Question

Six lead-acid types 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?

Answer 1

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 = `("nE")/("R" + "nr")`

= `(6 xx 2)/(8.5 + 6 xx 0.015)`

= `12/8.59`

= 1.39 A

Terminal voltage, V = IR = 1.39 × 8.5 = 11.87 A

Therefore, the current drawn from the supply is 1.39 A and the terminal voltage is 11.87 A.