Current Electricity Charging Discharging Of Capacitors Question 315
Question: Two resistances are connected in two gaps of a metre bridge. The balance point is 20 cm from the zero end. A resistance of 15 ohms is connected in series with the smaller of the two. The null point shifts to 40 cm. The value of the smaller resistance in ohms is
[KCET 2005]
Options:
A) 3
B) 6
C) 9
D) 12
Show Answer
Answer:
Correct Answer: C
Solution:
Let S be larger and R be smaller resistance connected in two gaps of meter bridge. \ $ S=( \frac{100-l}{l} )R=\frac{100-20}{20}R=4R $ …..(i) When $ 15\Omega $ resistance is added to resistance R, then $ S=( \frac{100-40}{40} )(R+15)=\frac{6}{4}(R+15) $ …. (ii) From equations (i) and (ii) $ R=9\Omega $
BETA
