This problem requires applying Kirchhoff's loop rule. Kirchhoff's loop rule states that the sum of all voltage changes around any closed loop must equal zero, reflecting energy conservation - a charge returning to its starting point must have the same energy. Starting at the battery's negative terminal and traversing clockwise, we gain +12V crossing from negative to positive terminal, then lose voltage across each resistor: -I(3.0) across the 3.0Ω resistor and -I(5.0) across the 5.0Ω resistor. Choice D incorrectly sets the sum equal to 12 instead of 0, showing a misconception that the loop equation equals the battery voltage. The correct strategy is to traverse the loop consistently, sum all voltage changes (rises positive, drops negative), and set the total equal to zero.

This problem applies Kirchhoff's loop rule. Kirchhoff's loop rule ensures that the total voltage change around any closed path equals zero, conserving energy in the circuit. Traversing clockwise with clockwise current I but crossing the 3V battery from + to - gives -3V (opposite to the standard - to + convention), while both resistors traversed with current contribute -2I and -4I respectively. The equation is -3 - 2I - 4I = 0. Choice B (+3 - 2I - 4I = 0) uses the wrong sign for the battery, forgetting that crossing from + to - gives negative voltage. Always apply sign conventions consistently: crossing a battery from + to - yields negative voltage.

This problem requires applying Kirchhoff's loop rule. Kirchhoff's loop rule ensures that the algebraic sum of potential differences around a closed loop equals zero, conserving energy. Traversing counterclockwise with counterclockwise current I: the 9V battery is crossed from - to +, giving +9V; the 3V battery is crossed from + to -, giving -3V; the 6Ω resistor is traversed with the current, giving -6I. The equation is +9 - 3 - 6I = 0. Choice D (+9 - 3 + 6I = 0) incorrectly uses +6I for the resistor term, revealing the misconception of confusing the sign when traversing with versus against current. Remember to use negative voltage drops for resistors when traversing in the direction of current flow.

This question involves applying Kirchhoff's loop rule. Kirchhoff's loop rule ensures conservation of energy by requiring that the sum of all voltage changes around a closed loop equals zero, meaning a charge returns to its starting point with the same potential energy. Traversing clockwise with clockwise current, we cross the battery from negative to positive (+10V), then traverse both resistors in the direction of current, experiencing drops of -I(4.0) and -I(2.0). Choice D incorrectly sets the equation equal to 12 instead of 0, suggesting confusion about what the loop rule represents - it's an energy balance, not a calculation of total voltage. The key strategy is to sum all voltage changes algebraically (rises positive, drops negative) and set the total to zero, not to any other value.

This problem tests Kirchhoff's loop rule. Kirchhoff's loop rule states that the algebraic sum of voltage changes around a closed loop must equal zero, ensuring energy is conserved. Traversing clockwise with clockwise current I: both batteries are crossed from - to +, contributing +4.5V and +1.5V respectively since they aid each other; the 3Ω resistor traversed with current contributes -3I. The equation is +4.5 + 1.5 - 3I = 0. Choice A (+4.5 - 1.5 - 3I = 0) incorrectly subtracts the second battery voltage, misunderstanding that aiding batteries have the same sign. When batteries aid each other in a loop, their voltages add with the same sign according to your traversal convention.

This question tests understanding of Kirchhoff's loop rule. Kirchhoff's loop rule states that the sum of all potential differences around a closed loop must be zero, ensuring that electric potential energy is conserved as charge completes a circuit. Traversing counterclockwise with counterclockwise current, we cross the battery from negative to positive (+15V), then move with the current through both resistors, experiencing voltage drops of -I(2.0) and -I(6.0). Choice C incorrectly uses positive signs for the resistor terms, revealing a sign convention error when the traverse direction matches the current direction. The reliable approach is to establish your traverse direction, then consistently apply signs: +ε for batteries traversed from - to +, and -IR for resistors traversed in the direction of current flow.

This question tests Kirchhoff's loop rule. Kirchhoff's loop rule requires that the sum of all voltage changes around any closed loop equals zero, reflecting the conservation of energy for charges moving through the circuit. Traversing clockwise, we encounter the 14V battery from negative to positive (+14V), drop voltage across R₁ = 4Ω (-4I), cross the 4V battery from negative to positive (+4V), and drop voltage across R₂ = 6Ω (-6I). The complete equation is: 14 - 4I + 4 - 6I = 0, which simplifies to 14 + 4 - I(4+6) = 0. Choice A incorrectly subtracts the 4V battery voltage, demonstrating the misconception of assuming all batteries after the first must be subtracted. Always determine signs based on traversal direction: negative to positive gives a positive voltage change regardless of battery position.

This question involves Kirchhoff's loop rule. Kirchhoff's loop rule requires that the algebraic sum of voltage changes around any closed loop equals zero, reflecting energy conservation. Starting at the positive terminal and traversing clockwise: we immediately cross the battery from + to - (-15V), then cross R₁ along current (-2.0I), then cross R₂ along current (-7.0I). The equation is -15 - 2.0I - 7.0I = 0. Choice B (+15 - 2.0I - 7.0I = 0) uses the wrong sign for the battery, showing the misconception that battery voltage is always positive. Remember that the sign depends on traversal direction: crossing from + to - gives -ℰ.

This question applies Kirchhoff's loop rule. Kirchhoff's loop rule states that the sum of voltage changes around any closed loop must equal zero, ensuring energy conservation in circuits. Traversing clockwise: ℰ₁ is crossed from + to - (-2.0V), ℰ₂ is crossed from - to + (+7.0V), and the resistor is crossed along current (-10I). The equation is -2.0 + 7.0 - 10I = 0. Choice D (-2.0 + 7.0 + 10I = 0) incorrectly uses +10I for the resistor term, showing the misconception that sign depends only on current direction, not traversal direction. Remember that crossing a resistor in the current direction always gives -IR in the loop equation.

This question applies Kirchhoff's loop rule. Kirchhoff's loop rule states that the algebraic sum of voltage changes around any closed loop equals zero, ensuring energy conservation in circuits. Starting at the battery's negative terminal and traversing counterclockwise (same as current direction): we cross the battery from - to + (+6.0V), then cross R₁ along the current (-2.0I), then cross R₂ along the current (-1.0I). The equation is +6.0 - 2.0I - 1.0I = 0. Choice D (+6.0 - 2.0I = 0) omits R₂, showing the misconception of incomplete loop traversal. To correctly apply the loop rule, traverse the entire loop and account for every component, using consistent sign conventions throughout.