question_answer

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.)            

A) 6.67mA   

B) 3.12mA   

C) 1.12mA

D) 5.14mA

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) \[1V-0V=\left( 10mA \right){{R}_{A}}\] \[{{R}_{A}}=100\Omega \]                                            ?.(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)) \[\frac{(5/3)-1}{100}=6.67\,mA\]