Question

A dry cell of emf 1.5V 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 ammter reading settles to a steady value of 2.0A. What is the steady
(a) rate of chemical energy consumption of the cell,
(b) rate of energy dissipation inside the cell,
(c) rate of energy dissipation inside the resistor,
(d) power output of the source?

Answer 1

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`