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 battery becomes 10 I. The value of ‘n’ is

• A. 10
• B. 20
• C. 11
• D. 9

Answer 1

The Correct Option is A

To solve this problem, we need to analyze both configurations of resistors and then determine the number of resistors n that satisfy the conditions given.

1. Series Configuration:
   • When n resistors, each of resistance R, are connected in series, the total resistance of these resistors is nR.
   • The total resistance in the circuit, considering the internal resistance R of the battery, is (nR + R) = (n + 1)R.
   • According to Ohm's Law, the current I drawn by the battery is given by: I = \frac{E}{(n+1)R}
2. Parallel Configuration:
   • When the same resistors are connected in parallel, the effective resistance R_{\text{parallel}} is given by: \frac{1}{R_{\text{parallel}}} = \frac{1}{R} + \frac{1}{R} + ... + \frac{1}{R} \quad (n \text{ times})
     Simplifying, we get: R_{\text{parallel}} = \frac{R}{n}
   • The total resistance in this circuit becomes \frac{R}{n} + R.
   • The new current I_{\text{new}} from the battery under this configuration is: I_{\text{new}} = \frac{E}{\left(\frac{R}{n} + R\right)} = \frac{E}{\frac{R + nR}{n}} = \frac{nE}{(n+1)R}
3. Given Condition:

   According to the problem, the current drawn when resistors are in parallel is 10 times that of when they are in series.

   I_{\text{new}} = 10 \times I
4. Equating the Currents:
   • Substitute the expressions for I_{\text{new}} and I from above: \frac{nE}{(n+1)R} = 10 \times \frac{E}{(n+1)R}
   • Simplify by canceling E and (n+1)R: n = 10
   Hence, the value of n is 10.

Therefore, the correct answer is 10.