Question

10 resistors, each of resistance R are connected in series to a battery of emf E and negligible internal resistance. Then those are connected in parallel to the same battery, the current is increased n times. The value of n is:

• A. 100
• B. 1
• C. 1000
• D. 10

Answer 1

The Correct Option is A

To find the value of \(n\), the factor by which current increases when the resistors are connected in parallel compared to series, let's analyze both configurations. When resistors are connected in series, the total resistance \(R_s\) is:

\[R_s = 10R\] 

The current \(I_s\) through the series circuit connected to a battery of emf \(E\) is given by Ohm's Law:

\[I_s = \frac{E}{R_s} = \frac{E}{10R}\]

For resistors connected in parallel, the total resistance \(R_p\) is:

\[\frac{1}{R_p} = \frac{1}{R} + \frac{1}{R} + \cdots + \frac{1}{R} = \frac{10}{R}\]

Simplifying, we get:

\[R_p = \frac{R}{10}\]

The current \(I_p\) through the parallel circuit is:

\[I_p = \frac{E}{R_p} = \frac{E}{\frac{R}{10}} = \frac{10E}{R}\]

The ratio of currents \(n\) is:

\[n = \frac{I_p}{I_s} = \frac{\frac{10E}{R}}{\frac{E}{10R}} = 100\]

Therefore, the value of \(n\) is 100.

Answer 2

\(I_s=\frac{E}{10R}\)     ……….(1)

\(I_p=\frac{E}{\frac{R}{10}}=\frac{10E}{R}\)    ……..(2)

According to eq. (1) and (2)

\(n=\frac{I_p}{I_s}=100\)

\(\Rightarrow n=100\)

So, the correct answer is option (A): 100