Question

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

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

Answer 1

Answer: (A)

$ T.P.D ( V )= E$ - Ir (Remember it )
$V=E-\left(\frac{E}{R+r}\right) r=\frac{E R}{(R+r)}$
from given conditions $ E=2.2$ & when $ R=5$
then $TPD\, V =1.8\, V$
therefore $1.8=\frac{2.2 \times 5}{5+ r }$
$ \Rightarrow r =\frac{10}{9} \Omega$