Question

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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 1

Answer:

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In series combination of resistors, current I is given by

• $I = E/E + nR$

Whereas, in parallel combination current 10 I is given by

• $E/R + R/n = 10 I$
• $===> E/R+R/n = 10 (E/R+nR)$

Now, according to problem,

• $=== > 1+n/1+1/n =10$
• $===>10 = (1+n/n+1)n$
• $===> n = 10$

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Answer 2

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In series combination of resistors, current I is given by

• I = E/E + nRI=E/E+nR

Whereas, in parallel combination current 10 I is given by

E/R + R/n = 10 IE/R+R/n=10I

=== > E/R+R/n = 10 (E/R+nR)===>E/R+R/n=10(E/R+nR)

Now, according to problem,

=== > 1+n/1+1/n =10===>1+n/1+1/n=10

=== > 10 = (1+n/n+1)n===>10=(1+n/n+1)n

=== > n = 10===>n=10