JEE Main 27-Aug-2021 Shift 1 Solved Paper

Given, number of cells, n = 5
Internal resistance of each cell, r = 1Ω
Emf of each cell, e = 5 V
When all cells are connected in parallel as shown below. Potential will remain same as, VP=5V
Net resistance in parallel combination will be given as   Rp=R+

1

5

(∵ r = 1 Ω) ...(i)
When all cells are connected in series as shown below The net potential will increase as cells are connected in series,
VS=5+5+5+5+5=25V
The net resistance of circuit will be equivalent of sum of all resistances as all are connected in series.
Rs=r+r+r+r+r+R=5r+R
Rs=5+R ( ∵ r = 1Ω) ...(ii)
By Ohm’s law, current flowing in circuit is given as
I=

V

R

As current in both series and parallel combination is same,

Vp

Rp

=

Vs

Rs

⇒

5

R+ 1 5

=

25

5+R

[From Eqs. (i) and (ii)]
⇒25+5R=25R+5
⇒ R = 1 Ω