The correct option is
C 2235 ΩThe given circuit posses mirror symmetry about a perpendicular axis joining
AB, as shown below
The current distribution will be such that the mirror image will have same current as that of object element.
From the distribution of current, we infer that connection at junction
′O′ is useless, hence it can be removed.
After junction removal :
⇒ For the two set of rectangular arrangement of resistors, two
r (in series), one
r and again two
r (in series) are in parallel combination.
1req=12r+1r+12r=1+2+12r
req=2r4+r2
Simillarly for the triangular arrangements, two
′r′ (being in series) are in parallel with third resistance
′r′.
⇒1req=12r+1r=1+22r
req=2r3
Now redrawing the simplified circuit,
The upper and lower branches can be replaced by
r1=2r3+2r3+r2
r1=4r+4r+3r6=11r6
The three resistances are in parallel about
A &
B,
1Req=1(11r/6)+1(11r/6)+1(2r)
1Req=611r+611r+12r
1Req=12+12+1122r
⇒Req=22r35
(∵r=1 Ω)
⇒Req=2235 Ω
Why this question?
Tip: In problems involving complex circuit and resistances are arranged to form geometrical shape, then always think to identity the parallel or perpendicular axis of symmetry about the required terminal/end. This will help to remove unwanted junction, as well as gives ''insight to points with same potential''. |