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 $R=5\Omega$ then determine internal resistance of cell is :-

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

Answer 1

The correct option is A $\frac{10}{9}\Omega$

When the circuit is open , the terminal voltage is equal to the emf i.e $E=2.2V$.
Current $I=E/(R+r)$
Now terminal voltage $V=E-IR=E-ER/(R+r)$
or $V(R+r)=ER$
or $r=(E/V-1)R=(2.2/1.8-1)5=(11/9-1)5=10/9\Omega$