1.

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 observed that the current is increased 10 times. What is n?

Answer»

When n resistors are in series, \(I=\frac{E}{R+nR};\)

When n resistors are in parallel, \(\frac{E}{R\,+\,\frac{R}{n}}=10I\)

\(\frac{1+n}{1+\frac{1}{n}}=10\) 

\(\Rightarrow\,\,\frac{1+n}{n+1}n=10\)  \(\Rightarrow\,n=10.\)



Discussion

No Comment Found