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First a set of n equal resistors of R each are connected in series to a battery of emf E and internal resistance R. A current I is observed to flow. Then the n resistors are connected in parallel to the same battery. It is observed that the current is increased 10 times. What is 'n'? |
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Answer» SOLUTION :When n RESISTORS of value R are connected in SERIES with cell with emf `epsilon` and internal resistance R then current . `I = (E)/(R + nR) = (E)/(R (1 + n)) ""` .... (1) Instead n resistor of R value are connected in parallel then current, 10 I = `(E)/(R + (R)/(n))` `= (nE)/(R (n + 1)) "" ` ... (2) `therefore 10 I = n ((E)/(R (n + 1) )) ` `therefore 10 I = nI "" therefore n = 10 ` |
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