Question: In Fig. 27-72, the ideal batteries have emfs${\epsilon }_1=20.0V,{\epsilon }_2=10.0V,and{\epsilon }_3=5.0V$, and the resistances are each$2.00\Omega$. What are the (a) size and (b) direction (left or right) of current$i_1$? (c) Does battery 1 supply or absorb energy, and (d) what is its power? (e) Does battery 2 supply or absorb energy, and (f) what is its power? (g) Does battery 3 supply or absorb energy, and (h) what is its power?

• a)Emf of ideal battery 1,${\epsilon }_1=20.0V$.
• b)Emf of ideal battery 2,${\epsilon }_2=10.0V$.
• c)Emf of battery 3,${\epsilon }_3=5.0V$.
• d)The value of each resistance,$R=2.0\Omega$.

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. 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,

$R_{eq}=\sum _1^n{R}_1$ (ii)

The equivalent resistance for a parallel combination,

$R_{eq}=\sum _1^n\frac{1}{R_1}$ (iii)

Kirchhoff’s voltage law,

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

The power of a battery,

P=IV (v)

Kirchhoff’s junction rule,

$l_h=l_{out}$ (vi)

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

First, the bottom parallel pair of resistances is reduced to one single resistance using equation (iii) as follows:

$$\begin{gathered} R\text{'}=\frac{R^2}{2R} \\ =\frac{R}{2} \\ =\frac{2.0\Omega }{2} \\ =1.0\Omega \end{gathered}$$

Now, the equivalent resistance of the bottom resistors with the resistor in series on the left of the circuit can be given using equation (ii) as follows:

$$\begin{gathered} R\text{'}=R+R\text{'} \\ =2.0\Omega +1.0\Omega \\ =3.0\Omega \end{gathered}$$

It is clear from the figure that the current through the equivalent resistance is $i_1$. Now,the loop rule is employed in a chosen path that includes R'' and all the batteries (proceeding clockwise.)

Thus, using equation (i) in the loop rule of equation (iv), the size of the current can be given as follows: (assuming$i_1$ goes leftward through R'')
$$\begin{gathered} {\epsilon }_3+{\epsilon }_1-{\epsilon }_2-i_{1}R\text{'}\text{'}=0 \\ 5.0V+20.0V-10.0V-i_1\left(3.0\Omega \right)=0 \\ {i}_1=\frac{15V}{3.0\Omega } \\ =5.0A \end{gathered}$$

Hence, the value of the current is $5.0A.$.

Since the sign of the current in part (a) is positive, this implies that our assumption of the current being in a leftward direction is correct.

Hence, the direction of the current is leftwards.

Since the current through battery 1 (${\epsilon }_1=20.0V$) is “forward”, battery 1 is supplying energy.
Hence, battery 1 supplies energy.

The rate of energy supplied by battery 1 is its power that is given using the given data in equation (v) as follows:

$$\begin{gathered} P=(5.0A)(20.0V) \\ =100W \end{gathered}$$

Hence, the value of the power is 100 W .

Now, the equivalent resistance of the parallel pair of resistors attached with battery 2 can be given using equation (iii) as follows:

$$\begin{gathered} R\text{'}\text{'}\text{'}=\frac{R^2}{2R} \\ =\frac{R}{2} \\ =\frac{2.0\Omega }{2} \\ =1.0\Omega \end{gathered}$$

Now, the amount of current through equivalent resistance is given using equation (i) as follows:

$$\begin{gathered} i\text{'}=\frac{10.0V}{1.0\Omega } \\ =10.0A,Downward \end{gathered}$$

But for the current to be in the upward direction for battery 2, the current can be given using equation (vi) as:

$$\begin{gathered} i=i\text{'}-i_1 \\ =10.0A-5.0A \\ =5.0A,upward \end{gathered}$$

Hence, battery 2 supplies energy.

The rate of energy supplied by battery 2 is its power that is given using the given data in equation (v) as follows:

$$\begin{gathered} P=\left(5.0A\right)\left(10.0V\right) \\ =50W \end{gathered}$$

Hence, the value of the power is 50W .

Now, the equivalent resistance of the pair of resistors attached with battery 3 can be given using equation (iii) as follows:

$$\begin{gathered} \frac{1}{R\text{'}\text{'}\text{'}\text{'}}=\frac{1}{R}+\frac{1}{R}+\frac{1}{2R} \\ =\frac{2+2+1}{2R} \\ =\frac{5}{2R} \\ R\text{'}\text{'}\text{'}\text{'}=\frac{2R}{5} \\ =\frac{2\left(2.0\Omega \right)}{5} \\ =0.8\Omega \end{gathered}$$

Now, the amount of current through equivalent resistance is given using equation (i) as follows:

$$\begin{gathered} i\text{'}\text{'}=\frac{5.0V}{0.8\Omega } \\ =6.25A,Download \end{gathered}$$

But for the current to be in the upward direction for battery 3 using the junction rule of equation (vi), the current can be given as:

$$\begin{gathered} i\text{'}\text{'}\text{'}=i\text{'}\text{'}+i_1 \\ =6.25A+5.0A \\ =11.25Aupward \end{gathered}$$

Hence, battery 3 supplies energy.

The rate of energy supplied by battery 1 is its power that is given using the given data in equation (v) as follows:

$$\begin{gathered} P=(11.25A)(5.0V) \\ =56.25W \\ \approx 56.3W \end{gathered}$$

Hence, the value of the power is 56.3W.