Question

The ammeter in FIGURE P28.52 reads$3.0\mathrm{A}$. Find ${\mathrm{I}}_1,{\mathrm{I}}_2,$and $\mathrm{\epsilon }.$

Short answer

The value of

$$\begin{gathered} {\mathrm{I}}_1=5\mathrm{A} \\ {\mathrm{I}}_2=8\mathrm{A} \\ \mathrm{\epsilon }=14\mathrm{V}. \end{gathered}$$

Step-by-step solution

Given information

We have given,

Ammeter reading = $3\mathrm{A}$

We have to find the value of${\mathrm{I}}_1,{\mathrm{I}}_2,\mathrm{\epsilon }.$

Simplify

Using the Kirchhoff voltage law in the upper loop

$$\begin{gathered} -9\mathrm{V}+3\Omega \times {\mathrm{I}}_1=2\Omega \times 3\mathrm{A} \\ {\mathrm{I}}_1=\frac{9+6}{3} \\ {\mathrm{I}}_1=5\mathrm{A} \end{gathered}$$

Using Kirchhoff current rule at the node and substitute the value of${\mathrm{I}}_1.$

localid="1649416745087" $$\begin{gathered} {\mathrm{I}}_2={\mathrm{I}}_1+3\mathrm{A} \\ {\mathrm{I}}_2=5\mathrm{A}+3\mathrm{A}=8\mathrm{A} \end{gathered}$$

Using Kirchhoff voltage rule in the lower loop

localid="1649416765380" $$\begin{gathered} (2\Omega \times 3\mathrm{A})-\mathrm{\epsilon }+(1\Omega \times {\mathrm{I}}_2)=0 \\ \mathrm{\epsilon }=6+8 \\ \mathrm{\epsilon }=14\mathrm{V} \end{gathered}$$