A) 10V
B) 15V
C) 18V
D) 20V
Correct Answer: A
Solution :
According to Kirchhoff?s junction law \[{{I}_{CO}}+{{I}_{OB}}+{{I}_{AO}}={{I}_{OD}}\] \[\Rightarrow \]\[\frac{{{V}_{C}}-{{V}_{O}}}{R}+\frac{{{V}_{B}}-{{V}_{O}}}{R}+\frac{{{V}_{A}}-{{V}_{O}}}{R}=\frac{{{V}_{O}}-{{V}_{D}}}{R}\] \[\Rightarrow \]\[3\left( \frac{{{V}_{A}}-{{V}_{O}}}{R} \right)=\frac{{{V}_{O}}-{{V}_{A}}+{{V}_{A}}-{{V}_{D}}}{R}\] \[\Rightarrow \]\[3\left[ \frac{{{V}_{A}}-{{V}_{O}}}{R} \right]=-\left[ \frac{{{V}_{A}}-{{V}_{O}}}{R} \right]+\left[ \frac{{{V}_{A}}-{{V}_{D}}}{R} \right]\] \[\Rightarrow \]\[4\left[ \frac{{{V}_{A}}-{{V}_{O}}}{R} \right]=\frac{{{V}_{A}}-{{V}_{D}}}{R}\] \[\Rightarrow \]\[4({{V}_{A}}-{{V}_{O}})=40\] \[\Rightarrow \]\[{{V}_{A}}-{{V}_{O}}=10V\]You need to login to perform this action.
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