Write any two factors on which internal resistance of a cell depends . The reading on a high resistance voltmeter , when a cell is connected across it , is 2.0 V . When the terminals of the cell are also connected to a resistance of ` 3 Omega` as shown in the circuit , the voltmeter reading drops to 1.5 V . Find the internal resistance of the cell.

Explanation

Factors effecting internal resistance of a cell :
Numerical : Let r be the internal resistance of the cell
`therefore " " I = (2.0)/(r) … [because I = (E)/(r)]`
When the terminals of the cell are connected to a resistance of `3 Omega` then
` I = ((1.5)/(3r))/(3 + r) [(1)/(R) = (1)/(R_(1)) + (1)/(R_(2)) therefore R = (R_(1) R_(2))/(R_(1) + R_(2))]`
`I = 1.5 xx ((3 + r)/(3r))`
Solving (i) and (ii) , we get
`(2.0)/(r) = 1.5 xx ((3 + r)/(3r)) implies (2.0)/(1.5) = (3 + r)/(3)`
`implies (4)/(3) = (3 +r )/(3) implies 3 + r = 5 therefore r = 1 Omega`.