Q. 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 rate of 2A. Find (i) chemical energy consumption of the cell (ii) energy dissipation inside the cell (iii) energy dissipation inside the resistor (iv) power output of source is
A) (i) 3 W (ii) 0.4 W (iii) 2.6 W (iv) 2.6 W
B) (i) 0.4 W (ii) 3 W (iii) 2.6 W (iv) 2.6 W
C) (i) 2.6 W (ii) 0.4 W (iii) 9 W (iv) 1 W
D) None of the above
Explanation
(a) (i) Rate of chemical energy consumption
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 1.5 \times 2 =$3 W
(ii) Rate of energy dissipation inside the cell
$\ = 2 \times 2 \times 0.1 = 0.4$ W
(iii) Rate of energy dissipation inside the resistor
$= (3 - 0.4)$ W = 2.6 W
(iv) Power output of source $= (3 - 0.4)$ W = 2.6 W
$*\ $ $EI$ represents rate of chemical energy consumption of the cell.
$*I^{2}r$ represents the rate of energy dissipation inside the cell.
$*(EI - I^{2}r)$ represents the power output of the source of emf.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.