Question

First a set of n equal resistors of r each are connected in series to a battery of emf e and internal resistance r. a current i is observed to flow. then the n resistors are connected in parallel to the same battery. it is observwd that the current is increased 10 times. what is n

Answer 1

I don't know...

Sryy

But I need pts.

Answer 2

When n resistors are connected in series,
equivalent resistance, Rs=nr .
Total resistance of the circuit Rts =nr+ r= r(n+1)
Current Is=e/{r(n+1)}
When n resistors are connected in parallel,
Equivalent resistance Rp = r/n
Total resistance, Rtp = r/n + r = (n+1)r/n
Current Ip = en/{(n+1)r}
But, Ip=10Is
=> en/{(n+1)r} = 10[e/{r(n+1)}]
=> n = 10