GATE Papers >> EEE >> 2016 >> Question No 65

Question No. 65 EEE | GATE 2016

In the circuit shown below, the node voltage VA is ___________ V.


Answer : 11.25 : 11.50


Solution of Question No 65 of GATE 2016 EEE Paper

Applying KCL at node A, we get

$ \begin{array}{l}\frac{{\mathrm V}_\mathrm A}5+\frac{{\mathrm V}_\mathrm A-10}{10}+\frac{{\mathrm V}_\mathrm A+10{\mathrm I}_1}5=5\\\mathrm{So},\;2\;{\mathrm V}_\mathrm A+{\mathrm V}_\mathrm A-10+2\;{\mathrm V}_\mathrm A+20\;{\mathrm I}_1=5\\\;\;\;\;\;\;\;5\;{\mathrm V}_\mathrm A+20\;{\mathrm I}_1=60\\\mathrm{Since},\;\;\;\;\;\;{\mathrm I}_1=\frac{{\mathrm V}_\mathrm A-10}{10}\\\mathrm{So},\;5\;{\mathrm V}_\mathrm A+2{\mathrm V}_\mathrm A-20=60\\\;\;\;\;\;\;\;\;\;7\;{\mathrm V}_\mathrm A=80\\\;\;\;\;\;\;\;\;\;\;{\mathrm V}_\mathrm A=\frac{80}7=11.42\end{array} $

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