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Current Electricity question

2018 · Q120
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Current Electricity question

2018 · Q120

NEETPhysicsCurrent ElectricityMCQ+4 / −1
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 10I. The value of n is
  1. A
    10
  2. B
    11
  3. C
    20
  4. D
    9
View written solutionFree

Correct answer: A

n series grouping equivalent resistance

Rseries = nR

In parallel grouping equivalent resistance

Rparallel = Rn{R \over n}nR​

Current drawn from a battery when n resistors are connected in series is

I = EnR+R{E \over {nR + R}}nR+RE​ ......(i)

Current drawn from same battery when n resistors are connected in parallel is

10I = ERn+R{E \over {{R \over n} + R}}nR​+RE​ ......(ii)

On dividing eqn. (ii) by (i),

10 = (n+1)R(1n+1)R{{\left( {n + 1} \right)R} \over {\left( {{1 \over n} + 1} \right)R}}(n1​+1)R(n+1)R​

Solving we get, n = 10

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