Current Electricity Charging Discharging Of Capacitors Question 287
Question: To convert a 800 mV range milli voltmeter of resistance 40 W into a galvanometer of 100 mA range, the resistance to be connected as shunt is
[CBSE PMT 2002]
Options:
A) 10 W
B) 20 W
C) 30 W
D) 40 W
Show Answer
Answer:
Correct Answer: A
Solution:
$ \frac{i}{i _{g}}=1+\frac{G}{S} $
$ \Rightarrow \frac{i.G}{V _{g}}=1+\frac{G}{S}\Rightarrow \frac{100\times {{10}^{-3}}\times 40}{800\times {{10}^{-3}}}=1+\frac{40}{S} $
Therefore $ S=10\Omega $ .
BETA
