(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 is 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?

Answer –

According to the question statement,

(a) Emf of the secondary cells is ε = 2.0 V

N is the number of secondary cells  = 6

Then total EMF is given by –

E = nε = 6 x 2

E = 12 V

r  = 0.015 Ω is the internal resistance of the secondary cells

And the resistance to which the secondary cells are connected is given by  R = 8.5 Ω
Total resistance in circuit can be determined by –

R_total = nr + R = 6 × 0.015 + 8.5

R_tota    =  8.59Ω

Similarly, current drawn from the supply can be calculated by –

I = E / R_total =12 / 8.59

I = 1.4 A

Using Ohm’s law, the terminal voltage can be calculated as –

V = IR = 1.4×8.5

V = 11.9 V

(b) Emf of the secondary cell is ε = 1.9 V

Internal resistance is r = 380 Ω
Maximum current drawn from the cell,

I =ε/r = 1.9/380

I = 0.005A.

It is given that the current required to start a motor is 100 Amp. Here, the current produced is 0.005 A, so the motor of the car cannot be started with this current.