Question

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.

Short answer

The EMF Ԑ is 8.60 V.

Step-by-step solution

Concept Introduction

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

Application of Kirchoff’s law

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.