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)) \[\frac{(5/3)-1}{100}=6.67\,mA\]You need to login to perform this action.
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