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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# RUHANIKA
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# RUHANIKA
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Hlo friend
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Here is ur answer :
In series combination of resistors, current I is given by
I = E/E + nR
Whereas, in parallel combination current 10 I is given by
E/R + R/n = 10 I
====> E/R+R/n = 10 (E/R+nR)
Now, according to problem, 1+n/1+1/n =10
===>10 = (1+n/n+1)n
===> n = 10
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_________________________________
Here is ur answer :
In series combination of resistors, current I is given by
I = E/E + nR
Whereas, in parallel combination current 10 I is given by
E/R + R/n = 10 I
====> E/R+R/n = 10 (E/R+nR)
Now, according to problem, 1+n/1+1/n =10
===>10 = (1+n/n+1)n
===> n = 10
HOPE IT HELPS YOU.. PLS MARK ME AS BRAINLIEST...
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