Question

⚡Six lead acid type of secondary cells each of emf 2.0V and internal resistance 0.015 ohm are joined in series to provide a supply to a resistance of 8.5ohm. What are the current drawn from the supply and its terminal voltage ?​

Answer 1

Given,

Emf of the cell , E = 2.0V

No of cells , n= 6

internal resistance = 0.015 ohm.

External resistance = 8.5 ohm.

As per the question, cells are connected in series.

thus I = $\frac{nE}{R+nr}$

= $\frac{6*2.0}{8.5+6*0.015}$

=$\frac{12}{8.59}$

= 1.4 A.

By ohms law, V= IR

V = $1.4*8.5$

    = 11.9 V.

thus, terminal voltage V = 11.9V

Answer 2

$\huge\sf\underline{♡Answer♡}$

Emf of secondary cells, E = 2.0 V (Given)

No. of secondary cells, n = 6 (Given)

External resistance, R = 8.5 ohm (Given)

Internal resistance of cell, r = 0.015 ohm (Given)

Current drawn from the supply,

$I = \frac{nE}{R + nr}$

$\frac{6 \times 2.0}{8.5 + 6 \times 0.015}$

$= 1.4 \: A$

Now,

terminal voltage of the supply is,

V = IR (By ohm's law)

= 1.4 × 8.5

= 11.9 V