This question applies Kirchhoff's junction rule. The junction rule embodies charge conservation: at any junction, the total current flowing in must equal the total current flowing out, as charge cannot accumulate at a point. Here, currents I_A = 0.40 A and I_B = 0.70 A enter junction J, providing total entering current of 1.10 A. Currents I_C and I_D = 0.50 A leave the junction, so their sum must equal 1.10 A. Choice C incorrectly adds all currents regardless of direction, demonstrating a fundamental misunderstanding of how current conservation works at junctions. Applying the rule: I_A + I_B = I_C + I_D, which gives I_C = I_A + I_B - I_D. Remember: always separate entering from leaving currents before applying the junction rule.

This problem tests understanding of Kirchhoff's junction rule. At any junction in a circuit, the total current entering must equal the total current leaving, which reflects conservation of charge—charge cannot accumulate at a junction. Here, currents I₁ = 2.0 A and I₂ = 1.5 A enter junction J, giving a total entering current of 3.5 A. Currents I₃ and I₄ = 0.5 A leave the junction, so the total leaving current must also be 3.5 A. Choice C incorrectly assumes all currents are equal, which represents a misconception that current is the same everywhere in a circuit. Setting entering currents equal to leaving currents: I₁ + I₂ = I₃ + I₄, which gives I₃ = I₁ + I₂ - I₄. Remember: at any junction, sum of currents in equals sum of currents out.

This problem involves Kirchhoff's junction rule. At a junction, charge conservation requires that the rate of charge entering (total current in) equals the rate of charge leaving (total current out), preventing charge accumulation. In this scenario, currents I₁ = 3.0 A and I₄ = 0.5 A enter junction J, giving total entering current of 3.5 A. Currents I₂ = 1.0 A and I₃ leave the junction, so their sum must also equal 3.5 A. Choice B incorrectly subtracts entering currents from leaving currents, revealing confusion about the direction-dependent nature of the junction rule. Setting entering equal to leaving: I₁ + I₄ = I₂ + I₃, which rearranges to I₃ = I₁ + I₄ - I₂. Key strategy: carefully track current directions and balance entering versus leaving currents.

This question tests Kirchhoff's junction rule. The junction rule ensures charge conservation by requiring that the total current entering any junction equals the total current leaving it—charge cannot build up or disappear at a junction. Here, currents I₃ = 0.9 A and I₄ enter junction J, while currents I₁ = 1.2 A and I₂ = 0.6 A leave, giving a total leaving current of 1.8 A. Therefore, the total entering current must also be 1.8 A. Choice C incorrectly adds all currents together, showing a misconception that fails to distinguish between entering and leaving currents. Applying conservation: I₃ + I₄ = I₁ + I₂, which gives I₄ = I₁ + I₂ - I₃. Strategy: list entering currents on one side and leaving currents on the other, then solve.

This question tests Kirchhoff's junction rule. The junction rule ensures charge conservation at circuit junctions: the total current entering must equal the total current leaving, preventing charge accumulation. Here, currents I_A = 1.8 A and I_B = 0.4 A enter junction J, giving total entering current of 2.2 A. Currents I_C = 0.9 A and I_D leave the junction, so their sum must also equal 2.2 A. Choice C incorrectly adds all currents without considering direction, a common error when students don't recognize that entering and leaving currents play opposite roles. Applying the rule: I_A + I_B = I_C + I_D, which gives I_D = I_A + I_B - I_C. Strategy: list all entering currents, all leaving currents, then set their sums equal.

This question applies Kirchhoff's junction rule. Kirchhoff's junction rule states that the algebraic sum of currents at any junction must be zero, reflecting the conservation of electric charge. Here, $I_1 = 1.2$ A enters, while $I_2 = 0.4$ A, $I_3 = 0.5$ A, and $I_4$ all leave the junction. Setting up the conservation equation: current in = current out gives us $1.2 = 0.4 + 0.5 + I_4$, which simplifies to $1.2 = 0.9 + I_4$, yielding $I_4 = 0.3$ A. Choice B ($I_4 = 2.1$ A) likely results from adding all currents instead of balancing inflow and outflow. Remember: at any junction, the sum of entering currents must equal the sum of leaving currents.

This question applies Kirchhoff's junction rule to three currents. Kirchhoff's junction rule states that charge is conserved at junctions—current flowing in must equal current flowing out. At junction J, currents I₁ = 0.30 A and I₂ = 0.10 A enter, while I₃ leaves upward. Therefore: I₁ + I₂ = I₃, giving 0.30 + 0.10 = I₃, so I₃ = 0.40 A. Choice B (I₃ = I₁ - I₂) incorrectly subtracts the two entering currents, misunderstanding that both contribute to the outgoing current. Remember: at any junction, add all entering currents and set equal to the sum of leaving currents.

This problem requires applying Kirchhoff's junction rule. Kirchhoff's junction rule states that the total current entering a junction must equal the total current leaving the junction, which reflects conservation of charge. At junction J, currents I₁ = 2.0 A and I₂ = 1.5 A enter, while currents I₃ and I₄ = 0.5 A leave. Setting current in equal to current out: I₁ + I₂ = I₃ + I₄, which gives 2.0 + 1.5 = I₃ + 0.5, so I₃ = 3.0 A. Choice C incorrectly assumes all currents are equal, which violates the physical constraint that charge cannot accumulate at a junction. Remember: at any junction, sum of currents in equals sum of currents out.

This problem applies Kirchhoff's junction rule. The junction rule reflects charge conservation: at any junction, the sum of entering currents must equal the sum of leaving currents, as charge cannot accumulate at a point. In this scenario, only I₁ = 0.60 A enters junction J, while currents I₂ = 0.10 A, I₃, and I₄ = 0.20 A all leave the junction. The total leaving current must equal 0.60 A. Choice B incorrectly adds all currents together, showing confusion about the directional nature of current flow at junctions. Setting entering equal to leaving: I₁ = I₂ + I₃ + I₄, which rearranges to I₃ = I₁ - I₂ - I₄. Remember: always separate currents by direction (in vs. out) before applying the junction rule.

This problem tests Kirchhoff's junction rule. At any junction, the total current entering must equal the total current leaving, which expresses conservation of charge—charge cannot accumulate at a junction. Here, currents $I_1 = 2.0$ A and $I_2 = 1.5$ A enter the junction, while $I_3$ and $I_4 = 0.5$ A leave. Setting current in equal to current out: $I_1 + I_2 = I_3 + I_4$, which gives $2.0 + 1.5 = I_3 + 0.5$, so $I_3 = 3.0$ A. Choice D incorrectly assumes all currents are equal, ignoring that different branches can carry different currents. Remember: at any junction, sum of currents in equals sum of currents out.