In Fig. 27-84, ${\mathit{R}}_{\mathbf{1}}\mathbf{=}{\mathit{R}}_{\mathbf{2}}\mathbf{=}\mathbf{2}\mathbf{.0}\mathbf{}\mathit{\Omega }$,${\mathit{R}}_{\mathbf{3}}\mathbf{=}\mathbf{4}\mathbf{.0}\mathbf{}\mathit{\Omega }$ ,${\mathit{R}}_{\mathbf{4}}\mathbf{=}\mathbf{3}\mathbf{.0}\mathbf{}\mathit{\Omega }$ ,${\mathit{R}}_{\mathbf{5}}\mathbf{=}\mathbf{1}\mathbf{.0}\mathbf{}\mathit{\Omega }$ , and${\mathit{R}}_{\mathbf{6}}\mathbf{=}{\mathit{R}}_{\mathbf{7}}\mathbf{=}{\mathit{R}}_{\mathbf{8}}\mathbf{=}\mathbf{8}\mathbf{.0}\mathbf{}\mathit{\Omega }$ , and the ideal batteries have emfs${\mathit{e}}_{\mathbf{1}}\mathbf{=}\mathbf{16}\mathbf{}\mathit{V}$ and${\mathit{e}}_{\mathbf{2}}\mathbf{=}\mathbf{8}\mathbf{.0}\mathbf{}\mathit{V}$ .What are the

(a) size and

(b) direction (up or down) of current i₁ and the

(c) size and

(d) direction of current i₂? What is the energy transfer rate in

(e) battery 1and

(f) battery 2? Is energy being supplied or absorbed in

(g) battery 1 and

(h) battery 2?

Consider the given values of the resistance as:

$$\begin{gathered} R_1=R_2=2.0\Omega \\ {R}_3=4.0\Omega \\ {R}_4=3.0\Omega \\ {R}_5=1.0\Omega \\ {R}_6=R_7=R_8=8.0\Omega \end{gathered}$$

Consider the given values of the voltage source as:

$$\begin{gathered} e_1=16V \\ {e}_2=8.0V \end{gathered}$$

Consider the concept of Kirchhoff’s voltage law and current law to find the unknown currents and their direction. By applying KVL to both loops, determine two different equations.

Consider the formulas:

According to KVL, $\sum V=0$

According to KCL, $\sum I_{in}=\sum I_{out}$

Solve for the equivalent of combination $R_7\&R_8$.

$\begin{array}{c}{R}_{78}=\frac{R_7\times R_8}{R_7+R_8}\\ =\frac{8.0\times 8.0}{8.0+8.0}\\ =4.0\Omega \end{array}$

Solve for the equivalent of combination $R_1,R_2,R_7\&R_8$.

$\begin{array}{c}{R}_{1278}=R_1+R_2+R_{78}\\ =2.0+2.0+4.0\\ =8.0\Omega \end{array}$

Solve for the equivalent of combination $R_{1278}\&R_6$.

$\begin{array}{c}{R}_{12678}=\frac{8.0\times 8.0}{8.0+8.0}\\ =4.0\Omega \end{array}$

Solve for the equivalent of combination .$R_{12678}\&R_3$

$R=4.0+4.0=8.0\Omega$

Consider the equivalent circuit is obtained as follows:

Applying KVL to first loop,

${\epsilon }_1-(i_1+i_4)R=0$ ….. (1)

Applying KVL to second loop,

$$\begin{gathered} {\epsilon }_1-{\epsilon }_2+i_4(R_4+R_5)=0 \\ 16-8+i_4(3+1)=0 \\ 8+4i_4=0 \end{gathered}$$ …(2)

Solve further as:

$i_4=-2A$

Substitute the values and solve as:

$$\begin{gathered} 16-(i_1-2)\times 8=0 \\ {i}_1=4.0A \end{gathered}$$

Size of current $i_1$is $4.0A$.

The direction of current will be upwards as suggested from the equivalent circuit.

Solve for the value of the current as$i_2$:

$i=i_1+i_4$

Substitute the values and solve as:

$\begin{array}{c}i=-2+4\\ =2A\end{array}$

Consider the equivalent circuit for the given condition as:

Aplying KVL to above loop,

$i{R}_3+i_6{R}_6-{\epsilon }_1=0$

Substitute the values and solve as:

$(2\times 4)+(i_6\times 8)-16=0$

Hence,

$i_6=1\text{\hspace{0.17em}A}$

Applying KCL in the circuit and solve as:

$\begin{array}{c}{i}_3=i-i_6\\ =2-1\\ =1\text{\hspace{0.17em}A}\end{array}$

Now, from the current division rule solve as:

$\begin{array}{c}{i}_2=i_3\times \frac{R_7}{R_7+R_8}\\ =1\times \frac{8}{8+8}\\ =0.50A\end{array}$

The value of thecurrent i₂is $0.5A$

The direction of current $i_2$ as directed from the equivalent circuit is in the downward.

Rate of energy transfer through battery 1

$P={\epsilon }_1{i}_1$

Substitute the values and solve as:

$\begin{array}{c}P=16\times 4\\ =64W\end{array}$

The energy transfer rate in Battery 1 is$P=64W$.

Consider the formula for the power:

$P={\epsilon }_2{i}_4$

Substitute the values and solve as:

$\begin{array}{c}P=8.0\times 2.0\\ =16W\end{array}$

The energy transfer rate in Battery 2 is$16W$

Energy is supplied by battery1 as the current direction is from positive side to the negative terminal.

Energy is absorbed by battery 2 as the current flows from negative terminal to the positive terminal of the battery.