Current Electricity Charging Discharging Of Capacitors Question 238
Question: A 100 V voltmeter of internal resistance $ 20,k\Omega $ in series with a high resistance R is connected to a 110 V line. The voltmeter reads 5 V, the value of R is
[MP PET 1999]
Options:
A) $ 210,k\Omega $
B) $ 315,k\Omega $
C) $ 420,k\Omega $
D) $ 440,k\Omega $
Show Answer
Answer:
Correct Answer: C
Solution:
Here $ i=\frac{110}{20\times 10^{3}+R} $
$ \because $
$ V=iR $
Therefore $ 5=( \frac{110}{20\times 10^{3}+R} )\times 20\times 10^{3} $
Therefore $ 10^{5}+5R=22\times 10^{5} $
Therefore $ R=21\times \frac{10^{5}}{5}=420,K\Omega $
BETA
