Current Electricity Charging Discharging Of Capacitors Question 723
Question: A 9 V battery with internal resistance of 0.512 is connected across an infinite network as shown in the figure. All ammeters $ A _{1},A _{2},A _{3} $ and voltmeter V are ideal. Choose correct statement.
Options:
A) Reading of $ A{ _{1}} $ is 2 A
B) Reading of $ A{ _{1}} $ is 18 A
C) Reading of V is 9 V
D) Reading of V is 7 V
Show Answer
Answer:
Correct Answer: A
Solution:
[a] The given circuit can be redrawn as, as 4 Q and $ x,\Omega $ are parallel
$ x’=\frac{1}{4}+\frac{1}{x}=\frac{( 4+x )}{4x}\text{ }x’=\frac{4x}{4+x} $ &
$ 1\Omega $ and $ 1\Omega $ are also parallel $ x’=2,\Omega $
$ \frac{4-V _{1}}{100}=\frac{V _{1}-1}{100}+\frac{V _{1}-0}{100} $
$ V _{P}-V _{Q}=4-\frac{1}{3}\times 3=3volt $
Now equivalent resistance of circuit $ x=\frac{4x}{4+x}+2=\frac{8+6x}{4+x} $
$ 4x+x^{2}=8+6x\Rightarrow x^{2}-2x-8=0 $
$ x=\frac{2\pm \sqrt{4-4( 1 )( -8 )}}{2}=\frac{2\pm \sqrt{36}}{2}=4\Omega $ Reading of Ammeter $ A _{1}=\frac{V}{( R+r )} $
$ A _{1}=\frac{9}{4+0.5}=2\text{ Ampere} $
BETA
