question_answer

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.  

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

Correct Answer: A

Solution :

[a] The given circuit can be redrawn as, as 4 Q and \[x\,\Omega \] are parallel \[x'=\frac{1}{4}+\frac{1}{x}=\frac{\left( 4+x \right)}{4x}\text{    }x'=\frac{4x}{4+x}\] & \[1\Omega \] 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 \[x=\frac{4x}{4+x}+2=\frac{8+6x}{4+x}\] \[4x+{{x}^{2}}=8+6x\Rightarrow {{x}^{2}}-2x-8=0\] \[x=\frac{2\pm \sqrt{4-4\left( 1 \right)\left( -8 \right)}}{2}=\frac{2\pm \sqrt{36}}{2}=4\Omega \] Reading of Ammeter \[{{A}_{1}}=\frac{V}{\left( R+r \right)}\] \[{{A}_{1}}=\frac{9}{4+0.5}=2\text{ Ampere}\]