The internal resistance of a battery is relatively small whe

The internal resistance of a battery is relatively small when the battery is new but increases as the battery ages. When a new 12.0-V battery is attached to a 100 load, the potential difference across the load is 11.9 V. After the circuit has operated for a while, the potential difference across the load is 11.1 V . By how much has the internal resistance of the battery changed?

R_{used batt} / R_{new batt} = ?

Solution

Given E = 12 volt

Load resistance R = 100 ohm

new :

Potential difference V = 11.9 volt

Let the internal resistance be r.

We know E = V + ir

             12 = 11.9 + ir

            ir = 12 - 11.9

               = 0.1       --------------( 1)

Used :

E = 12 volt

V \' = 11.1 volt

Internal resistance r \' = ?

We know E = V \' + ir \'

         12 = 11.1 + i r \'

         ir \' = 12-11.1

             = 0.9             --------( 2)

Equation(2)/ equation(1) ==>

ir \' / ir = 0.9 / 0.1

   r \' / r = 9

i.e., R_{used batt} / R_{new batt} = 9