Question

For a cell Terminal potential difference is $2.2V$ when circuit is open and reduces to $1.8V$ when cell is connected to a resistance of $R=5\Omega$. Determine internal resistance of cell ($r$)

• A. $\frac{10}{9}\Omega$
• B. $\frac{9}{10}\Omega$
• C. $\frac{11}{9}\Omega$
• D. $\frac{5}{9}\Omega$

Answer 1

Answer: (A)

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

Let $Ri=$ internal resistance of the battery
Let $Vo=$open circuit voltage of the battery $=2.2V$
Let $Vl=$ loaded voltage of the battery $=1.8V$
Let $Rl=$resistance of the load $=5\Omega$

$Vl=\frac{\left(Rl\right)\times \left(Vo\right)}{Rl+Ri}$

$Rl+Ri=\frac{\left(Rl\right)\left(Vo\right)}{Vl}$

$Ri=Rl(\frac{Vo}{Vl}-1)$

$Ri=5(\frac{2.2}{1.8}-1)$

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