Question

The p.d. between the terminal of a cell in an open circuit was $2.2V$. When it was measured across a resistor of $5$ it was found to be $1.8V$. The internal resistance of the cell is

• A. $\frac{10}{9}\Omega$
• B. $\frac{9}{10}\Omega$
• C. $\frac{12}{7}$
• D. $\frac{7}{12}$

Answer 1

The correct option is A

$\frac{10}{9}\Omega$

Step 1: Give data:

Open circuit emf of the cell is $E=2.2V$

When measured across the $5\Omega$ resistor the Voltage becomes $V=1.8V$

Step 2: Calculating the internal resistance,

We have a relation

$V=E-IR ...... \left(i\right)$

Also, we know

$l=\frac{E}{R+r}$

Where $r$ = internal resistance putting in ($i$)

$V=E - \left(\frac{E}{R+r}\right)r$

Now we substitute the given values

$$\begin{gathered} 1.8=2.2-\left(\frac{2.2}{5+r}\right)r \\ 1.8= \frac{2.2(5+r)-2.2r}{5+r} \\ 1.8=\frac{2.2\times 5}{5+r} \\ 5+r=\frac{11}{1.8} \\ r=\frac{11}{1.8}-5 \\ r=\frac{10}{9} \end{gathered}$$

Thus, the correct answer is option A,$\frac{10}{9}\Omega$.