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 battery becomes 10 I. The value of 'n' is

CONCEPT:

As we know that the relation between the EMF 'E' and internal resistance 'R' is written as;

E = I(r+R) ----(1)

Here E is the 'EMF', r is the internal resistance and R is the load resistance.

Resistance in series- When n numbers of resistors are in series then the total resistance is written as;

\({R_s} = {R_1} + {R_2} + .... + {R_n}\)

Here, R₁, R_2, .....R_n is denoted as resistance and R_s is the resistance in series.

Resistance in parallel- When n numbers of resistors are in parallel then the total resistance is written as;

\(\frac{1}{{{R_p}}} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}} + ..... + \frac{1}{{{R_n}}}\)

Here, R1, R2, .....Rn is denoted as resistance and R_p is the resistance in parallel.

CALCULATION:

In the question, the n number of internal resistance,r = R

Using equation (1) we get;

E = I(nR+R)

⇒ \(I = \frac{E}{{nR + R}}\)      ----(2)

Now, the n number of resistance are connected parallel and the current drawn is 10I we have;

\(10I = \frac{E}{{\frac{R}{n} + R}}\)      ----(3)

Now, on dividing the equation (3) by (2), we get;

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

\(10\left( {\frac{1}{n} + 1} \right) = (n+1)\\ 10\left( {\frac{n+1}{n} } \right) = (n+1)\)

⇒ n = 10

After solving the equation, n = 10

Hence, option 1) is the correct answer.