91Ó°ÊÓ

$\mathrm{Î\mu }=1.\text{20V}$

R is a variable resistor.

The graph of the electric potentials V between the terminals of each battery as a function of R is given

Write an expression for the current through the circuit by applying Kirchhoff’s voltage law to the given circuit. Using this, write two equations for the terminal voltages of the batteries. Solving these simultaneous equations, get the values of internal resistances.

Kirchhoff's loop rule states that the sum of all the electric potential differences around a loop is zero.

Formulae are as follow:

$V=I.r$

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

Let internal resistances of the battery 1 and 2 be r₁ and r₂ respectively and the current through the circuit be I.

Appling Kirchhoff’s voltage law to the given circuit gives,

$$\begin{gathered} \mathrm{Î\mu }−I{r}_2+\mathrm{Î\mu }−I{r}_1−IR=0 \\ 1.20−I{r}_2+1.20−I{r}_1−IR=0 \\ I{r}_2+I{r}_1+IR=2.40 \\ I=\frac{2.40}{r_2+r_1+R} \end{gathered}$$

The terminal voltage of the battery 1 is

role="math" localid="1662657633899" $$\begin{gathered} V_1=\mathrm{Î\mu }−I{r}_1 \\ ∴V_2=1.2−\frac{2.40r_2}{r_2+r_1+R} \end{gathered}$$

Terminal voltage of battery 1 is

$\begin{array}{rcl}{V}_2& =& \mathrm{Î\mu }−I{r}_2\\ ∴V_2& =& 1.2−\frac{2.40r_2}{r_2+r_1+R}\end{array}$

From the graph, we can infer that at$R=0.1\text{Ω},V_1=0.4\text{Vand}{V}_2=\text{0V.}$

$$\begin{gathered} 0.4=1.2−\frac{2.40r_1}{r_2+r_1+0.1} \\ 0=1.2−\frac{2.40r_1}{r_2+r_1+0.1} \end{gathered}$$

Solving these simultaneous equations give,

$$\begin{gathered} r_1=0.20\text{Ω} \\ {r}_2=0.30\text{Ω} \end{gathered}$$

Hence, the internal resistance of the battery 1 and battery 2 is$r_1=0.20\text{Ω}$,$r_2=0.30\text{Ω}$

Therefore, the internal resistances of the batteries can be found using Kirchhoff’s law and the graph between the voltages of the batteries vs the variable external resistance.