Current Electricity Charging Discharging Of Capacitors Question 745
Question: In the figure ammeter $ A _{1} $ reads a current of 10mA, while the voltmeter reads a potential difference of 3V. What does ammeter $ A _{2} $ in mA read? The ammeters are identical, the internal resistance of the battery is negligible. (Consider all ammeters and voltmeters as non- ideal.)
Options:
A) 6.67mA
B) 3.12mA
C) 1.12mA
D) 5.14mA
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ R _{A}= $ resistance of ammeter $ \frac{4-V _{1}}{100}=\frac{V _{1}-1}{R _{A}}+\frac{V _{1}-0}{100} $ ?.(i) $ 1V-0V=( 10mA )R _{A} $
$ R _{A}=100\Omega $ ?.(ii)
$ \frac{4-V _{1}}{100}=\frac{V _{1}-1}{100}+\frac{V _{1}-0}{100} $ [By using eq. (1) and (2)]
$ V _{1}=(5/3),V $
$ \frac{v _{1}-1}{R _{A}},= $ (Current in ammeter (II))
$ \frac{(5/3)-1}{100}=6.67,mA $
BETA
