Ncert Exercises & Solutions

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'?

Hint: Use the concept of series and parallel combination of resistors.

Step 1:

In a series combination of resistors, the current I is given by,

I = \frac{E}{R + nR}

Step 2:

whereas in parallel combination current is given by,

10 I = \frac{E}{R + \frac{R}{n}}

Step 3:

Now, according to the problem,

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

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