AP Physics 2: Algebra-Based Practice Quiz: Kirchhoff's Junction Rule

Correct Answer: B

As stated in the provided content, Kirchhoff's junction rule is a consequence of the conservation of electric charge. This means that charge cannot be created or destroyed at a junction.

Correct Answer: B

According to Kirchhoff's Junction Rule, the total current entering must equal the total current exiting ($\sum I_{in}=\sum I_{out}$). Therefore, 7 A (in) = 3 A (out) + I_2 (out). Solving for I_2 gives 4 A.

Correct Answer: A

The rule states that the sum of currents entering a junction must equal the sum of currents exiting it ($\sum I_{in}=\sum I_{out}$). In this case, I_1 and I_2 are entering, and I_3 is exiting, so I_1 + I_2 = I_3.

Correct Answer: B

Using the equation $\sum I_{in}=\sum I_{out}$, the sum of incoming currents is 2 A + 5 A = 7 A. The sum of outgoing currents is 1 A + I_x. Setting them equal gives 7 A = 1 A + I_x. Solving for I_x yields 6 A.

Correct Answer: C

This is the definition provided in the content. The rule states that the total amount of charge entering a junction per unit time (which is current) must equal the total amount of charge exiting that junction per unit time.

Correct Answer: B

Applying Kirchhoff's Junction Rule, $\sum I_{in}=\sum I_{out}$. The total incoming current is 10 A. The total outgoing current is the sum of the currents in the three paths: 4 A + 2 A + I_3. Therefore, 10 A = 6 A + I_3, which means I_3 = 4 A.

Correct Answer: C

The total current in is 5 A + 3 A = 8 A, while the total current out is 9 A. Since $\sum I_{in}$ does not equal $\sum I_{out}$, this violates Kirchhoff's Junction Rule. The rule itself is based on the conservation of electric charge, so this observation implies charge is not being conserved.

Correct Answer: C

The provided content explicitly states the relevant equation for Kirchhoff's Junction Rule is $\sum I_{in}=\sum I_{out}$, which means the sum of currents flowing into a junction equals the sum of currents flowing out.

Correct Answer: B

First, sum the known incoming currents: 3 A (B) + 2 A (C) = 5 A. The known outgoing current is 8 A (A). According to the rule $\sum I_{in}=\sum I_{out}$, we have 5 A = 8 A + I_D (assuming I_D is out) OR 5 A + I_D = 8 A (assuming I_D is in). Let's assume I_D is an outgoing current. The equation is 5 A = 8 A + I_D, which gives I_D = -3 A. The negative sign means our assumption was wrong, and the current is actually 3 A flowing in. Let's try the other way: assume I_D is an incoming current. The equation is 5 A + I_D = 8 A. This gives I_D = 3 A. Since the result is positive, our assumption was correct. The current is 3 A flowing in. Wait, let me re-calculate. $\sum I_{in} = 3 A + 2 A = 5 A$. $\sum I_{out} = 8 A$. To balance, we need more current flowing IN. Let's set up the equation: $\sum I_{in} = \sum I_{out}$. Let's say $I_D$ is the unknown. $I_B + I_C + ... = I_A + ...$. So, $3A + 2A = 8A$. This is not balanced. We need to add $I_D$. If $I_D$ is IN: $3A + 2A + I_D = 8A ightarrow 5A + I_D = 8A ightarrow I_D = 3A$ flowing IN. If $I_D$ is OUT: $3A + 2A = 8A + I_D ightarrow 5A = 8A + I_D ightarrow I_D = -3A$. A negative outgoing current is the same as a positive incoming current. So, the current is 3 A flowing in. My initial correct answer was A. Let me re-read the question and options. Ah, I made a mistake in my final check. The answer is 3 A flowing in. Let's re-write the explanation. Total known IN: 3 A + 2 A = 5 A. Total known OUT: 8 A. To satisfy $\sum I_{in}=\sum I_{out}$, the two sides must be equal. Since the OUT side is larger, the unknown current in Wire D must be flowing IN. The equation becomes: (3 A + 2 A + I_D) = 8 A. This simplifies to 5 A + I_D = 8 A, so I_D = 3 A. The direction is flowing in.