Current Electricity Charging Discharging Of Capacitors Question 276
Question: A voltmeter has a range 0-V with a series resistance R. With a series resistance 2R, the range is 0-V¢. The correct relation between V and V¢ is
[CPMT 2001]
Options:
A) $ {V}’=2V $
B) $ {V}’>2V $
C) $ {V}’»2V $
D) $ V’<2V $
Show Answer
Answer:
Correct Answer: D
Solution:
For conversion of galvanometer (of resistances) into voltmeter, a resistance R is connected in series. \ $ i _{g}=\frac{V _{1}}{R+G} $ and $ i _{g}=\frac{V _{2}}{2R+G} $
Therefore $ \frac{V _{1}}{R+G}=\frac{V _{2}}{2R+G} $
Therefore $ \frac{V _{2}}{V _{1}}=\frac{2R+G}{R+G}=\frac{2(R+G)-G}{(R+G)} $
$ =2-\frac{G}{(R+G)} $
Therefore $ V _{2}=2V _{1}-\frac{V _{1}G}{(R+G)} $
Therefore $ V _{2}<2V _{1} $
BETA
