A dry cell of emf `1.5 V` and internal resistance `0.10 Omega` is connected across a resistor in series with a very low resistance ammeter. When the circuit is switched on , the ammeter reading settles to a steady value of `2.0 A`.
(i) What is the steady rate of chemical energy consumption of the cell ?
(ii) What is the steady rate of energy dissipation inside the cell ?
(ii) What is the steady rate of energy dissipation inside the resistor ?
(iv) What is the steady power out put of the source?

Explanation

Suppose a resitance R is connected in series, then
Current through the circuit `I=(E)/(R+r)rArr2=(1.5)/(R+0.10)rArr2R+0.2=1.5rArr2R=1.3rArrR=0.65Omega`
(a) Rate of chemical energy consumption actual (Power) of the cell `P=El=1.5xx2=3W`
(b) Rate of energy dissipation (Power dissipated) inside the cell `=I^(2)r=(2)^(2)xx0.1=0.4W`
(c) Rate of energy dissipation (power dissipated) inside the resistor `=I^(2)R=(2)^ (2)xx0.65=2.6W`
(d) Power output of the source=Actual power of the cell -power dissipated inside the cell
`=3-0.4=2.6W`