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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)

20

(B)

11

(C)

10

(D)

9

NEET - Medical UG-2018 Medium

Answer & Solution:

\(\large E=emf\\ \text{~series combination of n equal resistor} \\ R_{eq}=nR\\R_{total}=(n+1)R\\ \Large {I_1}={E\over R_{total}}={E\over (n+1)R} \\ \text{ ~parallel combination of n equal resistor} \\ \Large {1\over R_eq}={n\over R} \\ \Large R_{eq}={R\over n} \\ \Large R_{total}={R\over n}+R={Rn\over (n+1)}\\ \Large {I_2}={E(n+1)\over Rn} \\It \text{ is given} \\I_2=10I_1 \\ \Large {E(n+1)\over Rn} =10×{E\over (n+1)R} \\n=10 \)

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