This question involves Kirchhoff's loop rule. Kirchhoff's loop rule requires the sum of all potential differences around a closed loop to equal zero, representing energy conservation. Traversing clockwise with clockwise current I: the 20V battery crossed from - to + gives +20V, the 8V battery crossed from + to - (opposing) gives -8V, and the 5Ω resistor traversed with current gives -5I. The equation is +20 - 8 - 5I = 0. Choice A (+20 + 8 - 5I = 0) incorrectly adds both battery voltages, missing that opposing batteries subtract. When batteries oppose in a circuit, pay careful attention to their polarities relative to your traversal direction.

This question involves Kirchhoff's loop rule. Kirchhoff's loop rule ensures that the total energy change around any closed path is zero, reflecting conservation of energy in electrical circuits. Traversing clockwise, we cross the 6V battery from negative to positive (+6V), then encounter three resistors in series: R₁ = 1Ω (-1I), R₂ = 2Ω (-2I), and R₃ = 3Ω (-3I), all producing voltage drops since we traverse in the current direction. The complete loop equation is: 6 - 1I - 2I - 3I = 0, which simplifies to 6 - I(1+2+3) = 0. Choice D incorrectly treats the resistors as if they have different currents, revealing the misconception that series resistors can have different currents flowing through them. Remember that in a series circuit, the same current flows through all components.

This problem requires applying Kirchhoff's loop rule. Kirchhoff's loop rule ensures that the total voltage change around any closed loop equals zero, conserving energy in the circuit. Traversing clockwise with clockwise current, we cross the 18V battery from negative to positive (+18V), then R₁ = 3Ω with the current (-3I), and R₂ = 6Ω with the current (-6I). The complete equation is 18 - 3I - 6I = 0. Choice D (18 - 6I = 0) omits the first resistor, suggesting a misconception that resistors in series can be selectively ignored. To avoid errors, methodically account for every component in the loop using consistent sign conventions.

This question requires applying Kirchhoff's loop rule. Kirchhoff's loop rule states that the total voltage change around any closed loop must be zero, ensuring energy conservation. Traversing counterclockwise with counterclockwise current, both batteries are crossed from negative to positive (ℰ₁: +5V, ℰ₂: +7V), and the 4Ω resistor is traversed with the current (-4I). The equation becomes 5 + 7 - 4I = 0. Choice B (5 - 7 - 4I = 0) incorrectly subtracts the second battery voltage, indicating confusion about applying sign conventions when both batteries aid the current. Remember that the sign depends on how you traverse each component, not on their relative values.

This problem tests understanding of Kirchhoff's loop rule. Kirchhoff's loop rule ensures that the algebraic sum of all voltage changes around a closed loop equals zero, conserving energy. Traversing clockwise with clockwise current, we cross the 15V battery from negative to positive (+15V), then encounter three resistors in series, each producing a voltage drop: R₁ (-1I), R₂ (-2I), and R₃ (-3I). The equation can be written as 15 - 1I - 2I - 3I = 0 or 15 - (1+2+3)I = 0. Choice C (15 - 1I - 2I = 0) omits the third resistor, revealing a misconception about accounting for all components. Always include every element in the loop, and remember that series resistors can be combined algebraically.

This problem tests understanding of Kirchhoff's loop rule. Kirchhoff's loop rule ensures that the algebraic sum of voltage changes around a closed loop equals zero, maintaining energy conservation. Traversing clockwise, we cross ℰ₁ = 9V from negative to positive (+9V), then ℰ₂ = 3V from positive to negative (-3V), and finally the resistor R = 6Ω with the current (-6I). The correct equation is 9 - 3 - 6I = 0. Choice B (9 + 3 - 6I = 0) incorrectly adds both battery voltages, revealing a misconception about sign conventions when traversing batteries in different directions. Remember to carefully track the polarity of each battery relative to your traversal direction.

This question involves applying Kirchhoff's loop rule. Kirchhoff's loop rule states that the sum of all voltage changes around a closed loop must be zero, reflecting the conservation of energy. When traversing counterclockwise with a counterclockwise current, we encounter the 6V battery from negative to positive (+6V), then pass through R₁ = 1Ω with the current (-1I or simply -I), and through R₂ = 5Ω with the current (-5I). The equation becomes 6 - I - 5I = 0. Choice D (6 + I + 5I = 0) incorrectly uses positive signs for the resistor terms, indicating a sign convention error when current and traversal directions match. Always use negative signs for voltage drops across resistors when moving with the current direction.

This problem involves applying Kirchhoff's loop rule. Kirchhoff's loop rule ensures that the sum of all voltage changes around a closed loop equals zero, reflecting energy conservation in electrical circuits. Traversing clockwise with clockwise current, we cross the 8V battery from negative to positive (+8V), then R₁ = 2Ω with the current (-2I), and R₂ = 2Ω with the current (-2I). The equation is 8 - 2I - 2I = 0. Choice B (8 - 2I + 2I = 0) incorrectly uses opposite signs for identical resistors, suggesting a misconception that resistors can have different sign conventions in the same loop. Apply the same sign convention consistently: voltage drops are negative when moving with the current.

This question tests Kirchhoff's loop rule application. Kirchhoff's loop rule states that the algebraic sum of voltage changes around a closed loop must equal zero, ensuring energy conservation. Traversing clockwise, we encounter the 10V battery from negative to positive (+10V), the 4V battery from positive to negative (-4V), and the 5Ω resistor with the current (-5I). The equation becomes 10 - 4 - 5I = 0. Choice B (10 + 4 - 5I = 0) incorrectly adds both battery voltages, revealing confusion about sign conventions when batteries oppose each other. Always determine each battery's contribution based on how you traverse it, not on their relative orientations.

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. When traversing the loop clockwise, we encounter the battery from negative to positive terminal (+12V), then pass through R₁ = 4Ω with the current (-4I), and finally through R₂ = 2Ω with the current (-2I). The equation becomes 12 - 4I - 2I = 0. Choice C incorrectly omits the second resistor, suggesting a misconception that only one resistor contributes to the voltage drop. To solve these problems correctly, always traverse the entire loop systematically, applying consistent sign conventions for each component.