Question

Three identical cells, each of $4\mathrm{V}$ and internal resistance $r$, are connected in series to a $6-\mathrm{ohm}$ resistor. If the current flowing in the circuit is $2\mathrm{A}$. The internal resistance of each cell is?

Answer 1

Step 1: Given parameters

Current in the circuit, $\mathrm{I}=2\mathrm{A}$

Total emf $$\begin{gathered} =3\times 4 \\ =12\mathrm{V} \end{gathered}$$

$\therefore \mathrm{E}=\frac{\mathrm{W}}{\mathrm{Q}}$

Here,

E is the electromotive force.

W is the work done.

Q is the charge.

Total resistance $=6+3\mathrm{r}$

Step 2: To find

We have to find the internal resistance of each cell

Step 3: Calculation

By using Ohm's law.

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

By substituting the values, we get

$$\begin{gathered} 2=\frac{12}{6+3\mathrm{r}} \\ 2(6+3\mathrm{r})=12 \\ 12+6\mathrm{r}=12 \\ 6\mathrm{r}=12-12 \\ \mathrm{r}=\frac{0}{6} \\ \mathrm{r}=0 \end{gathered}$$

Therefore, the internal resistance of each cell is $0\mathrm{\Omega }$.