Question: In Fig. 27-77, the ideal batteries have emfs ${\epsilon }_1=12.0Vand{\epsilon }_2=4.0V$, and the resistances are each$4.00\Omega$. What are the (a) size and (b) direction (up or down) of$i_1$and the (c) size and (d) direction of$i_2$? (e) Does battery 1 supply or absorb energy, and (f) what is its energy transfer rate? (g) Does battery 2 supply or absorb energy, and (h) what is its energy transfer rate?

• a)Emf of ideal battery 1,${\epsilon }_1=12.0V.$ .
• b)Emf of ideal battery 2,${\epsilon }_2=4.0V$.
• c) The value of each resistance,$R=4.0\Omega$ .

In the given problem, the current flowing through the entire circuit and the given batteries is given by Kirchhoff’s voltage law. If an amount of current is passing through the positive terminal of the device, then the device is

supplying power to an external circuit; and if the current enters a device at its positive terminal, then that device is absorbing or being supplied with power.

Formulae:

The voltage equation using Ohm’s law,

V =IR (i)

The equivalent resistance for a series combination,

$R=\underset{1}{\overset{n}{\sum R_l}}$ (ii)

The equivalent resistance for a parallel combination,

$R_{eq}=\underset{1}{\overset{n}{\sum \frac{1}{R_l}}}$ (iii)

Kirchhoff’s voltage law,

$\sum _{Closedloop}V=0$ (iv)

The power of a battery,

P = IV (v)

Kirchhoff’s junction rule,

$I_h=l_{out}$ (vi)

Here $l$ is the current V is the voltage, and R is the resistance.

From the lower left loop, the current value$i_1$ can be given using equation (i) as follows:
$$\begin{gathered} i_1=\frac{{\epsilon }_1}{R} \\ =\frac{12.0V}{4.0\Omega } \\ =3.0A \end{gathered}$$

Hence, the size of the current is$3.0A$ .

From the given figure and battery 1 position, the current$i_1$ direction is found to be downward.

Hence, the direction of the current is downward.

Applying the loop rule of equation (iv) to the tall rectangular lop in the center of the figure, the current value from the voltage equation can be given using equation (i) as follows: (proceeding clockwise)

$$\begin{gathered} {\epsilon }_2+(+i_{1}R)+(-i_{2}R)+\left(-\frac{i_2}{2}R\right)+\left(-i_{2}R\right)=0 \\ {\epsilon }_2+i_{1}R+\left(-\frac{5i_2}{2}R\right)=0 \\ {i}_2=\frac{2\left({\epsilon }_2+i_{1}R\right)}{5R} \end{gathered}$$

Substitute the values in the above expression, and we get,

$$\begin{gathered} i_2=\frac{2\left(4.0V+(3.0A)\times 4.0\Omega \right)}{5\left(4.0\Omega \right)} \\ {i}_2=1.60A \end{gathered}$$

Hence, the size of the current is 1.60 A.

From the assumption of part (c) calculations, the current$i_2$ direction is found to be downward.
Hence, the direction of the current is downward.

Battery 1 is supplying this power since the current is in the "forward" direction through the battery.

Hence, battery 1 supplies energy.

Applying the junction rule of equation (vi), the current through battery 1 can be given as follows:

$$\begin{gathered} i=i_1+i_2 \\ =3.0A+1.60A \\ =4.60A \end{gathered}$$

Now, the rate of energy transferred by the battery to the circuit can be given using equation (v) as follows:

P = (4.60 A) (12.0 V)

=55. 2 W

Hence, the value of the energy transfer rate is 55.2 W.

Battery 2 is supplying this power since the current is in the "forward" direction through the battery.

Hence, battery 1 supplies energy.

Now, the rate of energy transferred by battery 2 is given using the given data in equation (v) as follows:

P = (1.60 A) (4.0 V)
= 6.40 W

Hence, the value of the energy transfer rate is 6.40 W .