Worked exercise — two generators, two loops

Problem

A circuit has two generators $\mathcal{E}_1 = 12\;\text{V}$, $\mathcal{E}_2 = 5\;\text{V}$ and two resistors $R_1 = 100\;\Omega$, $R_2 = 50\;\Omega$. The currents $i_1$ and $i_2$ flow through two loops that share resistor $R_2$. Find $i_1$, $i_2$ and the total current $i$.

Two loops sharing the central branch with $R_2$.

Solution

Setup. Traversing the two loops in the conventional direction, Kirchhoff’s second law gives: $\text{Loop 1:}\quad -R_2\,i_1 + R_1\,i_2 + \mathcal{E}_1 = 0$ $\text{Loop 2:}\quad -R_2\,i_1 + \mathcal{E}_2 = 0$

From the second equation $i_1$ follows at once: $i_1 = \frac{\mathcal{E}_2}{R_2} = \frac{5}{50} = 0{,}10\;\text{A}$

Substituting into the first: $-50\cdot 0{,}10 + 100\,i_2 + 12 = 0 \;\Rightarrow\; 100\,i_2 = -7 \;\Rightarrow\; i_2 = -\frac{7}{100} = -0{,}07\;\text{A}$ The negative sign indicates that $i_2$ flows in the direction opposite to the one assumed.

Total current at the node: $i = i_1 + i_2 = 0{,}10 - 0{,}07 = 0{,}03\;\text{A}$

Check (loop 1): $-50\cdot 0{,}10 + 100\cdot(-0{,}07) + 12 = -5 - 7 + 12 = 0$. Correct.

$\ev{i_1 = 0{,}10\;\text{A},\quad i_2 = -0{,}07\;\text{A},\quad i = 0{,}03\;\text{A}}$

Links

Topics: Electric circuits Concepts: Kirchhoff’s laws Skills: Applying Kirchhoff’s laws