(a) Six lead-acid type of secondary cells each of emf 2.0 V and internal resistance 0.015 Ω are joined in series to provide a supply to a resistance of 8.5 Ω. What are the current drawn from the supply and its terminal voltage?
(b) 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?
(a) Number of secondary cells, n = 6
Emf of each secondary cell, E = 2.0 V
Internal resistance of each cell, r = 0.015 Ω
series resistor is connected to the combination of cells.
Resistance of the resistor, R = 8.5 Ω
Current drawn from the supply = I, which is given by the relation,
$I=\frac{n E}{R+n r}$
$=\frac{6 \times 2}{8.5+6 \times 0.015}$
$=\frac{12}{8.59}=1.39 \mathrm{~A}$
Terminal voltage, V = IR = 1.39 × 8.5 = 11.87 A
Therefore, the current drawn from the supply is 1.39 A and terminal voltage is
11.87 A.
(b) After a long use, emf of the secondary cell, E = 1.9 V
Internal resistance of the cell, r = 380 Ω
Hence, maximum current $=\frac{E}{r}=\frac{1.9}{380}=0.005 \mathrm{~A}$
Therefore, the maximum current drawn from the cell is 0.005 A. Since a large current is required to start the motor of a car, the cell cannot be used to start a motor.