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 than that at Q.
D. `I_2 = 2A`

Answer

A, B, C, D

Explanation

(a,b,c,d)
The given circuit is an extension of wheatstone bridge,
therefore points P and Q are at the same potential and point
S and T are also at the same potential. Therefore no current
passes through PQ and ST and the circuit reduces to as
shown

`I_1 = 12/4` `[:. R_(eq) = (6xx12)/(6+12)]`
=3A
`:. I_2 = 3[12/(6+12)] = 2A`
As P and Q are equipotential and potential at S is less than the potential at P (potential drops across a
resistance as current passes through it), therefore `V_S lt V_Q`.