TS EAMCET 4 May 2018 Shift 2 Solved Paper

Consider the following figure of Wheatstone bridge. Applying Kirchhoff’s voltage law in loop ABCA,
−3Ri1−4R(i1−i2)+2Ri2=0
7i1=6i2
i1=

6

7

i2
Also,
i1+i2=i
Therefore,
(

6

7

+1)i2=i
i2=

7

13

i
Therefore,
i1=

6

13

i
In loop ABDA,
3Ri1+2Ri2≈V
So
3R (

6

13

i)+2R(

7

13

i)=iReq
Req=

32

13

R
Therefore,
Req=

32

13

R
5=

32

13

R
R≈2Ω