Worked exercise — three-branch circuit with two sources

Problem

A circuit has a node A (top) and a node B (bottom), connected by three branches in parallel: branch 1 with $\mathcal{E}_1 = 10\;\text{V}$ and $R_1 = 10\;\Omega$; branch 2 with $R_2 = 20\;\Omega$; branch 3 with $\mathcal{E}_2 = 20\;\text{V}$ and $R_3 = 30\;\Omega$. Find all the currents $i_1$, $i_2$, $i_3$.

Three branches in parallel between nodes A and B.

Solution

Node A (currents $i_1$, $i_2$ entering, $i_3$ leaving): $i_1 + i_2 - i_3 = 0 \quad\Rightarrow\quad i_3 = i_1 + i_2$

Loops (currents leaving A): $\mathcal{E}_1 - R_1\,i_1 - R_2\,i_2 = 0 \;\Rightarrow\; 10 - 10\,i_1 - 20\,i_2 = 0$ $R_2\,i_2 - \mathcal{E}_2 + R_3\,i_3 = 0 \;\Rightarrow\; 20\,i_2 - 20 + 30\,i_3 = 0$

Substituting $i_3 = i_1 + i_2$: $10\,i_1 + 20\,i_2 = 10 \qquad 20\,i_2 + 30\,(i_1 + i_2) = 20 \;\Rightarrow\; 30\,i_1 + 50\,i_2 = 20$

From the first, $i_1 = 1 - 2\,i_2$. Substituting into the second: $30(1 - 2i_2) + 50\,i_2 = 20 \;\Rightarrow\; 30 - 60\,i_2 + 50\,i_2 = 20 \;\Rightarrow\; -10\,i_2 = -10 \;\Rightarrow\; i_2 = 1\;\text{A}$

So: $i_1 = 1 - 2\cdot 1 = -1\;\text{A} \qquad i_3 = i_1 + i_2 = -1 + 1 = 0\;\text{A}$ The negative sign on $i_1$ indicates that the source $\mathcal{E}_1$ is traversed in the direction opposite to the one assumed.

$\ev{i_1 = -1\;\text{A},\quad i_2 = 1\;\text{A},\quad i_3 = 0\;\text{A}}$

Links

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