Current Electricity Charging Discharging Of Capacitors Question 203
Question: In Wheatstone’s bridge $ P=9 $ ohm, $ Q=11 $ ohm, $ R=4 $ ohm and $ S=6 $ ohm. How much resistance must be put in parallel to the resistance $ S $ to balance the bridge
[DPMT 1999]
Options:
A) 24 ohm
B) $ \frac{44}{9} $ ohm
C) 26.4 ohm
D) 18.7 ohm
Show Answer
Answer:
Correct Answer: C
Solution:
$ \frac{P}{Q}=\frac{R}{{{S}’}} $ (For balancing bridge)
Therefore $ {S}’=\frac{4\times 11}{9}=\frac{44}{9} $
Therefore $ \frac{1}{{{S}’}}=\frac{1}{r}+\frac{1}{6} $
Therefore $ \frac{9}{44}-\frac{1}{6}=\frac{1}{r} $
Therefore $ r=\frac{132}{5}=26.4,\Omega $
BETA
