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).You need to login to perform this action.
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