AP Physics C- E&M- 11.7 Kirchhoff’s Junction Rule - Exam Style questions - FRQs- New Syllabus

This problem analyzes multi-loop configurations containing resistors, batteries, and eventually reactive components like capacitors and inductors under steady-state conditions. By carefully establishing equations using Kirchhoff’s laws and applying parallel/series simplification, we can systematically solve each dynamic state of the circuit.

(a)i. Using the junction rule at the top middle node: $I_1 + I_3 = I_2$
Applying the loop rule to the left loop (clockwise): $6.0 – 150I_1 – 200I_2 = 0$
Applying the loop rule to the right loop (counterclockwise): $6.0 – 100I_3 – 200I_2 = 0$

(a)ii. We solve the system of linear equations for the target branch current:
From the loop equations, we isolate $I_1$ and $I_3$:
$I_1 = \frac{6.0 – 200I_2}{150}$
$I_3 = \frac{6.0 – 200I_2}{100}$
Substitute these into the junction rule:
$\frac{6.0 – 200I_2}{150} + \frac{6.0 – 200I_2}{100} = I_2$
Multiply through by the common denominator 300:
$2(6.0 – 200I_2) + 3(6.0 – 200I_2) = 300I_2$
$12.0 – 400I_2 + 18.0 – 600I_2 = 300I_2$
$30.0 = 1300I_2$
$I_2 = \frac{30.0}{1300} \approx 0.023\text{ A}$
$\boxed{I_2 = 0.023\text{ A}}$

(a)iii. The electric power dissipated by a resistor is calculated via Joule’s law:
$P = I_2^2 R = (0.0231\text{ A})^2 \times 200\text{ }\Omega$
$P = 0.000533 \times 200 \approx 0.107\text{ W}$
$\boxed{P = 0.107\text{ W}}$

(b) In Figure 2, the $200\text{ }\Omega$ and the right branch are in parallel. Let’s find the voltage across the right branch components:
The voltmeter reads $4.4\text{ V}$ across the $200\text{ }\Omega$ resistor, which means the combined parallel section of $200\text{ }\Omega$ and the $(100\text{ }\Omega + 50\text{ }\Omega)$ branch also has a total potential drop of $4.4\text{ V}$.
The current running through the rightmost branch containing the $50\text{ }\Omega$ resistor is:
$I_{\text{right}} = \frac{V_{\text{parallel}}}{100 + 50} = \frac{4.4\text{ V}}{150\text{ }\Omega}$
$I_{\text{right}} \approx 0.0293\text{ A}$
$\boxed{I_{50\Omega} = 0.029\text{ A}}$

(c) To determine the total source electromotive force, we find the total current coming from the battery:
The current through the central $200\text{ }\Omega$ resistor is $I_{\text{center}} = \frac{4.4\text{ V}}{200\text{ }\Omega} = 0.022\text{ A}$.
Therefore, total current from the battery is $I_{\text{total}} = I_{\text{center}} + I_{\text{right}} = 0.022\text{ A} + 0.0293\text{ A} = 0.0513\text{ A}$.
Applying Kirchhoff’s loop rule from the battery through the main path:
$\mathcal{E} = I_{\text{total}} \times 150 + V_{\text{parallel}}$
$\mathcal{E} = (0.0513\text{ A} \times 150\text{ }\Omega) + 4.4\text{ V} = 7.7\text{ V} + 4.4\text{ V} = 12.1\text{ V}$
$\boxed{\mathcal{E} = 12.1\text{ V}}$

(d)i. In a steady-state DC circuit, a capacitor behaves as an open circuit (infinite resistance):
Since no current flows into the middle branch, the circuit reduces to a simple single-loop path with the $150\text{ }\Omega$, $100\text{ }\Omega$, and $50\text{ }\Omega$ resistors connected in series.
$R_{\text{eq}} = 150 + 100 + 50 = 300\text{ }\Omega$
$I = \frac{\mathcal{E}}{R_{\text{eq}}} = \frac{12.1\text{ V}}{300\text{ }\Omega} \approx 0.040\text{ A}$
$\boxed{I_{50\Omega} = 0.040\text{ A}}$

(d)ii. Correct choice:
$\boxed{\text{Greater than}}$
In a steady state under constant DC voltage conditions, an ideal inductor acts as a short circuit with zero resistance. This bypasses the right-hand branch completely, causing all current to flow exclusively through the central loop and leaving zero current through the $50\text{ }\Omega$ resistor, which is less than the current found in part (b).