Question

# For the resistance network shown in the figure, choose the correct option(s).

A
The current through PQ is zero.
B
I1=3A.
C
The potential at S is less than that at Q.
D
I2=2A.

Solution

## The correct options are A The current through PQ is zero. B The potential at S is less than that at Q. C $$I_{1} = 3A.$$ D $$I_{2} = 2A.$$Node  P  and  Q  are  equipotential  and  node  S  and  T  are  equipotential due to formation of wheatstone bridge across themThus,  no  current  passes  through  PQ  and  ST$$\displaystyle I_{1}=\dfrac{12}{4}= 3A$$$$\displaystyle I_{2}=I_{1}(\dfrac{12}{6+12})= 2A$$now point P and Q are Equipotential and point S and T are equipotential.from resistor PS current flows from P to S hence voltage of point P is more than voltage of point S , also P is equipotential with Q that's  why Q also has more potential than S.Physics

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