(i) Six lead-acid type of secondary cells each of emf 2 V and internal resistance 0.015 Ω are joined in series to provide a supply to a resistance of 8.5Ω. What are the current drawn from the supply and its terminal voltage?
(ii) A secondary cell after long use has an emf of 1.9 V and a large internal resistance of 380 Ω. What maximum current can be drawn from the cell?
Could the cell drive the starting motor of a car?
(i) Six cells are joined in series
Emf of each cell,E=2v
Number of cells n=6
Total emf of circuit = nxE=6x2=12 V
Internal resistance of each cell, r =.015 Ω
Total internal resistance = nx r=6x0.015=0.09 Ω
External load, R = 8.5 Ω
Current in the circuit,
I = nE/nr+R = 12/0.09 + 8.5 = 1.4A
The terminal voltage of battery,
V = IR = 1.4 x 8.5=11.9 V
(ii) Emf of cell,E = 1.9 V
Internal resistance of cell, r =380 Ω
Maximum current can be drawn from the cell, if there is zero external resistance.Therefore,
Now, we see that the maximum current drawn from the cell is very low, thus the cell cannot be used to drive the starting motor of a car as the current required for this purpose is approximately 100 A for few records.