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+[] =r+ Rp=10+ (∵ 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= As,
ip=20is =20 ⇒= [∵Vp=Vs=e=20V] ⇒= ⇒ 20n = 400
n = 20
Thus, the value of number of resistances n is 20.