A dry cell of emf `1.6V` and internal resistance of `0.10 Omega` is connected to a resistor of resistance `R omega`. If the current drawn the cell is `2A`, then (i) What is the voltage drop across R ? (ii) What is the rate of energy dissipation in the resistor ?

Answer

(i) `1.4V`(ii)`2.0W`

Explanation

Here `epsilon = 1.6 V, r = 0 .10 Omega 1 = 2A`
As` I = (epsilon)/(R+ r ) or R + r = (epsilon)/(l) = (1.6)/(2) = 0.8 Omega `
or ` R = 0.8 - 0.1 = 0.70 Omega`
(i) voltage drop across `R`
`V = IR = 2 xx 0.70 = 1.4V`
(ii) Rate of energy dissipation inside the resistor
`= VI = 1.4 xx 2.0 = 2.0W`