Question

In Fig. 27-81, the ideal batteries have emfs ${\mathit{\epsilon }}_{\mathbf{1}}\mathbf{=}\mathbf{20}\mathbf{}\mathit{V}$,${\mathit{\epsilon }}_{\mathbf{2}}\mathbf{=}\mathbf{10}\mathbf{}\mathbf{\text{\hspace{0.17em}V}}$ ,${\mathit{\epsilon }}_{\mathbf{3}}\mathbf{=}\mathbf{5}$ , and${\mathit{\epsilon }}_{\mathbf{4}}\mathbf{=}\mathbf{5}\mathbf{}\mathbf{\text{\hspace{0.17em}V}}$ , and the resistances are each$\mathbf{2}\mathbf{.00}\mathbf{}\mathit{\Omega }$ . What are the

(a) size and

(b) direction (left or right) of currenti₁and the

(c) size and

(d) direction of current?(This can be answered with only mental calculation.) (e) At what rate is energy being transferred in battery 4, and

(f) is the energy being supplied or absorbed by the battery?

Short answer

a)Size of current$i_1=7.50A$

b)Direction of the current i₁Leftward.

c) Size of the current $i_2=10A$

d) Direction of the current $i_1$Leftward.

e) Rate of the energy being transferred in battery 4 is $P=87.5W$

f) Energy is being supplied by the battery

Step-by-step solution

Determine the given quantities

$$\begin{gathered} {\epsilon }_1=20V \\ {\epsilon }_2=10V \\ {\epsilon }_3=5 \\ {\epsilon }_4=5V \end{gathered}$$

Determine the concept of Ohm’s Law

According to Ohm’s law, the current through a conductor placed between two pointsis directly proportional to the voltage induced across the points. That is,

$\mathit{I}\mathbf{=}\frac{\mathbf{V}}{\mathbf{R}}$

Here, $\mathit{I}\mathbf{-}\mathit{c}\mathit{u}\mathit{r}\mathit{r}\mathit{e}\mathit{n}\mathit{t}\mathbf{,}\mathbf{}\mathit{R}\mathbf{-}\mathit{r}\mathit{e}\mathit{s}\mathit{i}\mathit{s}\mathit{t}\mathit{a}\mathit{n}\mathit{c}\mathit{e}$

The rate at which energy transfers is as follows:

$P=i\epsilon$

Here,$\mathit{P}\mathbf{-}\mathit{p}\mathit{o}\mathit{w}\mathit{e}\mathit{r}\mathbf{,}\mathbf{}\mathit{i}\mathbf{-}\mathit{c}\mathit{u}\mathit{r}\mathit{r}\mathit{e}\mathit{n}\mathit{t}\mathbf{,}\mathbf{}\mathit{\epsilon }\mathbf{-}\mathit{v}\mathit{o}\mathit{l}\mathit{t}\mathit{a}\mathit{g}\mathit{e}\mathbf{}\mathit{i}\mathit{n}\mathit{d}\mathit{u}\mathit{c}\mathit{e}\mathit{d}\mathbf{}\mathbf{.}$

Step 3:(a)Determine the magnitude ofcurrent i1

By applying KVL,

$\begin{array}{l}{\epsilon }_1+{\epsilon }_3+{\epsilon }_4=i_1(R+R)\\ 20+5+5=i_1(2+2)\end{array}$

Solve further as:

$i_1=7.50A$

Magnitude of current $i_1=7.50A$.

Step 4:(b) Determine the direction of the current i1

Current $\left(i_1\right)$as suggested from the diagram is leftwards this is because the current flows from the positive terminal of the battery to the negative terminal.

Step 5:(c) Determine the size of current i2

By applying KVL to bottom circuit,

$$\begin{gathered} i_1(R+R)-i_{2}R-i_2\left(\frac{R}{2}\right)=0 \\ 2i_{1}R-1.5i_{2}R=0 \end{gathered}$$

Solve further as:

$\begin{array}{c}{i}_2=\frac{2i_1}{1.5}\\ =10A\end{array}$

Size of the current $i_2=10A$

Step 6:(d) Determine the direction of the current i2

Current$\left(i_2\right)$is directed towards leftward direction. As observed from the figure the current will be directed from the positive to the negative terminal.

Step 7:(e) Determine the rate of energy transfer from battery 4

Rate of energy transfer in battery 4:

$P=(i_1+i_2){\epsilon }_4$

Substitute the values and solve as:

$\begin{array}{c}P=(7.5+10)\times 5\\ =87.5W\end{array}$

Rate of the energy being transferred in battery 4 is $P=87.5W$

Step 8:(f) Determine if the energy is being supplied or absorbed

Here, energy is being supplied by the battery as the current is in the forward direction from the battery and the current flows from the negative terminal to the positive terminal.