Question

The number of dry cells, each of e.m.f. 1.5 volt and internal resistance 0.5 ohm that must be joined in series with a resistance of 20 ohm so as to send a current of 0.6 ampere through the circuit is
(a) 2
(b) 8
(c) 10
(d) 12​

Answer 1

Answer: Option C

Explanation: In series

i = nE/nr+R

0.6 = n×1.5/n×0.5+20

0.3n + 12 = 1.5n

1.2n = 12

n = 12/1.2

n = 10

Answer 2

Answer:

Option (c). The number of required dry cells is 10.

Explanation:

The equation connecting current, emf, internal resistance, and external resistance is given by the equation

i = E / (r + R)

Where i is the current, E and r are the emf and internal resistance of the cell, and R is the external resistance.

If many cells with the same emf and internal resistance are connected in series the above equation for current will become:

i = nE / (nr + R)

Where n is the number of cells.

∴ n = -iR / (ir - E)

We have,

Current, i = 0.6 A

Internal resistance, r = 0.5 Ω

emf of each cell, E = 1.5 V

External resistance, R = 20 Ω

∴ n = -(0.6 × 20) / ([0.6 × 0.5] - 1.5)

  n = -12 / -1.2

  n = 10