91Ó°ÊÓ

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

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,

$\mathrm{Î\mu }={\mathrm{Î\mu }}_1+{\mathrm{Î\mu }}_2$

The total resistance in the circuit is,

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

It is given that,

$\mathrm{Î\mu }=V+Ir$

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

$\mathrm{Î\mu }=Ir$

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

$∴{\mathrm{Î\mu }}_1=I{r}_1…………………………………………………………………………………………\mathrm{..1})$

According to Ohm’s law,

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

In this case,

$I=\frac{\mathrm{Î\mu }}{R_{total}}=\frac{{\mathrm{Î\mu }}_1+{\mathrm{Î\mu }}_2}{r_1+r_2+R}………………………………………………………………………….2)$

From the equations 1) and 2),

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

$$\begin{gathered} ∴R=\frac{{\mathrm{Î\mu }}_2{r}_1−{\mathrm{Î\mu }}_1{r}_2}{{\mathrm{Î\mu }}_1} \\ ∴R=\frac{12\left(0.016\right)−12\left(0.012\right)}{12} \\ ∴R=0.004\text{Ω} \end{gathered}$$

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.