question_answer

In the figure ammeter \[{{A}_{1}}\]reads a current of 10mA, while the voltmeter reads a potential difference of 3 V What does ammeter \[{{A}_{2}}\]in mA read? The am- meters are identical, the internal resistance of the battery is negligible. (Consider all ammeters and voltmeters as non-ideal.)

A)  6.57 mA                     

B)  3.12 mA         

C)     1.12 mA         

D)                5.14 mA                     

Correct Answer: A

Solution :

\[{{R}_{A}}=\]resistance of ammeter             \[\frac{4-{{V}_{1}}}{100}=\frac{{{V}_{1}}-1}{{{R}_{A}}}+\frac{{{V}_{1}}-0}{100}\] \[1V-0V=(10\,mA){{R}_{A}}\] \[{{R}_{A}}=100\,\Omega \]                                                (2) \[\frac{4-{{V}_{1}}}{100}=\frac{{{V}_{1}}-1}{100}+\frac{{{V}_{1}}-0}{100}\][By using eq.(1) and (2)] \[{{V}_{1}}=(5/3)V\] \[\frac{{{V}_{1}}-1}{{{R}_{A}}}=\](Current in ammeter)                       (ii) \[\frac{5/3-1}{10}=6.67\,mA\] Hence, the correction option is (a).