Question

5. emf of a cell is 2.2V. when resistance R = 512 is connected in
series, potential drop across the cell becomes 1.8volt. Value of
internal resistance of the cell is:​

Answer 1

Answer:

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=10/9Ω

solution