For the resistance network shown in the figure, choose the correct options (s)

Options

A. The current through `PQ` is zero
B. `I_(1) = 3A`
C. The potential, at `S` is less then at `Q`
D. `I_(2) = 2A`

Answer

a,b,c,d

Explanation

If a plate mirror is placed along a line which is perpendicular bisector of line `AB`, then the left part of the circuit is exact image of right part of the circuit so current in AP = current in SB and current circuit the point `P` and `Q` are at the same potential. Also the point `S` and `T` are also as the same potential Therefore no current flow through arms `PQ and ST`. The equivalent circuit will be
The equivalent resistance of the circuit
`R = ((2+2+2) xx (4+4+4))/((2+2+2) + (4+4+4)) = 4 Omega`
Pot diff. area `A` and `B = 12 V`
`:. I_(1) = (12)/(4) = 3A`

Since current is flowing from P to S therefore
`V_(P) gt V_(S)` As `V_(P) = V_(Q)`, so `V_(Q) gt V_(S) gt V_(S)` or `V_(S) lt V_(Q)`