First a set of n equal resistors of resistance 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?
Answers
Answered by
0
in series total resistance is nR +R
:.current = E/(n + 1)R
in parallel total resistance is n/R +R
:.current = E/n +R^2/R = ER/n +R^2
:.ER/(n+R^2)/E/(n +1)R = 10
=>R^2(n + 1)/(n +R^2) = 10
=>nR^2+ R^2 = 10n + 10 R^2
=>nR^2+R^2-10R^2 = 10n
=>nR^2-9R^2 =10n
=>R^2(n-R)/n = 10
now find the answer yourself
:.current = E/(n + 1)R
in parallel total resistance is n/R +R
:.current = E/n +R^2/R = ER/n +R^2
:.ER/(n+R^2)/E/(n +1)R = 10
=>R^2(n + 1)/(n +R^2) = 10
=>nR^2+ R^2 = 10n + 10 R^2
=>nR^2+R^2-10R^2 = 10n
=>nR^2-9R^2 =10n
=>R^2(n-R)/n = 10
now find the answer yourself
Answered by
0
Explanation:
I = E / R+ nr
=E / R +nR = E/ R(n+1)
Given when the resistors are parallel arranged , I' = 10I
I' = E / R+ (R/n) = 10I
E / R+ (R/n) = 10 x E/ R(n+1)
10[ R(1 + 1/n )] = R (n+1 )
TIP :- For exam if binomial is hard just substitute each options for fast answering
OR
10 + 10/n = n +1
n = 10
Hope you understand
This is Habel Sabu ....... Isn't it the Brainliest
Similar questions