Current Electricity Ques 41
- For the resistance network shown in the figure, choose the correct option(s).
(2012)
(a) The current through $P Q$ is zero
(b) $I _1=3 $ $A$
(c) The potential at $S$ is less than that at $Q$
(d) $I _2=2 $ $A$
Show Answer
Answer:
Correct Answer: 41.(a,b,c,d)
Solution:
Formula:
- Due to symmetry on upper side and lower side, points $P$ and $Q$ are at same potentials. Similarly, points $S$ and $T$ are at same potentials. Therefore, the simple circuit can be drawn as shown below
$I _2 =\frac{12}{2+2+2}=2 A $
$I _3 =\frac{12}{4+4+4}=1 A $
$I _1 =I _2+I _3=3 A $
$I _{P Q} =0 $
$V _P =V _Q$
Potential drop (from left to right) across each resistance is
$ \frac{12}{3} =4 V $
$\therefore \quad V _{M S} =2 \times 4=8 V $
$V _{N Q} =1 \times 4=4 V \text { or } V _S<V _Q$
BETA
