Question

A set of ‘n’ equal resistors, of value \('R'\) each, are connected in series to a battery of emf ‘E’ and internal resistance ‘R' The current drawn is \(I\) . Now, the ‘n’ resistors are connected in parallel to the same battery. Then the current drawn from the battery becomes \(10 \,I\) . The value of ‘n’ is

• A. 9
• B. 10
• C. 20
• D. 11

Answer 1

The Correct Option is B

$I = \frac{E}{nR + R}$ .....(i)
$ 10\,I = \frac{E}{\frac{R}{n} + R} $ ....(ii)
Dividing (ii) by (i),
$10 = \frac{\left(n+1\right)R}{\left(\frac{1}{n} + 1\right)R} $
After solving the equation, $n = 10$

Answer 2

In the given problem, we are given two cases of series combination and parallel combination. In the first case, we have 

′n′ equal resistors have resistance ′R′ each is connected in series. Also, the internal resistance of the battery is present in series. Therefore, the total resistance in this series of combination 

R_S will be:

R_s = nR+R

⇒R_s = R(n+1)

Thus, using Ohm’s law, let’s consider the emf voltage in the circuit as 

E. Therefore, the current I is as:

I=\(\frac{E}{R(n+1)}\)----equation 1

In the second case, we have 

′n′ equal resistors have resistance ′R′ each is connected in parallel. Here, the internal resistance of the battery is in series. Therefore, the total resistance in this series of combination 

Rp will be:

R_p =R_n+R

Here, 

R/n is the effective resistance for ′n′ equal resistors having resistance ′R′ each is connected in parallel. This is obtained using the formula for effective resistance in parallel combination as follows:

\(\frac{1}{R_p}\)=\(\frac{1}{R}\)+\(\frac{1}{R}\)+\(\frac{1}{R}\)+...+\(\frac{1}{R}\)

⇒\(\frac{1}{R_p}\)=\(\frac{1}{R_(n)}\)

∴

R_p =\(\frac{R}{n}\)

And the internal resistance is in series with this parallel combination thus the total resistance will be 

R_p = \(\frac{R}{n}\)+R

The current using Ohm’s law will be:

10I=\(\frac{V}{\frac{R}{n}+R}\)

⇒ 10I=\(\frac{nE}{R(1+n)}\)

Taking the ratio of this value with equation 1 we get

\(\frac{10I}{I}\)=\(\frac{\frac{nE}{R(1+n)}}{\frac{E}{R(n+1)}}\)

⇒ n=10

Therefore, the value of 

n=10

Thus, B is the correct option.

 

Answer 3

I = nR+\(\frac{R}{E}\) ……….(1)

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10I= \(\frac{E}{\frac{R}{N}}\)+R …………(2)

 

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Dividing (ii) by (i),

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

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After solving the equation, n=10

So, The correct answer will be Option (B) 10.