What must the emf Ԑ in Fig P26.60 be in order for the current through the 7.00 Ω resistor to be 1.80 A? Each emf source has negligible internal resistance.

Kirchoff’s law state that at any junction the algebraic sum of all currents should be always equal to zero.

The voltage around a loop equals the sum of every voltage drop in the same loop for any closed network and equals zero

Consider the following circuit, where ${\epsilon }^{\text{'}}=24.0\hspace{0.33em}V,\hspace{0.33em}{R}_1=2.00\hspace{0.33em}\Omega ,\hspace{0.33em}{R}_2=3.00\hspace{0.33em}\Omega ,\hspace{0.33em}and\hspace{0.33em}{R}_3=7.00\hspace{0.33em}\Omega$we need to find the emf $\epsilon$, such that the current flowing in localid="1668332814111" $R_2=\frac{3R_1}{2}and{R}_3=\frac{7R_1}{2}$is = 1.80 A. First apply the loop rule to the left loop(clockwise), to get

${\epsilon }^{\text{'}}-\epsilon +I_1{R}_1-I_2{R}_2=0$ (1)

And to the right loop (also clockwise) to get,

$\epsilon -I_3{R}_3-I_1{R}_1=0$ (2)

And from the junction rule, we can write,

$I_1+I_2=I_3$ (3)

To simplify the problem, we will use the substitutions $R_2=\frac{3R_1}{2}and{R}_3=\frac{7R_1}{2}$

$\left\{\begin{array}{l}{\epsilon }^{\mathrm{\prime }}-\epsilon +{\mathrm{I}}_1{\mathrm{R}}_1-\frac{3{\mathrm{I}}_2{\mathrm{R}}_1}{2}=0\\ \epsilon -\frac{7{\mathrm{I}}_3{\mathrm{R}}_1}{2}-{\mathrm{I}}_1{\mathrm{R}}_1=0\\ {\mathrm{I}}_1+{\mathrm{I}}_2={\mathrm{I}}_3\end{array}.\right.$……………………...(4)

Substitute the third equation in (4) into the first one to eliminate $l_2$

$\begin{array}{r}{\epsilon }^{\mathrm{\prime }}-\epsilon +I_1{R}_1-\frac{3R_1}{2}\left(I_3-I_1\right)=0\\ {\epsilon }^{\mathrm{\prime }}-\epsilon +\frac{5I_1{R}_1}{2}-\frac{3I_3{R}_1}{2}=0\end{array}$(5)

Multiply the second equation in (4) by factor 5/2 so we get,

$\frac{5\epsilon }{2}-\frac{35l_3{R}_1}{4}-\frac{5l_1{R}_1}{2}=0$ (6)

Adding equations (5) and (6), we get,

$$\begin{gathered} {\epsilon }^{\mathrm{\prime }}-\epsilon +\frac{5I_1{R}_1}{2}-\frac{3I_3{R}_1}{2}+\frac{5\epsilon }{2}-\frac{35I_3{R}_1}{4}-\frac{5I_1{R}_1}{2}=0 \\ {\epsilon }^{\mathrm{\prime }}+\frac{3\epsilon }{2}-\frac{{41}_3{R}_1}{4}=0 \\ \epsilon =\frac{{41}_3{R}_1}{6}-\frac{2{\epsilon }^{\mathrm{\prime }}}{3} \end{gathered}$$

Substitute with given data in the above expression, and we get,

$$\begin{gathered} \epsilon =\frac{4{\mathbb{1}}_3{\mathrm{R}}_1}{6}-\frac{2{\epsilon }^{\mathrm{\prime }}}{3} \\ =\frac{41\times 1.80\mathrm{A}\times 2.00\mathrm{\Omega }}{6}-\frac{2\times 24.0\mathrm{V}}{3} \\ =8.60\mathrm{V} \end{gathered}$$

Hence, the EMF is 8.60 V.