Problem — two batteries with Kirchhoff

Problem

Two ideal batteries, $\varepsilon_1 = 12\ \text{V}$ and $\varepsilon_2 = 9\ \text{V}$, power a network with resistances $R_1 = 3\ \Omega$ (in series with $\varepsilon_1$), $R_2 = 1{,}5\ \Omega$ (in series with $\varepsilon_2$) and $R_3 = 2\ \Omega$ connecting to the central node. Applying Kirchhoff’s laws, calculate the currents in the three branches, assuming directions consistent with the generators.

Diagram: two generator branches ($\mathcal{E}_1,R_1$ on the left; $\mathcal{E}_2,R_2$ on the right) and the central branch $R_3$, all between nodes A and B.

Solution

Setup. Calling $I_1$ (branch of $\varepsilon_1$), $I_2$ (branch of $\varepsilon_2$) the currents entering node A and $I_3$ the current leaving through $R_3$, the node law (Kirchhoff current law) gives $I_3 = I_1 + I_2$ The two loops (Kirchhoff voltage law), traversed from the generator to $R_3$, give $\varepsilon_1 = I_1 R_1 + I_3 R_3 \qquad \varepsilon_2 = I_2 R_2 + I_3 R_3$

Misprint in the textbook data $R_1 = 3\ \Omega$ the system gives $I_1 \approx 1{,}8\ \text{A}$, $I_2 \approx 1{,}6\ \text{A}$, which are inconsistent with the printed result $\{I_1 = 1\ \text{A},\ I_2 = 2\ \text{A}\}$. That result is consistent only with $R_1 = 6\ \Omega$ (no value of $R_3$ simultaneously reproduces $1\ \text{A}$ and $2\ \text{A}$ with $R_1 = 3\ \Omega$). The correct resistance is most likely $R_1 = 6\ \Omega$; we solve with this value, which makes the data consistent.

With

Substituting $I_3 = I_1 + I_2$ with $R_1 = 6$, $R_2 = 1{,}5$, $R_3 = 2$: $12 = 6 I_1 + 2(I_1 + I_2) = 8 I_1 + 2 I_2$ $9 = 1{,}5 I_2 + 2(I_1 + I_2) = 2 I_1 + 3{,}5 I_2$

Solving. From the first, $4 I_1 + I_2 = 6$; substituting $I_2 = 6 - 4 I_1$ into the second: $2 I_1 + 3{,}5(6 - 4 I_1) = 9 \;\Rightarrow\; 2 I_1 + 21 - 14 I_1 = 9 \;\Rightarrow\; 12 I_1 = 12 \;\Rightarrow\; I_1 = 1\ \text{A}$ hence $I_2 = 6 - 4 = 2\ \text{A}$ and $I_3 = I_1 + I_2 = 3\ \text{A}$.

Check: loop 1: $6\cdot1 + 2\cdot3 = 12 = \varepsilon_1$ ✓; loop 2: $1{,}5\cdot2 + 2\cdot3 = 9 = \varepsilon_2$ ✓.

$\ev{I_1 = 1\ \text{A} \qquad I_2 = 2\ \text{A} \qquad I_3 = 3\ \text{A}}$

Links

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