NEET Class neet Answered
current electricity
Asked by azazsindhiyasin | 18 May, 2022, 04:43: PM
Expert Answer
Given network of resistors is redrawn as hown in right side of figure.
Let us connect this network to a battery of EMF V . Let i be the total current drawn from battery.
Current distribution that is considered in the network is as shown in figure.
Let us apply Kirchoff's voltage law to the loop ABCDEA
(2r) i1 + r i5 - r i3 = 0 or 2 i1 + i5 - i3 = 0 .......................... ( 1 )
Let us apply Kirchoff's voltage law to the loop CHDGC
(2r) i6 - r i5 - r i5 = 0 or i6 = i5 .......................(2)
Let us apply Kirchoff's voltage law to the loop DGFED
r i3 + r i3 - (2r) i4 = 0 or i3 = i4 ..........................(3)
At node E , if we apply Kirchoff's current law , we get
i2 = i3 + i4 ..........................(4)
Using eqn.(3) and (4) , we get i2 = 2 i4 ........................ (5)
At node C , if we apply Kirchoff's current law , we get
i1 = i5 + i6 .............................(6)
Using eqn.(2) and eqn.(6) , we get
i1 = 2 i5 .............................. ( 7 )
using eqn.(7) and eqn.(3) , we get i1 from eqn.(1) as follows
2 i1 + (1/2) i1 - i4 = 0 or i1 = (2/5) i4 .................... (8)
Hence total Current i from battery is given as
i = i1 + i2 = (2/5) i4 + 2 i4 = (12/5) i4.......................(9)
As we see from figure, resitor 2r that is between E and F is directly connected across battery.
Hence we get , (2r) i4 = V or i4 = ( V / 2r ) ......................(10)
Hence from eqn.(9) and eqn.(10) , we get
i = (12/5) ( V / 2r ) = (6/5) ( V / r )
Hence we get , V / i = (5/6) r
Hence equivalent resistance of network = (5/6) r
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