Question

Each of the six real batteries in Fig. 27-68 has an emf of$\mathbf{\text{20V}}$and a resistance of$\mathbf{4}\mathbf{.}\mathbf{0}\mathit{\Omega }$. (a) What is the current through the (external) resistance$\mathbf{R}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{\Omega }$? (b) What is the potential difference across each battery? (c) What is the power of each battery? (d) At what rate does each battery transfer energy to internal thermal energy?

Short answer

a) The current through the external resistance is$6.0A$.

b) The potential difference across each battery is$8.0V$.

c) The power of each battery is $60.0W$.

d) The rate at which the battery transferred energy to the internal thermal energy is $36.0W$.

Step-by-step solution

The given data

a) Emf of each battery,$\epsilon =20.0\text{V}$

b) Resistance of the internal resistor of the batteries,$r=4.0\Omega$

c) External resistance,$R=4.0\Omega$

Understanding the concept of current and potential difference

Here, the battery with their emf value has different potential differences due to their internal resistances. Now, if the batteries are considered resistors, then the combination shows that the two given lines of the batteries with three in each line are in series combination, thus having the same current value. Thus, using Ohm's law and loop's concept, the required value of the current through batteries and external resistance can be calculated. With direct relation of emf and current to the power of the battery, it can be determined and the rate of energy conversion involves the dissipation process of charge.

Formula:

Kirchhoff’s voltage equation, $\sum _{closedloop}V=0$ (i)

The voltage equation according to Ohm’s law, $V=IR$ (ii)

The power of a battery, $P=i\epsilon$ (iii)

The rate at which the thermal energy is generated, $P=i^{2}R$ (iv)

Calculation of the current through the external resistance

(a)

From symmetry we see that the current through the top set of batteries$\left(i\right)$is the same as the current through the second set.

Thus, the current through the external resistance is given by:

$i_R=2i................\left(a\right)$

Now, considering the loop rule that is equation (i), for the outer loop, the current through each battery can be given using equation (ii) as:

$$\begin{gathered} 3\left(\epsilon -ir\right)-\left(2i\right)R=0 \\ 3\epsilon -i\left(3r+2R\right)=0 \\ i=\frac{3\epsilon }{3r+2R} \\ =\frac{3\left(20.0V\right)}{3\left(4.0\Omega \right)+2\left(4.0\Omega \right)} \\ =3.0A \end{gathered}$$

Now, the current through the external resistance is given using equation (a) as follows:

$$\begin{gathered} i_R=2\left(3.0\text{A}\right) \\ =6.0\text{A} \end{gathered}$$

Hence, the value of the current is6.0A.

Calculation of the potential difference through each battery

(b)

The potential difference across each battery can be given using the given data and equation (ii) as follows:

$$\begin{gathered} V_{battery}=\epsilon -ir \\ =\left(20.0\text{V}\right)-\left(3.0\text{A}\right)\left(4.0\Omega \right) \\ =8.0V \end{gathered}$$

Hence, the value of the potential difference is$8.0V$.

Calculation of the power of the battery

(c)

The power of each battery is given using the given data in equation (iii) as follows:

$$\begin{gathered} P=\left(3.0\text{A}\right)\left(20.0\text{V}\right) \\ =60.0\text{W} \end{gathered}$$

Hence, the value of the power is$60.0W$.

Calculation of the rate of thermal energy conversion

(d)

The rate at which the battery transferred energy to the internal thermal energy can be calculated using the given data that is of the battery’s internal resistance and current through it in equation (iv) as follows:

$$\begin{gathered} P={\left(3.0\text{A}\right)}^2\left(4.0\Omega \right) \\ =36.0W \end{gathered}$$

Hence, the value of the power is36.0W.