(a) Six lead-acid type of secondary cells each of emf `2.0` V and internal resistance `0.015 Omega` are joined in series to provide a supply to a resistance of `8.5Omega`. What are the current drawn from the supply and its terminal voltage ?
(b) A secondary cells after long use has an emf of `1.9` V and a large internal resistance of `380 Omega`. What maximum current can be drawn from the cell ? Could the cell drive the starting motor of a car ?

Explanation

(a) Number of secondary cells, `n = 6`
Emf of each secondary cell, `E = 1.0 V`
Internal resistance of each cell, `r = 0.015 Omega`
series resistor is connected to the combination of cells
Resistance of the resistor, `R = 8.5 Omega`
Current drawn from the supply `= I`, which is given by the relation,
`I = (nE)/(R+nr)`
`= (6xx2)/(8.5 + 6 xx 0.015)`
`=(12)/(8.59)=1.38 A`
Terminal voltage, `V = IR = 1.39 xx 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 = 1.9V`
Internal resistance of the cell, `r = 380 Omega`
Hence, maximum current `= (E)/(r)=(1.9)/(380)=0.005 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.