A) 4.0V
B) 3.0V
C) 2.5V
D) 2.0V
Correct Answer: C
Solution :
\[I=\frac{6-2}{20+5+R}\] \[\Rightarrow \] \[I=\frac{4}{25+R}\] \[\therefore \] \[0.1=\frac{4}{25+R}\] \[\Rightarrow \] \[40=25+R\] \[\Rightarrow \] \[R=15\,\Omega \] The current of 0.1 A must flow from\[x\]to y through R. So, \[{{V}_{x}}+6-(0.1)(15)-2-{{V}_{y}}=0\] \[\Rightarrow \] \[{{V}_{x}}+6-1.5-2-{{V}_{y}}=0\] \[\therefore \] \[{{V}_{x}}-{{V}_{y}}=2.5V\]
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