**Question:**

(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?

**Solution:**

(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.

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