 terms. This means loop C shares no components with the other loops - it's electrically isolated. The correct answer is B. Choice A is wrong because loop C doesn't share resistors with loops A or B - there are no common current terms. Choice C incorrectly describes loop C's topology; while it does contain resistors (3Ω and 9Ω) and a battery, they're not necessarily arranged as "two resistors in series with a battery between them" - the equation shows −3I3+6−9I3-3I_3 + 6 - 9I_3, indicating the battery could be between the resistors or at either end. Choice D makes assumptions about the total component count that can't be verified from these three equations alone. Study tip: In circuit analysis, shared current terms between loop equations always indicate shared components. Missing current variables between loops signal electrical isolation.

Question 9

A student applies Kirchhoff's loop rule and writes: 36−9I−15I−12=036 - 9I - 15I - 12 = 0. After solving, they get I=1I = 1 A. They then realize they need to check if their assumed current direction was correct. Which method provides the most reliable verification?

  1. Verify that all resistor voltage drops are positive when calculated using V=IRV = IR
  2. Check that the total power delivered by batteries equals total power dissipated by resistors (correct answer)
  3. Confirm that all batteries are delivering positive power to the circuit
  4. Verify that Kirchhoff's current law is satisfied at each junction in the circuit
  5. Check that the battery voltages add up correctly around the complete loop
Explanation: When analyzing circuits with Kirchhoff's laws, determining the correct current direction is crucial since the sign of your calculated current tells you whether your assumption was right. A positive result means your assumed direction was correct; negative means the actual current flows opposite to your assumption. The most reliable verification is checking that total power delivered by batteries equals total power dissipated by resistors (option B). This fundamental energy conservation principle must always hold in any circuit. Calculate battery power using P=VIP = VI (positive when delivering power) and resistor power using P=I2RP = I^2R. If these totals match, your current direction and magnitude are correct. Option A is flawed because resistor voltage drops calculated with V=IRV = IR are always positive regardless of current direction - the formula gives magnitude, not the actual voltage drop direction relative to your assumed current flow. Option C fails because not all batteries necessarily deliver positive power. Some batteries might be charging (receiving power) rather than discharging, especially in circuits with multiple voltage sources. Option D is problematic because Kirchhoff's current law deals with current magnitudes at junctions, not directions around loops. While KCL must be satisfied, it doesn't directly verify whether your assumed current direction in the loop equation was correct. Study tip: Always verify circuit solutions using energy conservation - power in must equal power out. This catches both calculation errors and incorrect current direction assumptions, making it the most comprehensive check for circuit analysis problems.

Question 10

A complex circuit has four loops. A student correctly identifies that three independent loop equations are needed to solve for all currents. When writing the third loop equation, which principle should guide the student's choice of loop path?

  1. Choose the loop with the most resistors to maximize the number of terms in the equation
  2. Choose the loop that includes all remaining circuit elements not covered by the first two equations
  3. Choose any loop that is linearly independent from the first two loop equations written (correct answer)
  4. Choose the outer perimeter loop to ensure all batteries are included in the analysis
  5. Choose the loop with the smallest total resistance to minimize computational complexity
Explanation: When analyzing complex circuits with multiple loops, you need to understand Kirchhoff's voltage law and the concept of linear independence in circuit equations. For a circuit with four loops, you need exactly three independent equations because the number of independent loop equations equals the number of loops minus one. The key insight is that your third loop equation must provide new, independent information that cannot be derived from combining your first two equations. This is what makes option C correct - you need any loop that is linearly independent from your previous equations. Linear independence means the third equation contains unique information that helps solve for unknown currents. Option A is wrong because having more resistors doesn't make an equation more useful or independent. A complex equation with many terms might actually be a combination of simpler loops you've already analyzed. Option B incorrectly assumes you must include every circuit element - this isn't necessary and might lead to a dependent equation that doesn't add new information. Option D is flawed because the outer perimeter loop might be dependent on your inner loops, and including all batteries doesn't guarantee independence. The mathematical requirement is linear independence, not physical completeness or complexity. Your third equation should provide a new constraint that, combined with the first two, allows you to solve the system of equations uniquely. Study tip: When writing loop equations, always check that each new loop you choose crosses at least one branch that wasn't fully characterized by your previous equations. This helps ensure linear independence.

Question 11

When applying Kirchhoff's loop rule to the circuit below, a student writes the equation 18−3I1−6I1−9(I1+I2)=018 - 3I_1 - 6I_1 - 9(I_1 + I_2) = 0 for one of the loops. Which statement about this equation is correct?

  1. The equation represents the outer loop and correctly accounts for all current interactions
  2. The equation has an error because currents I₁ and I₂ should be subtracted, not added, in the shared resistor
  3. The equation correctly represents a loop where currents I₁ and I₂ flow in the same direction through the 9Ω resistor (correct answer)
  4. The student incorrectly combined the 3Ω and 6Ω resistors before applying the loop rule
  5. The equation is missing a battery term because the voltage sum doesn't match the resistance pattern
Explanation: The term 9(I₁ + I₂) indicates that in the 9Ω resistor, both currents I₁ and I₂ flow in the same direction (both clockwise or both counterclockwise), so their effects add. This is correct when two loop currents flow the same way through a shared resistor. Choice A assumes it's the outer loop without justification. Choice B incorrectly suggests currents should always be subtracted in shared resistors. Choice D is wrong because separate resistance terms (3I₁ and 6I₁) indicate separate resistors, not combined ones. Choice E is wrong because one 18V battery can balance the total resistance drop.

Question 12

In the circuit shown, if you traverse the left loop counterclockwise starting from point A, what is the correct loop equation?

  1. −10+4I1+2(I1+I2)=0-10 + 4I_1 + 2(I_1 + I_2) = 0 (correct answer)
  2. +10−4I1−2(I1+I2)=0+10 - 4I_1 - 2(I_1 + I_2) = 0
  3. −10−4I1−2(I1−I2)=0-10 - 4I_1 - 2(I_1 - I_2) = 0
  4. +10+4I1+2(I1+I2)=0+10 + 4I_1 + 2(I_1 + I_2) = 0
  5. −10+4I1+2(I1−I2)=0-10 + 4I_1 + 2(I_1 - I_2) = 0
Explanation: Traversing the left loop counterclockwise from point A: we encounter the 10V battery from + to -, giving -10V; then the 4Ω resistor opposite to current I₁ direction, giving +4I₁; then the 2Ω resistor where both I₁ and I₂ flow downward (same direction), giving +2(I₁ + I₂). Choice B has wrong battery sign. Choice C has wrong current combination in the shared resistor. Choice D has wrong battery sign and wrong resistor signs. Choice E has wrong current combination in the shared resistor.

Question 13

In the circuit shown, what is the loop equation for the right-most loop if traversed clockwise starting from the top right corner?

  1. −12+5I3+3(I3+I2)=0-12 + 5I_3 + 3(I_3 + I_2) = 0
  2. +12−5I3−3(I3−I2)=0+12 - 5I_3 - 3(I_3 - I_2) = 0 (correct answer)
  3. −12+5I3+3(I3−I2)=0-12 + 5I_3 + 3(I_3 - I_2) = 0
  4. +12+5I3+3(I3+I2)=0+12 + 5I_3 + 3(I_3 + I_2) = 0
  5. +12−5I3−3I3=0+12 - 5I_3 - 3I_3 = 0
Explanation: Traversing the right loop clockwise from top right: +12V from battery (- to +), -5I₃ through 5Ω resistor (voltage drop), -3(I₃ - I₂) through shared 3Ω resistor where I₃ flows down and I₂ flows up, making net current (I₃ - I₂) downward. Choice A has wrong battery sign and wrong current combination. Choice C has wrong battery sign. Choice D has wrong signs for both resistor drops. Choice E ignores that I₂ also flows through the 3Ω resistor.

Question 14

In the circuit shown, a student wants to write a loop equation that includes the 10V battery, the 3Ω resistor, and the 7Ω resistor, but avoids the 5Ω resistor entirely. What is the correct equation for this path?

  1. 10−3I1−7I2=010 - 3I_1 - 7I_2 = 0
  2. 10−3I1−7(I1+I2)=010 - 3I_1 - 7(I_1 + I_2) = 0
  3. 10−3(I1−I2)−7I2=010 - 3(I_1 - I_2) - 7I_2 = 0 (correct answer)
  4. 10−3I1+7I2=010 - 3I_1 + 7I_2 = 0
  5. −10+3I1+7I2=0-10 + 3I_1 + 7I_2 = 0
Explanation: The path described goes through the 10V battery (+10V), then the 3Ω resistor where current I₁ flows one way and I₂ flows the opposite way (giving net current I₁ - I₂), then the 7Ω resistor where only I₂ flows. This gives: +10 - 3(I₁ - I₂) - 7I₂ = 0. Choice A ignores that both currents flow through the 3Ω resistor. Choice B has wrong current combination in the 7Ω resistor. Choice D has wrong sign for the 7Ω resistor. Choice E has wrong signs for the battery and both resistors.

Question 15

In the circuit shown, what loop equation results from applying Kirchhoff's loop rule to the path that goes through the 20V battery, the 5Ω resistor, and the 10V battery (avoiding the 8Ω resistor)?

  1. 20−5I1−10=020 - 5I_1 - 10 = 0
  2. 20−5I1+10=020 - 5I_1 + 10 = 0 (correct answer)
  3. 20−5(I1−I2)−10=020 - 5(I_1 - I_2) - 10 = 0
  4. 20+5I1−10=020 + 5I_1 - 10 = 0
  5. −20+5I1+10=0-20 + 5I_1 + 10 = 0
Explanation: Following the specified path: +20V from the 20V battery (traversed from - to +), -5I₁ through the 5Ω resistor (voltage drop in direction of current), then +10V from the 10V battery (traversed from - to +). Both batteries aid each other in this loop. Choice A has wrong sign for the 10V battery. Choice C incorrectly includes I₂ when this path avoids the branch containing I₂. Choice D has wrong sign for the resistor drop. Choice E has wrong signs for both batteries.

Question 16

Consider the circuit shown. A student writes the equation 20−4I1−6I1−8(I1−I2)+12=020 - 4I_1 - 6I_1 - 8(I_1 - I_2) + 12 = 0 for one of the loops. What is the most likely error in this equation?

  1. The student traversed through a battery in the wrong direction
  2. The student included too many resistor terms in a single loop equation
  3. The student incorrectly combined currents in the shared resistor term
  4. The student included two batteries in one loop when they should be in separate loops (correct answer)
  5. The student used the wrong sign convention for the shared resistor
Explanation: The equation shows both +20V and +12V battery terms, suggesting the student tried to include two batteries in one loop equation. In most circuit topologies, batteries are in separate loops, and including both would require traversing through other loops. The equation has too many resistance terms for a typical single loop. Choice A is wrong because both batteries have positive signs, suggesting consistent traversal direction. Choice B is possible but less likely than the battery issue. Choice C is wrong because (I₁ - I₂) is a standard form for shared resistors. Choice E is wrong because the shared resistor sign appears consistent with the traversal direction.

Question 17

A student applies Kirchhoff's loop rule to a circuit containing three resistors and two batteries. When writing the loop equation for a clockwise path, which statement about voltage drops and rises is correct?

  1. Voltage drops across resistors are negative when current flows clockwise through them
  2. Voltage rises occur when traversing a battery from positive to negative terminal
  3. Voltage drops across resistors are positive when the assumed current direction matches the loop direction
  4. Voltage rises occur when traversing a battery from negative to positive terminal (correct answer)
Explanation: When applying Kirchhoff's loop rule, a voltage rise occurs when traversing a battery from its negative terminal to its positive terminal, regardless of loop direction. This represents the battery doing work on the charges. Choice A is incorrect because voltage drops are positive when current flows in the loop direction. Choice B is backwards - traversing from positive to negative is a voltage drop. Choice C is incorrect because voltage drops are represented as positive terms (+IR+IR) when current direction matches loop direction.

Question 18

When applying Kirchhoff's loop rule to find the current in a circuit with multiple loops, a student obtains the equation 24−8I1−12I2=024 - 8I_1 - 12I_2 = 0 for one loop. If the student had chosen to traverse this same loop in the opposite direction, what equation would result?

  1. 24−8I1−12I2=024 - 8I_1 - 12I_2 = 0 (identical equation)
  2. −24+8I1+12I2=0-24 + 8I_1 + 12I_2 = 0 (all terms negated) (correct answer)
  3. 24+8I1+12I2=024 + 8I_1 + 12I_2 = 0 (resistor terms change sign)
  4. −24−8I1−12I2=0-24 - 8I_1 - 12I_2 = 0 (battery term changes sign)
Explanation: When traversing a loop in the opposite direction, all voltage drops become voltage rises and vice versa, so every term in the loop equation changes sign. This gives −24+8I1+12I2=0-24 + 8I_1 + 12I_2 = 0, which is mathematically equivalent to the original equation. Choice A is incorrect because the direction matters for signs. Choice C incorrectly assumes only resistor terms change. Choice D incorrectly assumes only the battery term changes.

Question 19

A student applies Kirchhoff's loop rule to determine the power dissipated in a circuit element. After solving the loop equations, the student finds that the current through a particular resistor is negative. What is the correct interpretation of this result?

  1. The resistor is actually supplying power to the circuit rather than dissipating power
  2. An error was made in the calculation since current through a resistor cannot be negative
  3. The actual current direction is opposite to what was initially assumed, but power dissipation is still I2RI^2R (correct answer)
  4. The negative sign indicates the resistor is storing energy rather than dissipating it
Explanation: A negative current simply means the actual current flows opposite to the assumed direction. Since power dissipation in a resistor is P=I2RP = I^2R, and the square of current is always positive, the power calculation remains valid regardless of current direction. Choice A is incorrect because resistors always dissipate power. Choice B is wrong - negative currents are valid solutions indicating direction. Choice D is incorrect because resistors cannot store energy.

Question 20

Consider the circuit shown where currents I₁, I₂, and I₃ are the three loop currents. If Kirchhoff's loop rule is applied to the middle loop (containing the 15V battery and both 6Ω resistors), what is the correct equation?

  1. 15−6(I2−I1)−6(I2−I3)=015 - 6(I_2 - I_1) - 6(I_2 - I_3) = 0 (correct answer)
  2. 15−6I2−6I2=015 - 6I_2 - 6I_2 = 0
  3. 15−6(I1+I2)−6(I2+I3)=015 - 6(I_1 + I_2) - 6(I_2 + I_3) = 0
  4. 15+6(I2−I1)+6(I2−I3)=015 + 6(I_2 - I_1) + 6(I_2 - I_3) = 0
  5. 15−6(I2+I1)−6(I2−I3)=015 - 6(I_2 + I_1) - 6(I_2 - I_3) = 0
Explanation: For the middle loop containing the 15V battery: +15V from the battery, then through the left 6Ω resistor where current I₂ flows clockwise and I₁ flows clockwise in the adjacent loop, making the net current (I₂ - I₁), giving -6(I₂ - I₁). Similarly, through the right 6Ω resistor, current I₂ flows clockwise and I₃ flows clockwise in the adjacent loop, making net current (I₂ - I₃), giving -6(I₂ - I₃). Choice B ignores interaction between loops. Choice C has wrong current combinations. Choice D has wrong signs for voltage drops. Choice E mixes current combinations incorrectly.