Question

Find the currents flowing in the circuit in Figure 21.47.

Short answer

The currents flowing in the circuit

$$\begin{gathered} {\mathrm{I}}_1=0.38\mathrm{A} \\ {\mathrm{I}}_2=-0.81\mathrm{A} \\ {\mathrm{I}}_3=1.2\mathrm{A} \end{gathered}$$

Step-by-step solution

Concept Introduction

Currentis a flow of electrical charge carriers, usually electrons or electron-deficient atoms. The common symbol for current is the uppercase letter I.

Explanation

Apply the Kirchhoff's law:

$$\begin{gathered} {\mathrm{I}}_1‐{\mathrm{I}}_2‐{\mathrm{I}}_3=0 \\ 18.0\mathrm{V}+0.5{\mathrm{I}}_1+20{\mathrm{I}}_1+6{\mathrm{I}}_3+3.0\mathrm{V}+0.25{\mathrm{I}}_3+15{\mathrm{I}}_1=0 \\ 3.0\mathrm{V}‐6{\mathrm{I}}_3+8{\mathrm{I}}_2‐12.0\mathrm{V}+0.5{\mathrm{I}}_2+0.75{\mathrm{I}}_2+24.0\mathrm{V}‐0.25{\mathrm{I}}_3=0 \end{gathered}$$

Therefore, by solving those equations, we get:

$$\begin{gathered} {\mathrm{I}}_1=0.38\mathrm{A} \\ {\mathrm{I}}_2=-0.81\mathrm{A} \\ {\mathrm{I}}_3=1.2\mathrm{A} \end{gathered}$$