Current Electricity Charging Discharging Of Capacitors Question 253
Question: We have a galvanometer of resistance $ 25,\Omega $ . It is shunted by a $ 2.5,\Omega $ wire. The part of total current that flows through the galvanometer is given as
[AFMC 1998; MH CET 1999; Pb. PMT 2002]
Options:
A) $ \frac{I}{I _{0}}=\frac{1}{11} $
B) $ \frac{I}{I _{0}}=\frac{1}{10} $
C) $ \frac{I}{I _{0}}=\frac{3}{11} $
D) $ \frac{I}{I _{0}}=\frac{4}{11} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \frac{i}{i _{g}}=\frac{G+S}{S} $
Therefore $ \frac{i _{g}}{i}=\frac{S}{G+S}=\frac{2.5}{27.5}=\frac{1}{11} $
BETA
