Consider the circuit in Figure21.53, and suppose that the emfs are unknown and the currents are given to be${\mathbf{I}}_{\mathbf{1}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{A}\mathbf{}\mathbf{,}\mathbf{}{\mathbf{I}}_{\mathbf{2}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{A}\mathbf{}\mathbf{,}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{I}}_{\mathbf{3}}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{A}$. (a) Could you find the emfs? (b) What is wrong with the assumptions?

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

(a)

Formula Used:

According to Kirchhoff's loop rule, the total voltage differences across all circuit elements must be zero in any closed electrical loop.

Calculation:

We can use Kirchhoff's loop rule to solve the loop abcdea.

$‐{\mathrm{I}}_2{\mathrm{R}}_2+{\mathrm{E}}_1‐{\mathrm{I}}_2{\mathrm{r}}_1‐{\mathrm{I}}_1{\mathrm{R}}_1=0$

We have

$$\begin{gathered} {\mathrm{I}}_1=5.00\mathrm{A} \\ {\mathrm{I}}_2=3.00\mathrm{A} \\ {\mathrm{r}}_2=0.5\mathrm{\Omega } \\ {\mathrm{R}}_1=6\mathrm{\Omega } \\ {\mathrm{R}}_2=2.5\mathrm{\Omega } \end{gathered}$$

Substituting these values in the above expression, we have

$$\begin{gathered} ‐\left(3\mathrm{A}\right)\left(2.5\mathrm{\Omega }\right)+{\mathrm{E}}_1‐\left(3\mathrm{A}\right)\left(5\mathrm{A}\right)\left(6\mathrm{\Omega }\right)=0 \\ {\mathrm{E}}_1‐39.0\mathrm{V}=0 \\ {\mathrm{E}}_1=39.0\mathrm{V} \end{gathered}$$

Using Kirchhoff's loop rule to solve the loop aefgha, we get

${\mathrm{I}}_1{\mathrm{R}}_1+{\mathrm{I}}_3{\mathrm{R}}_3+{\mathrm{I}}_3{\mathrm{r}}_2‐{\mathrm{E}}_2=0$

We have

role="math" localid="1656391767491" $$\begin{gathered} {\mathrm{I}}_1=5.00\mathrm{A} \\ {\mathrm{I}}_2=3.00\mathrm{A} \\ {\mathrm{r}}_2=0.5\mathrm{\Omega } \\ {\mathrm{R}}_1=6\mathrm{\Omega } \\ {\mathrm{R}}_3=1.5\mathrm{\Omega } \end{gathered}$$

Substituting the values, we have

$$\begin{gathered} \left(5\mathrm{A}\right)\left(6\mathrm{\Omega }\right)+\left(‐2\mathrm{A}\right)\left(1.5\mathrm{\Omega }\right)+\left(‐2\mathrm{A}\right)\left(0.5\mathrm{\Omega }\right)‐{\mathrm{E}}_2=0 \\ 30.0\mathrm{V}‐3.0\mathrm{V}‐1.0\mathrm{V}-{\mathrm{E}}_2=0 \\ 26.0\mathrm{V}-{\mathrm{E}}_2=0 \\ {\mathrm{E}}_2=26.0\mathrm{V} \end{gathered}$$

Conclusion:

Therefore, the value of ${\mathrm{E}}_1=39.0\mathrm{V}\mathrm{and}{\mathrm{E}}_2=26.0\mathrm{V}$

(b)

Given:

The currents are

$$\begin{gathered} {\mathrm{I}}_1=5.00\mathrm{A} \\ {\mathrm{I}}_2=3.00\mathrm{A} \\ {\mathrm{I}}_3=2.00\mathrm{A} \end{gathered}$$

Formula Used:

According to Kirchhoff's loop rule, in any closed electrical loop, the sum of voltage differences across all of the circuit elements must be zero

Calculation:

Applying the junction rule for point a

${\mathrm{I}}_1={\mathrm{I}}_2+{\mathrm{I}}_3$

Since ${\mathrm{I}}_1$is the sum of ${\mathrm{I}}_2$and ${\mathrm{I}}_3$

Thus, ${\mathrm{I}}_3$should be +2 A instead of -2 A

Or, reverse the current ${\mathrm{I}}_1$to make the assumption true.