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 ?

Explanation

Here `epsilon = 1.5 V, r = 0.20 Omega, I = 1.5A`
As `I = epsilon/(R + r) or R + r = epsilon/I`
or `R = epsilon/I - r = 1.5/1.5 - 0.2 = 0.8 Omega`
(a) Voltage drop across R,
`V = IR = 1.5 xx 0.8 = 1.2 V`
(b) Rate of energy dissipation inside the resistor
`= I^(2)R = (1.5)^(2) xx 0.8 = 1.8W`