SSC JE Electrical Engineering 22 Jan 2018 Evening Paper

Question : 118 of 200

Marks: +1, -0

What will be the voltage (in V) at node A of the given circuit diagram?

Solution:

At Node A, applying kcL

{V_A-12}/2+V_A/1+{V_A-24}/2=0

⇒V_A - 12 + 2V_A + V_A - 24 = 0

⇒V_A=36/4=9V

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