In Fig. $\mathbf{\text{27-33,}}$Battery1 has emf V and internal resistance$\mathbf{r}\mathbf{1}\mathbf{\text{=0.016}}$and battery 2has emf V and internal resistance$\mathbf{r}\mathbf{2}\mathbf{=}\mathbf{\text{0.012.}}$The batteries are connected in series with an external resistance R.

(a) What R-value makes the terminal-to-terminal potential difference of one of the batteries zero?

(b) Which battery is that?

$$\begin{gathered} {\epsilon }_1=\text{12.0V} \\ {r}_1=0.016\text{Ω} \\ {\epsilon }_2=\text{12.0V} \\ {r}_2=0.012\text{Ω} \end{gathered}$$

Write an expression for the total current through the circuit using Ohm’s law. write an expression for the emf in the circuit whenterminal-to-terminal potential difference of the battery having high internal resistance. From these two expressions, get the R value.

Ohm's law states that the current through a conductor between two points is directly proportional to the voltage across the two points.

Formulae are as follow:

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

Where, I is current, V is voltage, R is resistance.

The total emf in the circuit is,

$\epsilon ={\epsilon }_1+{\epsilon }_2$

The total resistance in the circuit is,

$R_{total}=r_1+r_2+R$

It is given that,

$\epsilon =V+Ir$

If the terminal-to-terminal potential difference becomes zero, then,

$\epsilon =Ir$

Since$r_1>r_2$,, the terminal-to-terminal potential difference across battery 1 is zero.

$\therefore {\epsilon }_1=I{r}_1\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \mathrm{..1})$

According to Ohm’s law,

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

In this case,

$I=\frac{\epsilon }{R_{total}}=\frac{{\epsilon }_1+{\epsilon }_2}{r_1+r_2+R}\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots .2)$

From the equations 1) and 2),

$$\begin{gathered} \frac{{\epsilon }_1}{r_1}=\frac{{\epsilon }_1+{\epsilon }_2}{r_1+r_2+R} \\ {\epsilon }_1(r_1+r_2+R)=r_1({\epsilon }_1+{\epsilon }_2) \end{gathered}$$

$$\begin{gathered} \therefore R=\frac{{\epsilon }_2{r}_1-{\epsilon }_1{r}_2}{{\epsilon }_1} \\ \therefore R=\frac{12\left(0.016\right)-12\left(0.012\right)}{12} \\ \therefore R=0.004\text{Ω} \end{gathered}$$

Hence, the R value which makes terminal-to-terminal potential difference of one of the batteries$0.004\text{Ω}$

From part a), we can conclude that, across battery 1, the terminal-to-terminal potential difference is zero.

Hence, across battery 1 the terminal-to-terminal potential difference is zero.

Therefore, the external resistances in the circuit which makes the terminal-to-terminal potential difference of one of the batteries zero can be determined using Ohm’s law.