Problem — Kirchhoff with two loops

Problem

Two loops share a resistor $R_c = 2\ \Omega$. Loop 1 contains a battery $\mathcal{E}_1 = 9\ \text{V}$ and a resistor $R_1 = 3\ \Omega$; loop 2 contains a battery $\mathcal{E}_2 = 6\ \text{V}$ and a resistor $R_2 = 4\ \Omega$. The polarities are oriented towards the common node. Determine the currents $I_1$ and $I_2$ in the two loops.

The two loops share the central resistor $R_c$ between nodes $A$ and $B$.

Solution

Setup — call $I_1$ the current that loop 1 pushes towards the common node $A$, and $I_2$ the one pushed by loop 2. Both batteries are oriented towards the common node, so in the shared resistor the two currents add up: by the junction rule, the current in $R_c$ is $I_1 + I_2$ (from $A$ towards $B$).

Loop law — going round each loop (battery $\to$ its own resistor $\to$ shared resistor): $\mathcal{E}_1 = R_1 I_1 + R_c (I_1 + I_2)$ $\mathcal{E}_2 = R_2 I_2 + R_c (I_1 + I_2)$

Substituting the values ($R_c = 2$): $9 = 3 I_1 + 2(I_1 + I_2) = 5 I_1 + 2 I_2$ $6 = 4 I_2 + 2(I_1 + I_2) = 2 I_1 + 6 I_2$

Solving — multiply the first by $3$ and subtract the second to eliminate $I_2$: $15 I_1 + 6 I_2 = 27 \quad\Rightarrow\quad (15 - 2) I_1 = 27 - 6 \quad\Rightarrow\quad 13 I_1 = 21$ $I_1 = \frac{21}{13} \approx 1{,}62\ \text{A}$

From the first equation: $I_2 = \frac{9 - 5 I_1}{2} = \frac{9 - 5\cdot 1{,}615}{2} \approx 0{,}46\ \text{A}$

The current in the shared resistor is $I_1 + I_2 \approx 2{,}08\ \text{A}$.

$\ev{I_1 \approx 1{,}62\ \text{A}, \quad I_2 \approx 0{,}46\ \text{A}}$

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Topics: Circuiti elettrici Concepts: Leggi di Kirchhoff Skills: Applicazione delle leggi di Kirchhoff