Current Electricity Part 2

Given, value of each resistance, R = 10 Ω
Emf of battery, e = 20V
Internal resistance of battery, r = 10 Ω
Current in parallel connection is 20 times current in series combination, ip=20is.
Net resistance in parallel combination will be given as
Rp=r+[

1

1 R + 1 R + 1 R ....+n

]
=r+

R

n

Rp=10+

10

n

(∵ r = 10 Ω and R = 10 Ω )...(i)
In series combination,
The net resistance of circuit will be equivalent to sum of all resistances as all are connected in series.
Rs=[R+R+.....+n]+r=nR+r
Rs=10n+10 (∵ r = 10 Ω and R = 10 Ω)...(ii)
By Ohm’s law, current flowing in circuit is given as
i=

V

R

As, ip=20is

Vp

Rp

=20

Vs

Rs

⇒

20

10+ 10 n

=

20×20

10n+10

[∵Vp=Vs=e=20V]
⇒

20n

10n+10

=

400

10n+10

⇒ 20n = 400
n = 20
Thus, the value of number of resistances n is 20.